3.374 \(\int \frac {x (1+c^2 x^2)^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx\)

Optimal. Leaf size=245 \[ -\frac {5 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{64 b c^2}-\frac {9 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2}-\frac {\sinh \left (\frac {7 a}{b}\right ) \text {Chi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{64 b c^2}+\frac {9 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2}+\frac {\cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{64 b c^2} \]

[Out]

5/64*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b/c^2+9/64*cosh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b/c^2+5/64*cosh(5*
a/b)*Shi(5*(a+b*arcsinh(c*x))/b)/b/c^2+1/64*cosh(7*a/b)*Shi(7*(a+b*arcsinh(c*x))/b)/b/c^2-5/64*Chi((a+b*arcsin
h(c*x))/b)*sinh(a/b)/b/c^2-9/64*Chi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b/c^2-5/64*Chi(5*(a+b*arcsinh(c*x))/b)
*sinh(5*a/b)/b/c^2-1/64*Chi(7*(a+b*arcsinh(c*x))/b)*sinh(7*a/b)/b/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.49, antiderivative size = 241, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5779, 5448, 3303, 3298, 3301} \[ -\frac {5 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{64 b c^2}-\frac {9 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{64 b c^2}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{64 b c^2}-\frac {\sinh \left (\frac {7 a}{b}\right ) \text {Chi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {9 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {\cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right )}{64 b c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(-5*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b])/(64*b*c^2) - (9*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]]*Sinh[(3
*a)/b])/(64*b*c^2) - (5*CoshIntegral[(5*a)/b + 5*ArcSinh[c*x]]*Sinh[(5*a)/b])/(64*b*c^2) - (CoshIntegral[(7*a)
/b + 7*ArcSinh[c*x]]*Sinh[(7*a)/b])/(64*b*c^2) + (5*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(64*b*c^2) + (
9*Cosh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(64*b*c^2) + (5*Cosh[(5*a)/b]*SinhIntegral[(5*a)/b + 5
*ArcSinh[c*x]])/(64*b*c^2) + (Cosh[(7*a)/b]*SinhIntegral[(7*a)/b + 7*ArcSinh[c*x]])/(64*b*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 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 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {x \left (1+c^2 x^2\right )^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cosh ^6(x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {5 \sinh (x)}{64 (a+b x)}+\frac {9 \sinh (3 x)}{64 (a+b x)}+\frac {5 \sinh (5 x)}{64 (a+b x)}+\frac {\sinh (7 x)}{64 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sinh (7 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {5 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {5 \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {9 \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}\\ &=\frac {\left (5 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {\left (9 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {\left (5 \cosh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}+\frac {\cosh \left (\frac {7 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}-\frac {\left (5 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}-\frac {\left (9 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}-\frac {\left (5 \sinh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}-\frac {\sinh \left (\frac {7 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{64 c^2}\\ &=-\frac {5 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{64 b c^2}-\frac {9 \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{64 b c^2}-\frac {5 \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {5 a}{b}\right )}{64 b c^2}-\frac {\text {Chi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {7 a}{b}\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {9 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{64 b c^2}+\frac {\cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right )}{64 b c^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.80, size = 180, normalized size = 0.73 \[ \frac {-5 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-9 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-5 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac {7 a}{b}\right ) \text {Chi}\left (7 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+5 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+9 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+5 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (7 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{64 b c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(-5*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] - 9*CoshIntegral[3*(a/b + ArcSinh[c*x])]*Sinh[(3*a)/b] - 5*Cosh
Integral[5*(a/b + ArcSinh[c*x])]*Sinh[(5*a)/b] - CoshIntegral[7*(a/b + ArcSinh[c*x])]*Sinh[(7*a)/b] + 5*Cosh[a
/b]*SinhIntegral[a/b + ArcSinh[c*x]] + 9*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] + 5*Cosh[(5*a)/b]*
SinhIntegral[5*(a/b + ArcSinh[c*x])] + Cosh[(7*a)/b]*SinhIntegral[7*(a/b + ArcSinh[c*x])])/(64*b*c^2)

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{5} + 2 \, c^{2} x^{3} + x\right )} \sqrt {c^{2} x^{2} + 1}}{b \operatorname {arsinh}\left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((c^4*x^5 + 2*c^2*x^3 + x)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a), x)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.42, size = 238, normalized size = 0.97 \[ \frac {{\mathrm e}^{\frac {7 a}{b}} \Ei \left (1, 7 \arcsinh \left (c x \right )+\frac {7 a}{b}\right )}{128 c^{2} b}+\frac {5 \,{\mathrm e}^{\frac {5 a}{b}} \Ei \left (1, 5 \arcsinh \left (c x \right )+\frac {5 a}{b}\right )}{128 c^{2} b}+\frac {9 \,{\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{128 c^{2} b}+\frac {5 \,{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{128 c^{2} b}-\frac {5 \,{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{128 c^{2} b}-\frac {9 \,{\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{128 c^{2} b}-\frac {5 \,{\mathrm e}^{-\frac {5 a}{b}} \Ei \left (1, -5 \arcsinh \left (c x \right )-\frac {5 a}{b}\right )}{128 c^{2} b}-\frac {{\mathrm e}^{-\frac {7 a}{b}} \Ei \left (1, -7 \arcsinh \left (c x \right )-\frac {7 a}{b}\right )}{128 c^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x)

[Out]

1/128/c^2/b*exp(7*a/b)*Ei(1,7*arcsinh(c*x)+7*a/b)+5/128/c^2/b*exp(5*a/b)*Ei(1,5*arcsinh(c*x)+5*a/b)+9/128/c^2/
b*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)+5/128/c^2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-5/128/c^2/b*exp(-a/b)*Ei(1
,-arcsinh(c*x)-a/b)-9/128/c^2/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-5/128/c^2/b*exp(-5*a/b)*Ei(1,-5*arcsin
h(c*x)-5*a/b)-1/128/c^2/b*exp(-7*a/b)*Ei(1,-7*arcsinh(c*x)-7*a/b)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

integrate((c^2*x^2 + 1)^(5/2)*x/(b*arcsinh(c*x) + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x\,{\left (c^2\,x^2+1\right )}^{5/2}}{a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c^2*x^2 + 1)^(5/2))/(a + b*asinh(c*x)),x)

[Out]

int((x*(c^2*x^2 + 1)^(5/2))/(a + b*asinh(c*x)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{a + b \operatorname {asinh}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c**2*x**2+1)**(5/2)/(a+b*asinh(c*x)),x)

[Out]

Integral(x*(c**2*x**2 + 1)**(5/2)/(a + b*asinh(c*x)), x)

________________________________________________________________________________________