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*(a^2*x^2+1)^(5/2)/a/arcsinh(a*x)+5/8*c^2*Shi(arcsinh(a*x))/a+15/16*c^2*Shi(3*arcsinh(a*x))/a+5/16*c^2*Shi
(5*arcsinh(a*x))/a

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Rubi [A]  time = 0.17, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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 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 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])

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*}

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Mathematica [A]  time = 0.40, size = 69, normalized size = 0.90 \[ \frac {c^2 \left (-16 \left (a^2 x^2+1\right )^{5/2}+10 \sinh ^{-1}(a x) \text {Shi}\left (\sinh ^{-1}(a x)\right )+15 \sinh ^{-1}(a x) \text {Shi}\left (3 \sinh ^{-1}(a x)\right )+5 \sinh ^{-1}(a x) \text {Shi}\left (5 \sinh ^{-1}(a x)\right )\right )}{16 a \sinh ^{-1}(a x)} \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}{\operatorname {arsinh}\left (a x\right )^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} c x^{2} + c\right )}^{2}}{\operatorname {arsinh}\left (a x\right )^{2}}\,{d x} \]

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)

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maple [A]  time = 0.07, size = 84, normalized size = 1.09 \[ \frac {c^{2} \left (10 \Shi \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+15 \Shi \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+5 \Shi \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-10 \sqrt {a^{2} x^{2}+1}-5 \cosh \left (3 \arcsinh \left (a x \right )\right )-\cosh \left (5 \arcsinh \left (a x \right )\right )\right )}{16 a \arcsinh \left (a x \right )} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{7} c^{2} x^{7} + 3 \, a^{5} c^{2} x^{5} + 3 \, a^{3} c^{2} x^{3} + a c^{2} x + {\left (a^{6} c^{2} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} x^{2} + 1}}{{\left (a^{3} x^{2} + \sqrt {a^{2} x^{2} + 1} a^{2} x + a\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )} + \int \frac {5 \, a^{8} c^{2} x^{8} + 16 \, a^{6} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{4} + 8 \, a^{2} c^{2} x^{2} + {\left (5 \, a^{6} c^{2} x^{6} + 9 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} - c^{2}\right )} {\left (a^{2} x^{2} + 1\right )} + c^{2} + 5 \, {\left (2 \, a^{7} c^{2} x^{7} + 5 \, a^{5} c^{2} x^{5} + 4 \, a^{3} c^{2} x^{3} + a c^{2} x\right )} \sqrt {a^{2} x^{2} + 1}}{{\left (a^{4} x^{4} + {\left (a^{2} x^{2} + 1\right )} a^{2} x^{2} + 2 \, a^{2} x^{2} + 2 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {a^{2} x^{2} + 1} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,a^2\,x^2+c\right )}^2}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + a^2*c*x^2)^2/asinh(a*x)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \frac {2 a^{2} x^{2}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx\right ) \]

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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