3.5.38 \(\int \frac {x^2}{\sqrt {1+c^2 x^2} (a+b \sinh ^{-1}(c x))^2} \, dx\) [438]

Optimal. Leaf size=79 \[ -\frac {x^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {\text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right ) \sinh \left (\frac {2 a}{b}\right )}{b^2 c^3}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c^3} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5818, 5780, 5556, 12, 3384, 3379, 3382} \begin {gather*} -\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c^3}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c^3}-\frac {x^2}{b c \left (a+b \sinh ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2/(b*c*(a + b*ArcSinh[c*x]))) - (CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b]*Sinh[(2*a)/b])/(b^2*c^3) + (Cosh
[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(b^2*c^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3384

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

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Mathematica [A]
time = 0.27, size = 70, normalized size = 0.89 \begin {gather*} \frac {-\frac {b c^2 x^2}{a+b \sinh ^{-1}(c x)}-\text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right ) \sinh \left (\frac {2 a}{b}\right )+\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*c^2*x^2)/(a + b*ArcSinh[c*x])) - CoshIntegral[2*(a/b + ArcSinh[c*x])]*Sinh[(2*a)/b] + Cosh[(2*a)/b]*Sinh
Integral[2*(a/b + ArcSinh[c*x])])/(b^2*c^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(79)=158\).
time = 5.74, size = 192, normalized size = 2.43

method	result	size
default	\(\frac {1}{2 c^{3} \left (a +b \arcsinh \left (c x \right )\right ) b}-\frac {2 c^{2} x^{2}-2 \sqrt {c^{2} x^{2}+1}\, c x +1}{4 c^{3} \left (a +b \arcsinh \left (c x \right )\right ) b}+\frac {{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{2 c^{3} b^{2}}-\frac {2 b \,c^{2} x^{2}+2 x b c \sqrt {c^{2} x^{2}+1}+2 \,{\mathrm e}^{-\frac {2 a}{b}} \arcsinh \left (c x \right ) \expIntegral \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right ) b +2 \,{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right ) a +b}{4 c^{3} b^{2} \left (a +b \arcsinh \left (c x \right )\right )}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c^3/(a+b*arcsinh(c*x))/b-1/4*(2*c^2*x^2-2*(c^2*x^2+1)^(1/2)*c*x+1)/c^3/(a+b*arcsinh(c*x))/b+1/2/c^3/b^2*ex
p(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-1/4/c^3/b^2*(2*b*c^2*x^2+2*x*b*c*(c^2*x^2+1)^(1/2)+2*exp(-2*a/b)*arcsinh(c
*x)*Ei(1,-2*arcsinh(c*x)-2*a/b)*b+2*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*a+b)/(a+b*arcsinh(c*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^2/(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-(c^3*x^5 + c*x^3 + (c^2*x^4 + x^2)*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((2*c^5*x^6 + 5*c^3*x^4 + 3*c*x^2 + (2*c^3*x^4 + c*x^2)*(c^2*x^2 + 1) + 2*(2*c^4*x^5 + 3*c^2*x^3
+ x)*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))*log(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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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/(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^2/(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)
*arcsinh(c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^2/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*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}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {c^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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