3.166 \(\int \frac {(d \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=260 \[ -\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a f}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3550, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a f}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {d \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

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

Rule 617

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 628

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 1162

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 1165

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 1168

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 3534

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 3550

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

Rubi steps

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

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Mathematica [A]  time = 0.93, size = 150, normalized size = 0.58 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) d \sqrt {\sin (2 (e+f x))} \csc (e+f x) \sqrt {d \tan (e+f x)} \left ((1+i) \sqrt {\sin (2 (e+f x))} \sec (e+f x)+(1+2 i) (\tan (e+f x)-i) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+(2+i) (\tan (e+f x)-i) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{a f (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*d*Csc[e + f*x]*Sqrt[Sin[2*(e + f*x)]]*Sqrt[d*Tan[e + f*x]]*((1 + I)*Sec[e + f*x]*Sqrt[Sin[2*(e +
f*x)]] + (1 + 2*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*(-I + Tan[e + f*x]) + (2 + I)*Log[Cos[e + f*x] + Sin[e
+ f*x] + Sqrt[Sin[2*(e + f*x)]]]*(-I + Tan[e + f*x])))/(a*f*(-I + Tan[e + f*x]))

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fricas [B]  time = 0.70, size = 521, normalized size = 2.00 \[ -\frac {{\left (a f \sqrt {-\frac {i \, d^{3}}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 \, d^{3}}{4 \, a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - a f \sqrt {-\frac {i \, d^{3}}{4 \, a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 \, d^{3}}{4 \, a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - a f \sqrt {\frac {i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 \, d^{3}}{a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + a f \sqrt {\frac {i \, d^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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 \, d^{3}}{a^{2} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + {\left (d e^{\left (2 i \, f x + 2 i \, e\right )} + d\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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.99, size = 177, normalized size = 0.68 \[ -\frac {1}{2} \, d {\left (\frac {i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {i \, \sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 111, normalized size = 0.43 \[ \frac {i d^{2} \sqrt {d \tan \left (f x +e \right )}}{2 f a \left (d \tan \left (f x +e \right )-i d \right )}-\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{f a \sqrt {-i d}}-\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/f/a*d^2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)-I*d)-I/f/a*d^2/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d
)^(1/2))-1/2*I/f/a*d^2/(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 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [B]  time = 5.89, size = 137, normalized size = 0.53 \[ -\mathrm {atan}\left (\frac {2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}}}{d^2}\right )\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{4\,a^2\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}}}{d^2}\right )\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,2{}\mathrm {i}-\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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