3.4.93 \(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [393]

Optimal. Leaf size=307 \[ \frac {3003 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16384 \sqrt {2} a^{7/2} d}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d} \]

[Out]

3003/32768*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(7/2)/d*2^(1/2)+1001/8192*I*co
s(d*x+c)/a^3/d/(a+I*a*tan(d*x+c))^(1/2)+429/5120*I*cos(d*x+c)^3/a^3/d/(a+I*a*tan(d*x+c))^(1/2)-3003/16384*I*co
s(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^4/d-1001/10240*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/a^4/d+1/10*I*cos(d*
x+c)^3/d/(a+I*a*tan(d*x+c))^(7/2)+13/160*I*cos(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(5/2)+143/1920*I*cos(d*x+c)^3/a
^2/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.36, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 9, 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 {3003 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16384 \sqrt {2} a^{7/2} d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((3003*I)/16384)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(7/2)*d) +
((I/10)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((13*I)/160)*Cos[c + d*x]^3)/(a*d*(a + I*a*Tan[c +
d*x])^(5/2)) + (((143*I)/1920)*Cos[c + d*x]^3)/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((1001*I)/8192)*Cos[c
+ d*x])/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((429*I)/5120)*Cos[c + d*x]^3)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]]
) - (((3003*I)/16384)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) - (((1001*I)/10240)*Cos[c + d*x]^3*Sqrt
[a + I*a*Tan[c + d*x]])/(a^4*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))^{7/2}} \, dx &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx}{20 a}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{320 a^2}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 \int \frac {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{1280 a^3}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {3003 \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{10240 a^4}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}+\frac {1001 \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4096 a^3}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}+\frac {3003 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{16384 a^4}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}+\frac {3003 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{32768 a^3}\\ &=\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}+\frac {(3003 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 )}{16384 a^3 d}\\ &=\frac {3003 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16384 \sqrt {2} a^{7/2} d}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d}\\ \end {align*}

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Mathematica [A]
time = 2.83, size = 175, normalized size = 0.57 \begin {gather*} -\frac {\left (42140+20048 e^{-2 i (c+d x)}+71190 e^{2 i (c+d x)}+5856 e^{-4 i (c+d x)}-48640 e^{4 i (c+d x)}+768 e^{-6 i (c+d x)}-2560 e^{6 i (c+d x)}+\frac {90090 e^{4 i (c+d x)} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) \sec ^3(c+d x)}{491520 a^3 d (-i+\tan (c+d x))^3 \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])^(7/2),x]

[Out]

-1/491520*((42140 + 20048/E^((2*I)*(c + d*x)) + 71190*E^((2*I)*(c + d*x)) + 5856/E^((4*I)*(c + d*x)) - 48640*E
^((4*I)*(c + d*x)) + 768/E^((6*I)*(c + d*x)) - 2560*E^((6*I)*(c + d*x)) + (90090*E^((4*I)*(c + d*x))*ArcTanh[S
qrt[1 + E^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)*(c + d*x))])*Sec[c + d*x]^3)/(a^3*d*(-I + Tan[c + d*x])^3*Sqr
t[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.93, size = 427, normalized size = 1.39

method	result	size
default	\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (786432 i \left (\cos ^{11}\left (d x +c \right )\right )+786432 \left (\cos ^{10}\left (d x +c \right )\right ) \sin \left (d x +c \right )-466944 i \left (\cos ^{9}\left (d x +c \right )\right )-73728 \sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )+5120 i \left (\cos ^{7}\left (d x +c \right )\right )+66560 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )+9152 i \left (\cos ^{5}\left (d x +c \right )\right )+45045 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+82368 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+45045 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )+45045 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+24024 i \left (\cos ^{3}\left (d x +c \right )\right )+120120 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-180180 i \cos \left (d x +c \right )\right )}{983040 d \,a^{4}}\)	\(427\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/983040/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(786432*I*cos(d*x+c)^11+786432*cos(d*x+c)^10*sin(d*x
+c)-466944*I*cos(d*x+c)^9-73728*sin(d*x+c)*cos(d*x+c)^8+5120*I*cos(d*x+c)^7+66560*sin(d*x+c)*cos(d*x+c)^6+9152
*I*cos(d*x+c)^5+45045*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)+82368*sin(d*x+c)*cos(d*x+c)^4+45045*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)+45045*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)+24024*I*cos(d*x+c)^3+120120*cos(d*x+c
)^2*sin(d*x+c)-180180*I*cos(d*x+c))/a^4

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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. 5821 vs. \(2 (236) = 472\).
time = 0.81, size = 5821, normalized size = 18.96 \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))^(7/2),x, algorithm="maxima")

[Out]

-1/1966080*(40*(cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + sin(1/5*arctan2(sin(10*d*x + 10*c
), cos(10*d*x + 10*c)))^2 + 2*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 1)^(3/4)*((79*(-I*sqr
t(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))
^2 + 79*(-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(1
0*d*x + 10*c)))^2 + 837*(-I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 - I*sqrt(2)*sin
(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 - 2*I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(
10*d*x + 10*c))) - I*sqrt(2))*cos(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 158*(-I*sqrt(2)*cos(1
0*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 837*(sq
rt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + sqrt(2)*sin(1/5*arctan2(sin(10*d*x + 10*c),
cos(10*d*x + 10*c)))^2 + 2*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + sqrt(2))*sin(4/
5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 79*I*sqrt(2)*cos(10*d*x + 10*c) - 79*sqrt(2)*sin(10*d*x +
10*c))*cos(7/2*arctan2(sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))), cos(1/5*arctan2(sin(10*d*x +
10*c), cos(10*d*x + 10*c))) + 1)) + (-49*I*sqrt(2)*cos(10*d*x + 10*c) - 1155*I*sqrt(2)*cos(4/5*arctan2(sin(10
*d*x + 10*c), cos(10*d*x + 10*c))) + 3264*I*sqrt(2)*cos(3/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) -
624*I*sqrt(2)*cos(2/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 49*sqrt(2)*sin(10*d*x + 10*c) - 1155
*sqrt(2)*sin(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 3264*sqrt(2)*sin(3/5*arctan2(sin(10*d*x +
10*c), cos(10*d*x + 10*c))) - 624*sqrt(2)*sin(2/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 128*I*sqr
t(2))*cos(3/2*arctan2(sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))), cos(1/5*arctan2(sin(10*d*x + 1
0*c), cos(10*d*x + 10*c))) + 1)) + (79*(sqrt(2)*cos(10*d*x + 10*c) - I*sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arc
tan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 79*(sqrt(2)*cos(10*d*x + 10*c) - I*sqrt(2)*sin(10*d*x + 10*c
))*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 837*(sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c
), cos(10*d*x + 10*c)))^2 + sqrt(2)*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 2*sqrt(2)*cos
(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + sqrt(2))*cos(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*
x + 10*c))) + 158*(sqrt(2)*cos(10*d*x + 10*c) - I*sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*
c), cos(10*d*x + 10*c))) + 837*(-I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 - I*sqrt
(2)*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 - 2*I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c
), cos(10*d*x + 10*c))) - I*sqrt(2))*sin(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 79*sqrt(2)*cos
(10*d*x + 10*c) - 79*I*sqrt(2)*sin(10*d*x + 10*c))*sin(7/2*arctan2(sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*
d*x + 10*c))), cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 1)) + (49*sqrt(2)*cos(10*d*x + 10*c)
+ 1155*sqrt(2)*cos(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 3264*sqrt(2)*cos(3/5*arctan2(sin(10
*d*x + 10*c), cos(10*d*x + 10*c))) + 624*sqrt(2)*cos(2/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 49
*I*sqrt(2)*sin(10*d*x + 10*c) - 1155*I*sqrt(2)*sin(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 3264
*I*sqrt(2)*sin(3/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 624*I*sqrt(2)*sin(2/5*arctan2(sin(10*d*x
+ 10*c), cos(10*d*x + 10*c))) - 128*sqrt(2))*sin(3/2*arctan2(sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x +
10*c))), cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 1)))*sqrt(a) + 4*(cos(1/5*arctan2(sin(10*
d*x + 10*c), cos(10*d*x + 10*c)))^2 + sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 2*cos(1/5*a
rctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 1)^(1/4)*(105*((-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(
10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^4 + (-I*sqrt(2)*cos(10*d*x + 10*c) -
sqrt(2)*sin(10*d*x + 10*c))*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^4 + 4*(-I*sqrt(2)*cos(10*
d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^3 + 6*(-I*s
qrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)
))^2 + 2*((-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*x + 10*c), cos
(10*d*x + 10*c)))^2 + 2*(-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10*d*
x + 10*c), cos(10*d*x + 10*c))) - I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*sin(1/5*arctan2(s
in(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 4*(-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos
(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 1...

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Fricas [A]
time = 0.45, size = 322, normalized size = 1.05 \begin {gather*} \frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-\frac {3003 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8192 \, a^{3} d}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-\frac {3003 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8192 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1280 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 25600 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 11275 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 56665 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 31094 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 12952 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3312 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 384 i\right )}\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{491520 \, a^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/491520*(-45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c)*log(-3003/8192*(sqrt(2)*sqrt(1/2)*(
I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x - I*
c)/(a^3*d)) + 45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c)*log(-3003/8192*(sqrt(2)*sqrt(1/2
)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x
- I*c)/(a^3*d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1280*I*e^(14*I*d*x + 14*I*c) - 25600*I*e^(12*I*d
*x + 12*I*c) + 11275*I*e^(10*I*d*x + 10*I*c) + 56665*I*e^(8*I*d*x + 8*I*c) + 31094*I*e^(6*I*d*x + 6*I*c) + 129
52*I*e^(4*I*d*x + 4*I*c) + 3312*I*e^(2*I*d*x + 2*I*c) + 384*I))*e^(-10*I*d*x - 10*I*c)/(a^4*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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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))^(7/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^3/(I*a*tan(d*x + c) + a)^(7/2), 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 )}^{7/2}} \,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)^(7/2),x)

[Out]

int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(7/2), x)

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