3.321 \(\int (a-i b \sin ^{-1}(1+i d x^2))^4 \, dx\)

Optimal. Leaf size=153 \[ -\frac{192 b^3 \sqrt{d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )}{d x}+48 b^2 x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2-\frac{8 b \sqrt{d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^3}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4+384 b^4 x \]

[Out]

384*b^4*x - (192*b^3*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2]))/(d*x) + 48*b^2*x*(a - I*b*Arc
Sin[1 + I*d*x^2])^2 - (8*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2])^3)/(d*x) + x*(a - I*b*Ar
cSin[1 + I*d*x^2])^4

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Rubi [A]  time = 0.0318743, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4814, 8} \[ -\frac{192 b^3 \sqrt{d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )}{d x}+48 b^2 x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2-\frac{8 b \sqrt{d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^3}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4+384 b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a - I*b*ArcSin[1 + I*d*x^2])^4,x]

[Out]

384*b^4*x - (192*b^3*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2]))/(d*x) + 48*b^2*x*(a - I*b*Arc
Sin[1 + I*d*x^2])^2 - (8*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2])^3)/(d*x) + x*(a - I*b*Ar
cSin[1 + I*d*x^2])^4

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 )^4 \, dx &=-\frac{8 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 )^3}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4+\left (48 b^2\right ) \int \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2 \, dx\\ &=-\frac{192 b^3 \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}+48 b^2 x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2-\frac{8 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 )^3}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4+\left (384 b^4\right ) \int 1 \, dx\\ &=384 b^4 x-\frac{192 b^3 \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}+48 b^2 x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2-\frac{8 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 )^3}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4\\ \end{align*}

Mathematica [A]  time = 0.116773, size = 149, normalized size = 0.97 \[ 48 b^2 \left (-\frac{4 b \sqrt{d x^2 \left (d x^2-2 i\right )} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^2+8 b^2 x\right )+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^4-\frac{8 b \sqrt{d x^2 \left (d x^2-2 i\right )} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^3}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - I*b*ArcSin[1 + I*d*x^2])^4,x]

[Out]

(-8*b*Sqrt[d*x^2*(-2*I + d*x^2)]*(a - I*b*ArcSin[1 + I*d*x^2])^3)/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^4 +
48*b^2*(8*b^2*x - (4*b*Sqrt[d*x^2*(-2*I + d*x^2)]*(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 = 0.125, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( -i+d{x}^{2} \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(-I+d*x^2))^4,x)

[Out]

int((a+b*arcsinh(-I+d*x^2))^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{4} x \log \left (d x^{2} + \sqrt{d x^{2} - 2 i} \sqrt{d} x - i\right )^{4} + 4 \,{\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^{3} b + a^{4} x + \int \frac{{\left (4 \,{\left (a b^{3} d^{2} - 2 \, b^{4} d^{2}\right )} x^{4} - 8 \, a b^{3} +{\left (-12 i \, a b^{3} d + 16 i \, b^{4} d\right )} x^{2} +{\left (4 \,{\left (a b^{3} d^{\frac{3}{2}} - 2 \, b^{4} d^{\frac{3}{2}}\right )} x^{3} +{\left (-8 i \, a b^{3} \sqrt{d} + 8 i \, b^{4} \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 )^{3} +{\left (6 \, a^{2} b^{2} d^{2} x^{4} - 18 i \, a^{2} b^{2} d x^{2} - 12 \, a^{2} b^{2} + 6 \,{\left (a^{2} b^{2} d^{\frac{3}{2}} x^{3} - 2 i \, a^{2} b^{2} \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 )^{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))^4,x, algorithm="maxima")

[Out]

b^4*x*log(d*x^2 + sqrt(d*x^2 - 2*I)*sqrt(d)*x - I)^4 + 4*(x*arcsinh(d*x^2 - I) - 2*(d^(3/2)*x^2 - 2*I*sqrt(d))
/(sqrt(d*x^2 - 2*I)*d))*a^3*b + a^4*x + integrate(((4*(a*b^3*d^2 - 2*b^4*d^2)*x^4 - 8*a*b^3 + (-12*I*a*b^3*d +
16*I*b^4*d)*x^2 + (4*(a*b^3*d^(3/2) - 2*b^4*d^(3/2))*x^3 + (-8*I*a*b^3*sqrt(d) + 8*I*b^4*sqrt(d))*x)*sqrt(d*x
^2 - 2*I))*log(d*x^2 + sqrt(d*x^2 - 2*I)*sqrt(d)*x - I)^3 + (6*a^2*b^2*d^2*x^4 - 18*I*a^2*b^2*d*x^2 - 12*a^2*b
^2 + 6*(a^2*b^2*d^(3/2)*x^3 - 2*I*a^2*b^2*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)*sqrt(d*x^2 - 2*I) - 2), x)

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Fricas [B]  time = 2.79762, size = 648, normalized size = 4.24 \begin{align*} \frac{b^{4} d x^{2} \log \left (d x^{2} + \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} - i\right )^{4} +{\left (a^{4} + 48 \, a^{2} b^{2} + 384 \, b^{4}\right )} d x^{2} + 4 \,{\left (a b^{3} d x^{2} - 2 \, \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} b^{4}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} - i\right )^{3} - 6 \,{\left (4 \, \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} a b^{3} -{\left (a^{2} b^{2} + 8 \, b^{4}\right )} d x^{2}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} - i\right )^{2} + 4 \,{\left ({\left (a^{3} b + 24 \, a b^{3}\right )} d x^{2} - 6 \, \sqrt{d^{2} x^{4} - 2 i \, d x^{2}}{\left (a^{2} b^{2} + 8 \, b^{4}\right )}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} - 2 i \, d x^{2}} - i\right ) - 8 \, \sqrt{d^{2} x^{4} - 2 i \, d x^{2}}{\left (a^{3} b + 24 \, a 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))^4,x, algorithm="fricas")

[Out]

(b^4*d*x^2*log(d*x^2 + sqrt(d^2*x^4 - 2*I*d*x^2) - I)^4 + (a^4 + 48*a^2*b^2 + 384*b^4)*d*x^2 + 4*(a*b^3*d*x^2
- 2*sqrt(d^2*x^4 - 2*I*d*x^2)*b^4)*log(d*x^2 + sqrt(d^2*x^4 - 2*I*d*x^2) - I)^3 - 6*(4*sqrt(d^2*x^4 - 2*I*d*x^
2)*a*b^3 - (a^2*b^2 + 8*b^4)*d*x^2)*log(d*x^2 + sqrt(d^2*x^4 - 2*I*d*x^2) - I)^2 + 4*((a^3*b + 24*a*b^3)*d*x^2
- 6*sqrt(d^2*x^4 - 2*I*d*x^2)*(a^2*b^2 + 8*b^4))*log(d*x^2 + sqrt(d^2*x^4 - 2*I*d*x^2) - I) - 8*sqrt(d^2*x^4
- 2*I*d*x^2)*(a^3*b + 24*a*b^3))/(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))**4,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 )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(-I+d*x^2))^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x^2 - I) + a)^4, x)