3.6.71 \(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx\) [571]

Optimal. Leaf size=272 \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]

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

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Rubi [A]
time = 0.25, antiderivative size = 272, 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 = {3647, 3711, 3609, 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 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{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 {(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 {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3,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) + (2*b*(3*a^2 - b^2)*Sqrt[Tan[c + d*x]])/d + (8*a*b^2*Tan[c + d*x]^(3/2))/(5*d)
+ (2*b^2*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))/(5*d)

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 3609

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

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 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.83, size = 120, normalized size = 0.44 \begin {gather*} \frac {-5 \sqrt [4]{-1} (i a+b)^3 \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-5 (-1)^{3/4} (a+i b)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 a^2-5 b^2+5 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(-1)^(1/4)*(I*a + b)^3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 5*(-1)^(3/4)*(a + I*b)^3*ArcTanh[(-1)^(3/4)
*Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c + d*x]]*(15*a^2 - 5*b^2 + 5*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2))/(5*d
)

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Maple [A]
time = 0.06, size = 250, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(250\)
default	\(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/5*b^3*tan(d*x+c)^(5/2)+2*a*b^2*tan(d*x+c)^(3/2)+6*a^2*b*tan(d*x+c)^(1/2)-2*b^3*tan(d*x+c)^(1/2)+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*arc
tan(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 = 238, normalized size = 0.88 \begin {gather*} \frac {8 \, b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 40 \, a b^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 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 ) + 40 \, {\left (3 \, a^{2} b - b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{20 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/20*(8*b^3*tan(d*x + c)^(5/2) + 40*a*b^2*tan(d*x + c)^(3/2) + 10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arct
an(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 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)))) - 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) + 40*(3*a^2*b - b^3)*sqrt(tan(d*x + c)))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7179 vs. \(2 (234) = 468\).
time = 3.74, size = 7179, normalized size = 26.39 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/20*(20*sqrt(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^1
0 + 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 + 2
55*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 + 603*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)*sq
rt((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 - 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(((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 + 1611*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 + 15*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 - 1322*a^12*b^9 - 732*a^10*b^11 + 1038*a^8*b^13 + 2
80*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 + 2
55*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^1
9*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 + 15*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 + 5424*a^14*b^22 + 20420*a^12*b^24 + 15240*a^10*b^2
6 + 5556*a^8*b^28 + 848*a^6*b^30 - 39*a^4*b^32 - 18*a^2*b^34 + b^36))*cos(d*x + c)^2 + 20*sqrt(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^1
0*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 - 5
40*a^14*b^10 + 540*a^10*b^14 + 603*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...

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sqrt(tan(c + d*x)), 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(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 6.09, size = 1742, normalized size = 6.40 \begin {gather*} \frac {2\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {2\,b^3}{d}-\frac {6\,a^2\,b}{d}\right )+\frac {2\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{d}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}\,1{}\mathrm {i}}{-\frac {16\,\left (-a^9+6\,a^5\,b^4+8\,a^3\,b^6+3\,a\,b^8\right )}{d^3}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}}\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 {i}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}\,1{}\mathrm {i}}{-\frac {16\,\left (-a^9+6\,a^5\,b^4+8\,a^3\,b^6+3\,a\,b^8\right )}{d^3}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\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}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\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}}}\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 {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^3,x)

[Out]

(2*b^3*tan(c + d*x)^(5/2))/(5*d) - atan((((8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4
- 15*a^4*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)*1i - ((8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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)*1i)/(((8*(4*b^3*d^2 - 12
*a^2*b*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))/d^
3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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) - (16*(3*a*b^8 - a^9 + 8*a^3*b^6 + 6*a^5*b^4))/d^3 + (
(8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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)))*((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)*2i - atan((((8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 - (16*tan(c + d*x
)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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)*1i - ((8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^
4 - 15*a^4*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)*1i)/(((8*(4*b^3*d^2 - 12*a^2*b*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))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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) - (16*(3*a*b^8 - a^9 +
8*a^3*b^6 + 6*a^5*b^4))/d^3 + ((8*(4*b^3*d^2 - 12*a^2*b*d^2)*((6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*1
5i - 20*a^3*b^3 - a^4*b^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*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)))*((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)*2i - tan(c + d*x)^
(1/2)*((2*b^3)/d - (6*a^2*b)/d) + (2*a*b^2*tan(c + d*x)^(3/2))/d

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