3.4.31 \(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [331]

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

[Out]

11/128*I*a^(7/2)*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d*2^(1/2)+11/96*I*a^4*cos(d*
x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-11/64*I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-11/120*I*a^3*cos(d*x+c)^3*(a
+I*a*tan(d*x+c))^(1/2)/d-11/140*I*a^2*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2)/d-11/126*I*a*cos(d*x+c)^7*(a+I*a*t
an(d*x+c))^(5/2)/d-1/9*I*cos(d*x+c)^9*(a+I*a*tan(d*x+c))^(7/2)/d

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Rubi [A]
time = 0.33, antiderivative size = 268, normalized size of antiderivative = 1.00, 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 = {3578, 3583, 3571, 3570, 212} \begin {gather*} \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 ^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^2 \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{140 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} \end {gather*}

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 212

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

Rule 3570

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 3571

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 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 3583

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]

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 ) \text {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*}

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Mathematica [A]
time = 3.50, size = 188, normalized size = 0.70 \begin {gather*} -\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 (-315+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 )\right )}{20160 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/20160*I)*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)) + 7
034*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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1603 vs. \(2 (219 ) = 438\).
time = 2.09, size = 1604, normalized size = 5.99

method	result	size
default	\(\text {Expression too large to display}\)	\(1604\)

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,method=_RETURNVERBOSE)

[Out]

-1/10321920/d*(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)^9-9175040*cos(d*x+c)^17*sin(d*x+c)+4587520*cos(d*x+c)^16*sin(d*x+c)+983040*cos(d*x+c)
^15*sin(d*x+c)-3465*cos(d*x+c)^8*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-27720*cos(d*x+c)^7*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arcta
n(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-97020*cos(d*x+c)^6*sin(d*x+c)*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-194040*cos(d*x+c)^5*sin(d*x
+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-242
550*cos(d*x+c)^4*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*2^(1/2))*2^(1/2)-194040*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3465*I*cos(d*x+c)^8*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17
/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+27720*I*cos(d*x+c)
^7*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)/cos(d*x+c)*2^(1/2))*2^(1/2)+97020*I*cos(d*x+c)^6*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/
2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+194040*I*cos(d*x+c)^5*sin(d*x+c)
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*
2^(1/2))*2^(1/2)+242550*I*cos(d*x+c)^4*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+194040*I*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(
1/2)+97020*I*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+27720*I*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(17/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+655360*sin(
d*x+c)*cos(d*x+c)^14-1774080*cos(d*x+c)^9*sin(d*x+c)+811008*sin(d*x+c)*cos(d*x+c)^12-720896*cos(d*x+c)^13*sin(
d*x+c)-97020*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*2^(1/2))*2^(1/2)-27720*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*arctan(1/2*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3465*I*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*sin(d*x+c)-946176*sin(d*x+c)*cos(d*x
+c)^11+1182720*cos(d*x+c)^10*sin(d*x+c)-3465*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(17/2)*2^(1/2)*arctan(1/2*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c))*(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*a^3

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.57, size = 314, normalized size = 1.17 \begin {gather*} -\frac {{\left (3465 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {11 \, {\left (-i \, a^{4} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\right ) - 3465 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {11 \, {\left (-i \, a^{4} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{40320 \, d} \end {gather*}

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^(2*I*d*x + 2*I*c)*log(-11/32*(-I*a^4 + sqrt(2)*sqrt(1/2)*sqrt(-a^7
/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 3465*sqrt(1/2)*sqrt
(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c)*log(-11/32*(-I*a^4 - sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c)
+ d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 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) - 1798*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^(-2*I*d*x -
2*I*c)/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (c+d\,x\right )}^9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(7/2), x)

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