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

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

[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)

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Rubi [A]  time = 0.371763, antiderivative size = 292, normalized size of antiderivative = 1., 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 = {3557, 3596, 21, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{i \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 \sqrt{2} a^3 f}-\frac{i \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{8 \sqrt{2} a^3 f}+\frac{i \sqrt{d} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}-\frac{i \sqrt{d} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{16 \sqrt{2} a^3 f}+\frac{i \sqrt{d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2}+\frac{i \sqrt{d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]

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 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]

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 3476

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 329

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 211

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 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 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 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 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])

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 ) \operatorname{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 ) \operatorname{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) \operatorname{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) \operatorname{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 ) \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 )}{16 \sqrt{2} a^3 f}+\frac{\left (i \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 )}{16 \sqrt{2} a^3 f}-\frac{(i d) \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 2.38863, size = 225, normalized size = 0.77 \[ -\frac{d (\cos (3 (e+f x))-i \sin (3 (e+f x))) \left (-3 i \sin (e+f x)-3 i \sin (3 (e+f x))+\cos (e+f x)-\cos (3 (e+f x))+6 \tan ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) \sqrt{i \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x)))+6 \sqrt{i \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))\right )}{48 a^3 f \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)))]]*(Co
s[3*(e + f*x)] + I*Sin[3*(e + f*x)])*Sqrt[I*Tan[e + f*x]]))/(48*a^3*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]  time = 0.065, size = 141, normalized size = 0.5 \begin{align*}{\frac{-{\frac{i}{12}}{d}^{2}}{f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{3}}{4\,f{a}^{3} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{3}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{d}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{d}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

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)

[Out]

-1/12*I/f/a^3*d^2/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e))^(3/2)-1/4/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3*(d*tan(f*x+e)
)^(1/2)-1/8/f/a^3*d/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/8/f/a^3*d/(I*d)^(1/2)*arctan((d*t
an(f*x+e))^(1/2)/(I*d)^(1/2))

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

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

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Fricas [B]  time = 2.37205, size = 1542, normalized size = 5.28 \begin{align*} \text{result too large to display} \end{align*}

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(((16*I*a^3*f*e^(2*I*f*x + 2*I*e) + 16*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)) - 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(((-16*I*a^3*f*e^(2
*I*f*x + 2*I*e) - 16*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)) - 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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [A]  time = 1.20242, size = 278, normalized size = 0.95 \begin{align*} \frac{1}{24} \, d^{4}{\left (\frac{3 \, \sqrt{2} \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^{3} d^{\frac{7}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{3 \, \sqrt{2} \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^{3} d^{\frac{7}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} - \frac{2 \,{\left (i \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 3 \, \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d^{2} f}\right )} \end{align*}

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*d^4*(3*sqrt(2)*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^3*d^(7/2)*f*(I*d/sqrt(d^2) + 1)) - 3*sqrt(2)*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^3*d^(7/2)*f*(-I*d/sqrt(d^2) + 1)) - 2*(I*sqrt(d*tan(f*x + e))*d*tan(f*
x + e) + 3*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^3*a^3*d^2*f))