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

Optimal. Leaf size=93 \[ -\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}-\frac {x^2 \left (c^2 x^2+1\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5777, 5669, 5448, 12, 3303, 3298, 3301} \[ -\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}-\frac {x^2 \left (c^2 x^2+1\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^2*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) - (CoshIntegral[(4*a)/b + 4*ArcSinh[c*x]]*Sinh[(4*a)/b])/(2*
b^2*c^3) + (Cosh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c*x]])/(2*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 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 5448

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 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p -
1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] - Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x],
x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

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 \left (1+c^2 x^2\right )}{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 {(4 c) \int \frac {x^3}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^2 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {4 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 (a+b x)} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {4 \operatorname {Subst}\left (\int \left (-\frac {\sinh (2 x)}{4 (a+b x)}+\frac {\sinh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}\\ &=-\frac {x^2 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\cosh \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}\\ &=-\frac {x^2 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {\text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {4 a}{b}\right )}{2 b^2 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 82, normalized size = 0.88 \[ \frac {-\frac {2 b c^2 x^2 \left (c^2 x^2+1\right )}{a+b \sinh ^{-1}(c x)}-\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{2 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} x^{2} + 1} x^{2}}{b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.28, size = 248, normalized size = 2.67 \[ \frac {1}{8 c^{3} \left (a +b \arcsinh \left (c x \right )\right ) b}-\frac {8 c^{4} x^{4}-8 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+8 c^{2} x^{2}-4 c x \sqrt {c^{2} x^{2}+1}+1}{16 c^{3} \left (a +b \arcsinh \left (c x \right )\right ) b}+\frac {{\mathrm e}^{\frac {4 a}{b}} \Ei \left (1, 4 \arcsinh \left (c x \right )+\frac {4 a}{b}\right )}{4 c^{3} b^{2}}-\frac {8 x^{4} b \,c^{4}+8 \sqrt {c^{2} x^{2}+1}\, x^{3} b \,c^{3}+8 x^{2} b \,c^{2}+4 b c \sqrt {c^{2} x^{2}+1}\, x +4 \arcsinh \left (c x \right ) \Ei \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {4 a}{b}} b +4 \Ei \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {4 a}{b}} a +b}{16 c^{3} b^{2} \left (a +b \arcsinh \left (c x \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c^3/(a+b*arcsinh(c*x))/b-1/16*(8*c^4*x^4-8*c^3*x^3*(c^2*x^2+1)^(1/2)+8*c^2*x^2-4*c*x*(c^2*x^2+1)^(1/2)+1)/
c^3/(a+b*arcsinh(c*x))/b+1/4/c^3/b^2*exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)-1/16/c^3/b^2*(8*x^4*b*c^4+8*(c^2*x^
2+1)^(1/2)*x^3*b*c^3+8*x^2*b*c^2+4*b*c*(c^2*x^2+1)^(1/2)*x+4*arcsinh(c*x)*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a
/b)*b+4*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a/b)*a+b)/(a+b*arcsinh(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (c^{2} x^{4} + x^{2}\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (c^{3} x^{5} + c x^{3}\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} a b c^{2} x + a b c + {\left (b^{2} c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )} + \int \frac {{\left (4 \, c^{3} x^{4} + c x^{2}\right )} {\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}} + 2 \, {\left (4 \, c^{4} x^{5} + 4 \, c^{2} x^{3} + x\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (4 \, c^{5} x^{6} + 7 \, c^{3} x^{4} + 3 \, c x^{2}\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c + {\left (b^{2} c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \, {\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c^2*x^4 + x^2)*(c^2*x^2 + 1) + (c^3*x^5 + c*x^3)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sqrt(c^2*x^2 + 1)*a*b*c^
2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x^2 + 1))) + integrate(((
4*c^3*x^4 + c*x^2)*(c^2*x^2 + 1)^(3/2) + 2*(4*c^4*x^5 + 4*c^2*x^3 + x)*(c^2*x^2 + 1) + (4*c^5*x^6 + 7*c^3*x^4
+ 3*c*x^2)*sqrt(c^2*x^2 + 1))/(a*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^5*x^4
+ (c^2*x^2 + 1)*b^2*c^3*x^2 + 2*b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log(c*x +
sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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