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

Optimal. Leaf size=190 \[ \frac{(2+2 i) a^{3/2} B (2 a+3 i b) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{2 (-1)^{3/4} a^{5/2} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a B (a+3 i b) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.72894, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3593, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{(2+2 i) a^{3/2} B (2 a+3 i b) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{2 (-1)^{3/4} a^{5/2} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a B (a+3 i b) \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1]

Rule 3601

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

Rule 3544

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

Rule 205

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

Rule 3599

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

Rule 63

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^{5/2} \left (\frac{3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{3}{2} (a+3 i b) B+\frac{3}{2} i a B \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4}{3} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{9}{4} a (i a-2 b) B-\frac{3}{4} a^2 B \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-(i a B) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx+(2 a (2 i a-3 b) B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{\left (i a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (4 a^3 (2 a+3 i b) B\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{(2+2 i) a^{3/2} (2 a+3 i b) B \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{\left (2 i a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{(2+2 i) a^{3/2} (2 a+3 i b) B \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{\left (2 i a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac{2 (-1)^{3/4} a^{5/2} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{(2+2 i) a^{3/2} (2 a+3 i b) B \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (a+3 i b) B \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [B]  time = 10.2212, size = 485, normalized size = 2.55 \[ \frac{\cos ^3(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} \left (\frac{3 b B}{2 a}+B \tan (c+d x)\right ) \left (\csc (c) (2 \cos (2 c)-2 i \sin (2 c)) \csc (c+d x) (2 a \sin (d x)+7 i b \sin (d x))-i \csc (c) (2 \cos (2 c)-2 i \sin (2 c)) (-2 i a \cos (c)+i b \sin (c)+7 b \cos (c))+(-2 b \cos (2 c)+2 i b \sin (2 c)) \csc ^2(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (2 a \sin (c+d x)+3 b \cos (c+d x))}-\frac{2 \sqrt{2} e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left ((6 b-4 i a) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+i \sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \left (\frac{3 b B}{2 a}+B \tan (c+d x)\right )}{d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (2 a \sin (c+d x)+3 b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(((-4*I)*a + 6*b
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + I*Sqrt[2]*a*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt
[-1 + E^((2*I)*(c + d*x))]])*(a + I*a*Tan[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/(d*E^((2*I)*c)*Sqr
t[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*S
in[d*x])^(5/2)*(3*b*Cos[c + d*x] + 2*a*Sin[c + d*x])) + (Cos[c + d*x]^3*((-I)*Csc[c]*((-2*I)*a*Cos[c] + 7*b*Co
s[c] + I*b*Sin[c])*(2*Cos[2*c] - (2*I)*Sin[2*c]) + Csc[c + d*x]^2*(-2*b*Cos[2*c] + (2*I)*b*Sin[2*c]) + Csc[c]*
Csc[c + d*x]*(2*Cos[2*c] - (2*I)*Sin[2*c])*(2*a*Sin[d*x] + (7*I)*b*Sin[d*x]))*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[
c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(3*b*Cos[c + d*x] + 2*a*Sin[c +
d*x]))

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Maple [B]  time = 0.076, size = 551, normalized size = 2.9 \begin{align*}{\frac{aB}{2\,d} \left ( -i\sqrt{ia}\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-2\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}{a}^{2}+2\,i\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-14\,i\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}b+\sqrt{ia}\sqrt{2}\ln \left ( -{\frac{1}{\tan \left ( dx+c \right ) +i} \left ( -2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }+ia-3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-2\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}a-4\,\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\tan \left ( dx+c \right ) a-2\,b\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia} \right ) \sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*B*a*(-I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*t
an(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-2*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a^2+2*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a-14*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*b+(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1
+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-2*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^2*a-4*(I*a)^(1/2)*(-I*a)^(1/
2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)*a-2*b*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-
I*a)^(1/2))*(a*(1+I*tan(d*x+c)))^(1/2)/tan(d*x+c)^(3/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*
a)^(1/2)

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

[Out]

Timed out

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Fricas [B]  time = 1.96486, size = 2410, normalized size = 12.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*(4*B*a*b*e^(2*I*d*x + 2*I*c) + 4*I*B*a^2 - 12*B*a*b + (-4*I*B*a^2 + 16*B*a*b)*e^(4*I*d*x + 4*I*c)
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*
c) - (d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((32*I*B^2*a^5 - 96*B^2*a^4*b - 72*I*B^2*a^3*b^
2)/d^2)*log((sqrt(2)*(-4*I*B*a^2 + 6*B*a*b + (-4*I*B*a^2 + 6*B*a*b)*e^(2*I*d*x + 2*I*c))*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + d*sqrt((32*I*B^2*a
^5 - 96*B^2*a^4*b - 72*I*B^2*a^3*b^2)/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(-4*I*B*a^2 + 6*B*a*b)) +
(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((32*I*B^2*a^5 - 96*B^2*a^4*b - 72*I*B^2*a^3*b^2)/d
^2)*log((sqrt(2)*(-4*I*B*a^2 + 6*B*a*b + (-4*I*B*a^2 + 6*B*a*b)*e^(2*I*d*x + 2*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - d*sqrt((32*I*B^2*a^5 -
96*B^2*a^4*b - 72*I*B^2*a^3*b^2)/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(-4*I*B*a^2 + 6*B*a*b)) + sqr
t(4*I*B^2*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*
c) + B*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(
I*d*x + I*c) + I*sqrt(4*I*B^2*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(B*a^2)) - sqrt(4*I*B^2*a^5
/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*c) + B*a^2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)
- I*sqrt(4*I*B^2*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(B*a^2)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e
^(2*I*d*x + 2*I*c) + 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((a+I*a*tan(d*x+c))**(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.63756, size = 258, normalized size = 1.36 \begin{align*} -\frac{-3 i \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} b - 2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} -{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}}{2 \,{\left (\left (i + 1\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} - \left (5 i + 5\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a + \left (9 i + 9\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - \left (7 i + 7\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + \left (2 i + 2\right ) \, a^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-3*I*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^3*a^2*b - 2*((I*a*tan(d*x + c) + a
)^3*a^2 - (I*a*tan(d*x + c) + a)^2*a^3)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a))/(((I
+ 1)*(I*a*tan(d*x + c) + a)^4 - (5*I + 5)*(I*a*tan(d*x + c) + a)^3*a + (9*I + 9)*(I*a*tan(d*x + c) + a)^2*a^2
- (7*I + 7)*(I*a*tan(d*x + c) + a)*a^3 + (2*I + 2)*a^4)*d)