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

Optimal. Leaf size=94 \[ -\frac{c^3 \left (a^2 x^2+1\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{35 c^3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac{63 c^3 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac{35 c^3 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac{7 c^3 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \]

[Out]

-((c^3*(1 + a^2*x^2)^(7/2))/(a*ArcSinh[a*x])) + (35*c^3*SinhIntegral[ArcSinh[a*x]])/(64*a) + (63*c^3*SinhInteg
ral[3*ArcSinh[a*x]])/(64*a) + (35*c^3*SinhIntegral[5*ArcSinh[a*x]])/(64*a) + (7*c^3*SinhIntegral[7*ArcSinh[a*x
]])/(64*a)

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Rubi [A]  time = 0.183676, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5696, 5779, 5448, 3298} \[ -\frac{c^3 \left (a^2 x^2+1\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{35 c^3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac{63 c^3 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac{35 c^3 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac{7 c^3 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*(1 + a^2*x^2)^(7/2))/(a*ArcSinh[a*x])) + (35*c^3*SinhIntegral[ArcSinh[a*x]])/(64*a) + (63*c^3*SinhInteg
ral[3*ArcSinh[a*x]])/(64*a) + (35*c^3*SinhIntegral[5*ArcSinh[a*x]])/(64*a) + (7*c^3*SinhIntegral[7*ArcSinh[a*x
]])/(64*a)

Rule 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]
*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1),
x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

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 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]

Rubi steps

\begin{align*} \int \frac{\left (c+a^2 c x^2\right )^3}{\sinh ^{-1}(a x)^2} \, dx &=-\frac{c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\left (7 a c^3\right ) \int \frac{x \left (1+a^2 x^2\right )^{5/2}}{\sinh ^{-1}(a x)} \, dx\\ &=-\frac{c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{\left (7 c^3\right ) \operatorname{Subst}\left (\int \frac{\cosh ^6(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac{c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{\left (7 c^3\right ) \operatorname{Subst}\left (\int \left (\frac{5 \sinh (x)}{64 x}+\frac{9 \sinh (3 x)}{64 x}+\frac{5 \sinh (5 x)}{64 x}+\frac{\sinh (7 x)}{64 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac{c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{\left (7 c^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (7 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac{\left (35 c^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac{\left (35 c^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac{\left (63 c^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}\\ &=-\frac{c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac{35 c^3 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac{63 c^3 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac{35 c^3 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac{7 c^3 \text{Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a}\\ \end{align*}

Mathematica [A]  time = 0.48972, size = 82, normalized size = 0.87 \[ \frac{c^3 \left (-64 \left (a^2 x^2+1\right )^{7/2}+35 \sinh ^{-1}(a x) \text{Shi}\left (\sinh ^{-1}(a x)\right )+63 \sinh ^{-1}(a x) \text{Shi}\left (3 \sinh ^{-1}(a x)\right )+35 \sinh ^{-1}(a x) \text{Shi}\left (5 \sinh ^{-1}(a x)\right )+7 \sinh ^{-1}(a x) \text{Shi}\left (7 \sinh ^{-1}(a x)\right )\right )}{64 a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(-64*(1 + a^2*x^2)^(7/2) + 35*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] + 63*ArcSinh[a*x]*SinhIntegral[3*Ar
cSinh[a*x]] + 35*ArcSinh[a*x]*SinhIntegral[5*ArcSinh[a*x]] + 7*ArcSinh[a*x]*SinhIntegral[7*ArcSinh[a*x]]))/(64
*a*ArcSinh[a*x])

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Maple [A]  time = 0.052, size = 106, normalized size = 1.1 \begin{align*}{\frac{{c}^{3}}{64\,a{\it Arcsinh} \left ( ax \right ) } \left ( 35\,{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) +63\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) +35\,{\it Shi} \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) +7\,{\it Shi} \left ( 7\,{\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) -21\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) -7\,\cosh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) -\cosh \left ( 7\,{\it Arcsinh} \left ( ax \right ) \right ) -35\,\sqrt{{a}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/a*c^3*(35*Shi(arcsinh(a*x))*arcsinh(a*x)+63*Shi(3*arcsinh(a*x))*arcsinh(a*x)+35*Shi(5*arcsinh(a*x))*arcsi
nh(a*x)+7*Shi(7*arcsinh(a*x))*arcsinh(a*x)-21*cosh(3*arcsinh(a*x))-7*cosh(5*arcsinh(a*x))-cosh(7*arcsinh(a*x))
-35*(a^2*x^2+1)^(1/2))/arcsinh(a*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{9} c^{3} x^{9} + 4 \, a^{7} c^{3} x^{7} + 6 \, a^{5} c^{3} x^{5} + 4 \, a^{3} c^{3} x^{3} + a c^{3} x +{\left (a^{8} c^{3} x^{8} + 4 \, a^{6} c^{3} x^{6} + 6 \, a^{4} c^{3} x^{4} + 4 \, a^{2} c^{3} x^{2} + c^{3}\right )} \sqrt{a^{2} x^{2} + 1}}{{\left (a^{3} x^{2} + \sqrt{a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )} + \int \frac{7 \, a^{10} c^{3} x^{10} + 29 \, a^{8} c^{3} x^{8} + 46 \, a^{6} c^{3} x^{6} + 34 \, a^{4} c^{3} x^{4} + 11 \, a^{2} c^{3} x^{2} + c^{3} +{\left (7 \, a^{8} c^{3} x^{8} + 20 \, a^{6} c^{3} x^{6} + 18 \, a^{4} c^{3} x^{4} + 4 \, a^{2} c^{3} x^{2} - c^{3}\right )}{\left (a^{2} x^{2} + 1\right )} + 7 \,{\left (2 \, a^{9} c^{3} x^{9} + 7 \, a^{7} c^{3} x^{7} + 9 \, a^{5} c^{3} x^{5} + 5 \, a^{3} c^{3} x^{3} + a c^{3} x\right )} \sqrt{a^{2} x^{2} + 1}}{{\left (a^{4} x^{4} +{\left (a^{2} x^{2} + 1\right )} a^{2} x^{2} + 2 \, a^{2} x^{2} + 2 \,{\left (a^{3} x^{3} + a x\right )} \sqrt{a^{2} x^{2} + 1} + 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^9*c^3*x^9 + 4*a^7*c^3*x^7 + 6*a^5*c^3*x^5 + 4*a^3*c^3*x^3 + a*c^3*x + (a^8*c^3*x^8 + 4*a^6*c^3*x^6 + 6*a^4
*c^3*x^4 + 4*a^2*c^3*x^2 + c^3)*sqrt(a^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a^2
*x^2 + 1))) + integrate((7*a^10*c^3*x^10 + 29*a^8*c^3*x^8 + 46*a^6*c^3*x^6 + 34*a^4*c^3*x^4 + 11*a^2*c^3*x^2 +
c^3 + (7*a^8*c^3*x^8 + 20*a^6*c^3*x^6 + 18*a^4*c^3*x^4 + 4*a^2*c^3*x^2 - c^3)*(a^2*x^2 + 1) + 7*(2*a^9*c^3*x^
9 + 7*a^7*c^3*x^7 + 9*a^5*c^3*x^5 + 5*a^3*c^3*x^3 + a*c^3*x)*sqrt(a^2*x^2 + 1))/((a^4*x^4 + (a^2*x^2 + 1)*a^2*
x^2 + 2*a^2*x^2 + 2*(a^3*x^3 + a*x)*sqrt(a^2*x^2 + 1) + 1)*log(a*x + sqrt(a^2*x^2 + 1))), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}}{\operatorname{arsinh}\left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3)/arcsinh(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{3} \left (\int \frac{3 a^{2} x^{2}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx + \int \frac{3 a^{4} x^{4}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx + \int \frac{a^{6} x^{6}}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx + \int \frac{1}{\operatorname{asinh}^{2}{\left (a x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**3*(Integral(3*a**2*x**2/asinh(a*x)**2, x) + Integral(3*a**4*x**4/asinh(a*x)**2, x) + Integral(a**6*x**6/asi
nh(a*x)**2, x) + Integral(asinh(a*x)**(-2), x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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