3.4.52 \(\int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [352]

Optimal. Leaf size=248 \[ -\frac {99 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}+\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac {99 i}{256 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-99/512*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+99/256*I/a/d/(a+I*a*tan(d*x+
c))^(1/2)+99/224*I*a^2/d/(a+I*a*tan(d*x+c))^(7/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(7/2)-11
/16*I*a^3/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(7/2)+99/320*I*a/d/(a+I*a*tan(d*x+c))^(5/2)+33/128*I/d/(a+I*
a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 248, 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 = {3568, 44, 53, 65, 212} \begin {gather*} -\frac {99 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac {99 i}{256 a d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-99*I)/256)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) + (((99*I)/224)*a^2)
/(d*(a + I*a*Tan[c + d*x])^(7/2)) - ((I/4)*a^4)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(7/2)) - ((
(11*I)/16)*a^3)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(7/2)) + (((99*I)/320)*a)/(d*(a + I*a*Tan[c +
d*x])^(5/2)) + ((33*I)/128)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((99*I)/256)/(a*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))^{3/2}} \, dx &=-\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{9/2}} \, 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))^{7/2}}-\frac {\left (11 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{9/2}} \, 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))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}-\frac {\left (99 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}-\frac {\left (99 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}-\frac {(99 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac {(99 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac {99 i}{256 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(99 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{512 a d}\\ &=\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac {99 i}{256 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(99 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{256 a d}\\ &=-\frac {99 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} a^{3/2} d}+\frac {99 i a^2}{224 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {11 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {99 i a}{320 d (a+i a \tan (c+d x))^{5/2}}+\frac {33 i}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac {99 i}{256 a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.38, size = 168, normalized size = 0.68 \begin {gather*} -\frac {i e^{-7 i (c+d x)} \left (-40-328 e^{2 i (c+d x)}-1304 e^{4 i (c+d x)}-4584 e^{6 i (c+d x)}-2833 e^{8 i (c+d x)}+805 e^{10 i (c+d x)}+70 e^{12 i (c+d x)}+3465 e^{7 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sec (c+d x)}{17920 a 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])^(3/2),x]

[Out]

((-1/17920*I)*(-40 - 328*E^((2*I)*(c + d*x)) - 1304*E^((4*I)*(c + d*x)) - 4584*E^((6*I)*(c + d*x)) - 2833*E^((
8*I)*(c + d*x)) + 805*E^((10*I)*(c + d*x)) + 70*E^((12*I)*(c + d*x)) + 3465*E^((7*I)*(c + d*x))*Sqrt[1 + E^((2
*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x])/(a*d*E^((7*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 1.01, size = 389, normalized size = 1.57

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-10240 i \left (\cos ^{8}\left (d x +c \right )\right )-10240 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-512 i \left (\cos ^{6}\left (d x +c \right )\right )-5632 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+3465 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-1056 i \left (\cos ^{4}\left (d x +c \right )\right )+3465 i \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+3465 \sin \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-7392 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-4620 i \left (\cos ^{2}\left (d x +c \right )\right )-13860 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{35840 d \,a^{2}}\)	\(389\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/35840/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-10240*I*cos(d*x+c)^8-10240*sin(d*x+c)*cos(d*x+c)^7
-512*I*cos(d*x+c)^6-5632*sin(d*x+c)*cos(d*x+c)^5+3465*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*co
s(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)-1056*I*cos(
d*x+c)^4+3465*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))+3465*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)-7392*sin(d*x+c
)*cos(d*x+c)^3-4620*I*cos(d*x+c)^2-13860*sin(d*x+c)*cos(d*x+c))/a^2

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Maxima [A]
time = 0.50, size = 207, normalized size = 0.83 \begin {gather*} \frac {i \, {\left (\frac {3465 \, \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 )}{\sqrt {a}} + \frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} - 11550 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a + 7392 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 2112 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 1408 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 1280 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}}\right )}}{35840 \, 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))^(3/2),x, algorithm="maxima")

[Out]

1/35840*I*(3465*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*
x + c) + a)))/sqrt(a) + 4*(3465*(I*a*tan(d*x + c) + a)^5 - 11550*(I*a*tan(d*x + c) + a)^4*a + 7392*(I*a*tan(d*
x + c) + a)^3*a^2 + 2112*(I*a*tan(d*x + c) + a)^2*a^3 + 1408*(I*a*tan(d*x + c) + a)*a^4 + 1280*a^5)/((I*a*tan(
d*x + c) + a)^(11/2) - 4*(I*a*tan(d*x + c) + a)^(9/2)*a + 4*(I*a*tan(d*x + c) + a)^(7/2)*a^2))/(a*d)

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Fricas [A]
time = 0.39, size = 316, normalized size = 1.27 \begin {gather*} \frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} 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 (-70 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 805 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 2833 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 4584 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1304 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 328 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 40 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{17920 \, a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

1/17920*(-3465*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(7*I*d*x + 7*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*
d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c
)) + 3465*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(7*I*d*x + 7*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x
+ 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*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))*(-70*I*e^(12*I*d*x + 12*I*c) - 805*I*e^(10*I*d*x + 10*I*c) + 2833*I
*e^(8*I*d*x + 8*I*c) + 4584*I*e^(6*I*d*x + 6*I*c) + 1304*I*e^(4*I*d*x + 4*I*c) + 328*I*e^(2*I*d*x + 2*I*c) + 4
0*I))*e^(-7*I*d*x - 7*I*c)/(a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \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))**(3/2),x)

[Out]

Integral(cos(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(3/2), 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)^4/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4/(I*a*tan(d*x + c) + a)^(3/2), 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 )}^{3/2}} \,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)^(3/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(3/2), x)

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