3.148 \(\int \frac{\tan ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=307 \[ \frac{(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A-(5+7 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(2 A-(5+7 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(-B+i A) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]

[Out]

((2*A + (5 - 7*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - ((2*A + (5 - 7*I)*B)*ArcTan[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - ((2*A - (5 + 7*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]
+ Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + ((2*A - (5 + 7*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])
/(32*Sqrt[2]*a^3*d) + ((I*A - B)*Tan[c + d*x]^(5/2))/(6*d*(a + I*a*Tan[c + d*x])^3) + ((A + (4*I)*B)*Tan[c + d
*x]^(3/2))/(12*a*d*(a + I*a*Tan[c + d*x])^2) + (5*B*Sqrt[Tan[c + d*x]])/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.638102, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A-(5+7 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{(2 A-(5+7 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^3 d}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(-B+i A) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((2*A + (5 - 7*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - ((2*A + (5 - 7*I)*B)*ArcTan[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) - ((2*A - (5 + 7*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]
+ Tan[c + d*x]])/(32*Sqrt[2]*a^3*d) + ((2*A - (5 + 7*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])
/(32*Sqrt[2]*a^3*d) + ((I*A - B)*Tan[c + d*x]^(5/2))/(6*d*(a + I*a*Tan[c + d*x])^3) + ((A + (4*I)*B)*Tan[c + d
*x]^(3/2))/(12*a*d*(a + I*a*Tan[c + d*x])^2) + (5*B*Sqrt[Tan[c + d*x]])/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 3595

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

Rule 3534

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 1168

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} a (i A-B)-\frac{1}{2} a (A-11 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-3 a^2 (A+4 i B)-3 a^2 (i A+6 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \frac{15 a^3 B+3 a^3 (2 A-7 i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{15 a^3 B+3 a^3 (2 A-7 i B) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{24 a^6 d}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(2 A-(5+7 i) B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}-\frac{(2 A+(5-7 i) B) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{16 a^3 d}\\ &=\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{(2 A-(5+7 i) B) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 \sqrt{2} a^3 d}-\frac{(2 A-(5+7 i) B) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 \sqrt{2} a^3 d}-\frac{(2 A+(5-7 i) B) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}-\frac{(2 A+(5-7 i) B) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{32 a^3 d}\\ &=-\frac{(2 A-(5+7 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(2 A-(5+7 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{(2 A+(5-7 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}+\frac{(2 A+(5-7 i) B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}\\ &=\frac{(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A+(5-7 i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{16 \sqrt{2} a^3 d}-\frac{(2 A-(5+7 i) B) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(2 A-(5+7 i) B) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt{2} a^3 d}+\frac{(i A-B) \tan ^{\frac{5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(A+4 i B) \tan ^{\frac{3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac{5 B \sqrt{\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.59742, size = 254, normalized size = 0.83 \[ \frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac{4}{3} \sin (c+d x) (\cos (3 d x)-i \sin (3 d x)) ((A+19 i B) \sin (2 (c+d x))+3 (7 B-i A) \cos (2 (c+d x))+3 i A-6 B)+(\cos (3 c)+i \sin (3 c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left ((2 A+(5-7 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+(1-i) ((1+i) A+(1-6 i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^3*(((2*A + (5 - 7*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] + (1 - I)*
((1 + I)*A + (1 - 6*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(Cos[3*c] +
I*Sin[3*c])*Sqrt[Sin[2*(c + d*x)]] + (4*(Cos[3*d*x] - I*Sin[3*d*x])*Sin[c + d*x]*((3*I)*A - 6*B + 3*((-I)*A +
7*B)*Cos[2*(c + d*x)] + (A + (19*I)*B)*Sin[2*(c + d*x)]))/3)*(A + B*Tan[c + d*x]))/(32*d*(A*Cos[c + d*x] + B*S
in[c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3)

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Maple [A]  time = 0.057, size = 323, normalized size = 1.1 \begin{align*}{\frac{-{\frac{9\,i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{A}{4\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{19\,B}{12\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{i}{12}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{5\,i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{A}{4\,{a}^{3}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }+{\frac{{\frac{3\,i}{2}}B}{{a}^{3}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-{\frac{A}{4\,{a}^{3}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }+{\frac{{\frac{i}{4}}B}{{a}^{3}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

-9/8*I/d/a^3/(tan(d*x+c)-I)^3*B*tan(d*x+c)^(5/2)-1/4/d/a^3/(tan(d*x+c)-I)^3*A*tan(d*x+c)^(5/2)-19/12/d/a^3/(ta
n(d*x+c)-I)^3*B*tan(d*x+c)^(3/2)+1/12*I/d/a^3/(tan(d*x+c)-I)^3*tan(d*x+c)^(3/2)*A+5/8*I/d/a^3/(tan(d*x+c)-I)^3
*B*tan(d*x+c)^(1/2)-1/4/d/a^3/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))*A+3/2*I/d/a^3
/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))*B-1/4/d/a^3/(2^(1/2)+I*2^(1/2))*arctan(2*t
an(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A+1/4*I/d/a^3/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^
(1/2)))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.69918, size = 1778, normalized size = 5.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(2*((a^3*d*e^(2*I*d*x + 2*I*c) +
a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) +
(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2
))*e^(6*I*d*x + 6*I*c)*log(-2*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d
*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c
)/(I*A + B)) + 3*a^3*d*sqrt((-I*A^2 - 12*A*B + 36*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((a^3*d*e^(2*I
*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 12*A*B + 36
*I*B^2)/(a^6*d^2)) + A - 6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*a^3*d*sqrt((-I*A^2 - 12*A*B + 36*I*B^2)/(a^6
*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^
(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 12*A*B + 36*I*B^2)/(a^6*d^2)) - A + 6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)
) + 2*((-2*I*A + 20*B)*e^(6*I*d*x + 6*I*c) + (I*A + 14*B)*e^(4*I*d*x + 4*I*c) + (2*I*A - 5*B)*e^(2*I*d*x + 2*I
*c) - I*A + B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21076, size = 182, normalized size = 0.59 \begin{align*} -\frac{\left (i + 1\right ) \, \sqrt{2}{\left (A - 6 i \, B\right )} \arctan \left (\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac{\left (i - 1\right ) \, \sqrt{2}{\left (A - i \, B\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac{6 \, A \tan \left (d x + c\right )^{\frac{5}{2}} + 27 i \, B \tan \left (d x + c\right )^{\frac{5}{2}} - 2 i \, A \tan \left (d x + c\right )^{\frac{3}{2}} + 38 \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 15 i \, B \sqrt{\tan \left (d x + c\right )}}{24 \, a^{3} d{\left (\tan \left (d x + c\right ) - i\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-(1/16*I + 1/16)*sqrt(2)*(A - 6*I*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + (1/16*I - 1/16
)*sqrt(2)*(A - I*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) - 1/24*(6*A*tan(d*x + c)^(5/2) +
27*I*B*tan(d*x + c)^(5/2) - 2*I*A*tan(d*x + c)^(3/2) + 38*B*tan(d*x + c)^(3/2) - 15*I*B*sqrt(tan(d*x + c)))/(
a^3*d*(tan(d*x + c) - I)^3)