3.5.22 \(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) [422]

Optimal. Leaf size=256 \[ \frac {(a+b) B \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) B \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a^{5/2} B \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}-\frac {(a-b) B \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 B \sqrt {\tan (c+d x)}}{b d} \]

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {21, 3647, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} \frac {B (a+b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {B (a+b) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {B (a-b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {B (a-b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {2 a^{5/2} B \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{3/2} d \left (a^2+b^2\right )}+\frac {2 B \sqrt {\tan (c+d x)}}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^(5/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

((a + b)*B*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - ((a + b)*B*ArcTan[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*a^(5/2)*B*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(b^(3/2
)*(a^2 + b^2)*d) - ((a - b)*B*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) +
((a - b)*B*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + (2*B*Sqrt[Tan[c + d
*x]])/(b*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 211

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

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 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 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3734

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.14, size = 156, normalized size = 0.61 \begin {gather*} \frac {B \left (\sqrt [4]{-1} b^{3/2} (-i a+b) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 a^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt [4]{-1} b^{3/2} (i a+b) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 a^2 \sqrt {b} \sqrt {\tan (c+d x)}+2 b^{5/2} \sqrt {\tan (c+d x)}\right )}{b^{3/2} \left (a^2+b^2\right ) d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^(5/2)*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]
time = 0.09, size = 242, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {B \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b}+\frac {-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{3} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{b \left (a^{2}+b^{2}\right ) \sqrt {a b}}\right )}{d}\)	\(242\)
default	\(\frac {B \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b}+\frac {-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}-\frac {2 a^{3} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{b \left (a^{2}+b^{2}\right ) \sqrt {a b}}\right )}{d}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 188, normalized size = 0.73 \begin {gather*} -\frac {\frac {8 \, B a^{3} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt {a b}} + \frac {{\left (2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{a^{2} + b^{2}} - \frac {8 \, B \sqrt {\tan \left (d x + c\right )}}{b}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/4*(8*B*a^3*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2*b + b^3)*sqrt(a*b)) + (2*sqrt(2)*(a + b)*arctan(1/2
*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x +
c)))) - sqrt(2)*(a - b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a - b)*log(-sqrt(2)*sqr
t(tan(d*x + c)) + tan(d*x + c) + 1))*B/(a^2 + b^2) - 8*B*sqrt(tan(d*x + c))/b)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 4431 vs. \(2 (216) = 432\).
time = 11.98, size = 8974, normalized size = 35.05 \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)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(2)*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d^5*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b +
2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*(B^4/((a
^4 + 2*a^2*b^2 + b^4)*d^4))^(3/4)*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a
^2*b^6 + b^8)*d^4))*arctan(-((B^6*a^8 + 2*B^6*a^6*b^2 - 2*B^6*a^2*b^6 - B^6*b^8)*d^4*sqrt(B^4/((a^4 + 2*a^2*b^
2 + b^4)*d^4))*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))
- sqrt(2)*((a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*d^7*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt
((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (B^2*a^7 + 3*B^2
*a^5*b^2 + 3*B^2*a^3*b^4 + B^2*a*b^6)*d^5*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b
^4 + 4*a^2*b^6 + b^8)*d^4)))*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(
B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqrt(((B^4*a^6 - B^4*a^4*b^2 - B^4*a^
2*b^4 + B^4*b^6)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*cos(d*x + c) + sqrt(2)*((B^3*a^7 - B^3*a^5*b^2 -
B^3*a^3*b^4 + B^3*a*b^6)*d^3*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*cos(d*x + c) - (B^5*a^4*b - 2*B^5*a^2*b^3
+ B^5*b^5)*d*cos(d*x + c))*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B
^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*(B^4/(
(a^4 + 2*a^2*b^2 + b^4)*d^4))^(1/4) + (B^6*a^4 - 2*B^6*a^2*b^2 + B^6*b^4)*sin(d*x + c))/cos(d*x + c))*(B^4/((a
^4 + 2*a^2*b^2 + b^4)*d^4))^(3/4) - sqrt(2)*((B^3*a^10*b + 3*B^3*a^8*b^3 + 2*B^3*a^6*b^5 - 2*B^3*a^4*b^7 - 3*B
^3*a^2*b^9 - B^3*b^11)*d^7*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((
a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (B^5*a^9 + 2*B^5*a^7*b^2 - 2*B^5*a^3*b^6 - B^5*a*b^8)*d
^5*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))*sqrt((B^2*
a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^
2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))^(3/4))/(
B^10*a^4 - 2*B^10*a^2*b^2 + B^10*b^4)) + 4*sqrt(2)*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d^5*sqrt((B^2*a^4 + 2
*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 -
2*B^2*a^2*b^2 + B^2*b^4))*(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))^(3/4)*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/
((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4))*arctan(((B^6*a^8 + 2*B^6*a^6*b^2 - 2*B^6*a^2*b^6 - B^6*
b^8)*d^4*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 +
6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + sqrt(2)*((a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*d^7*sqrt(B^4/(
(a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8)*d^4)) - (B^2*a^7 + 3*B^2*a^5*b^2 + 3*B^2*a^3*b^4 + B^2*a*b^6)*d^5*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^
4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^
5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqr
t(((B^4*a^6 - B^4*a^4*b^2 - B^4*a^2*b^4 + B^4*b^6)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*cos(d*x + c) -
sqrt(2)*((B^3*a^7 - B^3*a^5*b^2 - B^3*a^3*b^4 + B^3*a*b^6)*d^3*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*cos(d*x
+ c) - (B^5*a^4*b - 2*B^5*a^2*b^3 + B^5*b^5)*d*cos(d*x + c))*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5
*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqrt
(sin(d*x + c)/cos(d*x + c))*(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))^(1/4) + (B^6*a^4 - 2*B^6*a^2*b^2 + B^6*b^4)*si
n(d*x + c))/cos(d*x + c))*(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))^(3/4) + sqrt(2)*((B^3*a^10*b + 3*B^3*a^8*b^3 + 2
*B^3*a^6*b^5 - 2*B^3*a^4*b^7 - 3*B^3*a^2*b^9 - B^3*b^11)*d^7*sqrt(B^4/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt((B^4
*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (B^5*a^9 + 2*B^5*a^7*
b^2 - 2*B^5*a^3*b^6 - B^5*a*b^8)*d^5*sqrt((B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 +
4*a^2*b^6 + b^8)*d^4)))*sqrt((B^2*a^4 + 2*B^2*a^2*b^2 + B^2*b^4 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d^2*sqrt(B^4/(
(a^4 + 2*a^2*b^2 + b^4)*d^4)))/(B^2*a^4 - 2*B^2*a^2*b^2 + B^2*b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*(B^4/((a^4
+ 2*a^2*b^2 + b^4)*d^4))^(3/4))/(B^10*a^4 - 2*B^10*a^2*b^2 + B^10*b^4)) + 2*B^5*a^2*sqrt(-a/b)*log(-(6*a*b*co
s(d*x + c)*sin(d*x + c) - (a^2 - b^2)*cos(d*x + c)^2 - b^2 - 4*(a*b*cos(d*x + c)^2 - b^2*cos(d*x + c)*sin(d*x
+ c))*sqrt(-a/b)*sqrt(sin(d*x + c)/cos(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^
2 + b^2)) - sqrt(2)*(2*(B^2*a^3*b^2 + B^2*a*b^4...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} B \int \frac {\tan ^{\frac {5}{2}}{\left (c + d x \right )}}{a + b \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)**(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)

[Out]

B*Integral(tan(c + d*x)**(5/2)/(a + b*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)^(5/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 35.55, size = 2500, normalized size = 9.77 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^(5/2)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^2,x)

[Out]

(log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)
^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2) + (768*B*a^3*b^3*(a^2 + b^2))/d)*((
4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))
^(1/2))/4 + (64*B^2*a^3*tan(c + d*x)^(1/2)*(2*a^8 + 15*b^8 - 17*a^2*b^6 + 51*a^4*b^4 + 21*a^6*b^2))/(d^2*(a^2
+ b^2)^2))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a
^2 + b^2)^4))^(1/2))/4 + (32*B^3*a^4*(4*a^8 + b^8 - 77*a^2*b^6 + 47*a^4*b^4 + 33*a^6*b^2))/(d^3*(a^2 + b^2)^3)
)*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)
^4))^(1/2))/4 + (16*B^4*a^4*tan(c + d*x)^(1/2)*(a^10 - 2*b^10 - 4*a^2*b^8 - 27*a^4*b^6 + 15*a^6*b^4 + 9*a^8*b^
2))/(b*d^4*(a^2 + b^2)^4))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^
5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (8*B^5*a^7*(a^6 + 10*b^6 + 27*a^2*b^4 + 10*a^4*b^2))/(b*d^5*(a^2 + b^
2)^4))*(((192*B^4*a^6*b^6*d^4 - 16*B^4*a^4*b^8*d^4 - 16*B^4*a^12*d^4 - 608*B^4*a^8*b^4*d^4 + 192*B^4*a^10*b^2*
d^4)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(a^8*d^4 + b^8*d^4 + 4*a^2*b^6*d^4 + 6*a^4*b^4*d^4 + 4*a^6
*b^2*d^4))^(1/2))/4 + (log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(-(4*(-B^4*a^4*d^4*(a^
4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2) + (768*B*a^3
*b^3*(a^2 + b^2))/d)*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d
^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (64*B^2*a^3*tan(c + d*x)^(1/2)*(2*a^8 + 15*b^8 - 17*a^2*b^6 + 51*a^4*b^4 +
21*a^6*b^2))/(d^2*(a^2 + b^2)^2))*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 +
16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (32*B^3*a^4*(4*a^8 + b^8 - 77*a^2*b^6 + 47*a^4*b^4 + 33*a^6*
b^2))/(d^3*(a^2 + b^2)^3))*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a
^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^4*a^4*tan(c + d*x)^(1/2)*(a^10 - 2*b^10 - 4*a^2*b^8 - 27*a^4*b
^6 + 15*a^6*b^4 + 9*a^8*b^2))/(b*d^4*(a^2 + b^2)^4))*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*
B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (8*B^5*a^7*(a^6 + 10*b^6 + 27*a^2*b^4 + 10
*a^4*b^2))/(b*d^5*(a^2 + b^2)^4))*(-((192*B^4*a^6*b^6*d^4 - 16*B^4*a^4*b^8*d^4 - 16*B^4*a^12*d^4 - 608*B^4*a^8
*b^4*d^4 + 192*B^4*a^10*b^2*d^4)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(a^8*d^4 + b^8*d^4 + 4*a^2*b^6
*d^4 + 6*a^4*b^4*d^4 + 4*a^6*b^2*d^4))^(1/2))/4 - log((8*B^5*a^7*(a^6 + 10*b^6 + 27*a^2*b^4 + 10*a^4*b^2))/(b*
d^5*(a^2 + b^2)^4) - ((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-B^4*a^4*d^4*(a^4 + b^4
- 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2) - (768*B*a^3*b^3*(a
^2 + b^2))/d)*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(d^4
*(a^2 + b^2)^4))^(1/2))/4 + (64*B^2*a^3*tan(c + d*x)^(1/2)*(2*a^8 + 15*b^8 - 17*a^2*b^6 + 51*a^4*b^4 + 21*a^6*
b^2))/(d^2*(a^2 + b^2)^2))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^
5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 - (32*B^3*a^4*(4*a^8 + b^8 - 77*a^2*b^6 + 47*a^4*b^4 + 33*a^6*b^2))/(d^
3*(a^2 + b^2)^3))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/
(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^4*a^4*tan(c + d*x)^(1/2)*(a^10 - 2*b^10 - 4*a^2*b^8 - 27*a^4*b^6 + 15*a^
6*b^4 + 9*a^8*b^2))/(b*d^4*(a^2 + b^2)^4))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*b^3
*d^2 - 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4)*(((192*B^4*a^6*b^6*d^4 - 16*B^4*a^4*b^8*d^4 - 16*B^4*a
^12*d^4 - 608*B^4*a^8*b^4*d^4 + 192*B^4*a^10*b^2*d^4)^(1/2) + 16*B^2*a^3*b^3*d^2 - 16*B^2*a^5*b*d^2)/(16*a^8*d
^4 + 16*b^8*d^4 + 64*a^2*b^6*d^4 + 96*a^4*b^4*d^4 + 64*a^6*b^2*d^4))^(1/2) - log((8*B^5*a^7*(a^6 + 10*b^6 + 27
*a^2*b^4 + 10*a^4*b^2))/(b*d^5*(a^2 + b^2)^4) - ((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(
-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4
))^(1/2) - (768*B*a^3*b^3*(a^2 + b^2))/d)*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3
*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (64*B^2*a^3*tan(c + d*x)^(1/2)*(2*a^8 + 15*b^8 - 17*a
^2*b^6 + 51*a^4*b^4 + 21*a^6*b^2))/(d^2*(a^2 + b^2)^2))*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) -
16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 - (32*B^3*a^4*(4*a^8 + b^8 - 77*a^2*b^6 +
47*a^4*b^4 + 33*a^6*b^2))/(d^3*(a^2 + b^2)^3))*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a
^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^4*a^4*tan(c + d*x)^(1/2)*(a^10 - 2*b^10 -
4*a^2*b^8 - 27*a^4*b^6 + 15*a^6*b^4 + 9*a^8*b^...

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