3.6.75 \(\int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [575]

Optimal. Leaf size=270 \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(
1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)
-1/4*(a-b)*(a^2+4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2*a*(a^2-3*b^2)/d/tan(d*x+c)^(1
/2)-8/5*a^2*b/d/tan(d*x+c)^(3/2)-2/5*a^2*(a+b*tan(d*x+c))/d/tan(d*x+c)^(5/2)

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Rubi [A]
time = 0.25, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3646, 3709, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(7/2),x]

[Out]

-(((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + ((a + b)*(a^2 - 4*a*b +
b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[T
an[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]
+ Tan[c + d*x]])/(2*Sqrt[2]*d) - (8*a^2*b)/(5*d*Tan[c + d*x]^(3/2)) + (2*a*(a^2 - 3*b^2))/(d*Sqrt[Tan[c + d*x]
]) - (2*a^2*(a + b*Tan[c + d*x]))/(5*d*Tan[c + d*x]^(5/2))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {6 a^2 b-\frac {5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {-\frac {5}{2} a \left (a^2-3 b^2\right )-\frac {5}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {-\frac {5}{2} b \left (3 a^2-b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 \text {Subst}\left (\int \frac {-\frac {5}{2} b \left (3 a^2-b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{5 d}\\ &=-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.52, size = 103, normalized size = 0.38 \begin {gather*} -\frac {2 \left (3 a \left (a^2-3 b^2\right ) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\tan ^2(c+d x)\right )+b \left (5 \left (3 a^2-b^2\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right ) \tan (c+d x)+b (9 a+5 b \tan (c+d x))\right )\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(7/2),x]

[Out]

(-2*(3*a*(a^2 - 3*b^2)*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] + b*(5*(3*a^2 - b^2)*Hypergeometric2F
1[-3/4, 1, 1/4, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(9*a + 5*b*Tan[c + d*x]))))/(15*d*Tan[c + d*x]^(5/2))

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Maple [A]
time = 0.06, size = 243, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 a \left (a^{2}-3 b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a^{2} b}{\tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(243\)
default	\(\frac {-\frac {2 a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 a \left (a^{2}-3 b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a^{2} b}{\tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^3/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/5*a^3/tan(d*x+c)^(5/2)+2*a*(a^2-3*b^2)/tan(d*x+c)^(1/2)-2*a^2*b/tan(d*x+c)^(3/2)+1/4*(-3*a^2*b+b^3)*2^
(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*
tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(a^3-3*a*b^2)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(
1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(
1/2)*tan(d*x+c)^(1/2))))

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Maxima [A]
time = 0.51, size = 235, normalized size = 0.87 \begin {gather*} \frac {10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (5 \, a^{2} b \tan \left (d x + c\right ) + a^{3} - 5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/20*(10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 10*sqr
t(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 5*sqrt(2)*(a^3 +
3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b
^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 8*(5*a^2*b*tan(d*x + c) + a^3 - 5*(a^3 - 3*a*b
^2)*tan(d*x + c)^2)/tan(d*x + c)^(5/2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7742 vs. \(2 (232) = 464\).
time = 3.33, size = 7742, normalized size = 28.67 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/20*(20*sqrt(2)*(d^5*cos(d*x + c)^4 - 2*d^5*cos(d*x + c)^2 + d^5)*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*
a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*sqrt((a^12 + 6*a^10*b^2 + 15
*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6
+ 255*a^4*b^8 - 30*a^2*b^10 + b^12))*((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 +
b^12)/d^4)^(3/4)*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4
)*arctan(-((a^24 - 6*a^22*b^2 - 84*a^20*b^4 - 322*a^18*b^6 - 603*a^16*b^8 - 540*a^14*b^10 + 540*a^10*b^14 + 60
3*a^8*b^16 + 322*a^6*b^18 + 84*a^4*b^20 + 6*a^2*b^22 - b^24)*d^4*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6
*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8
- 30*a^2*b^10 + b^12)/d^4) + sqrt(2)*((3*a^2*b - b^3)*d^7*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 +
15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*
a^2*b^10 + b^12)/d^4) + (a^9 - 6*a^5*b^4 - 8*a^3*b^6 - 3*a*b^8)*d^5*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 4
52*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^
4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a
^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 -
30*a^2*b^10 + b^12))*sqrt(((a^18 - 27*a^16*b^2 + 168*a^14*b^4 + 224*a^12*b^6 - 366*a^10*b^8 - 366*a^8*b^10 +
224*a^6*b^12 + 168*a^4*b^14 - 27*a^2*b^16 + b^18)*d^2*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a
^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*cos(d*x + c) + sqrt(2)*((a^15 - 33*a^13*b^2 + 345*a^11*b^4 - 1217*a^9*b^6 + 1
611*a^7*b^8 - 795*a^5*b^10 + 91*a^3*b^12 - 3*a*b^14)*d^3*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 1
5*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*cos(d*x + c) + (3*a^20*b - 82*a^18*b^3 + 531*a^16*b^5 + 504*a^14*b^7 - 132
2*a^12*b^9 - 732*a^10*b^11 + 1038*a^8*b^13 + 280*a^6*b^15 - 249*a^4*b^17 + 30*a^2*b^19 - b^21)*d*cos(d*x + c))
*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^5*b - 10*a^3*b^3
+ 3*a*b^5)*d^2*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4))/(a^12
- 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12))*sqrt(sin(d*x + c)/cos(d*x + c)
)*((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)^(1/4) + (a^24 - 24*a^22
*b^2 + 90*a^20*b^4 + 648*a^18*b^6 + 783*a^16*b^8 - 624*a^14*b^10 - 1748*a^12*b^12 - 624*a^10*b^14 + 783*a^8*b^
16 + 648*a^6*b^18 + 90*a^4*b^20 - 24*a^2*b^22 + b^24)*sin(d*x + c))/cos(d*x + c))*((a^12 + 6*a^10*b^2 + 15*a^8
*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)^(3/4) + sqrt(2)*((3*a^14*b - 37*a^12*b^3 - 69*a^10*b^
5 + 27*a^8*b^7 + 81*a^6*b^9 + 9*a^4*b^11 - 15*a^2*b^13 + b^15)*d^7*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a
^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b
^8 - 30*a^2*b^10 + b^12)/d^4) + (a^21 - 12*a^19*b^2 - 33*a^17*b^4 + 64*a^15*b^6 + 282*a^13*b^8 + 264*a^11*b^10
- 82*a^9*b^12 - 288*a^7*b^14 - 171*a^5*b^16 - 28*a^3*b^18 + 3*a*b^20)*d^5*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*
b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6
+ 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4
+ 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^
4*b^8 - 30*a^2*b^10 + b^12))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 1
5*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)^(3/4))/(a^36 - 18*a^34*b^2 - 39*a^32*b^4 + 848*a^30*b^6 + 5556*a^28*b^8 +
15240*a^26*b^10 + 20420*a^24*b^12 + 5424*a^22*b^14 - 25938*a^20*b^16 - 42988*a^18*b^18 - 25938*a^16*b^20 + 542
4*a^14*b^22 + 20420*a^12*b^24 + 15240*a^10*b^26 + 5556*a^8*b^28 + 848*a^6*b^30 - 39*a^4*b^32 - 18*a^2*b^34 + b
^36)) + 20*sqrt(2)*(d^5*cos(d*x + c)^4 - 2*d^5*cos(d*x + c)^2 + d^5)*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20
*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12 - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d^2*sqrt((a^12 + 6*a^10*b^2 + 1
5*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4))/(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6
+ 255*a^4*b^8 - 30*a^2*b^10 + b^12))*((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10
+ b^12)/d^4)^(3/4)*sqrt((a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^
4)*arctan(((a^24 - 6*a^22*b^2 - 84*a^20*b^4 - 322*a^18*b^6 - 603*a^16*b^8 - 540*a^14*b^10 + 540*a^10*b^14 + 60
3*a^8*b^16 + 322*a^6*b^18 + 84*a^4*b^20 + 6*a^2*b^22 - b^24)*d^4*sqrt((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6
*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)/d^4)*sqr...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**3/tan(d*x+c)**(7/2),x)

[Out]

Integral((a + b*tan(c + d*x))**3/tan(c + d*x)**(7/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 6.27, size = 1777, normalized size = 6.58 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {32\,a^6\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {3\,a\,b^5}{2\,d^2}+\frac {b^6\,1{}\mathrm {i}}{4\,d^2}-\frac {a^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a^5\,b}{2\,d^2}-\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}+\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2+a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2-a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2+a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2-a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2+b^9\,d^2\,16{}\mathrm {i}}-\frac {32\,b^6\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {3\,a\,b^5}{2\,d^2}+\frac {b^6\,1{}\mathrm {i}}{4\,d^2}-\frac {a^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a^5\,b}{2\,d^2}-\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}+\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2+a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2-a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2+a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2-a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2+b^9\,d^2\,16{}\mathrm {i}}+\frac {480\,a^2\,b^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {3\,a\,b^5}{2\,d^2}+\frac {b^6\,1{}\mathrm {i}}{4\,d^2}-\frac {a^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a^5\,b}{2\,d^2}-\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}+\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2+a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2-a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2+a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2-a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2+b^9\,d^2\,16{}\mathrm {i}}-\frac {480\,a^4\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {3\,a\,b^5}{2\,d^2}+\frac {b^6\,1{}\mathrm {i}}{4\,d^2}-\frac {a^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a^5\,b}{2\,d^2}-\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}+\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2+a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2-a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2+a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2-a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2+b^9\,d^2\,16{}\mathrm {i}}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}-2\,\mathrm {atanh}\left (\frac {32\,a^6\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^6\,1{}\mathrm {i}}{4\,d^2}-\frac {b^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a\,b^5}{2\,d^2}+\frac {3\,a^5\,b}{2\,d^2}+\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}-\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2-a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2+a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2-a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2+a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2-b^9\,d^2\,16{}\mathrm {i}}-\frac {32\,b^6\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^6\,1{}\mathrm {i}}{4\,d^2}-\frac {b^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a\,b^5}{2\,d^2}+\frac {3\,a^5\,b}{2\,d^2}+\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}-\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2-a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2+a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2-a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2+a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2-b^9\,d^2\,16{}\mathrm {i}}+\frac {480\,a^2\,b^4\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^6\,1{}\mathrm {i}}{4\,d^2}-\frac {b^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a\,b^5}{2\,d^2}+\frac {3\,a^5\,b}{2\,d^2}+\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}-\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2-a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2+a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2-a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2+a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2-b^9\,d^2\,16{}\mathrm {i}}-\frac {480\,a^4\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^6\,1{}\mathrm {i}}{4\,d^2}-\frac {b^6\,1{}\mathrm {i}}{4\,d^2}+\frac {3\,a\,b^5}{2\,d^2}+\frac {3\,a^5\,b}{2\,d^2}+\frac {a^2\,b^4\,15{}\mathrm {i}}{4\,d^2}-\frac {5\,a^3\,b^3}{d^2}-\frac {a^4\,b^2\,15{}\mathrm {i}}{4\,d^2}}}{16\,a^9\,d^2-a^8\,b\,d^2\,48{}\mathrm {i}-288\,a^7\,b^2\,d^2+a^6\,b^3\,d^2\,736{}\mathrm {i}+960\,a^5\,b^4\,d^2-a^4\,b^5\,d^2\,960{}\mathrm {i}-736\,a^3\,b^6\,d^2+a^2\,b^7\,d^2\,288{}\mathrm {i}+48\,a\,b^8\,d^2-b^9\,d^2\,16{}\mathrm {i}}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (6\,a\,b^2-2\,a^3\right )+\frac {2\,a^3}{5}+2\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^3/tan(c + d*x)^(7/2),x)

[Out]

- 2*atanh((32*a^6*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/
(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i +
48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b
^3*d^2*736i - 288*a^7*b^2*d^2) - (32*b^6*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^
5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a
^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i +
960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) + (480*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2)
- (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^
2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b
^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (480*a^4*b^2*d^3*tan(c + d
*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2)
- (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i -
a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2))
*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2) - 2*atanh((32*
a^6*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2
*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 -
a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i -
288*a^7*b^2*d^2) - (32*b^6*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (
3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 - b^9*d
^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^
2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) + (480*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*
d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2)
)^(1/2))/(16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b
^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (480*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((a^
6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)
/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*28
8i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2))*((6*a*b^5 + 6
*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2) - (tan(c + d*x)^2*(6*a*b^2 -
2*a^3) + (2*a^3)/5 + 2*a^2*b*tan(c + d*x))/(d*tan(c + d*x)^(5/2))

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