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

Optimal. Leaf size=121 \[ \frac {3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^4}-\frac {\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^4}+\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^4} \]

[Out]

-3/4*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b/c^4+1/4*cosh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b/c^4+3/4*Chi((a+b*
arcsinh(c*x))/b)*sinh(a/b)/b/c^4-1/4*Chi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b/c^4

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Rubi [A]  time = 0.39, antiderivative size = 117, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5779, 3312, 3303, 3298, 3301} \[ \frac {3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^4}-\frac {\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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 {x^3}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac {i \operatorname {Subst}\left (\int \left (\frac {3 i \sinh (x)}{4 (a+b x)}-\frac {i \sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^4}-\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 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 )}{4 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 )}{4 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 )}{4 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 )}{4 c^4}\\ &=\frac {3 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{4 b c^4}-\frac {\text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{4 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^4}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 92, normalized size = 0.76 \[ \frac {3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{4 b c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{3}}{a c^{2} x^{2} + {\left (b c^{2} x^{2} + b\right )} \operatorname {arsinh}\left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.23, size = 118, normalized size = 0.98 \[ \frac {{\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{8 c^{4} b}-\frac {3 \,{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{8 c^{4} b}+\frac {3 \,{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{8 c^{4} b}-\frac {{\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{8 c^{4} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c^4/b*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-3/8/c^4/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)+3/8/c^4/b*exp(-a/b)*
Ei(1,-arcsinh(c*x)-a/b)-1/8/c^4/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {c^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \sqrt {c^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/((a + b*asinh(c*x))*sqrt(c**2*x**2 + 1)), x)

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