3.331 \(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=268 \[ -\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}+\frac{11 i a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{64 \sqrt{2} d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d} \]

[Out]

(((11*I)/64)*a^(7/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((11
*I)/96)*a^4*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((11*I)/64)*a^3*Cos[c + d*x]*Sqrt[a + I*a*Tan[c +
d*x]])/d - (((11*I)/120)*a^3*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (((11*I)/140)*a^2*Cos[c + d*x]^5*(
a + I*a*Tan[c + d*x])^(3/2))/d - (((11*I)/126)*a*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2))/d - ((I/9)*Cos[c
+ d*x]^9*(a + I*a*Tan[c + d*x])^(7/2))/d

________________________________________________________________________________________

Rubi [A]  time = 0.411437, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3497, 3502, 3490, 3489, 206} \[ -\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 d}+\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}+\frac{11 i a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{64 \sqrt{2} d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((11*I)/64)*a^(7/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((11
*I)/96)*a^4*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((11*I)/64)*a^3*Cos[c + d*x]*Sqrt[a + I*a*Tan[c +
d*x]])/d - (((11*I)/120)*a^3*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (((11*I)/140)*a^2*Cos[c + d*x]^5*(
a + I*a*Tan[c + d*x])^(3/2))/d - (((11*I)/126)*a*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2))/d - ((I/9)*Cos[c
+ d*x]^9*(a + I*a*Tan[c + d*x])^(7/2))/d

Rule 3497

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

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3490

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

Rule 3489

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

Rule 206

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

Rubi steps

\begin{align*} \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{18} (11 a) \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{28} \left (11 a^2\right ) \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{40} \left (11 a^3\right ) \int \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{48} \left (11 a^4\right ) \int \frac{\cos (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{64} \left (11 a^3\right ) \int \cos (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 d}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{1}{128} \left (11 a^4\right ) \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 d}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}+\frac{\left (11 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{64 d}\\ &=\frac{11 i a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{64 \sqrt{2} d}+\frac{11 i a^4 \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 d}-\frac{11 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{120 d}-\frac{11 i a^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 d}-\frac{11 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{126 d}-\frac{i \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2}}{9 d}\\ \end{align*}

Mathematica [A]  time = 3.20505, size = 188, normalized size = 0.7 \[ -\frac{i a^3 e^{-3 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (4303 e^{2 i (c+d x)}+7034 e^{4 i (c+d x)}+3754 e^{6 i (c+d x)}+1798 e^{8 i (c+d x)}+530 e^{10 i (c+d x)}+70 e^{12 i (c+d x)}-3465 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-315\right )}{20160 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-I/20160)*a^3*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-315 + 4303*E^((2*I)*(c + d*x)) + 703
4*E^((4*I)*(c + d*x)) + 3754*E^((6*I)*(c + d*x)) + 1798*E^((8*I)*(c + d*x)) + 530*E^((10*I)*(c + d*x)) + 70*E^
((12*I)*(c + d*x)) - 3465*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x
))]]))/(Sqrt[2]*d*E^((3*I)*(c + d*x)))

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Maple [B]  time = 0.615, size = 1604, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

-1/10321920/d*a^3*(97020*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^6*2^(1/2)+194040*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17
/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^5*2^
(1/2)+242550*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+194040*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*arctanh(
1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+97020*
I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/co
s(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+27720*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*arctanh(1/2*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)*2^(1/2)+3465*I*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(17/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x
+c)*cos(d*x+c)^8*2^(1/2)+983040*sin(d*x+c)*cos(d*x+c)^15-3465*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)+655360*sin(d*x+c)*cos(d*x+c)^14-720896*sin(d
*x+c)*cos(d*x+c)^13-9175040*sin(d*x+c)*cos(d*x+c)^17+811008*sin(d*x+c)*cos(d*x+c)^12-1774080*sin(d*x+c)*cos(d*
x+c)^9-946176*sin(d*x+c)*cos(d*x+c)^11+1182720*sin(d*x+c)*cos(d*x+c)^10+27720*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(17/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^
7*2^(1/2)+3465*I*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)-3465*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^8*2^(1/2)-27720*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^7*2^(1/2)-97020*arctan(1/2*2^(1
/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^6*2^(1/2
)-194040*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*
x+c)*cos(d*x+c)^5*2^(1/2)-242550*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)-194040*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-97020*arctan(1/2*2^(1/2)*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)-27720*arctan(
1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(17/2)*sin(d*x+c)*cos(d*x+c)*
2^(1/2)+4587520*sin(d*x+c)*cos(d*x+c)^16+9175040*I*cos(d*x+c)^18-4587520*I*cos(d*x+c)^17-5570560*I*cos(d*x+c)^
16+1638400*I*cos(d*x+c)^15+65536*I*cos(d*x+c)^14+90112*I*cos(d*x+c)^13+135168*I*cos(d*x+c)^12+236544*I*cos(d*x
+c)^11+591360*I*cos(d*x+c)^10-1774080*I*cos(d*x+c)^9)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*
x+c)+cos(d*x+c)-1)/cos(d*x+c)^8

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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(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.25117, size = 1034, normalized size = 3.86 \begin{align*} -\frac{{\left (3465 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{{\left (22 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 11 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{11 \, a^{3}}\right ) - 3465 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{{\left (-22 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 11 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{11 \, a^{3}}\right ) - \sqrt{2}{\left (-70 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 530 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 1798 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 3754 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 7034 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4303 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 315 i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{40320 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/40320*(3465*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/11*(22*I*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(I*d
*x + I*c) + 11*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)/a^3) - 3465*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/11*(-22*I*sqrt(1/2)*sqrt(-a^7/d^
2)*d*e^(I*d*x + I*c) + 11*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)/a^3) - sqrt(2)*(-70*I*a^3*e^(12*I*d*x + 12*I*c) - 530*I*a^3*e^(10*I*d*x + 10*I*c) - 17
98*I*a^3*e^(8*I*d*x + 8*I*c) - 3754*I*a^3*e^(6*I*d*x + 6*I*c) - 7034*I*a^3*e^(4*I*d*x + 4*I*c) - 4303*I*a^3*e^
(2*I*d*x + 2*I*c) + 315*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-3*I*d*x - 3*I*c)/d

________________________________________________________________________________________

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(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [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(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out