3.4.72 \(\int \frac {x^3 (1+c^2 x^2)^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx\) [372]

Optimal. Leaf size=245 \[ \frac {3 \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{128 b c^4}+\frac {\text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right ) \sinh \left (\frac {3 a}{b}\right )}{32 b c^4}-\frac {3 \text {Chi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right ) \sinh \left (\frac {7 a}{b}\right )}{256 b c^4}-\frac {\text {Chi}\left (\frac {9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right ) \sinh \left (\frac {9 a}{b}\right )}{256 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{128 b c^4}-\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^4}+\frac {3 \cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}+\frac {\cosh \left (\frac {9 a}{b}\right ) \text {Shi}\left (\frac {9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4} \]

[Out]

-3/128*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b/c^4-1/32*cosh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b/c^4+3/256*cosh
(7*a/b)*Shi(7*(a+b*arcsinh(c*x))/b)/b/c^4+1/256*cosh(9*a/b)*Shi(9*(a+b*arcsinh(c*x))/b)/b/c^4+3/128*Chi((a+b*a
rcsinh(c*x))/b)*sinh(a/b)/b/c^4+1/32*Chi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b/c^4-3/256*Chi(7*(a+b*arcsinh(c*
x))/b)*sinh(7*a/b)/b/c^4-1/256*Chi(9*(a+b*arcsinh(c*x))/b)*sinh(9*a/b)/b/c^4

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Rubi [A]
time = 0.35, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5819, 5556, 3384, 3379, 3382} \begin {gather*} \frac {3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{128 b c^4}+\frac {\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^4}-\frac {3 \sinh \left (\frac {7 a}{b}\right ) \text {Chi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}-\frac {\sinh \left (\frac {9 a}{b}\right ) \text {Chi}\left (\frac {9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{128 b c^4}-\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^4}+\frac {3 \cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4}+\frac {\cosh \left (\frac {9 a}{b}\right ) \text {Shi}\left (\frac {9 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{256 b c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(3*CoshIntegral[(a + b*ArcSinh[c*x])/b]*Sinh[a/b])/(128*b*c^4) + (CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b]*Sin
h[(3*a)/b])/(32*b*c^4) - (3*CoshIntegral[(7*(a + b*ArcSinh[c*x]))/b]*Sinh[(7*a)/b])/(256*b*c^4) - (CoshIntegra
l[(9*(a + b*ArcSinh[c*x]))/b]*Sinh[(9*a)/b])/(256*b*c^4) - (3*Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/
(128*b*c^4) - (Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(32*b*c^4) + (3*Cosh[(7*a)/b]*SinhInteg
ral[(7*(a + b*ArcSinh[c*x]))/b])/(256*b*c^4) + (Cosh[(9*a)/b]*SinhIntegral[(9*(a + b*ArcSinh[c*x]))/b])/(256*b
*c^4)

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 5819

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (1+c^2 x^2\right )^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cosh ^6(x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {3 \sinh (x)}{128 (a+b x)}-\frac {\sinh (3 x)}{32 (a+b x)}+\frac {3 \sinh (7 x)}{256 (a+b x)}+\frac {\sinh (9 x)}{256 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4}\\ &=\frac {\text {Subst}\left (\int \frac {\sinh (9 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac {3 \text {Subst}\left (\int \frac {\sinh (7 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}-\frac {3 \text {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{128 c^4}-\frac {\text {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^4}\\ &=-\frac {\left (3 \cosh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{128 c^4}-\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^4}+\frac {\left (3 \cosh \left (\frac {7 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac {\cosh \left (\frac {9 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {9 a}{b}+9 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}+\frac {\left (3 \sinh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{128 c^4}+\frac {\sinh \left (\frac {3 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^4}-\frac {\left (3 \sinh \left (\frac {7 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {7 a}{b}+7 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}-\frac {\sinh \left (\frac {9 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {9 a}{b}+9 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{256 c^4}\\ &=\frac {3 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{128 b c^4}+\frac {\text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{32 b c^4}-\frac {3 \text {Chi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {7 a}{b}\right )}{256 b c^4}-\frac {\text {Chi}\left (\frac {9 a}{b}+9 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {9 a}{b}\right )}{256 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{128 b c^4}-\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{32 b c^4}+\frac {3 \cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (\frac {7 a}{b}+7 \sinh ^{-1}(c x)\right )}{256 b c^4}+\frac {\cosh \left (\frac {9 a}{b}\right ) \text {Shi}\left (\frac {9 a}{b}+9 \sinh ^{-1}(c x)\right )}{256 b c^4}\\ \end {align*}

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Mathematica [A]
time = 0.72, size = 180, normalized size = 0.73 \begin {gather*} \frac {6 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )+8 \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\right )-3 \text {Chi}\left (7 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right ) \sinh \left (\frac {7 a}{b}\right )-\text {Chi}\left (9 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right ) \sinh \left (\frac {9 a}{b}\right )-6 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-8 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+3 \cosh \left (\frac {7 a}{b}\right ) \text {Shi}\left (7 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {9 a}{b}\right ) \text {Shi}\left (9 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{256 b c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(1 + c^2*x^2)^(5/2))/(a + b*ArcSinh[c*x]),x]

[Out]

(6*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] + 8*CoshIntegral[3*(a/b + ArcSinh[c*x])]*Sinh[(3*a)/b] - 3*CoshI
ntegral[7*(a/b + ArcSinh[c*x])]*Sinh[(7*a)/b] - CoshIntegral[9*(a/b + ArcSinh[c*x])]*Sinh[(9*a)/b] - 6*Cosh[a/
b]*SinhIntegral[a/b + ArcSinh[c*x]] - 8*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] + 3*Cosh[(7*a)/b]*S
inhIntegral[7*(a/b + ArcSinh[c*x])] + Cosh[(9*a)/b]*SinhIntegral[9*(a/b + ArcSinh[c*x])])/(256*b*c^4)

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Maple [A]
time = 6.16, size = 238, normalized size = 0.97

method	result	size
default	\(\frac {{\mathrm e}^{\frac {9 a}{b}} \expIntegral \left (1, 9 \arcsinh \left (c x \right )+\frac {9 a}{b}\right )}{512 c^{4} b}+\frac {3 \,{\mathrm e}^{\frac {7 a}{b}} \expIntegral \left (1, 7 \arcsinh \left (c x \right )+\frac {7 a}{b}\right )}{512 c^{4} b}-\frac {{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{64 c^{4} b}-\frac {3 \,{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{256 c^{4} b}+\frac {3 \,{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{256 c^{4} b}+\frac {{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{64 c^{4} b}-\frac {3 \,{\mathrm e}^{-\frac {7 a}{b}} \expIntegral \left (1, -7 \arcsinh \left (c x \right )-\frac {7 a}{b}\right )}{512 c^{4} b}-\frac {{\mathrm e}^{-\frac {9 a}{b}} \expIntegral \left (1, -9 \arcsinh \left (c x \right )-\frac {9 a}{b}\right )}{512 c^{4} b}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/512/c^4/b*exp(9*a/b)*Ei(1,9*arcsinh(c*x)+9*a/b)+3/512/c^4/b*exp(7*a/b)*Ei(1,7*arcsinh(c*x)+7*a/b)-1/64/c^4/b
*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-3/256/c^4/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)+3/256/c^4/b*exp(-a/b)*Ei(1,
-arcsinh(c*x)-a/b)+1/64/c^4/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-3/512/c^4/b*exp(-7*a/b)*Ei(1,-7*arcsinh(
c*x)-7*a/b)-1/512/c^4/b*exp(-9*a/b)*Ei(1,-9*arcsinh(c*x)-9*a/b)

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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^3*(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^3/(b*arcsinh(c*x) + a), 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^3*(c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((c^4*x^7 + 2*c^2*x^5 + x^3)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{a + b \operatorname {asinh}{\left (c x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c**2*x**2+1)**(5/2)/(a+b*asinh(c*x)),x)

[Out]

Integral(x**3*(c**2*x**2 + 1)**(5/2)/(a + b*asinh(c*x)), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,{\left (c^2\,x^2+1\right )}^{5/2}}{a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(c^2*x^2 + 1)^(5/2))/(a + b*asinh(c*x)),x)

[Out]

int((x^3*(c^2*x^2 + 1)^(5/2))/(a + b*asinh(c*x)), x)

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