3.4.89 \(\int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \sinh ^{-1}(c x))} \, dx\) [389]

Optimal. Leaf size=144 \[ -\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b c^5}+\frac {3 \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b c^5} \]

[Out]

-1/2*Chi(2*(a+b*arcsinh(c*x))/b)*cosh(2*a/b)/b/c^5+1/8*Chi(4*(a+b*arcsinh(c*x))/b)*cosh(4*a/b)/b/c^5+3/8*ln(a+
b*arcsinh(c*x))/b/c^5+1/2*Shi(2*(a+b*arcsinh(c*x))/b)*sinh(2*a/b)/b/c^5-1/8*Shi(4*(a+b*arcsinh(c*x))/b)*sinh(4
*a/b)/b/c^5

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Rubi [A]
time = 0.22, antiderivative size = 144, 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 {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{8 b c^5}+\frac {3 \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])),x]

[Out]

-1/2*(Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(b*c^5) + (Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*
ArcSinh[c*x]))/b])/(8*b*c^5) + (3*Log[a + b*ArcSinh[c*x]])/(8*b*c^5) + (Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*A
rcSinh[c*x]))/b])/(2*b*c^5) - (Sinh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(8*b*c^5)

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^4}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx &=\frac {\text {Subst}\left (\int \frac {\sinh ^4(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^5}\\ &=\frac {\text {Subst}\left (\int \left (\frac {3}{8 (a+b x)}-\frac {\cosh (2 x)}{2 (a+b x)}+\frac {\cosh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5}\\ &=\frac {3 \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5}-\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5}\\ &=\frac {3 \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b c^5}-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5}\\ &=-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {3 \log \left (a+b \sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{8 b c^5}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 109, normalized size = 0.76 \begin {gather*} -\frac {4 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-3 \log \left (a+b \sinh ^{-1}(c x)\right )-4 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{8 b c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])),x]

[Out]

-1/8*(4*Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])] - Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcSinh[c*x])
] - 3*Log[a + b*ArcSinh[c*x]] - 4*Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])] + Sinh[(4*a)/b]*SinhInteg
ral[4*(a/b + ArcSinh[c*x])])/(b*c^5)

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Maple [A]
time = 10.38, size = 139, normalized size = 0.97

method	result	size
default	\(\frac {3 \ln \left (a +b \arcsinh \left (c x \right )\right )}{8 c^{5} b}-\frac {{\mathrm e}^{\frac {4 a}{b}} \expIntegral \left (1, 4 \arcsinh \left (c x \right )+\frac {4 a}{b}\right )}{16 c^{5} b}+\frac {{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{4 c^{5} b}+\frac {{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right )}{4 c^{5} b}-\frac {{\mathrm e}^{-\frac {4 a}{b}} \expIntegral \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right )}{16 c^{5} b}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

3/8*ln(a+b*arcsinh(c*x))/c^5/b-1/16/c^5/b*exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)+1/4/c^5/b*exp(2*a/b)*Ei(1,2*ar
csinh(c*x)+2*a/b)+1/4/c^5/b*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)-1/16/c^5/b*exp(-4*a/b)*Ei(1,-4*arcsinh(c*x
)-4*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^4/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/(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^4/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*x^2 + 1)*x^4/(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^{4}}{\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**4/(a+b*asinh(c*x))/(c**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/((a + b*asinh(c*x))*sqrt(c**2*x**2 + 1)), x)

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Giac [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^4/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/(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 \begin {gather*} \int \frac {x^4}{\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^4/((a + b*asinh(c*x))*(c^2*x^2 + 1)^(1/2)),x)

[Out]

int(x^4/((a + b*asinh(c*x))*(c^2*x^2 + 1)^(1/2)), x)

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