3.4.84 \(\int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [384]

Optimal. Leaf size=306 \[ -\frac {195 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} a^{7/2} d}+\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-195/2048*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(7/2)/d*2^(1/2)+195/1024*I/a^3/d/(a+I*a*ta
n(d*x+c))^(1/2)+195/352*I*a^2/d/(a+I*a*tan(d*x+c))^(11/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^
(11/2)-15/16*I*a^3/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(11/2)+65/192*I*a/d/(a+I*a*tan(d*x+c))^(9/2)+195/89
6*I/d/(a+I*a*tan(d*x+c))^(7/2)+39/256*I/a/d/(a+I*a*tan(d*x+c))^(5/2)+65/512*I/a^2/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.14, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \begin {gather*} -\frac {195 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} a^{7/2} d}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-195*I)/1024)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(7/2)*d) + (((195*I)/352)*a
^2)/(d*(a + I*a*Tan[c + d*x])^(11/2)) - ((I/4)*a^4)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(11/2))
- (((15*I)/16)*a^3)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(11/2)) + (((65*I)/192)*a)/(d*(a + I*a*T
an[c + d*x])^(9/2)) + ((195*I)/896)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + ((39*I)/256)/(a*d*(a + I*a*Tan[c + d*x]
)^(5/2)) + ((65*I)/512)/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((195*I)/1024)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]
])

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 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 3568

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]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{13/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {\left (15 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{13/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}-\frac {\left (195 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{13/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}-\frac {\left (195 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}-\frac {(195 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {(195 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}-\frac {(195 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{512 a d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {(195 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{1024 a^2 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {(195 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{2048 a^3 d}\\ &=\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {(195 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{1024 a^3 d}\\ &=-\frac {195 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} a^{7/2} d}+\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}-\frac {15 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.71, size = 202, normalized size = 0.66 \begin {gather*} -\frac {i e^{-13 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-168-1456 e^{2 i (c+d x)}-5728 e^{4 i (c+d x)}-13824 e^{6 i (c+d x)}-24688 e^{8 i (c+d x)}-54112 e^{10 i (c+d x)}+6699 e^{12 i (c+d x)}+462 e^{14 i (c+d x)}\right )+45045 e^{11 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sec ^3(c+d x)}{1892352 a^3 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-1/1892352*I)*(1 + E^((2*I)*(c + d*x)))^(5/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-168 - 1456*E^((2*I)*(c + d*x)
) - 5728*E^((4*I)*(c + d*x)) - 13824*E^((6*I)*(c + d*x)) - 24688*E^((8*I)*(c + d*x)) - 54112*E^((10*I)*(c + d*
x)) + 6699*E^((12*I)*(c + d*x)) + 462*E^((14*I)*(c + d*x))) + 45045*E^((11*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x
))])*Sec[c + d*x]^3)/(a^3*d*E^((13*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 1.18, size = 443, normalized size = 1.45

method	result	size
default	\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (688128 i \left (\cos ^{12}\left (d x +c \right )\right )+688128 \sin \left (d x +c \right ) \left (\cos ^{11}\left (d x +c \right )\right )-401408 i \left (\cos ^{10}\left (d x +c \right )\right )-57344 \left (\cos ^{9}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4096 i \left (\cos ^{8}\left (d x +c \right )\right )+61440 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+6656 i \left (\cos ^{6}\left (d x +c \right )\right )+73216 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+13728 i \left (\cos ^{4}\left (d x +c \right )\right )-45045 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+96096 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-45045 \sin \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-45045 i \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+60060 i \left (\cos ^{2}\left (d x +c \right )\right )+180180 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{946176 d \,a^{4}}\)	\(443\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/946176/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(688128*I*cos(d*x+c)^12+688128*sin(d*x+c)*cos(d*x+c)
^11-401408*I*cos(d*x+c)^10-57344*cos(d*x+c)^9*sin(d*x+c)+4096*I*cos(d*x+c)^8+61440*sin(d*x+c)*cos(d*x+c)^7+665
6*I*cos(d*x+c)^6+73216*sin(d*x+c)*cos(d*x+c)^5+13728*I*cos(d*x+c)^4-45045*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/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)+96096*sin(d*x+c)*cos(d*x+c)^3-45045*sin(d*x+c)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)-45045*I*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*2^(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*2^(1/2))+60060*I*cos(d*x+c)^2+180180*sin(d*x+c)*cos(d*x+c))/a^4

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Maxima [A]
time = 0.49, size = 249, normalized size = 0.81 \begin {gather*} \frac {i \, {\left (\frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} - 150150 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 96096 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 27456 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 18304 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 16640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} + 17920 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} + 21504 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{3} + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{4}} + \frac {45045 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}}\right )}}{946176 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/946176*I*(4*(45045*(I*a*tan(d*x + c) + a)^7 - 150150*(I*a*tan(d*x + c) + a)^6*a + 96096*(I*a*tan(d*x + c) +
a)^5*a^2 + 27456*(I*a*tan(d*x + c) + a)^4*a^3 + 18304*(I*a*tan(d*x + c) + a)^3*a^4 + 16640*(I*a*tan(d*x + c) +
a)^2*a^5 + 17920*(I*a*tan(d*x + c) + a)*a^6 + 21504*a^7)/((I*a*tan(d*x + c) + a)^(15/2)*a^2 - 4*(I*a*tan(d*x
+ c) + a)^(13/2)*a^3 + 4*(I*a*tan(d*x + c) + a)^(11/2)*a^4) + 45045*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*t
an(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2))/(a*d)

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Fricas [A]
time = 0.40, size = 338, normalized size = 1.10 \begin {gather*} \frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (11 i \, d x + 11 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a 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^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (11 i \, d x + 11 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a 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 (-462 i \, e^{\left (16 i \, d x + 16 i \, c\right )} - 7161 i \, e^{\left (14 i \, d x + 14 i \, c\right )} + 47413 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 78800 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 38512 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 19552 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 7184 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1624 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 168 i\right )}\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{473088 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/473088*(-45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(11*I*d*x + 11*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(
2*I*d*x + 2*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x -
I*c)) + 45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(11*I*d*x + 11*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2
*I*d*x + 2*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - a*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))*(-462*I*e^(16*I*d*x + 16*I*c) - 7161*I*e^(14*I*d*x + 14*I*c)
+ 47413*I*e^(12*I*d*x + 12*I*c) + 78800*I*e^(10*I*d*x + 10*I*c) + 38512*I*e^(8*I*d*x + 8*I*c) + 19552*I*e^(6*
I*d*x + 6*I*c) + 7184*I*e^(4*I*d*x + 4*I*c) + 1624*I*e^(2*I*d*x + 2*I*c) + 168*I))*e^(-11*I*d*x - 11*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)**4/(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)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4/(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 )}^4}{{\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)^4/(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(7/2), x)

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