3.1.85 \(\int \frac {x^2}{\sinh ^{-1}(a+b x)^2} \, dx\) [85]

Optimal. Leaf size=154 \[ -\frac {a^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}+\frac {2 a (a+b x) \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {1+(a+b x)^2}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {a^2 \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3} \]

[Out]

-2*a*Chi(2*arcsinh(b*x+a))/b^3-1/4*Shi(arcsinh(b*x+a))/b^3+a^2*Shi(arcsinh(b*x+a))/b^3+3/4*Shi(3*arcsinh(b*x+a
))/b^3-a^2*(1+(b*x+a)^2)^(1/2)/b^3/arcsinh(b*x+a)+2*a*(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^3/arcsinh(b*x+a)-(b*x+a)^2
*(1+(b*x+a)^2)^(1/2)/b^3/arcsinh(b*x+a)

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Rubi [A]
time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5859, 5829, 5773, 5819, 3379, 5778, 3382} \begin {gather*} \frac {a^2 \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {a^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {2 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac {\text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac {2 a (a+b x) \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac {(a+b x)^2 \sqrt {(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a + b*x]^2,x]

[Out]

-((a^2*Sqrt[1 + (a + b*x)^2])/(b^3*ArcSinh[a + b*x])) + (2*a*(a + b*x)*Sqrt[1 + (a + b*x)^2])/(b^3*ArcSinh[a +
b*x]) - ((a + b*x)^2*Sqrt[1 + (a + b*x)^2])/(b^3*ArcSinh[a + b*x]) - (2*a*CoshIntegral[2*ArcSinh[a + b*x]])/b
^3 - SinhIntegral[ArcSinh[a + b*x]]/(4*b^3) + (a^2*SinhIntegral[ArcSinh[a + b*x]])/b^3 + (3*SinhIntegral[3*Arc
Sinh[a + b*x]])/(4*b^3)

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 5773

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

Rule 5778

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^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -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]

Rule 5829

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5859

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

Rubi steps

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

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Mathematica [A]
time = 0.28, size = 83, normalized size = 0.54 \begin {gather*} \frac {-\frac {4 b^2 x^2 \sqrt {1+a^2+2 a b x+b^2 x^2}}{\sinh ^{-1}(a+b x)}-8 a \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )+\left (-1+4 a^2\right ) \text {Shi}\left (\sinh ^{-1}(a+b x)\right )+3 \text {Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a + b*x]^2,x]

[Out]

((-4*b^2*x^2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/ArcSinh[a + b*x] - 8*a*CoshIntegral[2*ArcSinh[a + b*x]] + (-1
+ 4*a^2)*SinhIntegral[ArcSinh[a + b*x]] + 3*SinhIntegral[3*ArcSinh[a + b*x]])/(4*b^3)

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Maple [A]
time = 4.12, size = 146, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {a \left (2 \hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sinh \left (2 \arcsinh \left (b x +a \right )\right )\right )}{\arcsinh \left (b x +a \right )}+\frac {\sqrt {1+\left (b x +a \right )^{2}}}{4 \arcsinh \left (b x +a \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right )}{4}-\frac {\cosh \left (3 \arcsinh \left (b x +a \right )\right )}{4 \arcsinh \left (b x +a \right )}+\frac {3 \hyperbolicSineIntegral \left (3 \arcsinh \left (b x +a \right )\right )}{4}+\frac {a^{2} \left (\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sqrt {1+\left (b x +a \right )^{2}}\right )}{\arcsinh \left (b x +a \right )}}{b^{3}}\)	\(146\)
default	\(\frac {-\frac {a \left (2 \hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sinh \left (2 \arcsinh \left (b x +a \right )\right )\right )}{\arcsinh \left (b x +a \right )}+\frac {\sqrt {1+\left (b x +a \right )^{2}}}{4 \arcsinh \left (b x +a \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right )}{4}-\frac {\cosh \left (3 \arcsinh \left (b x +a \right )\right )}{4 \arcsinh \left (b x +a \right )}+\frac {3 \hyperbolicSineIntegral \left (3 \arcsinh \left (b x +a \right )\right )}{4}+\frac {a^{2} \left (\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sqrt {1+\left (b x +a \right )^{2}}\right )}{\arcsinh \left (b x +a \right )}}{b^{3}}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b + b)*x^3 + (a^3 + a)*x^2 + (b^2*x^4 + 2*a*b*x^3 + (a^2 + 1)*x^2)*sqrt(b^2*x
^2 + 2*a*b*x + a^2 + 1))/((b^3*x^2 + 2*a*b^2*x + a^2*b + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*
log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate((3*b^5*x^6 + 14*a*b^4*x^5 + 2*(13*a^2*b^3 + 3*b^
3)*x^4 + 8*(3*a^3*b^2 + 2*a*b^2)*x^3 + (11*a^4*b + 14*a^2*b + 3*b)*x^2 + (3*b^3*x^4 + 8*a*b^2*x^3 + (7*a^2*b +
b)*x^2 + 2*(a^3 + a)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 2*(a^5 + 2*a^3 + a)*x + (6*b^4*x^5 + 22*a*b^3*x^4 + (
30*a^2*b^2 + 7*b^2)*x^3 + (18*a^3*b + 13*a*b)*x^2 + 2*(2*a^4 + 3*a^2 + 1)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)/((b^5*x^4 + 4*a*b^4*x^3 + a^4*b + 2*a^2*b + 2*(3*a^2*b^3 + b^3)*x^2 + (b^3*x^2 + 2*a*b^2*x + a^2*b)*(b^2*x^2
+ 2*a*b*x + a^2 + 1) + 4*(a^3*b^2 + a*b^2)*x + 2*(b^4*x^3 + 3*a*b^3*x^2 + a^3*b + a*b + (3*a^2*b^2 + b^2)*x)*
sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^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^2/arcsinh(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^2/arcsinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a + b*x)**2, 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^2/arcsinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^2/arcsinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/asinh(a + b*x)^2,x)

[Out]

int(x^2/asinh(a + b*x)^2, x)

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