3.10.90 \(\int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx\) [990]

Optimal. Leaf size=231 \[ \frac {63 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

63/256*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(5/2)/f*2^(1/2)-63/128*I/a^2/c^2/f/(c-I*c
*tan(f*x+e))^(1/2)-63/160*I/a^2/f/(c-I*c*tan(f*x+e))^(5/2)+1/4*I/a^2/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(
5/2)+9/16*I/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-21/64*I/a^2/c/f/(c-I*c*tan(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 9, 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 {63 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (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])^2*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(((63*I)/128)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*c^(5/2)*f) - ((63*I)/160)/(a
^2*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I/4)/(a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2)) + ((9*I
)/16)/(a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) - ((21*I)/64)/(a^2*c*f*(c - I*c*Tan[e + f*x])^
(3/2)) - ((63*I)/128)/(a^2*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))^2 (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\cos ^4(e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx}{a^2 c^2}\\ &=\frac {\left (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 )}{a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {\left (9 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 )}{8 a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(63 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f}\\ &=-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac {(63 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 f}\\ &=-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(63 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^2 c f}\\ &=-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(63 i) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^2 c^2 f}\\ &=-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {(63 i) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{128 a^2 c^2 f}\\ &=\frac {63 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 3.66, size = 182, normalized size = 0.79 \begin {gather*} \frac {\sec ^2(e+f x) (-i \cos (3 (e+f x))+\sin (3 (e+f x))) \left (315 e^{-i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-547 \cos (e+f x)+31 \cos (3 (e+f x))+2 \cos (5 (e+f x))-141 i \sin (e+f x)-159 i \sin (3 (e+f x))-18 i \sin (5 (e+f x))\right ) \sqrt {c-i c \tan (e+f x)}}{1280 a^2 c^3 f (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*((-I)*Cos[3*(e + f*x)] + Sin[3*(e + f*x)])*((315*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1
+ E^((2*I)*(e + f*x))]])/E^(I*(e + f*x)) - 547*Cos[e + f*x] + 31*Cos[3*(e + f*x)] + 2*Cos[5*(e + f*x)] - (141*
I)*Sin[e + f*x] - (159*I)*Sin[3*(e + f*x)] - (18*I)*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(1280*a^2*c^
3*f*(-I + Tan[e + f*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.34, size = 158, normalized size = 0.68

method	result	size
derivativedivides	\(-\frac {2 i c^{3} \left (-\frac {\frac {-\frac {15 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16}+\frac {17 c \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {63 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{5}}+\frac {3}{16 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{16 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{40 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\)	\(158\)
default	\(-\frac {2 i c^{3} \left (-\frac {\frac {-\frac {15 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16}+\frac {17 c \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {63 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{5}}+\frac {3}{16 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{16 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{40 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/f/a^2*c^3*(-1/16/c^5*(4*(-15/64*(c-I*c*tan(f*x+e))^(3/2)+17/32*c*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x
+e))^2+63/32*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))+3/16/c^5/(c-I*c*tan(f*x+e)
)^(1/2)+1/16/c^4/(c-I*c*tan(f*x+e))^(3/2)+1/40/c^3/(c-I*c*tan(f*x+e))^(5/2))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 214, normalized size = 0.93 \begin {gather*} -\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} - 1050 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c + 672 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} + 192 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} + 128 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{2} c - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} c^{2} + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c^{3}} + \frac {315 \, \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^{2} c^{\frac {3}{2}}}\right )}}{2560 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2560*I*(4*(315*(-I*c*tan(f*x + e) + c)^4 - 1050*(-I*c*tan(f*x + e) + c)^3*c + 672*(-I*c*tan(f*x + e) + c)^2
*c^2 + 192*(-I*c*tan(f*x + e) + c)*c^3 + 128*c^4)/((-I*c*tan(f*x + e) + c)^(9/2)*a^2*c - 4*(-I*c*tan(f*x + e)
+ c)^(7/2)*a^2*c^2 + 4*(-I*c*tan(f*x + e) + c)^(5/2)*a^2*c^3) + 315*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^2*c^(3/2)))/(c*f)

________________________________________________________________________________________

Fricas [A]
time = 1.50, size = 354, normalized size = 1.53 \begin {gather*} \frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 64 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 344 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 203 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 95 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 10 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{1280 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/1280*(-315*I*sqrt(1/2)*a^2*c^3*f*sqrt(1/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(-63/64*(sqrt(2)*sqrt(1/2)*(I*
a^2*c^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^5*f^2)) - I)*e^(-
I*f*x - I*e)/(a^2*c^2*f)) + 315*I*sqrt(1/2)*a^2*c^3*f*sqrt(1/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I*e)*log(-63/64*(sq
rt(2)*sqrt(1/2)*(-I*a^2*c^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4
*c^5*f^2)) - I)*e^(-I*f*x - I*e)/(a^2*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-8*I*e^(10*I*f*x +
10*I*e) - 64*I*e^(8*I*f*x + 8*I*e) - 344*I*e^(6*I*f*x + 6*I*e) - 203*I*e^(4*I*f*x + 4*I*e) + 95*I*e^(2*I*f*x +
2*I*e) + 10*I))*e^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 5.64, size = 208, normalized size = 0.90 \begin {gather*} -\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,21{}\mathrm {i}}{20\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{5\,a^2\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,105{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,63{}\mathrm {i}}{128\,a^2\,c^2\,f}+\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{10\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\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 )\,63{}\mathrm {i}}{256\,a^2\,{\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)^2*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

- (((c - c*tan(e + f*x)*1i)^2*21i)/(20*a^2*f) + (c^2*1i)/(5*a^2*f) - ((c - c*tan(e + f*x)*1i)^3*105i)/(64*a^2*
c*f) + ((c - c*tan(e + f*x)*1i)^4*63i)/(128*a^2*c^2*f) + (c*(c - c*tan(e + f*x)*1i)*3i)/(10*a^2*f))/((c - c*ta
n(e + f*x)*1i)^(9/2) - 4*c*(c - c*tan(e + f*x)*1i)^(7/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(5/2)) - (2^(1/2)*ata
n((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*63i)/(256*a^2*(-c)^(5/2)*f)

________________________________________________________________________________________