3.2.69 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx\) [169]

Optimal. Leaf size=287 \[ \frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}+\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \]

[Out]

(5/8+3/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(3/2)/f*2^(1/2)-(5/8+3/8*I)*arctan(1+2^(1/2)*(d
*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(3/2)/f*2^(1/2)+(-5/16+3/16*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)
*tan(f*x+e))/a/d^(3/2)/f*2^(1/2)+(5/16-3/16*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a/d
^(3/2)/f*2^(1/2)-5/2/a/d/f/(d*tan(f*x+e))^(1/2)+1/2/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))

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Rubi [A]
time = 0.21, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3633, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a d^{3/2} f}+\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a d^{3/2} f}-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5/4 + (3*I)/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*d^(3/2)*f) - ((5/4 + (3*I)/4)*
ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*d^(3/2)*f) - ((5/8 - (3*I)/8)*Log[Sqrt[d] + Sqr
t[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*d^(3/2)*f) + ((5/8 - (3*I)/8)*Log[Sqrt[d] + Sqrt
[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*d^(3/2)*f) - 5/(2*a*d*f*Sqrt[d*Tan[e + f*x]]) + 1
/(2*d*f*Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3633

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a
)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c
+ d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x
] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx &=\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac {\int \frac {-\frac {5 a d}{2}+\frac {3}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{2 a^2 d}\\ &=-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac {\int \frac {\frac {3}{2} i a d^2+\frac {5}{2} a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d^3}\\ &=-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac {\text {Subst}\left (\int \frac {\frac {3}{2} i a d^3+\frac {5}{2} a d^2 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d^3 f}\\ &=-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}--\frac {\left (\frac {5}{4}-\frac {3 i}{4}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d f}\\ &=-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{8}+\frac {3 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d f}-\frac {\left (\frac {5}{8}+\frac {3 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a d f}\\ &=-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}+\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}--\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}\\ &=\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}+\frac {\left (\frac {5}{8}-\frac {3 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.37, size = 155, normalized size = 0.54 \begin {gather*} \frac {16 i-20 \tan (e+f x)+(5+3 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) \sec (e+f x) \sqrt {\sin (2 (e+f x))} (-i+\tan (e+f x))+(5-3 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sec (e+f x) \sqrt {\sin (2 (e+f x))} (-i+\tan (e+f x))}{8 a d f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*I - 20*Tan[e + f*x] + (5 + 3*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sec[e + f*x]*Sqrt[Sin[2*(e + f*x)]]*(-
I + Tan[e + f*x]) + (5 - 3*I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sec[e + f*x]*Sqrt[Sin[
2*(e + f*x)]]*(-I + Tan[e + f*x]))/(8*a*d*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.14, size = 116, normalized size = 0.40

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{-i d +d \tan \left (f x +e \right )}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{3} \sqrt {i d}}\right )}{f a}\)	\(116\)
default	\(\frac {2 d^{2} \left (-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{-i d +d \tan \left (f x +e \right )}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{3} \sqrt {i d}}\right )}{f a}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a*d^2*(-1/d^3/(d*tan(f*x+e))^(1/2)+1/4/d^3*(-(d*tan(f*x+e))^(1/2)/(-I*d+d*tan(f*x+e))-4/(-I*d)^(1/2)*arcta
n((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/4/d^3/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 677 vs. \(2 (218) = 436\).
time = 0.38, size = 677, normalized size = 2.36 \begin {gather*} \frac {{\left (a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} - a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {i}{4 \, a^{2} d^{3} f^{2}}} \log \left (-2 \, {\left (2 \, {\left (i \, a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{4 \, a^{2} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - {\left (a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} - a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {i}{4 \, a^{2} d^{3} f^{2}}} \log \left (-2 \, {\left (2 \, {\left (-i \, a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{4 \, a^{2} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + {\left (a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} - a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {4 i}{a^{2} d^{3} f^{2}}} \log \left (\frac {{\left ({\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i}{a^{2} d^{3} f^{2}}} + 2\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d f}\right ) - {\left (a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} - a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {4 i}{a^{2} d^{3} f^{2}}} \log \left (-\frac {{\left ({\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i}{a^{2} d^{3} f^{2}}} - 2\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a d f}\right ) + \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-9 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{4 \, {\left (a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} - a d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt(1/4*I/(a^2*d^3*f^2))*log(-2*(2*(I*a*d^2*
f*e^(2*I*f*x + 2*I*e) + I*a*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/4*I
/(a^2*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - (a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2
*I*f*x + 2*I*e))*sqrt(1/4*I/(a^2*d^3*f^2))*log(-2*(2*(-I*a*d^2*f*e^(2*I*f*x + 2*I*e) - I*a*d^2*f)*sqrt((-I*d*e
^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/4*I/(a^2*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-
2*I*f*x - 2*I*e)) + (a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt(-4*I/(a^2*d^3*f^2))*log((
(a*d*f*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I
/(a^2*d^3*f^2)) + 2)*e^(-2*I*f*x - 2*I*e)/(a*d*f)) - (a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e
))*sqrt(-4*I/(a^2*d^3*f^2))*log(-((a*d*f*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e
^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I/(a^2*d^3*f^2)) - 2)*e^(-2*I*f*x - 2*I*e)/(a*d*f)) + sqrt((-I*d*e^(2*I*f*x +
2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(-9*I*e^(4*I*f*x + 4*I*e) - 8*I*e^(2*I*f*x + 2*I*e) + I))/(a*d^2*f*e
^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x) - I*(d*tan(e + f*x))**(3/2)), x)/a

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Giac [A]
time = 0.59, size = 202, normalized size = 0.70 \begin {gather*} -\frac {\frac {5 i \, d \tan \left (f x + e\right ) + 4 \, d}{{\left (i \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + \sqrt {d \tan \left (f x + e\right )} d\right )} a f} + \frac {4 i \, \sqrt {2} \arctan \left (\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {i \, \sqrt {2} \arctan \left (\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((5*I*d*tan(f*x + e) + 4*d)/((I*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + sqrt(d*tan(f*x + e))*d)*a*f) + 4*I*
sqrt(2)*arctan(8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a*sqrt
(d)*f*(I*d/sqrt(d^2) + 1)) + I*sqrt(2)*arctan(8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^(3/2) + 4*sqr
t(2)*sqrt(d^2)*sqrt(d)))/(a*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)))/d

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Mupad [B]
time = 6.06, size = 147, normalized size = 0.51 \begin {gather*} -\frac {-\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{2\,a\,f}+\frac {2{}\mathrm {i}}{a\,f}}{-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}+2\,\mathrm {atanh}\left (a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{a^2\,d^3\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{a^2\,d^3\,f^2}}+2\,\mathrm {atanh}\left (4\,a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{16\,a^2\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{16\,a^2\,d^3\,f^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atanh(a*d*f*(d*tan(e + f*x))^(1/2)*(-1i/(a^2*d^3*f^2))^(1/2))*(-1i/(a^2*d^3*f^2))^(1/2) - (2i/(a*f) - (5*tan
(e + f*x))/(2*a*f))/(d*(d*tan(e + f*x))^(1/2)*1i - (d*tan(e + f*x))^(3/2)) + 2*atanh(4*a*d*f*(d*tan(e + f*x))^
(1/2)*(1i/(16*a^2*d^3*f^2))^(1/2))*(1i/(16*a^2*d^3*f^2))^(1/2)

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