3.1.60 \(\int \frac {x^4}{\sinh ^{-1}(a x)^3} \, dx\) [60]

Optimal. Leaf size=97 \[ -\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {\text {Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac {27 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac {25 \text {Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5} \]

[Out]

-2*x^3/a^2/arcsinh(a*x)-5/2*x^5/arcsinh(a*x)+1/16*Chi(arcsinh(a*x))/a^5-27/32*Chi(3*arcsinh(a*x))/a^5+25/32*Ch
i(5*arcsinh(a*x))/a^5-1/2*x^4*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^2

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Rubi [A]
time = 0.25, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5779, 5818, 5780, 5556, 3382} \begin {gather*} \frac {\text {Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac {27 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac {25 \text {Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {x^4 \sqrt {a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac {5 x^5}{2 \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/ArcSinh[a*x]^3,x]

[Out]

-1/2*(x^4*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x]^2) - (2*x^3)/(a^2*ArcSinh[a*x]) - (5*x^5)/(2*ArcSinh[a*x]) + Cosh
Integral[ArcSinh[a*x]]/(16*a^5) - (27*CoshIntegral[3*ArcSinh[a*x]])/(32*a^5) + (25*CoshIntegral[5*ArcSinh[a*x]
])/(32*a^5)

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 5556

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 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 1)), Int[x^(m - 1)*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^
2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x]
/; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x^4}{\sinh ^{-1}(a x)^3} \, dx &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}+\frac {2 \int \frac {x^3}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx}{a}+\frac {1}{2} (5 a) \int \frac {x^5}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx\\ &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {25}{2} \int \frac {x^4}{\sinh ^{-1}(a x)} \, dx+\frac {6 \int \frac {x^2}{\sinh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {6 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}+\frac {25 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^4(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {6 \text {Subst}\left (\int \left (-\frac {\cosh (x)}{4 x}+\frac {\cosh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^5}+\frac {25 \text {Subst}\left (\int \left (\frac {\cosh (x)}{8 x}-\frac {3 \cosh (3 x)}{16 x}+\frac {\cosh (5 x)}{16 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {25 \text {Subst}\left (\int \frac {\cosh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}-\frac {3 \text {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}+\frac {3 \text {Subst}\left (\int \frac {\cosh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}+\frac {25 \text {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a^5}-\frac {75 \text {Subst}\left (\int \frac {\cosh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{32 a^5}\\ &=-\frac {x^4 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \sinh ^{-1}(a x)}-\frac {5 x^5}{2 \sinh ^{-1}(a x)}+\frac {\text {Chi}\left (\sinh ^{-1}(a x)\right )}{16 a^5}-\frac {27 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )}{32 a^5}+\frac {25 \text {Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 102, normalized size = 1.05 \begin {gather*} -\frac {16 a^4 x^4 \sqrt {1+a^2 x^2}+64 a^3 x^3 \sinh ^{-1}(a x)+80 a^5 x^5 \sinh ^{-1}(a x)-2 \sinh ^{-1}(a x)^2 \text {Chi}\left (\sinh ^{-1}(a x)\right )+27 \sinh ^{-1}(a x)^2 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )-25 \sinh ^{-1}(a x)^2 \text {Chi}\left (5 \sinh ^{-1}(a x)\right )}{32 a^5 \sinh ^{-1}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcSinh[a*x]^3,x]

[Out]

-1/32*(16*a^4*x^4*Sqrt[1 + a^2*x^2] + 64*a^3*x^3*ArcSinh[a*x] + 80*a^5*x^5*ArcSinh[a*x] - 2*ArcSinh[a*x]^2*Cos
hIntegral[ArcSinh[a*x]] + 27*ArcSinh[a*x]^2*CoshIntegral[3*ArcSinh[a*x]] - 25*ArcSinh[a*x]^2*CoshIntegral[5*Ar
cSinh[a*x]])/(a^5*ArcSinh[a*x]^2)

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Maple [A]
time = 1.56, size = 120, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {a^{2} x^{2}+1}}{16 \arcsinh \left (a x \right )^{2}}-\frac {a x}{16 \arcsinh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )}{16}+\frac {3 \cosh \left (3 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )^{2}}+\frac {9 \sinh \left (3 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )}-\frac {27 \hyperbolicCosineIntegral \left (3 \arcsinh \left (a x \right )\right )}{32}-\frac {\cosh \left (5 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )^{2}}-\frac {5 \sinh \left (5 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )}+\frac {25 \hyperbolicCosineIntegral \left (5 \arcsinh \left (a x \right )\right )}{32}}{a^{5}}\)	\(120\)
default	\(\frac {-\frac {\sqrt {a^{2} x^{2}+1}}{16 \arcsinh \left (a x \right )^{2}}-\frac {a x}{16 \arcsinh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )}{16}+\frac {3 \cosh \left (3 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )^{2}}+\frac {9 \sinh \left (3 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )}-\frac {27 \hyperbolicCosineIntegral \left (3 \arcsinh \left (a x \right )\right )}{32}-\frac {\cosh \left (5 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )^{2}}-\frac {5 \sinh \left (5 \arcsinh \left (a x \right )\right )}{32 \arcsinh \left (a x \right )}+\frac {25 \hyperbolicCosineIntegral \left (5 \arcsinh \left (a x \right )\right )}{32}}{a^{5}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsinh(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/a^5*(-1/16/arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)-1/16/arcsinh(a*x)*a*x+1/16*Chi(arcsinh(a*x))+3/32/arcsinh(a*x)^2
*cosh(3*arcsinh(a*x))+9/32/arcsinh(a*x)*sinh(3*arcsinh(a*x))-27/32*Chi(3*arcsinh(a*x))-1/32/arcsinh(a*x)^2*cos
h(5*arcsinh(a*x))-5/32/arcsinh(a*x)*sinh(5*arcsinh(a*x))+25/32*Chi(5*arcsinh(a*x)))

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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/arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^8*x^11 + 3*a^6*x^9 + 3*a^4*x^7 + a^2*x^5 + (a^5*x^8 + a^3*x^6)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^9 + 5*a^
4*x^7 + 2*a^2*x^5)*(a^2*x^2 + 1) + (5*a^8*x^11 + 15*a^6*x^9 + 15*a^4*x^7 + 5*a^2*x^5 + (5*a^5*x^8 + 8*a^3*x^6
+ 3*a*x^4)*(a^2*x^2 + 1)^(3/2) + (15*a^6*x^9 + 31*a^4*x^7 + 20*a^2*x^5 + 4*x^3)*(a^2*x^2 + 1) + (15*a^7*x^10 +
38*a^5*x^8 + 32*a^3*x^6 + 9*a*x^4)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^10 + 7*a^5*x^8
+ 5*a^3*x^6 + a*x^4)*sqrt(a^2*x^2 + 1))/((a^8*x^6 + 3*a^6*x^4 + (a^2*x^2 + 1)^(3/2)*a^5*x^3 + 3*a^4*x^2 + 3*(a
^6*x^4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 3*(a^7*x^5 + 2*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*
x^2 + 1))^2) + integrate(1/2*(25*a^10*x^12 + 100*a^8*x^10 + 150*a^6*x^8 + 100*a^4*x^6 + 25*a^2*x^4 + (25*a^6*x
^8 + 24*a^4*x^6 + 3*a^2*x^4)*(a^2*x^2 + 1)^2 + (100*a^7*x^9 + 172*a^5*x^7 + 87*a^3*x^5 + 12*a*x^3)*(a^2*x^2 +
1)^(3/2) + 3*(50*a^8*x^10 + 124*a^6*x^8 + 105*a^4*x^6 + 35*a^2*x^4 + 4*x^2)*(a^2*x^2 + 1) + (100*a^9*x^11 + 32
4*a^7*x^9 + 381*a^5*x^7 + 193*a^3*x^5 + 36*a*x^3)*sqrt(a^2*x^2 + 1))/((a^10*x^8 + 4*a^8*x^6 + (a^2*x^2 + 1)^2*
a^6*x^4 + 6*a^6*x^4 + 4*a^4*x^2 + 4*(a^7*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/2) + 6*(a^8*x^6 + 2*a^6*x^4 + a^4*x^2
)*(a^2*x^2 + 1) + a^2 + 4*(a^9*x^7 + 3*a^7*x^5 + 3*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2
+ 1))), 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/arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

integral(x^4/arcsinh(a*x)^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\operatorname {asinh}^{3}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/asinh(a*x)**3,x)

[Out]

Integral(x**4/asinh(a*x)**3, 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/arcsinh(a*x)^3,x, algorithm="giac")

[Out]

integrate(x^4/arcsinh(a*x)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\mathrm {asinh}\left (a\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/asinh(a*x)^3,x)

[Out]

int(x^4/asinh(a*x)^3, x)

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