3.70 \(\int \frac{x}{\sinh ^{-1}(a x)^4} \, dx\)
Optimal. Leaf size=95 \[ \frac{2 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{3 a^2}-\frac{2 x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)}-\frac{x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}-\frac{1}{6 a^2 \sinh ^{-1}(a x)^2}-\frac{x^2}{3 \sinh ^{-1}(a x)^2} \]
[Out]
-(x*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^3) - 1/(6*a^2*ArcSinh[a*x]^2) - x^2/(3*ArcSinh[a*x]^2) - (2*x*Sqrt[1
+ a^2*x^2])/(3*a*ArcSinh[a*x]) + (2*CoshIntegral[2*ArcSinh[a*x]])/(3*a^2)
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Rubi [A] time = 0.163369, antiderivative size = 95, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used =
{5667, 5774, 5665, 3301, 5675} \[ \frac{2 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{3 a^2}-\frac{2 x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)}-\frac{x \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}-\frac{1}{6 a^2 \sinh ^{-1}(a x)^2}-\frac{x^2}{3 \sinh ^{-1}(a x)^2} \]
Antiderivative was successfully verified.
[In]
Int[x/ArcSinh[a*x]^4,x]
[Out]
-(x*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^3) - 1/(6*a^2*ArcSinh[a*x]^2) - x^2/(3*ArcSinh[a*x]^2) - (2*x*Sqrt[1
+ a^2*x^2])/(3*a*ArcSinh[a*x]) + (2*CoshIntegral[2*ArcSinh[a*x]])/(3*a^2)
Rule 5667
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 5774
Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 0]
Rule 5665
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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]
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 5675
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]
Rubi steps
\begin{align*} \int \frac{x}{\sinh ^{-1}(a x)^4} \, dx &=-\frac{x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}+\frac{\int \frac{1}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3} \, dx}{3 a}+\frac{1}{3} (2 a) \int \frac{x^2}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3} \, dx\\ &=-\frac{x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{1}{6 a^2 \sinh ^{-1}(a x)^2}-\frac{x^2}{3 \sinh ^{-1}(a x)^2}+\frac{2}{3} \int \frac{x}{\sinh ^{-1}(a x)^2} \, dx\\ &=-\frac{x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{1}{6 a^2 \sinh ^{-1}(a x)^2}-\frac{x^2}{3 \sinh ^{-1}(a x)^2}-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{1}{6 a^2 \sinh ^{-1}(a x)^2}-\frac{x^2}{3 \sinh ^{-1}(a x)^2}-\frac{2 x \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)}+\frac{2 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{3 a^2}\\ \end{align*}
Mathematica [A] time = 0.115911, size = 84, normalized size = 0.88 \[ -\frac{2 a x \sqrt{a^2 x^2+1}+4 a x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2+\left (2 a^2 x^2+1\right ) \sinh ^{-1}(a x)-4 \sinh ^{-1}(a x)^3 \text{Chi}\left (2 \sinh ^{-1}(a x)\right )}{6 a^2 \sinh ^{-1}(a x)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[x/ArcSinh[a*x]^4,x]
[Out]
-(2*a*x*Sqrt[1 + a^2*x^2] + (1 + 2*a^2*x^2)*ArcSinh[a*x] + 4*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2 - 4*ArcSinh[
a*x]^3*CoshIntegral[2*ArcSinh[a*x]])/(6*a^2*ArcSinh[a*x]^3)
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Maple [A] time = 0.025, size = 60, normalized size = 0.6 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{6\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}}-{\frac{\cosh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{6\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\sinh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{3\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{2\,{\it Chi} \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{3}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/arcsinh(a*x)^4,x)
[Out]
1/a^2*(-1/6/arcsinh(a*x)^3*sinh(2*arcsinh(a*x))-1/6/arcsinh(a*x)^2*cosh(2*arcsinh(a*x))-1/3/arcsinh(a*x)*sinh(
2*arcsinh(a*x))+2/3*Chi(2*arcsinh(a*x)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/arcsinh(a*x)^4,x, algorithm="maxima")
[Out]
-1/6*(2*a^12*x^12 + 10*a^10*x^10 + 20*a^8*x^8 + 20*a^6*x^6 + 10*a^4*x^4 + 2*a^2*x^2 + 2*(a^7*x^7 + a^5*x^5)*(a
^2*x^2 + 1)^(5/2) + 2*(5*a^8*x^8 + 9*a^6*x^6 + 4*a^4*x^4)*(a^2*x^2 + 1)^2 + (4*a^12*x^12 + 20*a^10*x^10 + 40*a
^8*x^8 + 40*a^6*x^6 + 20*a^4*x^4 + 4*a^2*x^2 + 4*(a^7*x^7 + a^5*x^5)*(a^2*x^2 + 1)^(5/2) + (20*a^8*x^8 + 36*a^
6*x^6 + 16*a^4*x^4 - 3*a^2*x^2 - 3)*(a^2*x^2 + 1)^2 + (40*a^9*x^9 + 104*a^7*x^7 + 88*a^5*x^5 + 21*a^3*x^3 - 3*
a*x)*(a^2*x^2 + 1)^(3/2) + (40*a^10*x^10 + 136*a^8*x^8 + 168*a^6*x^6 + 91*a^4*x^4 + 22*a^2*x^2 + 3)*(a^2*x^2 +
1) + (20*a^11*x^11 + 84*a^9*x^9 + 136*a^7*x^7 + 107*a^5*x^5 + 42*a^3*x^3 + 7*a*x)*sqrt(a^2*x^2 + 1))*log(a*x
+ sqrt(a^2*x^2 + 1))^2 + 4*(5*a^9*x^9 + 13*a^7*x^7 + 11*a^5*x^5 + 3*a^3*x^3)*(a^2*x^2 + 1)^(3/2) + 4*(5*a^10*x
^10 + 17*a^8*x^8 + 21*a^6*x^6 + 11*a^4*x^4 + 2*a^2*x^2)*(a^2*x^2 + 1) + (2*a^12*x^12 + 10*a^10*x^10 + 20*a^8*x
^8 + 20*a^6*x^6 + 10*a^4*x^4 + 2*a^2*x^2 + 2*(a^7*x^7 + a^5*x^5)*(a^2*x^2 + 1)^(5/2) + (10*a^8*x^8 + 18*a^6*x^
6 + 9*a^4*x^4 + a^2*x^2)*(a^2*x^2 + 1)^2 + (20*a^9*x^9 + 52*a^7*x^7 + 47*a^5*x^5 + 17*a^3*x^3 + 2*a*x)*(a^2*x^
2 + 1)^(3/2) + (20*a^10*x^10 + 68*a^8*x^8 + 87*a^6*x^6 + 51*a^4*x^4 + 13*a^2*x^2 + 1)*(a^2*x^2 + 1) + (10*a^11
*x^11 + 42*a^9*x^9 + 69*a^7*x^7 + 55*a^5*x^5 + 21*a^3*x^3 + 3*a*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 +
1)) + 2*(5*a^11*x^11 + 21*a^9*x^9 + 34*a^7*x^7 + 26*a^5*x^5 + 9*a^3*x^3 + a*x)*sqrt(a^2*x^2 + 1))/((a^12*x^10
+ 5*a^10*x^8 + (a^2*x^2 + 1)^(5/2)*a^7*x^5 + 10*a^8*x^6 + 10*a^6*x^4 + 5*a^4*x^2 + 5*(a^8*x^6 + a^6*x^4)*(a^2
*x^2 + 1)^2 + 10*(a^9*x^7 + 2*a^7*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/2) + 10*(a^10*x^8 + 3*a^8*x^6 + 3*a^6*x^4 +
a^4*x^2)*(a^2*x^2 + 1) + a^2 + 5*(a^11*x^9 + 4*a^9*x^7 + 6*a^7*x^5 + 4*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log
(a*x + sqrt(a^2*x^2 + 1))^3) + integrate(1/6*(8*a^13*x^13 + 48*a^11*x^11 + 120*a^9*x^9 + 8*(a^2*x^2 + 1)^3*a^7
*x^7 + 160*a^7*x^7 + 120*a^5*x^5 + 48*a^3*x^3 + (48*a^8*x^8 + 48*a^6*x^6 + 4*a^4*x^4 + 12*a^2*x^2 + 15)*(a^2*x
^2 + 1)^(5/2) + 8*(15*a^9*x^9 + 30*a^7*x^7 + 17*a^5*x^5 + 5*a^3*x^3 + 3*a*x)*(a^2*x^2 + 1)^2 + 2*(80*a^10*x^10
+ 240*a^8*x^8 + 252*a^6*x^6 + 104*a^4*x^4 + 3*a^2*x^2 - 9)*(a^2*x^2 + 1)^(3/2) + 8*(15*a^11*x^11 + 60*a^9*x^9
+ 92*a^7*x^7 + 63*a^5*x^5 + 15*a^3*x^3 - a*x)*(a^2*x^2 + 1) + 8*a*x + (48*a^12*x^12 + 240*a^10*x^10 + 484*a^8
*x^8 + 484*a^6*x^6 + 243*a^4*x^4 + 58*a^2*x^2 + 7)*sqrt(a^2*x^2 + 1))/((a^13*x^12 + 6*a^11*x^10 + 15*a^9*x^8 +
(a^2*x^2 + 1)^3*a^7*x^6 + 20*a^7*x^6 + 15*a^5*x^4 + 6*a^3*x^2 + 6*(a^8*x^7 + a^6*x^5)*(a^2*x^2 + 1)^(5/2) + 1
5*(a^9*x^8 + 2*a^7*x^6 + a^5*x^4)*(a^2*x^2 + 1)^2 + 20*(a^10*x^9 + 3*a^8*x^7 + 3*a^6*x^5 + a^4*x^3)*(a^2*x^2 +
1)^(3/2) + 15*(a^11*x^10 + 4*a^9*x^8 + 6*a^7*x^6 + 4*a^5*x^4 + a^3*x^2)*(a^2*x^2 + 1) + 6*(a^12*x^11 + 5*a^10
*x^9 + 10*a^8*x^7 + 10*a^6*x^5 + 5*a^4*x^3 + a^2*x)*sqrt(a^2*x^2 + 1) + a)*log(a*x + sqrt(a^2*x^2 + 1))), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\operatorname{arsinh}\left (a x\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/arcsinh(a*x)^4,x, algorithm="fricas")
[Out]
integral(x/arcsinh(a*x)^4, x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{asinh}^{4}{\left (a x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/asinh(a*x)**4,x)
[Out]
Integral(x/asinh(a*x)**4, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{arsinh}\left (a x\right )^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/arcsinh(a*x)^4,x, algorithm="giac")
[Out]
integrate(x/arcsinh(a*x)^4, x)