3.28 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))}{d+c^2 d x^2} \, dx\)

Optimal. Leaf size=156 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}-\frac{b \left (c^2 x^2+1\right )^{3/2}}{9 c^5 d}+\frac{4 b \sqrt{c^2 x^2+1}}{3 c^5 d} \]

[Out]

(4*b*Sqrt[1 + c^2*x^2])/(3*c^5*d) - (b*(1 + c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSinh[c*x]))/(c^4*d) + (x^
3*(a + b*ArcSinh[c*x]))/(3*c^2*d) + (2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c^5*d) - (I*b*PolyLog[2,
(-I)*E^ArcSinh[c*x]])/(c^5*d) + (I*b*PolyLog[2, I*E^ArcSinh[c*x]])/(c^5*d)

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Rubi [A]  time = 0.240951, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5767, 5693, 4180, 2279, 2391, 261, 266, 43} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}-\frac{b \left (c^2 x^2+1\right )^{3/2}}{9 c^5 d}+\frac{4 b \sqrt{c^2 x^2+1}}{3 c^5 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2),x]

[Out]

(4*b*Sqrt[1 + c^2*x^2])/(3*c^5*d) - (b*(1 + c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSinh[c*x]))/(c^4*d) + (x^
3*(a + b*ArcSinh[c*x]))/(3*c^2*d) + (2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c^5*d) - (I*b*PolyLog[2,
(-I)*E^ArcSinh[c*x]])/(c^5*d) + (I*b*PolyLog[2, I*E^ArcSinh[c*x]])/(c^5*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 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}-\frac{\int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{c^2}-\frac{b \int \frac{x^3}{\sqrt{1+c^2 x^2}} \, dx}{3 c d}\\ &=-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{c^4}+\frac{b \int \frac{x}{\sqrt{1+c^2 x^2}} \, dx}{c^3 d}-\frac{b \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )}{6 c d}\\ &=\frac{b \sqrt{1+c^2 x^2}}{c^5 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}+\frac{\operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}-\frac{b \operatorname{Subst}\left (\int \left (-\frac{1}{c^2 \sqrt{1+c^2 x}}+\frac{\sqrt{1+c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{6 c d}\\ &=\frac{4 b \sqrt{1+c^2 x^2}}{3 c^5 d}-\frac{b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}+\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac{(i b) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}+\frac{(i b) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}\\ &=\frac{4 b \sqrt{1+c^2 x^2}}{3 c^5 d}-\frac{b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}+\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ &=\frac{4 b \sqrt{1+c^2 x^2}}{3 c^5 d}-\frac{b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d}+\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac{i b \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac{i b \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ \end{align*}

Mathematica [A]  time = 0.231707, size = 170, normalized size = 1.09 \[ \frac{-9 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+9 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+3 a c^3 x^3-9 a c x+9 a \tan ^{-1}(c x)-b c^2 x^2 \sqrt{c^2 x^2+1}+11 b \sqrt{c^2 x^2+1}+3 b c^3 x^3 \sinh ^{-1}(c x)-9 b c x \sinh ^{-1}(c x)+9 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-9 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{9 c^5 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2),x]

[Out]

(-9*a*c*x + 3*a*c^3*x^3 + 11*b*Sqrt[1 + c^2*x^2] - b*c^2*x^2*Sqrt[1 + c^2*x^2] - 9*b*c*x*ArcSinh[c*x] + 3*b*c^
3*x^3*ArcSinh[c*x] + 9*a*ArcTan[c*x] + (9*I)*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - (9*I)*b*ArcSinh[c*x]*L
og[1 + I*E^ArcSinh[c*x]] - (9*I)*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] + (9*I)*b*PolyLog[2, I*E^ArcSinh[c*x]])/(9*
c^5*d)

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Maple [A]  time = 0.158, size = 266, normalized size = 1.7 \begin{align*}{\frac{a{x}^{3}}{3\,{c}^{2}d}}-{\frac{ax}{{c}^{4}d}}+{\frac{a\arctan \left ( cx \right ) }{{c}^{5}d}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{3}}{3\,{c}^{2}d}}-{\frac{b{\it Arcsinh} \left ( cx \right ) x}{{c}^{4}d}}+{\frac{b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{{c}^{5}d}}-{\frac{b{x}^{2}}{9\,{c}^{3}d}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{11\,b}{9\,{c}^{5}d}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{b\arctan \left ( cx \right ) }{{c}^{5}d}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{b\arctan \left ( cx \right ) }{{c}^{5}d}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{ib}{{c}^{5}d}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{ib}{{c}^{5}d}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/c^2*a/d*x^3-1/c^4*a/d*x+1/c^5*a/d*arctan(c*x)+1/3/c^2*b/d*arcsinh(c*x)*x^3-1/c^4*b/d*arcsinh(c*x)*x+1/c^5*
b/d*arcsinh(c*x)*arctan(c*x)-1/9/c^3*b/d*x^2*(c^2*x^2+1)^(1/2)+11/9*b*(c^2*x^2+1)^(1/2)/c^5/d+1/c^5*b/d*arctan
(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/c^5*b/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I/c^5*b/d*
dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I/c^5*b/d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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^{4}}{c^{2} d x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

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