3.10.91 \(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx\) [991]

Optimal. Leaf size=274 \[ \frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

231/1024*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/c^(5/2)/f*2^(1/2)-231/512*I/a^3/c^2/f/(c-
I*c*tan(f*x+e))^(1/2)-231/640*I/a^3/f/(c-I*c*tan(f*x+e))^(5/2)+1/6*I/a^3/f/(1+I*tan(f*x+e))^3/(c-I*c*tan(f*x+e
))^(5/2)+11/48*I/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2)+33/64*I/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f
*x+e))^(5/2)-77/256*I/a^3/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.18, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3603, 3568, 44, 53, 65, 212} \begin {gather*} \frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((231*I)/512)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(5/2)*f) - ((231*I)/640)/
(a^3*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)) + ((1
1*I)/48)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + ((33*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])
*(c - I*c*Tan[e + f*x])^(5/2)) - ((77*I)/256)/(a^3*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((231*I)/512)/(a^3*c^2*
f*Sqrt[c - I*c*Tan[e + f*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]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {\int \cos ^6(e+f x) \sqrt {c-i c \tan (e+f x)} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {\left (11 i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {\left (33 i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(231 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(231 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(231 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(231 i) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{1024 a^3 c^2 f}\\ &=-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(231 i) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{512 a^3 c^2 f}\\ &=\frac {231 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 3.30, size = 171, normalized size = 0.62 \begin {gather*} -\frac {i e^{-6 i (e+f x)} \left (-40-350 e^{2 i (e+f x)}-1645 e^{4 i (e+f x)}+1433 e^{6 i (e+f x)}+3184 e^{8 i (e+f x)}+464 e^{10 i (e+f x)}+48 e^{12 i (e+f x)}-3465 e^{6 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )\right ) \sqrt {c-i c \tan (e+f x)}}{15360 a^3 c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

((-1/15360*I)*(-40 - 350*E^((2*I)*(e + f*x)) - 1645*E^((4*I)*(e + f*x)) + 1433*E^((6*I)*(e + f*x)) + 3184*E^((
8*I)*(e + f*x)) + 464*E^((10*I)*(e + f*x)) + 48*E^((12*I)*(e + f*x)) - 3465*E^((6*I)*(e + f*x))*Sqrt[1 + E^((2
*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*c^3*E^((6*I)*(e + f*x
))*f)

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Maple [A]
time = 0.30, size = 177, normalized size = 0.65

method	result	size
derivativedivides	\(\frac {2 i c^{4} \left (-\frac {5}{32 c^{6} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{5} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{80 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\frac {71 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {59 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {89 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {231 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{6}}\right )}{f \,a^{3}}\)	\(177\)
default	\(\frac {2 i c^{4} \left (-\frac {5}{32 c^{6} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{5} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{80 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\frac {71 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {59 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {89 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {231 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{6}}\right )}{f \,a^{3}}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/f/a^3*c^4*(-5/32/c^6/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^5/(c-I*c*tan(f*x+e))^(3/2)-1/80/c^4/(c-I*c*tan(f*x+e)
)^(5/2)+1/32/c^6*(8*(71/256*(c-I*c*tan(f*x+e))^(5/2)-59/48*c*(c-I*c*tan(f*x+e))^(3/2)+89/64*c^2*(c-I*c*tan(f*x
+e))^(1/2))/(c+I*c*tan(f*x+e))^3+231/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))
)

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Maxima [A]
time = 0.49, size = 255, normalized size = 0.93 \begin {gather*} -\frac {i \, {\left (\frac {4 \, {\left (3465 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{5} - 18480 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} c + 30492 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c^{2} - 12672 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{3} - 2816 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{4} - 1536 \, c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}} a^{3} c - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} c^{2} + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c^{3} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{4}} + \frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} c^{\frac {3}{2}}}\right )}}{30720 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/30720*I*(4*(3465*(-I*c*tan(f*x + e) + c)^5 - 18480*(-I*c*tan(f*x + e) + c)^4*c + 30492*(-I*c*tan(f*x + e) +
c)^3*c^2 - 12672*(-I*c*tan(f*x + e) + c)^2*c^3 - 2816*(-I*c*tan(f*x + e) + c)*c^4 - 1536*c^5)/((-I*c*tan(f*x
+ e) + c)^(11/2)*a^3*c - 6*(-I*c*tan(f*x + e) + c)^(9/2)*a^3*c^2 + 12*(-I*c*tan(f*x + e) + c)^(7/2)*a^3*c^3 -
8*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^4) + 3465*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(
sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a^3*c^(3/2)))/(c*f)

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Fricas [A]
time = 1.11, size = 366, normalized size = 1.34 \begin {gather*} \frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {231 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {231 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-48 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 464 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 3184 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 1433 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 1645 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 350 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15360*(-3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*log(-231/256*(sqrt(2)*sqrt(1/2)
*(I*a^3*c^2*f*e^(2*I*f*x + 2*I*e) + I*a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^5*f^2)) - I)*
e^(-I*f*x - I*e)/(a^3*c^2*f)) + 3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f^2))*e^(6*I*f*x + 6*I*e)*log(-231/
256*(sqrt(2)*sqrt(1/2)*(-I*a^3*c^2*f*e^(2*I*f*x + 2*I*e) - I*a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt
(1/(a^6*c^5*f^2)) - I)*e^(-I*f*x - I*e)/(a^3*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-48*I*e^(12*
I*f*x + 12*I*e) - 464*I*e^(10*I*f*x + 10*I*e) - 3184*I*e^(8*I*f*x + 8*I*e) - 1433*I*e^(6*I*f*x + 6*I*e) + 1645
*I*e^(4*I*f*x + 4*I*e) + 350*I*e^(2*I*f*x + 2*I*e) + 40*I))*e^(-6*I*f*x - 6*I*e)/(a^3*c^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} + 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

I*Integral(1/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5 + I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f
*x)**4 - 2*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 + 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x
)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x)/a**3

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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(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^(5/2)), x)

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Mupad [B]
time = 5.14, size = 255, normalized size = 0.93 \begin {gather*} -\frac {-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,2541{}\mathrm {i}}{640\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{5\,a^3\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,77{}\mathrm {i}}{32\,a^3\,c\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^5\,231{}\mathrm {i}}{512\,a^3\,c^2\,f}+\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,33{}\mathrm {i}}{20\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{30\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,231{}\mathrm {i}}{1024\,a^3\,{\left (-c\right )}^{5/2}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

- ((c^3*1i)/(5*a^3*f) - ((c - c*tan(e + f*x)*1i)^3*2541i)/(640*a^3*f) + ((c - c*tan(e + f*x)*1i)^4*77i)/(32*a^
3*c*f) - ((c - c*tan(e + f*x)*1i)^5*231i)/(512*a^3*c^2*f) + (c*(c - c*tan(e + f*x)*1i)^2*33i)/(20*a^3*f) + (c^
2*(c - c*tan(e + f*x)*1i)*11i)/(30*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(9/2) - (c - c*tan(e + f*x)*1i)^(11/2)
+ 8*c^3*(c - c*tan(e + f*x)*1i)^(5/2) - 12*c^2*(c - c*tan(e + f*x)*1i)^(7/2)) - (2^(1/2)*atan((2^(1/2)*(c - c
*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*231i)/(1024*a^3*(-c)^(5/2)*f)

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