3.4.90 \(\int \frac {x^3}{\sqrt {1+c^2 x^2} (a+b \sinh ^{-1}(c x))} \, dx\) [390]

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

[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 = 0.23, antiderivative size = 121, normalized size of antiderivative = 1.00, 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 = {5819, 3393, 3384, 3379, 3382} \begin {gather*} \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} \end {gather*}

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

Rule 3379

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 3382

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 3384

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 3393

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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
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 {\text {Subst}\left (\int \frac {\sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac {i \text {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 {\text {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^4}-\frac {3 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.16, size = 92, normalized size = 0.76 \begin {gather*} \frac {3 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\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} \end {gather*}

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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Maple [A]
time = 6.10, size = 118, normalized size = 0.98

method	result	size
default	\(\frac {{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{8 c^{4} b}-\frac {3 \,{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{8 c^{4} b}+\frac {3 \,{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{8 c^{4} b}-\frac {{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{8 c^{4} b}\)	\(118\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

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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