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

Optimal. Leaf size=85 \[ -\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c}+\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c}-\frac{c^2 x^2+1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.180853, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {5696, 5669, 5448, 12, 3303, 3298, 3301} \[ -\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b^2 c}+\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b^2 c}-\frac{c^2 x^2+1}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5696

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

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 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 12

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.16659, size = 73, normalized size = 0.86 \[ \frac{-\frac{b c^2 x^2+b}{a+b \sinh ^{-1}(c x)}-\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.126, size = 192, normalized size = 2.3 \begin{align*} -{\frac{1}{2\,bc \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) }}-{\frac{1}{4\,bc \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( 2\,{c}^{2}{x}^{2}-2\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }+{\frac{1}{2\,c{b}^{2}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( cx \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{1}{4\,c{b}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( 2\,{x}^{2}b{c}^{2}+2\,bc\sqrt{{c}^{2}{x}^{2}+1}x+2\,{\it Arcsinh} \left ( cx \right ){{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) b+2\,{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) a+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b/c/(a+b*arcsinh(c*x))-1/4*(2*c^2*x^2-2*c*x*(c^2*x^2+1)^(1/2)+1)/c/(a+b*arcsinh(c*x))/b+1/2/c/b^2*exp(2*a
/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-1/4/c/b^2*(2*x^2*b*c^2+2*b*c*(c^2*x^2+1)^(1/2)*x+2*arcsinh(c*x)*exp(-2*a/b)*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., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (c^{2} x^{2} + 1\right )}^{2} +{\left (c^{3} x^{3} + c x\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 (2 \, c^{2} x^{2} - 1\right )}{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 2 \,{\left (2 \, c^{3} x^{3} + c x\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (2 \, c^{4} x^{4} + 3 \, c^{2} x^{2} + 1\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{4} x^{4} +{\left (c^{2} x^{2} + 1\right )} a b c^{2} x^{2} + 2 \, a b c^{2} x^{2} + a b +{\left (b^{2} c^{4} x^{4} +{\left (c^{2} x^{2} + 1\right )} b^{2} c^{2} x^{2} + 2 \, b^{2} c^{2} x^{2} + b^{2} + 2 \,{\left (b^{2} c^{3} x^{3} + b^{2} c 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^{3} x^{3} + a b c x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} x^{2} + 1}}{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c^{2} x^{2} + 1}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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