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

Optimal. Leaf size=206 \[ -\frac{(-1)^{3/4} a^{5/2} (23 B+20 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \]

[Out]

-((-1)^(3/4)*a^(5/2)*((20*I)*A + 23*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(4*d) + ((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d - (a^2*(4*A - (7*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/2)*a*B*Sqrt[Tan[c +
d*x]]*(a + I*a*Tan[c + d*x])^(3/2))/d

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Rubi [A]  time = 0.711822, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3594, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{(-1)^{3/4} a^{5/2} (23 B+20 i A) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(3/4)*a^(5/2)*((20*I)*A + 23*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x
]]])/(4*d) + ((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*
x]]])/d - (a^2*(4*A - (7*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/2)*a*B*Sqrt[Tan[c +
d*x]]*(a + I*a*Tan[c + d*x])^(3/2))/d

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, 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} (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (4 A-i B)+\frac{1}{2} a (4 i A+7 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^2 (4 A-3 i B)+\frac{1}{4} a^2 (20 i A+23 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\left (4 a^2 (A-i B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx-\frac{1}{8} (a (20 A-23 i B)) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac{\left (a^3 (20 A-23 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{\left (8 a^4 (i A+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{(4+4 i) a^{5/2} (i A+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{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac{\left (a^3 (20 A-23 i B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=-\frac{(4+4 i) a^{5/2} (i A+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{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac{\left (a^3 (20 A-23 i 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 )}{4 d}\\ &=\frac{\sqrt [4]{-1} a^{5/2} (20 A-23 i B) \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 d}-\frac{(4+4 i) a^{5/2} (i A+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{a^2 (4 A-7 i B) \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}\\ \end{align*}

Mathematica [B]  time = 9.17287, size = 499, normalized size = 2.42 \[ \frac{\cos ^3(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left ((4 A-11 i B) \left (-\frac{1}{4} \cos (2 c)+\frac{1}{4} i \sin (2 c)\right )+\sec (c+d x) \left (-\frac{1}{2} B \sin (3 c+d x)-\frac{1}{2} i B \cos (3 c+d x)\right )\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (128 (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )-\sqrt{2} (20 A-23 i B) \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )\right ) (A+B \tan (c+d x))}{16 \sqrt{2} d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(128*(A - I*B)*Log[E^(I*(c
+ d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] - Sqrt[2]*(20*A - (23*I)*B)*(Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[
2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))
*Sqrt[-1 + E^((2*I)*(c + d*x))]]))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(16*Sqrt[2]*d*E^((2*I)*c
)*Sqrt[-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*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*((4*A - (11*I)*B)*(-Cos[2*c]/4 + (I/4
)*Sin[2*c]) + Sec[c + d*x]*((-I/2)*B*Cos[3*c + d*x] - (B*Sin[3*c + d*x])/2))*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c
+ d*x])^(5/2)*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.058, size = 564, normalized size = 2.7 \begin{align*}{\frac{{a}^{2}}{8\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{\tan \left ( dx+c \right ) } \left ( -9\,iB\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}a+18\,iB\sqrt{ia}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-4\,B\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 ) +8\,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 ) a+12\,A\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}a-8\,A\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia}+16\,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 ) a\sqrt{-ia}-8\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{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 ) }{\tan \left ( dx+c \right ) +i}} \right ) a+16\,\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 ) a\sqrt{-ia} \right ){\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)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x)

[Out]

1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^(1/2)*a^2*(-9*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a+18*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)-4*B*(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)+8*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*tan(d*x+c))/(tan(d*x+c)+I
))*a+12*A*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)*a-8*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)+16*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))*a*(-I*a)^(1/2)-8*(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))*a+16*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))*a*(-I*a)^(1/2))/(I*a)^(1/2)/(-I
*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.01015, size = 2363, normalized size = 11.47 \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)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/8*(2*sqrt(2)*((4*A - 11*I*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*A - 7*I*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) + sqrt((-400*I*A^2 - 920*A*B + 5
29*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((20*I*A + 23*B)*a^2*e^(2*I*d*x + 2*I*c) + (20*I*A
+ 23*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) + 2*I*sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I
*c)/((20*I*A + 23*B)*a^2)) - sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(
(sqrt(2)*((20*I*A + 23*B)*a^2*e^(2*I*d*x + 2*I*c) + (20*I*A + 23*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr
t((-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) - 2*I*sqrt((-400*I*A^2 - 920*A*B + 5
29*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((20*I*A + 23*B)*a^2)) - 4*sqrt((-32*I*A^2 - 64
*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4
*I*A + 4*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((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c
)/((4*I*A + 4*B)*a^2)) + 4*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt
(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*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((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/
d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^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)*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.6426, size = 212, normalized size = 1.03 \begin{align*} \frac{-\left (i - 1\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 )}^{3} a^{2} +{\left (-\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 6 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 4 i \, a^{3}\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)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

(-(I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^3*a^2 + (-(2*I + 2)*(I*a*tan(d*x +
c) + a)^3*a + (2*I + 2)*(I*a*tan(d*x + c) + a)^2*a^2)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d
*x + c) + a)*B)/((2*I*(I*a*tan(d*x + c) + a)^2*a - 6*I*(I*a*tan(d*x + c) + a)*a^2 + 4*I*a^3)*d)