3.1.86 \(\int \frac {x}{\sinh ^{-1}(a+b x)^2} \, dx\) [86]

Optimal. Leaf size=84 \[ \frac {a \sqrt {1+(a+b x)^2}}{b^2 \sinh ^{-1}(a+b x)}-\frac {(a+b x) \sqrt {1+(a+b x)^2}}{b^2 \sinh ^{-1}(a+b x)}+\frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac {a \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5859, 5829, 5773, 5819, 3379, 5778, 3382} \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac {a \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {a \sqrt {(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)}-\frac {(a+b x) \sqrt {(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a + b*x]^2,x]

[Out]

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

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

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Mathematica [A]
time = 0.12, size = 62, normalized size = 0.74 \begin {gather*} -\frac {b x \sqrt {1+(a+b x)^2}-\sinh ^{-1}(a+b x) \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )+a \sinh ^{-1}(a+b x) \text {Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^2 \sinh ^{-1}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSinh[a + b*x]^2,x]

[Out]

-((b*x*Sqrt[1 + (a + b*x)^2] - ArcSinh[a + b*x]*CoshIntegral[2*ArcSinh[a + b*x]] + a*ArcSinh[a + b*x]*SinhInte
gral[ArcSinh[a + b*x]])/(b^2*ArcSinh[a + b*x]))

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Maple [A]
time = 3.24, size = 73, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {-\frac {\sinh \left (2 \arcsinh \left (b x +a \right )\right )}{2 \arcsinh \left (b x +a \right )}+\hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right )-\frac {a \left (\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sqrt {1+\left (b x +a \right )^{2}}\right )}{\arcsinh \left (b x +a \right )}}{b^{2}}\)	\(73\)
default	\(\frac {-\frac {\sinh \left (2 \arcsinh \left (b x +a \right )\right )}{2 \arcsinh \left (b x +a \right )}+\hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right )-\frac {a \left (\hyperbolicSineIntegral \left (\arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-\sqrt {1+\left (b x +a \right )^{2}}\right )}{\arcsinh \left (b x +a \right )}}{b^{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(-1/2/arcsinh(b*x+a)*sinh(2*arcsinh(b*x+a))+Chi(2*arcsinh(b*x+a))-a*(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/arcsinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b + b)*x^2 + (a^3 + a)*x + (b^2*x^3 + 2*a*b*x^2 + (a^2 + 1)*x)*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((2*b^5*x^5 + 9*a*b^4*x^4 + a^5 + 4*(4*a^2*b^3 + b^3)
*x^3 + 2*a^3 + 2*(7*a^3*b^2 + 5*a*b^2)*x^2 + (2*b^3*x^3 + 5*a*b^2*x^2 + 4*a^2*b*x + a^3 + a)*(b^2*x^2 + 2*a*b*
x + a^2 + 1) + 2*(3*a^4*b + 4*a^2*b + b)*x + (4*b^4*x^4 + 14*a*b^3*x^3 + 2*a^4 + 2*(9*a^2*b^2 + 2*b^2)*x^2 + 3
*a^2 + (10*a^3*b + 7*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + a)/((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)*l
og(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/arcsinh(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x/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}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x/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}{{\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/asinh(a + b*x)^2,x)

[Out]

int(x/asinh(a + b*x)^2, x)

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