3.353 \(\int \frac{\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=321 \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}-\frac{715 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} a^{3/2} d}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac{715 i}{2048 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(((-715*I)/2048)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) + (((715*I)/1152)*
a^3)/(d*(a + I*a*Tan[c + d*x])^(9/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(9/2))
- (((5*I)/16)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(9/2)) - (((65*I)/64)*a^4)/(d*(a - I*a*T
an[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + (((715*I)/1792)*a^2)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((143*I)
/512)*a)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((715*I)/3072)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((715*I)/2048)/(
a*d*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.180061, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}-\frac{715 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} a^{3/2} d}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac{715 i}{2048 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-715*I)/2048)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) + (((715*I)/1152)*
a^3)/(d*(a + I*a*Tan[c + d*x])^(9/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(9/2))
- (((5*I)/16)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(9/2)) - (((65*I)/64)*a^4)/(d*(a - I*a*T
an[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + (((715*I)/1792)*a^2)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((143*I)
/512)*a)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((715*I)/3072)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((715*I)/2048)/(
a*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3487

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 \frac{\cos ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{\left (5 i a^6\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{\left (65 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac{\left (715 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac{\left (715 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}-\frac{\left (715 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{512 d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}-\frac{(715 i a) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{1024 d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}-\frac{(715 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{2048 d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac{715 i}{2048 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(715 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4096 a d}\\ &=\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac{715 i}{2048 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(715 i) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2048 a d}\\ &=-\frac{715 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2048 \sqrt{2} a^{3/2} d}+\frac{715 i a^3}{1152 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{9/2}}-\frac{5 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac{65 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{715 i a^2}{1792 d (a+i a \tan (c+d x))^{7/2}}+\frac{143 i a}{512 d (a+i a \tan (c+d x))^{5/2}}+\frac{715 i}{3072 d (a+i a \tan (c+d x))^{3/2}}+\frac{715 i}{2048 a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.76946, size = 203, normalized size = 0.63 \[ \frac{i e^{-8 i (c+d x)} \left (1136 e^{2 i (c+d x)}+5440 e^{4 i (c+d x)}+17344 e^{6 i (c+d x)}+57632 e^{8 i (c+d x)}+33301 e^{10 i (c+d x)}-13209 e^{12 i (c+d x)}-1974 e^{14 i (c+d x)}-168 e^{16 i (c+d x)}-45045 e^{9 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+112\right )}{129024 a d \left (1+e^{2 i (c+d x)}\right ) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((I/129024)*(112 + 1136*E^((2*I)*(c + d*x)) + 5440*E^((4*I)*(c + d*x)) + 17344*E^((6*I)*(c + d*x)) + 57632*E^(
(8*I)*(c + d*x)) + 33301*E^((10*I)*(c + d*x)) - 13209*E^((12*I)*(c + d*x)) - 1974*E^((14*I)*(c + d*x)) - 168*E
^((16*I)*(c + d*x)) - 45045*E^((9*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))]))/(a*d*
E^((8*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.648, size = 422, normalized size = 1.3 \begin{align*}{\frac{1}{516096\,{a}^{2}d}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 114688\,i \left ( \cos \left ( dx+c \right ) \right ) ^{10}+114688\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{9}+4096\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+61440\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+6656\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+73216\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +45045\,i\sqrt{2}\cos \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) +13728\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+45045\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) +45045\,\sqrt{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +96096\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +60060\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+180180\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/516096/d/a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(114688*I*cos(d*x+c)^10+114688*sin(d*x+c)*cos(d*
x+c)^9+4096*I*cos(d*x+c)^8+61440*sin(d*x+c)*cos(d*x+c)^7+6656*I*cos(d*x+c)^6+73216*cos(d*x+c)^5*sin(d*x+c)+450
45*I*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/si
n(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+13728*I*cos(d*x+c)^4+45045*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+450
45*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+96096*cos(d*x+c)^3*sin(d*x+c)+60060*I*cos(d*x+c)^2+180180*cos(d*x
+c)*sin(d*x+c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.26408, size = 1106, normalized size = 3.45 \begin{align*} \frac{{\left (-45045 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-168 i \, e^{\left (16 i \, d x + 16 i \, c\right )} - 1974 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 13209 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 33301 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 57632 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 17344 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 5440 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1136 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 112 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{258048 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/258048*(-45045*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(10*I*d*x + 10*I*c)*log((2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*
d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(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)) + 45045*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(10*I*d*x + 10*I*c)*log(-(2*sqrt(1/2)*a^2*d
*sqrt(1/(a^3*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(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)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-168*I*e^(16*I*d*x + 16*I*c) -
1974*I*e^(14*I*d*x + 14*I*c) - 13209*I*e^(12*I*d*x + 12*I*c) + 33301*I*e^(10*I*d*x + 10*I*c) + 57632*I*e^(8*I*
d*x + 8*I*c) + 17344*I*e^(6*I*d*x + 6*I*c) + 5440*I*e^(4*I*d*x + 4*I*c) + 1136*I*e^(2*I*d*x + 2*I*c) + 112*I)*
e^(I*d*x + I*c))*e^(-10*I*d*x - 10*I*c)/(a^2*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(cos(d*x+c)**6/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{6}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^6/(I*a*tan(d*x + c) + a)^(3/2), x)