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

Optimal. Leaf size=206 \[ -\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}+\frac{\cosh \left (\frac{6 a}{b}\right ) \text{Chi}\left (\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}-\frac{\sinh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3} \]

[Out]

-(Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(32*b*c^3) + (Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*A
rcSinh[c*x]))/b])/(16*b*c^3) + (Cosh[(6*a)/b]*CoshIntegral[(6*(a + b*ArcSinh[c*x]))/b])/(32*b*c^3) - Log[a + b
*ArcSinh[c*x]]/(16*b*c^3) + (Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(32*b*c^3) - (Sinh[(4*a)/
b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(16*b*c^3) - (Sinh[(6*a)/b]*SinhIntegral[(6*(a + b*ArcSinh[c*x]))
/b])/(32*b*c^3)

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Rubi [A]  time = 0.494725, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {5779, 5448, 3303, 3298, 3301} \[ -\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac{\cosh \left (\frac{6 a}{b}\right ) \text{Chi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}+\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\sinh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[(2*a)/b]*CoshIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(32*b*c^3) + (Cosh[(4*a)/b]*CoshIntegral[(4*a)/b + 4*A
rcSinh[c*x]])/(16*b*c^3) + (Cosh[(6*a)/b]*CoshIntegral[(6*a)/b + 6*ArcSinh[c*x]])/(32*b*c^3) - Log[a + b*ArcSi
nh[c*x]]/(16*b*c^3) + (Sinh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(32*b*c^3) - (Sinh[(4*a)/b]*SinhI
ntegral[(4*a)/b + 4*ArcSinh[c*x]])/(16*b*c^3) - (Sinh[(6*a)/b]*SinhIntegral[(6*a)/b + 6*ArcSinh[c*x]])/(32*b*c
^3)

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 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^2 \left (1+c^2 x^2\right )^{3/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{16 (a+b x)}-\frac{\cosh (2 x)}{32 (a+b x)}+\frac{\cosh (4 x)}{16 (a+b x)}+\frac{\cosh (6 x)}{32 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (6 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}\\ &=-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\cosh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}+\frac{\cosh \left (\frac{6 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac{\sinh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}-\frac{\sinh \left (\frac{6 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}\\ &=-\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac{\cosh \left (\frac{6 a}{b}\right ) \text{Chi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\sinh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}\\ \end{align*}

Mathematica [A]  time = 0.480728, size = 152, normalized size = 0.74 \[ \frac{-\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+2 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{6 a}{b}\right ) \text{Chi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac{6 a}{b}\right ) \text{Shi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \log \left (a+b \sinh ^{-1}(c x)\right )}{32 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])]) + 2*Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcSinh[c*x])]
+ Cosh[(6*a)/b]*CoshIntegral[6*(a/b + ArcSinh[c*x])] - 2*Log[a + b*ArcSinh[c*x]] + Sinh[(2*a)/b]*SinhIntegral[
2*(a/b + ArcSinh[c*x])] - 2*Sinh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - Sinh[(6*a)/b]*SinhIntegral[6*
(a/b + ArcSinh[c*x])])/(32*b*c^3)

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Maple [A]  time = 0.208, size = 199, normalized size = 1. \begin{align*} -{\frac{\ln \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) }{16\,b{c}^{3}}}-{\frac{1}{64\,b{c}^{3}}{{\rm e}^{6\,{\frac{a}{b}}}}{\it Ei} \left ( 1,6\,{\it Arcsinh} \left ( cx \right ) +6\,{\frac{a}{b}} \right ) }-{\frac{1}{32\,b{c}^{3}}{{\rm e}^{4\,{\frac{a}{b}}}}{\it Ei} \left ( 1,4\,{\it Arcsinh} \left ( cx \right ) +4\,{\frac{a}{b}} \right ) }+{\frac{1}{64\,b{c}^{3}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( cx \right ) +2\,{\frac{a}{b}} \right ) }+{\frac{1}{64\,b{c}^{3}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) }-{\frac{1}{32\,b{c}^{3}}{{\rm e}^{-4\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-4\,{\it Arcsinh} \left ( cx \right ) -4\,{\frac{a}{b}} \right ) }-{\frac{1}{64\,b{c}^{3}}{{\rm e}^{-6\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-6\,{\it Arcsinh} \left ( cx \right ) -6\,{\frac{a}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*ln(a+b*arcsinh(c*x))/b/c^3-1/64/c^3/b*exp(6*a/b)*Ei(1,6*arcsinh(c*x)+6*a/b)-1/32/c^3/b*exp(4*a/b)*Ei(1,4
*arcsinh(c*x)+4*a/b)+1/64/c^3/b*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)+1/64/c^3/b*exp(-2*a/b)*Ei(1,-2*arcsinh(c
*x)-2*a/b)-1/32/c^3/b*exp(-4*a/b)*Ei(1,-4*arcsinh(c*x)-4*a/b)-1/64/c^3/b*exp(-6*a/b)*Ei(1,-6*arcsinh(c*x)-6*a/
b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} x^{4} + x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{b \operatorname{arsinh}\left (c x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^2*x^4 + x^2)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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