3.175 \(\int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=299 \[ -\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

(-1/32+3/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^2/f*2^(1/2)+(1/32-3/32*I)*arctan(1+2^(
1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^2/f*2^(1/2)+(1/64+3/64*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)
+d^(1/2)*tan(f*x+e))*d^(1/2)/a^2/f*2^(1/2)-(1/64+3/64*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f
*x+e))*d^(1/2)/a^2/f*2^(1/2)+1/8*I*(d*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))+1/4*I*(d*tan(f*x+e))^(1/2)/f/(a
+I*a*tan(f*x+e))^2

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Rubi [A]  time = 0.38, antiderivative size = 299, 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 = {3557, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/16 + (3*I)/16)*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) + ((1/16 - (3*
I)/16)*Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) + ((1/32 + (3*I)/32)*Sqrt[d
]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) - ((1/32 + (3*I)/32)*Sqr
t[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + ((I/8)*Sqrt[d*Tan[e
+ f*x]])/(a^2*f*(1 + I*Tan[e + f*x])) + ((I/4)*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2)

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 3557

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(b*(a + b*Tan[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*m), x] + Dist[1/(4*a^2*m), Int[((a + b*Tan[e + f*
x])^(m + 1)*Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x])/Sqrt[c + d*Tan[e + f*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] && LtQ[m, 0] && IntegersQ[2
*m]

Rule 3596

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

Rubi steps

\begin {align*} \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {i a d-3 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{8 a^2}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {3 i a^2 d^2-a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^4 d}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\operatorname {Subst}\left (\int \frac {3 i a^2 d^3-a^2 d^2 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^4 d f}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}--\frac {\left (\left (\frac {1}{16}-\frac {3 i}{16}\right ) d\right ) \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}-\frac {\left (\left (\frac {1}{16}+\frac {3 i}{16}\right ) d\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}--\frac {\left (\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d}\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^2 f}--\frac {\left (\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d}\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^2 f}--\frac {\left (\left (\frac {1}{32}-\frac {3 i}{32}\right ) d\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^2 f}--\frac {\left (\left (\frac {1}{32}-\frac {3 i}{32}\right ) d\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^2 f}\\ &=\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}--\frac {\left (\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d}\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^2 f}-\frac {\left (\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d}\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^2 f}\\ &=-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {i \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 227, normalized size = 0.76 \[ -\frac {d \sec ^3(e+f x) \left (3 i \sin (e+f x)+3 i \sin (3 (e+f x))-\cos (e+f x)+\cos (3 (e+f x))-(3+i) \sqrt {\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))+(3-i) \sin ^{\frac {3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )-(1+3 i) \sqrt {\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 f (\tan (e+f x)-i)^2 \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(d*Sec[e + f*x]^3*(-Cos[e + f*x] + Cos[3*(e + f*x)] + (3*I)*Sin[e + f*x] - (1 + 3*I)*Cos[2*(e + f*x)]*Lo
g[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (3 - I)*Log[Cos[e + f*x] + Si
n[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) - (3 + I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sqrt
[Sin[2*(e + f*x)]]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + (3*I)*Sin[3*(e + f*x)]))/(a^2*f*Sqrt[d*Tan[e +
f*x]]*(-I + Tan[e + f*x])^2)

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fricas [B]  time = 0.54, size = 531, normalized size = 1.78 \[ \frac {{\left (4 \, a^{2} f \sqrt {\frac {i \, d}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left ({\left ({\left (8 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, a^{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 \, d}{16 \, a^{4} f^{2}}} - 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 ) - 4 \, a^{2} f \sqrt {\frac {i \, d}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left ({\left ({\left (-8 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{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 \, d}{16 \, a^{4} f^{2}}} - 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 ) + 4 \, a^{2} f \sqrt {-\frac {i \, d}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 \, d}{64 \, a^{4} f^{2}}} + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - 4 \, a^{2} f \sqrt {-\frac {i \, d}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 \, d}{64 \, a^{4} f^{2}}} - d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{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}} {\left (2 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*f*sqrt(1/16*I*d/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(((8*I*a^2*f*e^(2*I*f*x + 2*I*e) + 8*I*a^2*f)*sq
rt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I*d/(a^4*f^2)) - 2*I*d*e^(2*I*f*x + 2
*I*e))*e^(-2*I*f*x - 2*I*e)) - 4*a^2*f*sqrt(1/16*I*d/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(((-8*I*a^2*f*e^(2*I*f*
x + 2*I*e) - 8*I*a^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(1/16*I*d/(a^4*f^
2)) - 2*I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 4*a^2*f*sqrt(-1/64*I*d/(a^4*f^2))*e^(4*I*f*x + 4*I*e)
*log(1/8*(8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^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(-1/64*I*d/(a^4*f^2)) + d)*e^(-2*I*f*x - 2*I*e)/(a^2*f)) - 4*a^2*f*sqrt(-1/64*I*d/(a^4*f^2))*e^(4*I*f*x
+ 4*I*e)*log(-1/8*(8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^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(-1/64*I*d/(a^4*f^2)) - d)*e^(-2*I*f*x - 2*I*e)/(a^2*f)) + sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I
*d)/(e^(2*I*f*x + 2*I*e) + 1))*(2*I*e^(4*I*f*x + 4*I*e) + 3*I*e^(2*I*f*x + 2*I*e) + I))*e^(-4*I*f*x - 4*I*e)/(
a^2*f)

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giac [A]  time = 1.09, size = 205, normalized size = 0.69 \[ \frac {\frac {i \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 i \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {\sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 3 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 139, normalized size = 0.46 \[ \frac {d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}-\frac {3 i d^{2} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}-\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 f \,a^{2} \sqrt {-i d}}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 f \,a^{2} \sqrt {i d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f/a^2*d/(d*tan(f*x+e)-I*d)^2*(d*tan(f*x+e))^(3/2)-3/8*I/f/a^2*d^2/(d*tan(f*x+e)-I*d)^2*(d*tan(f*x+e))^(1/2
)-1/8/f/a^2*d/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/4/f/a^2*d/(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))^(1/2)/(a+I*a*tan(f*x+e))^2,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 = 5.20, size = 147, normalized size = 0.49 \[ \frac {-\frac {d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8\,a^2\,f}+\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,3{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}+\frac {\mathrm {atanh}\left (\frac {2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d\,1{}\mathrm {i}}{4}}}{d}\right )\,\sqrt {-\frac {d\,1{}\mathrm {i}}{4}}}{4\,a^2\,f}-\frac {\mathrm {atanh}\left (\frac {4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d\,1{}\mathrm {i}}{16}}}{d}\right )\,\sqrt {\frac {d\,1{}\mathrm {i}}{16}}}{a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2*(d*tan(e + f*x))^(1/2)*3i)/(8*a^2*f) - (d*(d*tan(e + f*x))^(3/2))/(8*a^2*f))/(d^2*tan(e + f*x)*2i + d^2
- d^2*tan(e + f*x)^2) + (atanh((2*(d*tan(e + f*x))^(1/2)*(-(d*1i)/4)^(1/2))/d)*(-(d*1i)/4)^(1/2))/(4*a^2*f) -
(atanh((4*(d*tan(e + f*x))^(1/2)*((d*1i)/16)^(1/2))/d)*((d*1i)/16)^(1/2))/(a^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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