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

Optimal. Leaf size=77 \[ -\frac{c^2 \left (a^2 x^2+1\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{5 c^2 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac{15 c^2 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac{5 c^2 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a} \]

[Out]

-((c^2*(1 + a^2*x^2)^(5/2))/(a*ArcSinh[a*x])) + (5*c^2*SinhIntegral[ArcSinh[a*x]])/(8*a) + (15*c^2*SinhIntegra
l[3*ArcSinh[a*x]])/(16*a) + (5*c^2*SinhIntegral[5*ArcSinh[a*x]])/(16*a)

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Rubi [A]  time = 0.169523, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5696, 5779, 5448, 3298} \[ -\frac{c^2 \left (a^2 x^2+1\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{5 c^2 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac{15 c^2 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac{5 c^2 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^2*(1 + a^2*x^2)^(5/2))/(a*ArcSinh[a*x])) + (5*c^2*SinhIntegral[ArcSinh[a*x]])/(8*a) + (15*c^2*SinhIntegra
l[3*ArcSinh[a*x]])/(16*a) + (5*c^2*SinhIntegral[5*ArcSinh[a*x]])/(16*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 )^2}{\sinh ^{-1}(a x)^2} \, dx &=-\frac{c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\left (5 a c^2\right ) \int \frac{x \left (1+a^2 x^2\right )^{3/2}}{\sinh ^{-1}(a x)} \, dx\\ &=-\frac{c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac{c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 x}+\frac{3 \sinh (3 x)}{16 x}+\frac{\sinh (5 x)}{16 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac{c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a}+\frac{\left (15 c^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}\\ &=-\frac{c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac{5 c^2 \text{Shi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac{15 c^2 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac{5 c^2 \text{Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.382716, size = 69, normalized size = 0.9 \[ \frac{c^2 \left (-16 \left (a^2 x^2+1\right )^{5/2}+10 \sinh ^{-1}(a x) \text{Shi}\left (\sinh ^{-1}(a x)\right )+15 \sinh ^{-1}(a x) \text{Shi}\left (3 \sinh ^{-1}(a x)\right )+5 \sinh ^{-1}(a x) \text{Shi}\left (5 \sinh ^{-1}(a x)\right )\right )}{16 a \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-16*(1 + a^2*x^2)^(5/2) + 10*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] + 15*ArcSinh[a*x]*SinhIntegral[3*Ar
cSinh[a*x]] + 5*ArcSinh[a*x]*SinhIntegral[5*ArcSinh[a*x]]))/(16*a*ArcSinh[a*x])

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Maple [A]  time = 0.033, size = 84, normalized size = 1.1 \begin{align*}{\frac{{c}^{2}}{16\,a{\it Arcsinh} \left ( ax \right ) } \left ( 10\,{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) +15\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) +5\,{\it Shi} \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ){\it Arcsinh} \left ( ax \right ) -10\,\sqrt{{a}^{2}{x}^{2}+1}-5\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) -\cosh \left ( 5\,{\it Arcsinh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a*c^2*(10*Shi(arcsinh(a*x))*arcsinh(a*x)+15*Shi(3*arcsinh(a*x))*arcsinh(a*x)+5*Shi(5*arcsinh(a*x))*arcsin
h(a*x)-10*(a^2*x^2+1)^(1/2)-5*cosh(3*arcsinh(a*x))-cosh(5*arcsinh(a*x)))/arcsinh(a*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{7} c^{2} x^{7} + 3 \, a^{5} c^{2} x^{5} + 3 \, a^{3} c^{2} x^{3} + a c^{2} x +{\left (a^{6} c^{2} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\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{5 \, a^{8} c^{2} x^{8} + 16 \, a^{6} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{4} + 8 \, a^{2} c^{2} x^{2} +{\left (5 \, a^{6} c^{2} x^{6} + 9 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} - c^{2}\right )}{\left (a^{2} x^{2} + 1\right )} + c^{2} + 5 \,{\left (2 \, a^{7} c^{2} x^{7} + 5 \, a^{5} c^{2} x^{5} + 4 \, a^{3} c^{2} x^{3} + a c^{2} 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)^2/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^7*c^2*x^7 + 3*a^5*c^2*x^5 + 3*a^3*c^2*x^3 + a*c^2*x + (a^6*c^2*x^6 + 3*a^4*c^2*x^4 + 3*a^2*c^2*x^2 + c^2)*
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((5*a^8*c
^2*x^8 + 16*a^6*c^2*x^6 + 18*a^4*c^2*x^4 + 8*a^2*c^2*x^2 + (5*a^6*c^2*x^6 + 9*a^4*c^2*x^4 + 3*a^2*c^2*x^2 - c^
2)*(a^2*x^2 + 1) + c^2 + 5*(2*a^7*c^2*x^7 + 5*a^5*c^2*x^5 + 4*a^3*c^2*x^3 + a*c^2*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^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}{\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)^2/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

c**2*(Integral(2*a**2*x**2/asinh(a*x)**2, x) + Integral(a**4*x**4/asinh(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 )}^{2}}{\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)^2/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

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