3.240 \(\int (a+b \cosh ^{-1}(1+d x^2))^4 \, dx\)

Optimal. Leaf size=145 \[ -\frac{192 b^3 \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )}{x \sqrt{d x^2} \sqrt{d x^2+2}}+48 b^2 x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2-\frac{8 b \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^3}{x \sqrt{d x^2} \sqrt{d x^2+2}}+x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^4+384 b^4 x \]

[Out]

384*b^4*x - (192*b^3*(2*x^2 + d*x^4)*(a + b*ArcCosh[1 + d*x^2]))/(x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]) + 48*b^2*x*(a
+ b*ArcCosh[1 + d*x^2])^2 - (8*b*(2*x^2 + d*x^4)*(a + b*ArcCosh[1 + d*x^2])^3)/(x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]
) + x*(a + b*ArcCosh[1 + d*x^2])^4

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Rubi [A]  time = 0.0360027, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5880, 8} \[ -\frac{192 b^3 \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )}{x \sqrt{d x^2} \sqrt{d x^2+2}}+48 b^2 x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^2-\frac{8 b \left (d x^4+2 x^2\right ) \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^3}{x \sqrt{d x^2} \sqrt{d x^2+2}}+x \left (a+b \cosh ^{-1}\left (d x^2+1\right )\right )^4+384 b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[1 + d*x^2])^4,x]

[Out]

384*b^4*x - (192*b^3*(2*x^2 + d*x^4)*(a + b*ArcCosh[1 + d*x^2]))/(x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]) + 48*b^2*x*(a
+ b*ArcCosh[1 + d*x^2])^2 - (8*b*(2*x^2 + d*x^4)*(a + b*ArcCosh[1 + d*x^2])^3)/(x*Sqrt[d*x^2]*Sqrt[2 + d*x^2]
) + x*(a + b*ArcCosh[1 + d*x^2])^4

Rule 5880

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcCosh[c + d*x^2])^n, x] +
(Dist[4*b^2*n*(n - 1), Int[(a + b*ArcCosh[c + d*x^2])^(n - 2), x], x] - Simp[(2*b*n*(2*c*d*x^2 + d^2*x^4)*(a +
b*ArcCosh[c + d*x^2])^(n - 1))/(d*x*Sqrt[-1 + c + d*x^2]*Sqrt[1 + c + d*x^2]), 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+b \cosh ^{-1}\left (1+d x^2\right )\right )^4 \, dx &=-\frac{8 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3}{x \sqrt{d x^2} \sqrt{2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^4+\left (48 b^2\right ) \int \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2 \, dx\\ &=-\frac{192 b^3 \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )}{x \sqrt{d x^2} \sqrt{2+d x^2}}+48 b^2 x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2-\frac{8 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3}{x \sqrt{d x^2} \sqrt{2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^4+\left (384 b^4\right ) \int 1 \, dx\\ &=384 b^4 x-\frac{192 b^3 \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )}{x \sqrt{d x^2} \sqrt{2+d x^2}}+48 b^2 x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^2-\frac{8 b \left (2 x^2+d x^4\right ) \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^3}{x \sqrt{d x^2} \sqrt{2+d x^2}}+x \left (a+b \cosh ^{-1}\left (1+d x^2\right )\right )^4\\ \end{align*}

Mathematica [A]  time = 0.220184, size = 264, normalized size = 1.82 \[ \frac{d x^2 \left (48 a^2 b^2+a^4+384 b^4\right )-8 a b \left (a^2+24 b^2\right ) \sqrt{d x^2} \sqrt{d x^2+2}+6 b^2 \cosh ^{-1}\left (d x^2+1\right )^2 \left (a^2 d x^2-4 a b \sqrt{d x^2} \sqrt{d x^2+2}+8 b^2 d x^2\right )+4 b \cosh ^{-1}\left (d x^2+1\right ) \left (-6 a^2 b \sqrt{d x^2} \sqrt{d x^2+2}+a^3 d x^2+24 a b^2 d x^2-48 b^3 \sqrt{d x^2} \sqrt{d x^2+2}\right )+4 b^3 \cosh ^{-1}\left (d x^2+1\right )^3 \left (a d x^2-2 b \sqrt{d x^2} \sqrt{d x^2+2}\right )+b^4 d x^2 \cosh ^{-1}\left (d x^2+1\right )^4}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 + 48*a^2*b^2 + 384*b^4)*d*x^2 - 8*a*b*(a^2 + 24*b^2)*Sqrt[d*x^2]*Sqrt[2 + d*x^2] + 4*b*(a^3*d*x^2 + 24*a
*b^2*d*x^2 - 6*a^2*b*Sqrt[d*x^2]*Sqrt[2 + d*x^2] - 48*b^3*Sqrt[d*x^2]*Sqrt[2 + d*x^2])*ArcCosh[1 + d*x^2] + 6*
b^2*(a^2*d*x^2 + 8*b^2*d*x^2 - 4*a*b*Sqrt[d*x^2]*Sqrt[2 + d*x^2])*ArcCosh[1 + d*x^2]^2 + 4*b^3*(a*d*x^2 - 2*b*
Sqrt[d*x^2]*Sqrt[2 + d*x^2])*ArcCosh[1 + d*x^2]^3 + b^4*d*x^2*ArcCosh[1 + d*x^2]^4)/(d*x)

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Maple [F]  time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arccosh} \left (d{x}^{2}+1\right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x^2+1))^4,x)

[Out]

int((a+b*arccosh(d*x^2+1))^4,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x^2+1))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.08347, size = 626, normalized size = 4.32 \begin{align*} \frac{b^{4} d x^{2} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 \, d x^{2}} + 1\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 \, d x^{2}} b^{4}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 \, d x^{2}} + 1\right )^{3} - 6 \,{\left (4 \, \sqrt{d^{2} x^{4} + 2 \, 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 \, d x^{2}} + 1\right )^{2} + 4 \,{\left ({\left (a^{3} b + 24 \, a b^{3}\right )} d x^{2} - 6 \, \sqrt{d^{2} x^{4} + 2 \, d x^{2}}{\left (a^{2} b^{2} + 8 \, b^{4}\right )}\right )} \log \left (d x^{2} + \sqrt{d^{2} x^{4} + 2 \, d x^{2}} + 1\right ) - 8 \, \sqrt{d^{2} x^{4} + 2 \, 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*arccosh(d*x^2+1))^4,x, algorithm="fricas")

[Out]

(b^4*d*x^2*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1)^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*d*x^2)*b^4)*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1)^3 - 6*(4*sqrt(d^2*x^4 + 2*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*d*x^2) + 1)^2 + 4*((a^3*b + 24*a*b^3)*d*x^2 - 6*sqrt(
d^2*x^4 + 2*d*x^2)*(a^2*b^2 + 8*b^4))*log(d*x^2 + sqrt(d^2*x^4 + 2*d*x^2) + 1) - 8*sqrt(d^2*x^4 + 2*d*x^2)*(a^
3*b + 24*a*b^3))/(d*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{acosh}{\left (d x^{2} + 1 \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x**2+1))**4,x)

[Out]

Integral((a + b*acosh(d*x**2 + 1))**4, x)

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Giac [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((a+b*arccosh(d*x^2+1))^4,x, algorithm="giac")

[Out]

Exception raised: RuntimeError