3.439 \(\int \frac {x}{\sqrt {1+c^2 x^2} (a+b \sinh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=73 \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac {x}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5774, 5657, 3303, 3298, 3301} \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac {x}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(b*c*(a + b*ArcSinh[c*x]))) + (Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(b^2*c^2) - (Sinh[a/b]*Sinh
Integral[(a + b*ArcSinh[c*x])/b])/(b^2*c^2)

Rule 3298

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 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 3303

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 5657

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 60, normalized size = 0.82 \[ \frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-\frac {b c x}{a+b \sinh ^{-1}(c x)}}{b^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*c*x)/(a + b*ArcSinh[c*x])) + Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] - Sinh[a/b]*SinhIntegral[a/b + A
rcSinh[c*x]])/(b^2*c^2)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} x^{2} + 1} x}{a^{2} c^{2} x^{2} + {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \operatorname {arsinh}\left (c x\right )^{2} + a^{2} + 2 \, {\left (a b c^{2} x^{2} + a b\right )} \operatorname {arsinh}\left (c x\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*x^2 + 1)*x/(a^2*c^2*x^2 + (b^2*c^2*x^2 + b^2)*arcsinh(c*x)^2 + a^2 + 2*(a*b*c^2*x^2 + a*b)*a
rcsinh(c*x)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*x) + a)^2), x)

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maple [B]  time = 0.08, size = 151, normalized size = 2.07 \[ -\frac {c x -\sqrt {c^{2} x^{2}+1}}{2 c^{2} b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{2 c^{2} b^{2}}-\frac {\arcsinh \left (c x \right ) {\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) b +{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) a +x b c +\sqrt {c^{2} x^{2}+1}\, b}{2 c^{2} b^{2} \left (a +b \arcsinh \left (c x \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*x-(c^2*x^2+1)^(1/2))/c^2/b/(a+b*arcsinh(c*x))-1/2/c^2/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-1/2/c^2/b^2*
(arcsinh(c*x)*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*b+exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*a+x*b*c+(c^2*x^2+1)^(1/2)*
b)/(a+b*arcsinh(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {c^{3} x^{4} + c x^{2} + {\left (c^{2} x^{3} + x\right )} \sqrt {c^{2} x^{2} + 1}}{{\left (c^{2} x^{2} + 1\right )} a b c^{2} x + {\left ({\left (c^{2} x^{2} + 1\right )} b^{2} c^{2} x + {\left (b^{2} c^{3} x^{2} + b^{2} c\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (a b c^{3} x^{2} + a b c\right )} \sqrt {c^{2} x^{2} + 1}} + \int \frac {c^{5} x^{5} + {\left (c^{2} x^{2} + 1\right )} c^{3} x^{3} + 3 \, c^{3} x^{3} + 2 \, c x + {\left (2 \, c^{4} x^{4} + 3 \, c^{2} x^{2} + 1\right )} \sqrt {c^{2} x^{2} + 1}}{{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b c^{3} x^{2} + 2 \, {\left (a b c^{4} x^{3} + a b c^{2} x\right )} {\left (c^{2} x^{2} + 1\right )} + {\left ({\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} c^{3} x^{2} + 2 \, {\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (b^{2} c^{5} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{2} c\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (a b c^{5} x^{4} + 2 \, a b c^{3} x^{2} + a b c\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^3*x^4 + c*x^2 + (c^2*x^3 + x)*sqrt(c^2*x^2 + 1))/((c^2*x^2 + 1)*a*b*c^2*x + ((c^2*x^2 + 1)*b^2*c^2*x + (b^
2*c^3*x^2 + b^2*c)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + (a*b*c^3*x^2 + a*b*c)*sqrt(c^2*x^2 + 1))
+ integrate((c^5*x^5 + (c^2*x^2 + 1)*c^3*x^3 + 3*c^3*x^3 + 2*c*x + (2*c^4*x^4 + 3*c^2*x^2 + 1)*sqrt(c^2*x^2 +
1))/((c^2*x^2 + 1)^(3/2)*a*b*c^3*x^2 + 2*(a*b*c^4*x^3 + a*b*c^2*x)*(c^2*x^2 + 1) + ((c^2*x^2 + 1)^(3/2)*b^2*c^
3*x^2 + 2*(b^2*c^4*x^3 + b^2*c^2*x)*(c^2*x^2 + 1) + (b^2*c^5*x^4 + 2*b^2*c^3*x^2 + b^2*c)*sqrt(c^2*x^2 + 1))*l
og(c*x + sqrt(c^2*x^2 + 1)) + (a*b*c^5*x^4 + 2*a*b*c^3*x^2 + a*b*c)*sqrt(c^2*x^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {c^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2} \sqrt {c^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*asinh(c*x))**2/(c**2*x**2+1)**(1/2),x)

[Out]

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

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