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

Optimal. Leaf size=76 \[ -\frac{4 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2+8 b^2 x \]

[Out]

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

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Rubi [A]  time = 0.01526, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4814, 8} \[ -\frac{4 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2+8 b^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2 \, dx &=-\frac{4 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 )}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2+\left (8 b^2\right ) \int 1 \, dx\\ &=8 b^2 x-\frac{4 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 )}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2\\ \end{align*}

Mathematica [A]  time = 0.0257147, size = 76, normalized size = 1. \[ -\frac{4 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^2+8 b^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \,{\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 b +{\left (x \log \left (d x^{2} + \sqrt{d x^{2} + 2 i} \sqrt{d} x + i\right )^{2} - \int \frac{{\left (4 \, d^{2} x^{4} + 8 i \, d x^{2} +{\left (4 \, d^{\frac{3}{2}} x^{3} + 4 i \, \sqrt{d} 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 )}{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}\right )} b^{2} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x*arcsinh(d*x^2 + I) - 2*(d^(3/2)*x^2 + 2*I*sqrt(d))/(sqrt(d*x^2 + 2*I)*d))*a*b + (x*log(d*x^2 + sqrt(d*x^2
+ 2*I)*sqrt(d)*x + I)^2 - integrate((4*d^2*x^4 + 8*I*d*x^2 + (4*d^(3/2)*x^3 + 4*I*sqrt(d)*x)*sqrt(d*x^2 + 2*I
))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I)/(d^2*x^4 + 3*I*d*x^2 + (d^(3/2)*x^3 + 2*I*sqrt(d)*x)*sqrt(d*x^
2 + 2*I) - 2), x))*b^2 + a^2*x

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Fricas [B]  time = 2.64473, size = 288, normalized size = 3.79 \begin{align*} \frac{b^{2} d x^{2} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )^{2} +{\left (a^{2} + 8 \, b^{2}\right )} d x^{2} - 4 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} a b + 2 \,{\left (a b d x^{2} - 2 \, \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} b^{2}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 i \, d x^{2}} + i\right )}{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**2,x)

[Out]

Exception raised: TypeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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