3.437 \(\int \frac{x^3}{\sqrt{1+c^2 x^2} (a+b \sinh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=142 \[ -\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{4 b^2 c^4}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^4}+\frac{3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{4 b^2 c^4}-\frac{3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-(x^3/(b*c*(a + b*ArcSinh[c*x]))) - (3*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(4*b^2*c^4) + (3*Cosh[(
3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b^2*c^4) + (3*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])
/b])/(4*b^2*c^4) - (3*Sinh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b^2*c^4)

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Rubi [A]  time = 0.366808, antiderivative size = 138, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5774, 5669, 5448, 3303, 3298, 3301} \[ -\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2),x]

[Out]

-(x^3/(b*c*(a + b*ArcSinh[c*x]))) - (3*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(4*b^2*c^4) + (3*Cosh[(3*a)
/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b^2*c^4) + (3*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(4*b^
2*c^4) - (3*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b^2*c^4)

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^3}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{3 \int \frac{x^2}{a+b \sinh ^{-1}(c x)} \, dx}{b c}\\ &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{\left (3 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac{\left (3 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac{\left (3 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}-\frac{\left (3 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.260961, size = 113, normalized size = 0.8 \[ \frac{3 \left (-\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2),x]

[Out]

-(x^3/(b*c*(a + b*ArcSinh[c*x]))) + (3*(-(Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]]) + Cosh[(3*a)/b]*CoshInte
gral[3*(a/b + ArcSinh[c*x])] + Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] - Sinh[(3*a)/b]*SinhIntegral[3*(a/b
+ ArcSinh[c*x])]))/(4*b^2*c^4)

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Maple [B]  time = 0.128, size = 364, normalized size = 2.6 \begin{align*} -{\frac{1}{8\,b{c}^{4} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( 4\,{c}^{3}{x}^{3}-4\,{c}^{2}{x}^{2}\sqrt{{c}^{2}{x}^{2}+1}+3\,cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{3}{8\,{c}^{4}{b}^{2}}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\it Arcsinh} \left ( cx \right ) +3\,{\frac{a}{b}} \right ) }+{\frac{3}{8\,b{c}^{4} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{3}{8\,{c}^{4}{b}^{2}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }+{\frac{3}{8\,{c}^{4}{b}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ({\it Arcsinh} \left ( cx \right ){\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ){{\rm e}^{-{\frac{a}{b}}}}b+{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ){{\rm e}^{-{\frac{a}{b}}}}a+xbc+\sqrt{{c}^{2}{x}^{2}+1}b \right ) }-{\frac{1}{8\,{c}^{4}{b}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( 4\,{x}^{3}b{c}^{3}+4\,\sqrt{{c}^{2}{x}^{2}+1}{x}^{2}b{c}^{2}+3\,{\it Arcsinh} \left ( cx \right ){{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( cx \right ) -3\,{\frac{a}{b}} \right ) b+3\,{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( cx \right ) -3\,{\frac{a}{b}} \right ) a+3\,xbc+\sqrt{{c}^{2}{x}^{2}+1}b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x)

[Out]

-1/8*(4*c^3*x^3-4*c^2*x^2*(c^2*x^2+1)^(1/2)+3*c*x-(c^2*x^2+1)^(1/2))/c^4/b/(a+b*arcsinh(c*x))-3/8/c^4/b^2*exp(
3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)+3/8*(c*x-(c^2*x^2+1)^(1/2))/c^4/b/(a+b*arcsinh(c*x))+3/8/c^4/b^2*exp(a/b)*Ei
(1,arcsinh(c*x)+a/b)+3/8/c^4/b^2*(arcsinh(c*x)*Ei(1,-arcsinh(c*x)-a/b)*exp(-a/b)*b+Ei(1,-arcsinh(c*x)-a/b)*exp
(-a/b)*a+x*b*c+(c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(c*x))-1/8/c^4/b^2*(4*x^3*b*c^3+4*(c^2*x^2+1)^(1/2)*x^2*b*c^2+
3*arcsinh(c*x)*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)*b+3*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)*a+3*x*b*c+(
c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(c*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{c^{3} x^{6} + c x^{4} +{\left (c^{2} x^{5} + x^{3}\right )} \sqrt{c^{2} x^{2} + 1}}{{\left (c^{2} x^{2} + 1\right )} a b c^{2} x +{\left ({\left (c^{2} x^{2} + 1\right )} b^{2} c^{2} x +{\left (b^{2} c^{3} x^{2} + b^{2} c\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (a b c^{3} x^{2} + a b c\right )} \sqrt{c^{2} x^{2} + 1}} + \int \frac{3 \, c^{5} x^{7} + 7 \, c^{3} x^{5} + 4 \, c x^{3} +{\left (3 \, c^{3} x^{5} + 2 \, c x^{3}\right )}{\left (c^{2} x^{2} + 1\right )} + 3 \,{\left (2 \, c^{4} x^{6} + 3 \, c^{2} x^{4} + x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} a b c^{3} x^{2} + 2 \,{\left (a b c^{4} x^{3} + a b c^{2} x\right )}{\left (c^{2} x^{2} + 1\right )} +{\left ({\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b^{2} c^{3} x^{2} + 2 \,{\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (b^{2} c^{5} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{2} c\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (a b c^{5} x^{4} + 2 \, a b c^{3} x^{2} + a b c\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-(c^3*x^6 + c*x^4 + (c^2*x^5 + x^3)*sqrt(c^2*x^2 + 1))/((c^2*x^2 + 1)*a*b*c^2*x + ((c^2*x^2 + 1)*b^2*c^2*x + (
b^2*c^3*x^2 + b^2*c)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + (a*b*c^3*x^2 + a*b*c)*sqrt(c^2*x^2 + 1)
) + integrate((3*c^5*x^7 + 7*c^3*x^5 + 4*c*x^3 + (3*c^3*x^5 + 2*c*x^3)*(c^2*x^2 + 1) + 3*(2*c^4*x^6 + 3*c^2*x^
4 + x^2)*sqrt(c^2*x^2 + 1))/((c^2*x^2 + 1)^(3/2)*a*b*c^3*x^2 + 2*(a*b*c^4*x^3 + a*b*c^2*x)*(c^2*x^2 + 1) + ((c
^2*x^2 + 1)^(3/2)*b^2*c^3*x^2 + 2*(b^2*c^4*x^3 + b^2*c^2*x)*(c^2*x^2 + 1) + (b^2*c^5*x^4 + 2*b^2*c^3*x^2 + b^2
*c)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + (a*b*c^5*x^4 + 2*a*b*c^3*x^2 + a*b*c)*sqrt(c^2*x^2 + 1))
, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{3}}{a^{2} c^{2} x^{2} +{\left (b^{2} c^{2} x^{2} + b^{2}\right )} \operatorname{arsinh}\left (c x\right )^{2} + a^{2} + 2 \,{\left (a b c^{2} x^{2} + a b\right )} \operatorname{arsinh}\left (c x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*x^2 + 1)*x^3/(a^2*c^2*x^2 + (b^2*c^2*x^2 + b^2)*arcsinh(c*x)^2 + a^2 + 2*(a*b*c^2*x^2 + a*b)
*arcsinh(c*x)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2} \sqrt{c^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+b*asinh(c*x))**2/(c**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3/((a + b*asinh(c*x))**2*sqrt(c**2*x**2 + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{c^{2} x^{2} + 1}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*x) + a)^2), x)