3.38 \(\int \frac{x^3 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=145 \[ \frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}-\frac{b x}{2 c^3 d^2 \sqrt{c^2 x^2+1}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2} \]

[Out]

-(b*x)/(2*c^3*d^2*Sqrt[1 + c^2*x^2]) + (b*ArcSinh[c*x])/(2*c^4*d^2) - (x^2*(a + b*ArcSinh[c*x]))/(2*c^2*d^2*(1
+ c^2*x^2)) - (a + b*ArcSinh[c*x])^2/(2*b*c^4*d^2) + ((a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c^4*
d^2) + (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*c^4*d^2)

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Rubi [A]  time = 0.190549, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5751, 5714, 3718, 2190, 2279, 2391, 288, 215} \[ \frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d^2}-\frac{b x}{2 c^3 d^2 \sqrt{c^2 x^2+1}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x)/(2*c^3*d^2*Sqrt[1 + c^2*x^2]) + (b*ArcSinh[c*x])/(2*c^4*d^2) - (x^2*(a + b*ArcSinh[c*x]))/(2*c^2*d^2*(1
+ c^2*x^2)) - (a + b*ArcSinh[c*x])^2/(2*b*c^4*d^2) + ((a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c^4*
d^2) + (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*c^4*d^2)

Rule 5751

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p
+ 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 5714

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac{b \int \frac{x^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}+\frac{\int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{c^2 d}\\ &=-\frac{b x}{2 c^3 d^2 \sqrt{1+c^2 x^2}}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}+\frac{b \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{2 c^3 d^2}\\ &=-\frac{b x}{2 c^3 d^2 \sqrt{1+c^2 x^2}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac{b x}{2 c^3 d^2 \sqrt{1+c^2 x^2}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d^2}\\ &=-\frac{b x}{2 c^3 d^2 \sqrt{1+c^2 x^2}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}-\frac{b \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}\\ &=-\frac{b x}{2 c^3 d^2 \sqrt{1+c^2 x^2}}+\frac{b \sinh ^{-1}(c x)}{2 c^4 d^2}-\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d^2}+\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d^2}+\frac{b \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d^2}\\ \end{align*}

Mathematica [C]  time = 0.218747, size = 241, normalized size = 1.66 \[ \frac{2 b \left (c^2 x^2+1\right ) \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+2 b \left (c^2 x^2+1\right ) \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+a c^2 x^2 \log \left (c^2 x^2+1\right )+a \log \left (c^2 x^2+1\right )+a-b c x \sqrt{c^2 x^2+1}-b c^2 x^2 \sinh ^{-1}(c x)^2+2 b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+2 b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-b \sinh ^{-1}(c x)^2+b \sinh ^{-1}(c x)+2 b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+2 b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{2 c^4 d^2 \left (c^2 x^2+1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a - b*c*x*Sqrt[1 + c^2*x^2] + b*ArcSinh[c*x] - b*ArcSinh[c*x]^2 - b*c^2*x^2*ArcSinh[c*x]^2 + 2*b*ArcSinh[c*x]
*Log[1 - I*E^ArcSinh[c*x]] + 2*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + 2*b*ArcSinh[c*x]*Log[1 + I*E
^ArcSinh[c*x]] + 2*b*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + a*Log[1 + c^2*x^2] + a*c^2*x^2*Log[1 + c
^2*x^2] + 2*b*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSinh[c*x]] + 2*b*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]])/
(2*c^4*d^2*(1 + c^2*x^2))

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Maple [A]  time = 0.141, size = 206, normalized size = 1.4 \begin{align*}{\frac{a}{2\,{c}^{4}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{4}{d}^{2}}}-{\frac{b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,{c}^{4}{d}^{2}}}-{\frac{bx}{2\,{c}^{3}{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{x}^{2}}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{2\,{c}^{4}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b}{2\,{c}^{4}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{c}^{4}{d}^{2}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{b}{2\,{c}^{4}{d}^{2}}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x)

[Out]

1/2/c^4*a/d^2/(c^2*x^2+1)+1/2/c^4*a/d^2*ln(c^2*x^2+1)-1/2/c^4*b/d^2*arcsinh(c*x)^2-1/2*b*x/c^3/d^2/(c^2*x^2+1)
^(1/2)+1/2/c^2*b/d^2*x^2/(c^2*x^2+1)+1/2/c^4*b/d^2*arcsinh(c*x)/(c^2*x^2+1)+1/2/c^4*b/d^2/(c^2*x^2+1)+1/c^4*b/
d^2*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{8} \, b{\left (\frac{{\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - 4 \,{\left ({\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right ) + 1\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2}{c^{6} d^{2} x^{2} + c^{4} d^{2}} + 8 \, \int \frac{{\left (c^{2} x^{2} + 1\right )} \log \left (c^{2} x^{2} + 1\right ) + 1}{2 \,{\left (c^{8} d^{2} x^{5} + 2 \, c^{6} d^{2} x^{3} + c^{4} d^{2} x +{\left (c^{7} d^{2} x^{4} + 2 \, c^{5} d^{2} x^{2} + c^{3} d^{2}\right )} \sqrt{c^{2} x^{2} + 1}\right )}}\,{d x}\right )} + \frac{1}{2} \, a{\left (\frac{1}{c^{6} d^{2} x^{2} + c^{4} d^{2}} + \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b*(((c^2*x^2 + 1)*log(c^2*x^2 + 1)^2 - 4*((c^2*x^2 + 1)*log(c^2*x^2 + 1) + 1)*log(c*x + sqrt(c^2*x^2 + 1)
) - 2)/(c^6*d^2*x^2 + c^4*d^2) + 8*integrate(1/2*((c^2*x^2 + 1)*log(c^2*x^2 + 1) + 1)/(c^8*d^2*x^5 + 2*c^6*d^2
*x^3 + c^4*d^2*x + (c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2)*sqrt(c^2*x^2 + 1)), x)) + 1/2*a*(1/(c^6*d^2*x^2 + c
^4*d^2) + log(c^2*x^2 + 1)/(c^4*d^2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \operatorname{arsinh}\left (c x\right ) + a x^{3}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{3}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac{b x^{3} \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**3/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*x**3*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1),
x))/d**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{3}}{{\left (c^{2} d x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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