3.368 \(\int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=277 \[ -\frac {143 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

-143/1024*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+143/512*I/a^2/d/(a+I*a*tan
(d*x+c))^(1/2)+143/288*I*a^2/d/(a+I*a*tan(d*x+c))^(9/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(9
/2)-13/16*I*a^3/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(9/2)+143/448*I*a/d/(a+I*a*tan(d*x+c))^(7/2)+143/640*I
/d/(a+I*a*tan(d*x+c))^(5/2)+143/768*I/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.16, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {143 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-143*I)/512)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) + (((143*I)/288)*a^
2)/(d*(a + I*a*Tan[c + d*x])^(9/2)) - ((I/4)*a^4)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(9/2)) -
(((13*I)/16)*a^3)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + (((143*I)/448)*a)/(d*(a + I*a*Tan[
c + d*x])^(7/2)) + ((143*I)/640)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((143*I)/768)/(a*d*(a + I*a*Tan[c + d*x])^
(3/2)) + ((143*I)/512)/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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])

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]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\left (i a^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{11/2}} \, 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))^{9/2}}-\frac {\left (13 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{11/2}} \, 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))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {\left (143 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {\left (143 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}-\frac {(143 i a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {(143 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(143 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{512 a d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(143 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{1024 a^2 d}\\ &=\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(143 i) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{512 a^2 d}\\ &=-\frac {143 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.71, size = 189, normalized size = 0.68 \[ -\frac {i e^{-10 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \left (-2200 e^{2 i (c+d x)}-7944 e^{4 i (c+d x)}-18808 e^{6 i (c+d x)}-50584 e^{8 i (c+d x)}+7875 e^{10 i (c+d x)}+630 e^{12 i (c+d x)}-280\right )+45045 e^{9 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{645120 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-1/645120*I)*(1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-280 - 2200*E^((2*I)*(c + d*x))
- 7944*E^((4*I)*(c + d*x)) - 18808*E^((6*I)*(c + d*x)) - 50584*E^((8*I)*(c + d*x)) + 7875*E^((10*I)*(c + d*x)
) + 630*E^((12*I)*(c + d*x))) + 45045*E^((9*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(a^2*d*E^(
(10*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.86, size = 327, normalized size = 1.18 \[ \frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-630 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 8505 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 42709 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 69392 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 26752 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10144 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2480 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 280 i\right )}\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{322560 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/322560*(-45045*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(9*I*d*x + 9*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*
I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I
*c)) + 45045*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(9*I*d*x + 9*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d
*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*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))*(-630*I*e^(14*I*d*x + 14*I*c) - 8505*I*e^(12*I*d*x + 12*I*c) + 4
2709*I*e^(10*I*d*x + 10*I*c) + 69392*I*e^(8*I*d*x + 8*I*c) + 26752*I*e^(6*I*d*x + 6*I*c) + 10144*I*e^(4*I*d*x
+ 4*I*c) + 2480*I*e^(2*I*d*x + 2*I*c) + 280*I))*e^(-9*I*d*x - 9*I*c)/(a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)

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maple [A]  time = 1.37, size = 422, normalized size = 1.52 \[ \frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (286720 i \left (\cos ^{10}\left (d x +c \right )\right )+286720 \sin \left (d x +c \right ) \left (\cos ^{9}\left (d x +c \right )\right )-81920 i \left (\cos ^{8}\left (d x +c \right )\right )+61440 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+6656 i \left (\cos ^{6}\left (d x +c \right )\right )+73216 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+13728 i \left (\cos ^{4}\left (d x +c \right )\right )+45045 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+45045 i \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+96096 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+45045 \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+60060 i \left (\cos ^{2}\left (d x +c \right )\right )+180180 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{645120 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/645120/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(286720*I*cos(d*x+c)^10+286720*sin(d*x+c)*cos(d*x+c)
^9-81920*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)+13728*
I*cos(d*x+c)^4+45045*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*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)+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)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+96096*
cos(d*x+c)^3*sin(d*x+c)+45045*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)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+60060*I*cos(d*x+c)^2+180180*cos(d*x+c)
*sin(d*x+c))/a^3

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maxima [A]  time = 0.56, size = 229, normalized size = 0.83 \[ \frac {i \, {\left (\frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} - 150150 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a + 96096 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + 27456 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} + 18304 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} + 16640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 17920 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}} + \frac {45045 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}}\right )}}{645120 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/645120*I*(4*(45045*(I*a*tan(d*x + c) + a)^6 - 150150*(I*a*tan(d*x + c) + a)^5*a + 96096*(I*a*tan(d*x + c) +
a)^4*a^2 + 27456*(I*a*tan(d*x + c) + a)^3*a^3 + 18304*(I*a*tan(d*x + c) + a)^2*a^4 + 16640*(I*a*tan(d*x + c) +
a)*a^5 + 17920*a^6)/((I*a*tan(d*x + c) + a)^(13/2)*a - 4*(I*a*tan(d*x + c) + a)^(11/2)*a^2 + 4*(I*a*tan(d*x +
c) + a)^(9/2)*a^3) + 45045*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqr
t(I*a*tan(d*x + c) + a)))/a^(3/2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(cos(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(5/2), x)

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