3.315 \(\int (a+i b \sin ^{-1}(1-i d x^2))^3 \, dx\)

Optimal. Leaf size=129 \[ 24 a b^2 x-\frac{6 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3-\frac{48 b^3 \sqrt{d^2 x^4+2 i d x^2}}{d x}+24 i b^3 x \sin ^{-1}\left (1-i d x^2\right ) \]

[Out]

24*a*b^2*x - (48*b^3*Sqrt[(2*I)*d*x^2 + d^2*x^4])/(d*x) + (24*I)*b^3*x*ArcSin[1 - I*d*x^2] - (6*b*Sqrt[(2*I)*d
*x^2 + d^2*x^4]*(a + I*b*ArcSin[1 - I*d*x^2])^2)/(d*x) + x*(a + I*b*ArcSin[1 - I*d*x^2])^3

________________________________________________________________________________________

Rubi [A]  time = 0.0639983, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4814, 4840, 12, 1588} \[ 24 a b^2 x-\frac{6 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3-\frac{48 b^3 \sqrt{d^2 x^4+2 i d x^2}}{d x}+24 i b^3 x \sin ^{-1}\left (1-i d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + I*b*ArcSin[1 - I*d*x^2])^3,x]

[Out]

24*a*b^2*x - (48*b^3*Sqrt[(2*I)*d*x^2 + d^2*x^4])/(d*x) + (24*I)*b^3*x*ArcSin[1 - I*d*x^2] - (6*b*Sqrt[(2*I)*d
*x^2 + d^2*x^4]*(a + I*b*ArcSin[1 - I*d*x^2])^2)/(d*x) + x*(a + I*b*ArcSin[1 - I*d*x^2])^3

Rule 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rule 4840

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3 \, dx &=-\frac{6 b \sqrt{2 i d x^2+d^2 x^4} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3+\left (24 b^2\right ) \int \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right ) \, dx\\ &=24 a b^2 x-\frac{6 b \sqrt{2 i d x^2+d^2 x^4} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3+\left (24 i b^3\right ) \int \sin ^{-1}\left (1-i d x^2\right ) \, dx\\ &=24 a b^2 x+24 i b^3 x \sin ^{-1}\left (1-i d x^2\right )-\frac{6 b \sqrt{2 i d x^2+d^2 x^4} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3-\left (24 i b^3\right ) \int -\frac{2 i d x^2}{\sqrt{2 i d x^2+d^2 x^4}} \, dx\\ &=24 a b^2 x+24 i b^3 x \sin ^{-1}\left (1-i d x^2\right )-\frac{6 b \sqrt{2 i d x^2+d^2 x^4} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3-\left (48 b^3 d\right ) \int \frac{x^2}{\sqrt{2 i d x^2+d^2 x^4}} \, dx\\ &=24 a b^2 x-\frac{48 b^3 \sqrt{2 i d x^2+d^2 x^4}}{d x}+24 i b^3 x \sin ^{-1}\left (1-i d x^2\right )-\frac{6 b \sqrt{2 i d x^2+d^2 x^4} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^3\\ \end{align*}

Mathematica [A]  time = 0.133959, size = 180, normalized size = 1.4 \[ \frac{a d x^2 \left (a^2+24 b^2\right )-6 b \left (a^2+8 b^2\right ) \sqrt{d x^2 \left (d x^2+2 i\right )}+3 i b \sin ^{-1}\left (1-i d x^2\right ) \left (a^2 d x^2-4 a b \sqrt{d x^2 \left (d x^2+2 i\right )}+8 b^2 d x^2\right )+3 b^2 \sin ^{-1}\left (1-i d x^2\right )^2 \left (-a d x^2+2 b \sqrt{d x^2 \left (d x^2+2 i\right )}\right )-i b^3 d x^2 \sin ^{-1}\left (1-i d x^2\right )^3}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*b*ArcSin[1 - I*d*x^2])^3,x]

[Out]

(a*(a^2 + 24*b^2)*d*x^2 - 6*b*(a^2 + 8*b^2)*Sqrt[d*x^2*(2*I + d*x^2)] + (3*I)*b*(a^2*d*x^2 + 8*b^2*d*x^2 - 4*a
*b*Sqrt[d*x^2*(2*I + d*x^2)])*ArcSin[1 - I*d*x^2] + 3*b^2*(-(a*d*x^2) + 2*b*Sqrt[d*x^2*(2*I + d*x^2)])*ArcSin[
1 - I*d*x^2]^2 - I*b^3*d*x^2*ArcSin[1 - I*d*x^2]^3)/(d*x)

________________________________________________________________________________________

Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(I+d*x^2))^3,x)

[Out]

int((a+b*arcsinh(I+d*x^2))^3,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{3} x \log \left (d x^{2} + \sqrt{d x^{2} + 2 i} \sqrt{d} x + i\right )^{3} + 3 \,{\left (x \operatorname{arsinh}\left (d x^{2} + i\right ) - \frac{2 \,{\left (d^{\frac{3}{2}} x^{2} + 2 i \, \sqrt{d}\right )}}{\sqrt{d x^{2} + 2 i} d}\right )} a^{2} b + a^{3} x + \int \frac{{\left (3 \,{\left (a b^{2} d^{2} - 2 \, b^{3} d^{2}\right )} x^{4} - 6 \, a b^{2} +{\left (9 i \, a b^{2} d - 12 i \, b^{3} d\right )} x^{2} +{\left (3 \,{\left (a b^{2} d^{\frac{3}{2}} - 2 \, b^{3} d^{\frac{3}{2}}\right )} x^{3} +{\left (6 i \, a b^{2} \sqrt{d} - 6 i \, b^{3} \sqrt{d}\right )} x\right )} \sqrt{d x^{2} + 2 i}\right )} \log \left (d x^{2} + \sqrt{d x^{2} + 2 i} \sqrt{d} x + i\right )^{2}}{d^{2} x^{4} + 3 i \, d x^{2} +{\left (d^{\frac{3}{2}} x^{3} + 2 i \, \sqrt{d} x\right )} \sqrt{d x^{2} + 2 i} - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(I+d*x^2))^3,x, algorithm="maxima")

[Out]

b^3*x*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I)^3 + 3*(x*arcsinh(d*x^2 + I) - 2*(d^(3/2)*x^2 + 2*I*sqrt(d))
/(sqrt(d*x^2 + 2*I)*d))*a^2*b + a^3*x + integrate((3*(a*b^2*d^2 - 2*b^3*d^2)*x^4 - 6*a*b^2 + (9*I*a*b^2*d - 12
*I*b^3*d)*x^2 + (3*(a*b^2*d^(3/2) - 2*b^3*d^(3/2))*x^3 + (6*I*a*b^2*sqrt(d) - 6*I*b^3*sqrt(d))*x)*sqrt(d*x^2 +
2*I))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I)^2/(d^2*x^4 + 3*I*d*x^2 + (d^(3/2)*x^3 + 2*I*sqrt(d)*x)*sqr
t(d*x^2 + 2*I) - 2), x)

________________________________________________________________________________________

Fricas [A]  time = 2.72941, size = 458, normalized size = 3.55 \begin{align*} \frac{b^{3} d x^{2} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )^{3} +{\left (a^{3} + 24 \, a b^{2}\right )} d x^{2} + 3 \,{\left (a b^{2} d x^{2} - 2 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} b^{3}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )^{2} + 3 \,{\left ({\left (a^{2} b + 8 \, b^{3}\right )} d x^{2} - 4 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} a b^{2}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right ) - 6 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}}{\left (a^{2} b + 8 \, b^{3}\right )}}{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(I+d*x^2))^3,x, algorithm="fricas")

[Out]

(b^3*d*x^2*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I)^3 + (a^3 + 24*a*b^2)*d*x^2 + 3*(a*b^2*d*x^2 - 2*sqrt(d^2
*x^4 + 2*I*d*x^2)*b^3)*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I)^2 + 3*((a^2*b + 8*b^3)*d*x^2 - 4*sqrt(d^2*x^
4 + 2*I*d*x^2)*a*b^2)*log(d*x^2 + sqrt(d^2*x^4 + 2*I*d*x^2) + I) - 6*sqrt(d^2*x^4 + 2*I*d*x^2)*(a^2*b + 8*b^3)
)/(d*x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(I+d*x**2))**3,x)

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(I+d*x^2))^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x^2 + I) + a)^3, x)