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

Optimal. Leaf size=135 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}-\frac{b x \sqrt{c^2 x^2+1}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.195492, antiderivative size = 135, 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 = {5767, 5714, 3718, 2190, 2279, 2391, 321, 215} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}-\frac{b x \sqrt{c^2 x^2+1}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5767

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)/(e*(m + 2*p + 1)), x] + (-Dist[(f^2*(m - 1))/(c^2
*(m + 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d
+ e*x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(
a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[m
, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 )}{d+c^2 d x^2} \, dx &=\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}-\frac{\int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{c^2}-\frac{b \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{2 c d}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{4 c^3 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}-\frac{\operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}+\frac{b \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{4 c^3 d}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\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}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ &=-\frac{b x \sqrt{1+c^2 x^2}}{4 c^3 d}+\frac{b \sinh ^{-1}(c x)}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac{b \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [A]  time = 0.201929, size = 181, normalized size = 1.34 \[ -\frac{4 b \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+4 b \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-2 a c^2 x^2+2 a \log \left (c^2 x^2+1\right )+b c x \sqrt{c^2 x^2+1}-2 b c^2 x^2 \sinh ^{-1}(c x)+4 b \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+4 b \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-2 b \sinh ^{-1}(c x)^2-b \sinh ^{-1}(c x)}{4 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.081, size = 161, normalized size = 1.2 \begin{align*}{\frac{a{x}^{2}}{2\,{c}^{2}d}}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{4}d}}+{\frac{b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,{c}^{4}d}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{2\,{c}^{2}d}}-{\frac{bx}{4\,{c}^{3}d}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{4\,{c}^{4}d}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{c}^{4}d}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{b}{2\,{c}^{4}d}{\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),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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^{2} d x^{2} + d}, 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),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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}}{c^{2} d x^{2} + d}\,{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),x, algorithm="giac")

[Out]

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