3.5.14 \(\int \frac {\sqrt {1+c^2 x^2}}{(a+b \sinh ^{-1}(c x))^2} \, dx\) [414]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 = {5790, 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}+\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 )} \end {gather*}

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*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)

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 5790

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[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)/(b*(n + 1)))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^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]

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 \text {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 \text {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 {\text {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 ) \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}-\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}\\ &=-\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 = 0.13, size = 73, normalized size = 0.86 \begin {gather*} \frac {-\frac {b+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} \end {gather*}

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)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(85)=170\).
time = 5.49, size = 192, normalized size = 2.26

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

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,method=_RETURNVERBOSE)

[Out]

-1/2/b/c/(a+b*arcsinh(c*x))-1/4*(2*c^2*x^2-2*(c^2*x^2+1)^(1/2)*c*x+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*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))

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________