3.226 \(\int \frac{\tan ^{\frac{7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=218 \[ \frac{2 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((1/8
- I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) - Tan[c + d*x]^(5
/2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/2)*Tan[c + d*x]^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (7*Sq
rt[Tan[c + d*x]])/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.714238, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3558, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{2 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(7/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + ((1/8
- I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) - Tan[c + d*x]^(5
/2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/2)*Tan[c + d*x]^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (7*Sq
rt[Tan[c + d*x]])/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

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 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{\tan ^{\frac{7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\tan ^{\frac{3}{2}}(c+d x) \left (-\frac{5 a}{2}+5 i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\tan (c+d x)} \left (-\frac{45 i a^2}{4}-15 a^2 \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{105 a^3}{8}-15 i a^3 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a^4}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac{i \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 )}{4 a d}\\ &=\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{a^2 d}\\ &=\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 \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 )}{a^2 d}\\ &=\frac{2 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{i \tan ^{\frac{3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 \sqrt{\tan (c+d x)}}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.58197, size = 241, normalized size = 1.11 \[ \frac{i e^{-6 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-8 e^{2 i (c+d x)}+48 e^{4 i (c+d x)}-41 e^{6 i (c+d x)}-5 e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+40 \sqrt{2} e^{5 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+1\right )}{20 \sqrt{2} a^3 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(7/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((I/20)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 - 8*E^((2*I)*(c + d*x)) + 48*E^((4*I)*(c +
d*x)) - 41*E^((6*I)*(c + d*x)) - 5*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/
Sqrt[-1 + E^((2*I)*(c + d*x))]] + 40*Sqrt[2]*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[
2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.043, size = 958, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/80/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(-5*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-320*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)*a-80*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)^4*a+460*I*(
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)-20*(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)^3*a-516*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)^2+30*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))*tan(d*x+c)^2*a-196*I*(
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)^3+480*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*t
an(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+320*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)^3*a+20*
(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)*a-5*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))*tan(d*x+c)^4*a-80*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)+140*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*
(I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)

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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)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 4.23925, size = 1883, normalized size = 8.64 \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)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/40*(10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(2*I*d*x +
2*I*c) + sqrt(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^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(6*I*d*x +
6*I*c)*log(-1/4*(4*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(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^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(
-I*d*x - I*c)) - 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(52/605*(4*sqrt(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^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x +
I*c) + (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + 10*a^3*d*sqrt
(-4*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(52/605*(4*sqrt(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^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c) - (3*a^3*d*e^(2*I*d*x +
2*I*c) - a^3*d)*sqrt(-4*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(41*e^(6*I*d*x + 6*I*c) + 34*e^(4*I*d*x + 4*I*c) -
6*e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*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)**(7/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError