3.4.77 \(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [377]

Optimal. Leaf size=270 \[ \frac {1155 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d} \]

[Out]

1155/8192*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+385/2048*I*cos(
d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+33/256*I*cos(d*x+c)^3/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1155/4096*I*cos(d*x
+c)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d-77/512*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/8*I*cos(d*x+c)^3/d/(
a+I*a*tan(d*x+c))^(5/2)+11/96*I*cos(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.30, antiderivative size = 270, 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 = {3583, 3578, 3571, 3570, 212} \begin {gather*} \frac {1155 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((1155*I)/4096)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + (
(I/8)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((11*I)/96)*Cos[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*
x])^(3/2)) + (((385*I)/2048)*Cos[c + d*x])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((33*I)/256)*Cos[c + d*x]^3)/
(a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (((1155*I)/4096)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d) - (((77
*I)/512)*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*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 \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{16 a}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {33 \int \frac {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{64 a^2}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {231 \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{512 a^3}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}+\frac {385 \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{1024 a^2}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}+\frac {1155 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{4096 a^3}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}+\frac {1155 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{8192 a^2}\\ &=\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}+\frac {(1155 i) \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 )}{4096 a^2 d}\\ &=\frac {1155 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 1.46, size = 165, normalized size = 0.61 \begin {gather*} \frac {i \sec ^3(c+d x) \left (-3325-3465 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 )-1605 \cos (2 (c+d x))+1800 \cos (4 (c+d x))+80 \cos (6 (c+d x))+1111 i \sin (2 (c+d x))+2552 i \sin (4 (c+d x))+176 i \sin (6 (c+d x))\right )}{24576 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/24576)*Sec[c + d*x]^3*(-3325 - 3465*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^(
(2*I)*(c + d*x))]] - 1605*Cos[2*(c + d*x)] + 1800*Cos[4*(c + d*x)] + 80*Cos[6*(c + d*x)] + (1111*I)*Sin[2*(c +
d*x)] + (2552*I)*Sin[4*(c + d*x)] + (176*I)*Sin[6*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[
c + d*x]])

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Maple [A]
time = 0.87, size = 400, normalized size = 1.48

method	result	size
default	\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (24576 i \left (\cos ^{9}\left (d x +c \right )\right )+24576 \sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )-7168 i \left (\cos ^{7}\left (d x +c \right )\right )+5120 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+704 i \left (\cos ^{5}\left (d x +c \right )\right )+6336 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+3465 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+3465 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )+3465 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+1848 i \left (\cos ^{3}\left (d x +c \right )\right )+9240 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-13860 i \cos \left (d x +c \right )\right )}{49152 d \,a^{3}}\)	\(400\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/49152/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(24576*I*cos(d*x+c)^9+24576*sin(d*x+c)*cos(d*x+c)^8-7
168*I*cos(d*x+c)^7+5120*sin(d*x+c)*cos(d*x+c)^6+704*I*cos(d*x+c)^5+6336*sin(d*x+c)*cos(d*x+c)^4+3465*I*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)+3465*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-
I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)+3465*I*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*2^(1/2)+1848*I*cos(d*x+c)^3+9240*cos(d*x+c)^2*sin(d*x+c)-13860*I*cos(d*x+c))/a^3

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3789 vs. \(2 (207) = 414\).
time = 0.74, size = 3789, normalized size = 14.03 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/98304*(4*(cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*
d*x + 8*c)))^2 + 2*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)^(3/4)*(15*((-I*sqrt(2)*cos(8*d*x
+ 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + (-I*sqrt(2)*cos(8*
d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(-I*sqrt(2)*
cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - I*sqrt(2)*
cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(7/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c
))), cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + (55*I*sqrt(2)*cos(8*d*x + 8*c) + 960*I*sqrt(
2)*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - 1296*I*sqrt(2)*cos(1/2*arctan2(sin(8*d*x + 8*c), cos
(8*d*x + 8*c))) + 55*sqrt(2)*sin(8*d*x + 8*c) + 960*sqrt(2)*sin(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)
)) - 1296*sqrt(2)*sin(1/2*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 128*I*sqrt(2))*cos(3/2*arctan2(sin(1/
4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + 1
5*((sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))
)^2 + (sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*
c)))^2 + 2*(sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x
+ 8*c))) + sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(7/2*arctan2(sin(1/4*arctan2(sin(8*d*x +
8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) - (55*sqrt(2)*cos(8*d*x
+ 8*c) + 960*sqrt(2)*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - 1296*sqrt(2)*cos(1/2*arctan2(sin(8
*d*x + 8*c), cos(8*d*x + 8*c))) - 55*I*sqrt(2)*sin(8*d*x + 8*c) - 960*I*sqrt(2)*sin(3/4*arctan2(sin(8*d*x + 8*
c), cos(8*d*x + 8*c))) + 1296*I*sqrt(2)*sin(1/2*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 128*sqrt(2))*si
n(3/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*
x + 8*c))) + 1)))*sqrt(a) + 4*(cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + sin(1/4*arctan2(sin(8*
d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)^(1/4)*((73*(-I*
sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 +
73*(-I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)
))^2 + 792*(-I*sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 - I*sqrt(2)*sin(1/4*arctan2(sin(
8*d*x + 8*c), cos(8*d*x + 8*c)))^2 - 2*I*sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - I*sqrt
(2))*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 146*(-I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d
*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - 792*(sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c
), cos(8*d*x + 8*c)))^2 + sqrt(2)*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*sqrt(2)*cos(1/4*a
rctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + sqrt(2))*sin(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) -
73*I*sqrt(2)*cos(8*d*x + 8*c) - 73*sqrt(2)*sin(8*d*x + 8*c))*cos(5/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c),
cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + 3*(-5*I*sqrt(2)*cos(8*d*x +
8*c) - 120*I*sqrt(2)*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 336*I*sqrt(2)*cos(1/2*arctan2(sin(
8*d*x + 8*c), cos(8*d*x + 8*c))) - 64*I*sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - 5*sqrt(
2)*sin(8*d*x + 8*c) - 120*sqrt(2)*sin(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 336*sqrt(2)*sin(1/2*a
rctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - 64*sqrt(2)*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))
+ 640*I*sqrt(2))*cos(1/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d
*x + 8*c), cos(8*d*x + 8*c))) + 1)) + (73*(sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arct
an2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 73*(sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(1/
4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 792*(sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x +
8*c)))^2 + sqrt(2)*sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(8*d*
x + 8*c), cos(8*d*x + 8*c))) + sqrt(2))*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 146*(sqrt(2)*co
s(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 792*(-I*sq
rt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 - I*sqrt(2)*sin(1/4*arctan2(sin(8*d*x + 8*c), cos
(8*d*x + 8*c)))^2 - 2*I*sqrt(2)*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - I*sqrt(2))*sin(3/4*arct
an2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 73*s...

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Fricas [A]
time = 0.39, size = 311, normalized size = 1.15 \begin {gather*} \frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-128 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 2176 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 247 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3325 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1358 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 376 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{24576 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24576*(-3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log(-1155/2048*(sqrt(2)*sqrt(1/2)*(I*a^
2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(
a^2*d)) + 3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log(-1155/2048*(sqrt(2)*sqrt(1/2)*(-I*a
^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/
(a^2*d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-128*I*e^(12*I*d*x + 12*I*c) - 2176*I*e^(10*I*d*x + 10*I
*c) + 247*I*e^(8*I*d*x + 8*I*c) + 3325*I*e^(6*I*d*x + 6*I*c) + 1358*I*e^(4*I*d*x + 4*I*c) + 376*I*e^(2*I*d*x +
2*I*c) + 48*I))*e^(-8*I*d*x - 8*I*c)/(a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(cos(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), x)

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

[Out]

integrate(cos(d*x + c)^3/(I*a*tan(d*x + c) + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(5/2), x)

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