3.2.83 \(\int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\) [183]

Optimal. Leaf size=292 \[ \frac {i \sqrt {d} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2} \]

[Out]

1/16*I*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^3/f*2^(1/2)-1/16*I*arctan(1+2^(1/2)*(d*tan(f*x
+e))^(1/2)/d^(1/2))*d^(1/2)/a^3/f*2^(1/2)+1/32*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^3/f*2^(1/2)-1/32*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^3/f*2^(1/2)+
1/6*I*(d*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^3+1/12*I*(d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2

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Rubi [A]
time = 0.25, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3638, 3677, 21, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {i \sqrt {d} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{8 \sqrt {2} a^3 f}+\frac {i \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((I/8)*Sqrt[d]*ArcTan[1 +
(Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) + ((I/16)*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x]
- Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) - ((I/16)*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[
2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + ((I/6)*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^3) + ((I/12
)*Sqrt[d*Tan[e + f*x]])/(a*f*(a + I*a*Tan[e + f*x])^2)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3638

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-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 3677

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))^3} \, dx &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {i a d-5 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \, dx}{12 a^2}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {\int \frac {6 i a^2 d^2-6 a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{48 a^4 d}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {(i d) \int \frac {1}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {\left (i d^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (e+f x)\right )}{8 a^3 f}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {\left (i d^2\right ) \text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^3 f}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {(i d) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}-\frac {(i d) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}\\ &=\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}+\frac {\left (i \sqrt {d}\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 )}{16 \sqrt {2} a^3 f}+\frac {\left (i \sqrt {d}\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 )}{16 \sqrt {2} a^3 f}-\frac {(i d) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}-\frac {(i d) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}\\ &=\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}-\frac {\left (i \sqrt {d}\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 )}{8 \sqrt {2} a^3 f}+\frac {\left (i \sqrt {d}\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 )}{8 \sqrt {2} a^3 f}\\ &=\frac {i \sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 2.63, size = 225, normalized size = 0.77 \begin {gather*} -\frac {d (\cos (3 (e+f x))-i \sin (3 (e+f x))) \left (\cos (e+f x)-\cos (3 (e+f x))-3 i \sin (e+f x)-3 i \sin (3 (e+f x))+6 \text {ArcTan}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {i \tan (e+f x)}+6 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {i \tan (e+f x)}\right )}{48 a^3 f \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(d*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)])*(Cos[e + f*x] - Cos[3*(e + f*x)] - (3*I)*Sin[e + f*x] - (3*I)
*Sin[3*(e + f*x)] + 6*ArcTan[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]]*(Cos[3*(e + f*x)] + I
*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]] + 6*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]
]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]]))/(a^3*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]
time = 0.21, size = 120, normalized size = 0.41

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(120\)
default	\(\frac {2 d^{4} \left (\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(120\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(1/16/d^3*((-2/3*I*d*(d*tan(f*x+e))^(3/2)-2*d^2*(d*tan(f*x+e))^(1/2))/(-I*d+d*tan(f*x+e))^3-1/(-I*
d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/16/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((d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,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. 575 vs. \(2 (226) = 452\).
time = 0.38, size = 575, normalized size = 1.97 \begin {gather*} -\frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\left (8 \, {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} 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^{6} 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 ) - 12 \, a^{3} f \sqrt {\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\left (8 \, {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} 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^{6} 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 ) - 12 \, a^{3} f \sqrt {-\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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^{6} f^{2}}} + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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^{6} f^{2}}} - d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} 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 (6 i \, f x + 6 i \, e\right )} + 5 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

-1/48*(12*a^3*f*sqrt(1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(8*(I*a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*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^6*f^2)) + I*d*e^(2*I*f*x +
2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(8*(-I*a^3*f*e^(2
*I*f*x + 2*I*e) - I*a^3*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^6
*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(-1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*
e)*log(1/8*(8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*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^6*f^2)) + d)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*f*sqrt(-1/64*I*d/(a^6*f^2))*e^(6*I*
f*x + 6*I*e)*log(-1/8*(8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*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^6*f^2)) - d)*e^(-2*I*f*x - 2*I*e)/(a^3*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^(6*I*f*x + 6*I*e) + 5*I*e^(4*I*f*x + 4*I*e) + 4*I*e^(2*I*f*x + 2*I*e)
+ I))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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

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))**3,x)

[Out]

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

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Giac [A]
time = 0.71, size = 207, normalized size = 0.71 \begin {gather*} \frac {\frac {3 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {8 \, \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^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {8 \, \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^{3} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} d^{4} \tan \left (f x + e\right ) + 3 \, \sqrt {d \tan \left (f x + e\right )} d^{4}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}}{24 \, d} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

1/24*(3*sqrt(2)*d^(3/2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqr
t(d)))/(a^3*f*(I*d/sqrt(d^2) + 1)) - 3*sqrt(2)*d^(3/2)*arctan(8*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^3*f*(-I*d/sqrt(d^2) + 1)) - 2*(I*sqrt(d*tan(f*x + e))*d^4*tan(f*x +
e) + 3*sqrt(d*tan(f*x + e))*d^4)/((d*tan(f*x + e) - I*d)^3*a^3*f))/d

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Mupad [B]
time = 4.18, size = 157, normalized size = 0.54 \begin {gather*} \frac {\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}+\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f} \end {gather*}

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)^3,x)

[Out]

((d^3*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^2*(d*tan(e + f*x))^(3/2)*1i)/(12*a^3*f))/(3*d^3*tan(e + f*x) - d^
3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3) - ((-1)^(1/4)*d^(1/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/
2))/d^(1/2)))/(8*a^3*f) - ((-1)^(1/4)*d^(1/2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(8*a^3*f)

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