3.4.65 \(\int \frac {x^2 (1+c^2 x^2)^{3/2}}{a+b \sinh ^{-1}(c x)} \, dx\) [365]

Optimal. Leaf size=206 \[ -\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3} \]

[Out]

-1/32*Chi(2*(a+b*arcsinh(c*x))/b)*cosh(2*a/b)/b/c^3+1/16*Chi(4*(a+b*arcsinh(c*x))/b)*cosh(4*a/b)/b/c^3+1/32*Ch
i(6*(a+b*arcsinh(c*x))/b)*cosh(6*a/b)/b/c^3-1/16*ln(a+b*arcsinh(c*x))/b/c^3+1/32*Shi(2*(a+b*arcsinh(c*x))/b)*s
inh(2*a/b)/b/c^3-1/16*Shi(4*(a+b*arcsinh(c*x))/b)*sinh(4*a/b)/b/c^3-1/32*Shi(6*(a+b*arcsinh(c*x))/b)*sinh(6*a/
b)/b/c^3

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Rubi [A]
time = 0.28, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, 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 {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^3}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(b*c^3) + (Cosh[(4*a)/b]*CoshIntegral[(4*(a + b
*ArcSinh[c*x]))/b])/(16*b*c^3) + (Cosh[(6*a)/b]*CoshIntegral[(6*(a + b*ArcSinh[c*x]))/b])/(32*b*c^3) - Log[a +
b*ArcSinh[c*x]]/(16*b*c^3) + (Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(32*b*c^3) - (Sinh[(4*a
)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(16*b*c^3) - (Sinh[(6*a)/b]*SinhIntegral[(6*(a + b*ArcSinh[c*x]
))/b])/(32*b*c^3)

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^2 \left (1+c^2 x^2\right )^{3/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cosh ^4(x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {1}{16 (a+b x)}-\frac {\cosh (2 x)}{32 (a+b x)}+\frac {\cosh (4 x)}{16 (a+b x)}+\frac {\cosh (6 x)}{32 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=-\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac {\text {Subst}\left (\int \frac {\cosh (6 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}\\ &=-\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac {\cosh \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 )}{32 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}+\frac {\sinh \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 )}{32 c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c^3}\\ &=-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac {\log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c^3}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c^3}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c^3}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 152, normalized size = 0.74 \begin {gather*} \frac {-\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+2 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \log \left (a+b \sinh ^{-1}(c x)\right )+\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{32 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])]) + 2*Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcSinh[c*x])]
+ Cosh[(6*a)/b]*CoshIntegral[6*(a/b + ArcSinh[c*x])] - 2*Log[a + b*ArcSinh[c*x]] + Sinh[(2*a)/b]*SinhIntegral[
2*(a/b + ArcSinh[c*x])] - 2*Sinh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - Sinh[(6*a)/b]*SinhIntegral[6*
(a/b + ArcSinh[c*x])])/(32*b*c^3)

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Maple [A]
time = 8.98, size = 199, normalized size = 0.97

method	result	size
default	\(-\frac {\ln \left (a +b \arcsinh \left (c x \right )\right )}{16 b \,c^{3}}-\frac {{\mathrm e}^{\frac {6 a}{b}} \expIntegral \left (1, 6 \arcsinh \left (c x \right )+\frac {6 a}{b}\right )}{64 c^{3} b}-\frac {{\mathrm e}^{\frac {4 a}{b}} \expIntegral \left (1, 4 \arcsinh \left (c x \right )+\frac {4 a}{b}\right )}{32 c^{3} b}+\frac {{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{64 c^{3} b}+\frac {{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right )}{64 c^{3} b}-\frac {{\mathrm e}^{-\frac {4 a}{b}} \expIntegral \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right )}{32 c^{3} b}-\frac {{\mathrm e}^{-\frac {6 a}{b}} \expIntegral \left (1, -6 \arcsinh \left (c x \right )-\frac {6 a}{b}\right )}{64 c^{3} b}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*ln(a+b*arcsinh(c*x))/b/c^3-1/64/c^3/b*exp(6*a/b)*Ei(1,6*arcsinh(c*x)+6*a/b)-1/32/c^3/b*exp(4*a/b)*Ei(1,4
*arcsinh(c*x)+4*a/b)+1/64/c^3/b*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)+1/64/c^3/b*exp(-2*a/b)*Ei(1,-2*arcsinh(c
*x)-2*a/b)-1/32/c^3/b*exp(-4*a/b)*Ei(1,-4*arcsinh(c*x)-4*a/b)-1/64/c^3/b*exp(-6*a/b)*Ei(1,-6*arcsinh(c*x)-6*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^2*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

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

[Out]

integral((c^2*x^4 + x^2)*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^{2} \left (c^{2} x^{2} + 1\right )^{\frac {3}{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**2*(c**2*x**2+1)**(3/2)/(a+b*asinh(c*x)),x)

[Out]

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

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

[Out]

integrate((c^2*x^2 + 1)^(3/2)*x^2/(b*arcsinh(c*x) + a), x)

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

[Out]

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

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