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

Optimal. Leaf size=204 \[ \frac{5 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^6}-\frac{15 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^6}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^6}+\frac{15 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^6}-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-(x^5/(b*c*(a + b*ArcSinh[c*x]))) + (5*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^6) - (15*Cosh[
(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^6) + (5*Cosh[(5*a)/b]*CoshIntegral[(5*(a + b*ArcS
inh[c*x]))/b])/(16*b^2*c^6) - (5*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^6) + (15*Sinh[(3*a)/
b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^6) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c*
x]))/b])/(16*b^2*c^6)

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Rubi [A]  time = 0.480139, antiderivative size = 200, normalized size of antiderivative = 0.98, number of steps used = 13, 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} \[ \frac{5 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^6}-\frac{15 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^6}+\frac{15 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^5/(b*c*(a + b*ArcSinh[c*x]))) + (5*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(8*b^2*c^6) - (15*Cosh[(3*a
)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2*c^6) + (5*Cosh[(5*a)/b]*CoshIntegral[(5*a)/b + 5*ArcSinh[
c*x]])/(16*b^2*c^6) - (5*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(8*b^2*c^6) + (15*Sinh[(3*a)/b]*SinhInteg
ral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2*c^6) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c*x]])/(16*b^2
*c^6)

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^5}{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \int \frac{x^4}{a+b \sinh ^{-1}(c x)} \, dx}{b c}\\ &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^4(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^6}\\ &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \left (\frac{\cosh (x)}{8 (a+b x)}-\frac{3 \cosh (3 x)}{16 (a+b x)}+\frac{\cosh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^6}\\ &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^6}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^6}-\frac{15 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^6}\\ &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\left (5 \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 )}{8 b c^6}-\frac{\left (15 \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 )}{16 b c^6}+\frac{\left (5 \cosh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^6}-\frac{\left (5 \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 )}{8 b c^6}+\frac{\left (15 \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 )}{16 b c^6}-\frac{\left (5 \sinh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^6}\\ &=-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{5 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^6}-\frac{15 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^6}+\frac{15 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^6}\\ \end{align*}

Mathematica [A]  time = 0.330466, size = 158, normalized size = 0.77 \[ \frac{5 \left (2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )\right )}{16 b^2 c^6}-\frac{x^5}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^5/(b*c*(a + b*ArcSinh[c*x]))) + (5*(2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] - 3*Cosh[(3*a)/b]*CoshInt
egral[3*(a/b + ArcSinh[c*x])] + Cosh[(5*a)/b]*CoshIntegral[5*(a/b + ArcSinh[c*x])] - 2*Sinh[a/b]*SinhIntegral[
a/b + ArcSinh[c*x]] + 3*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] - Sinh[(5*a)/b]*SinhIntegral[5*(a/b
+ ArcSinh[c*x])]))/(16*b^2*c^6)

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Maple [B]  time = 0.179, size = 633, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(16*c^5*x^5-16*c^4*x^4*(c^2*x^2+1)^(1/2)+20*c^3*x^3-12*c^2*x^2*(c^2*x^2+1)^(1/2)+5*c*x-(c^2*x^2+1)^(1/2)
)/c^6/b/(a+b*arcsinh(c*x))-5/32/c^6/b^2*exp(5*a/b)*Ei(1,5*arcsinh(c*x)+5*a/b)+5/32*(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^6/b/(a+b*arcsinh(c*x))+15/32/c^6/b^2*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a
/b)-5/16*(c*x-(c^2*x^2+1)^(1/2))/c^6/b/(a+b*arcsinh(c*x))-5/16/c^6/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-5/16/c^
6/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))+5/32/c^6/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*a
rcsinh(c*x))-1/32/c^6/b^2*(16*x^5*b*c^5+16*(c^2*x^2+1)^(1/2)*x^4*b*c^4+20*x^3*b*c^3+12*(c^2*x^2+1)^(1/2)*x^2*b
*c^2+5*arcsinh(c*x)*exp(-5*a/b)*Ei(1,-5*arcsinh(c*x)-5*a/b)*b+5*exp(-5*a/b)*Ei(1,-5*arcsinh(c*x)-5*a/b)*a+5*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^{8} + c x^{6} +{\left (c^{2} x^{7} + x^{5}\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{5 \, c^{5} x^{9} + 11 \, c^{3} x^{7} + 6 \, c x^{5} +{\left (5 \, c^{3} x^{7} + 4 \, c x^{5}\right )}{\left (c^{2} x^{2} + 1\right )} + 5 \,{\left (2 \, c^{4} x^{8} + 3 \, c^{2} x^{6} + x^{4}\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^5/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-(c^3*x^8 + c*x^6 + (c^2*x^7 + x^5)*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((5*c^5*x^9 + 11*c^3*x^7 + 6*c*x^5 + (5*c^3*x^7 + 4*c*x^5)*(c^2*x^2 + 1) + 5*(2*c^4*x^8 + 3*c^2*x
^6 + x^4)*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^{5}}{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^5/(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^5/(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^{5}}{\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**5/(a+b*asinh(c*x))**2/(c**2*x**2+1)**(1/2),x)

[Out]

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

[Out]

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