3.372 \(\int \frac{x^3 (1+c^2 x^2)^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx\)

Optimal. Leaf size=245 \[ \frac{3 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{128 b c^4}+\frac{\sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^4}-\frac{3 \sinh \left (\frac{7 a}{b}\right ) \text{Chi}\left (\frac{7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}-\frac{\sinh \left (\frac{9 a}{b}\right ) \text{Chi}\left (\frac{9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}-\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{128 b c^4}-\frac{\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^4}+\frac{3 \cosh \left (\frac{7 a}{b}\right ) \text{Shi}\left (\frac{7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}+\frac{\cosh \left (\frac{9 a}{b}\right ) \text{Shi}\left (\frac{9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4} \]

[Out]

(3*CoshIntegral[(a + b*ArcSinh[c*x])/b]*Sinh[a/b])/(128*b*c^4) + (CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b]*Sin
h[(3*a)/b])/(32*b*c^4) - (3*CoshIntegral[(7*(a + b*ArcSinh[c*x]))/b]*Sinh[(7*a)/b])/(256*b*c^4) - (CoshIntegra
l[(9*(a + b*ArcSinh[c*x]))/b]*Sinh[(9*a)/b])/(256*b*c^4) - (3*Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/
(128*b*c^4) - (Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(32*b*c^4) + (3*Cosh[(7*a)/b]*SinhInteg
ral[(7*(a + b*ArcSinh[c*x]))/b])/(256*b*c^4) + (Cosh[(9*a)/b]*SinhIntegral[(9*(a + b*ArcSinh[c*x]))/b])/(256*b
*c^4)

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Rubi [A]  time = 0.576533, antiderivative size = 241, normalized size of antiderivative = 0.98, number of steps used = 15, 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} \[ \frac{3 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{128 b c^4}+\frac{\sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{32 b c^4}-\frac{3 \sinh \left (\frac{7 a}{b}\right ) \text{Chi}\left (\frac{7 a}{b}+7 \sinh ^{-1}(c x)\right )}{256 b c^4}-\frac{\sinh \left (\frac{9 a}{b}\right ) \text{Chi}\left (\frac{9 a}{b}+9 \sinh ^{-1}(c x)\right )}{256 b c^4}-\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{128 b c^4}-\frac{\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{32 b c^4}+\frac{3 \cosh \left (\frac{7 a}{b}\right ) \text{Shi}\left (\frac{7 a}{b}+7 \sinh ^{-1}(c x)\right )}{256 b c^4}+\frac{\cosh \left (\frac{9 a}{b}\right ) \text{Shi}\left (\frac{9 a}{b}+9 \sinh ^{-1}(c x)\right )}{256 b c^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(3*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b])/(128*b*c^4) + (CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]]*Sinh[(3*a
)/b])/(32*b*c^4) - (3*CoshIntegral[(7*a)/b + 7*ArcSinh[c*x]]*Sinh[(7*a)/b])/(256*b*c^4) - (CoshIntegral[(9*a)/
b + 9*ArcSinh[c*x]]*Sinh[(9*a)/b])/(256*b*c^4) - (3*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(128*b*c^4) -
(Cosh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(32*b*c^4) + (3*Cosh[(7*a)/b]*SinhIntegral[(7*a)/b + 7*
ArcSinh[c*x]])/(256*b*c^4) + (Cosh[(9*a)/b]*SinhIntegral[(9*a)/b + 9*ArcSinh[c*x]])/(256*b*c^4)

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^3 \left (1+c^2 x^2\right )^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^6(x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{3 \sinh (x)}{128 (a+b x)}-\frac{\sinh (3 x)}{32 (a+b x)}+\frac{3 \sinh (7 x)}{256 (a+b x)}+\frac{\sinh (9 x)}{256 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (9 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (7 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{128 c^4}-\frac{\operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^4}\\ &=-\frac{\left (3 \cosh \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 )}{128 c^4}-\frac{\cosh \left (\frac{3 a}{b}\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 )}{32 c^4}+\frac{\left (3 \cosh \left (\frac{7 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac{\cosh \left (\frac{9 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{9 a}{b}+9 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac{\left (3 \sinh \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 )}{128 c^4}+\frac{\sinh \left (\frac{3 a}{b}\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 )}{32 c^4}-\frac{\left (3 \sinh \left (\frac{7 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}-\frac{\sinh \left (\frac{9 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{9 a}{b}+9 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}\\ &=\frac{3 \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac{a}{b}\right )}{128 b c^4}+\frac{\text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{3 a}{b}\right )}{32 b c^4}-\frac{3 \text{Chi}\left (\frac{7 a}{b}+7 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{7 a}{b}\right )}{256 b c^4}-\frac{\text{Chi}\left (\frac{9 a}{b}+9 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{9 a}{b}\right )}{256 b c^4}-\frac{3 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{128 b c^4}-\frac{\cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{32 b c^4}+\frac{3 \cosh \left (\frac{7 a}{b}\right ) \text{Shi}\left (\frac{7 a}{b}+7 \sinh ^{-1}(c x)\right )}{256 b c^4}+\frac{\cosh \left (\frac{9 a}{b}\right ) \text{Shi}\left (\frac{9 a}{b}+9 \sinh ^{-1}(c x)\right )}{256 b c^4}\\ \end{align*}

Mathematica [A]  time = 0.969224, size = 180, normalized size = 0.73 \[ \frac{6 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+8 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-3 \sinh \left (\frac{7 a}{b}\right ) \text{Chi}\left (7 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac{9 a}{b}\right ) \text{Chi}\left (9 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-6 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-8 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+3 \cosh \left (\frac{7 a}{b}\right ) \text{Shi}\left (7 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{9 a}{b}\right ) \text{Shi}\left (9 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{256 b c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(6*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] + 8*CoshIntegral[3*(a/b + ArcSinh[c*x])]*Sinh[(3*a)/b] - 3*CoshI
ntegral[7*(a/b + ArcSinh[c*x])]*Sinh[(7*a)/b] - CoshIntegral[9*(a/b + ArcSinh[c*x])]*Sinh[(9*a)/b] - 6*Cosh[a/
b]*SinhIntegral[a/b + ArcSinh[c*x]] - 8*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] + 3*Cosh[(7*a)/b]*S
inhIntegral[7*(a/b + ArcSinh[c*x])] + Cosh[(9*a)/b]*SinhIntegral[9*(a/b + ArcSinh[c*x])])/(256*b*c^4)

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Maple [A]  time = 0.35, size = 238, normalized size = 1. \begin{align*}{\frac{1}{512\,{c}^{4}b}{{\rm e}^{9\,{\frac{a}{b}}}}{\it Ei} \left ( 1,9\,{\it Arcsinh} \left ( cx \right ) +9\,{\frac{a}{b}} \right ) }+{\frac{3}{512\,{c}^{4}b}{{\rm e}^{7\,{\frac{a}{b}}}}{\it Ei} \left ( 1,7\,{\it Arcsinh} \left ( cx \right ) +7\,{\frac{a}{b}} \right ) }-{\frac{1}{64\,{c}^{4}b}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\it Arcsinh} \left ( cx \right ) +3\,{\frac{a}{b}} \right ) }-{\frac{3}{256\,{c}^{4}b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }+{\frac{3}{256\,{c}^{4}b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }+{\frac{1}{64\,{c}^{4}b}{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( cx \right ) -3\,{\frac{a}{b}} \right ) }-{\frac{3}{512\,{c}^{4}b}{{\rm e}^{-7\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-7\,{\it Arcsinh} \left ( cx \right ) -7\,{\frac{a}{b}} \right ) }-{\frac{1}{512\,{c}^{4}b}{{\rm e}^{-9\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-9\,{\it Arcsinh} \left ( cx \right ) -9\,{\frac{a}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x)

[Out]

1/512/c^4/b*exp(9*a/b)*Ei(1,9*arcsinh(c*x)+9*a/b)+3/512/c^4/b*exp(7*a/b)*Ei(1,7*arcsinh(c*x)+7*a/b)-1/64/c^4/b
*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-3/256/c^4/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)+3/256/c^4/b*exp(-a/b)*Ei(1,
-arcsinh(c*x)-a/b)+1/64/c^4/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-3/512/c^4/b*exp(-7*a/b)*Ei(1,-7*arcsinh(
c*x)-7*a/b)-1/512/c^4/b*exp(-9*a/b)*Ei(1,-9*arcsinh(c*x)-9*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{5}{2}} x^{3}}{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^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

integrate((c^2*x^2 + 1)^(5/2)*x^3/(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^{4} x^{7} + 2 \, c^{2} x^{5} + x^{3}\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^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((c^4*x^7 + 2*c^2*x^5 + x^3)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c**2*x**2+1)**(5/2)/(a+b*asinh(c*x)),x)

[Out]

Timed out

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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{5}{2}} x^{3}}{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^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

integrate((c^2*x^2 + 1)^(5/2)*x^3/(b*arcsinh(c*x) + a), x)