3.46 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^3} \, dx\)

Optimal. Leaf size=186 \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{3 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (c^2 x^2+1\right )}+\frac{3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}-\frac{5 b}{8 c^5 d^3 \sqrt{c^2 x^2+1}}+\frac{b}{12 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

b/(12*c^5*d^3*(1 + c^2*x^2)^(3/2)) - (5*b)/(8*c^5*d^3*Sqrt[1 + c^2*x^2]) - (x^3*(a + b*ArcSinh[c*x]))/(4*c^2*d
^3*(1 + c^2*x^2)^2) - (3*x*(a + b*ArcSinh[c*x]))/(8*c^4*d^3*(1 + c^2*x^2)) + (3*(a + b*ArcSinh[c*x])*ArcTan[E^
ArcSinh[c*x]])/(4*c^5*d^3) - (((3*I)/8)*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^5*d^3) + (((3*I)/8)*b*PolyLog[2,
I*E^ArcSinh[c*x]])/(c^5*d^3)

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Rubi [A]  time = 0.232901, antiderivative size = 186, 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 = {5751, 5693, 4180, 2279, 2391, 261, 266, 43} \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{3 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (c^2 x^2+1\right )}+\frac{3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}-\frac{5 b}{8 c^5 d^3 \sqrt{c^2 x^2+1}}+\frac{b}{12 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^3,x]

[Out]

b/(12*c^5*d^3*(1 + c^2*x^2)^(3/2)) - (5*b)/(8*c^5*d^3*Sqrt[1 + c^2*x^2]) - (x^3*(a + b*ArcSinh[c*x]))/(4*c^2*d
^3*(1 + c^2*x^2)^2) - (3*x*(a + b*ArcSinh[c*x]))/(8*c^4*d^3*(1 + c^2*x^2)) + (3*(a + b*ArcSinh[c*x])*ArcTan[E^
ArcSinh[c*x]])/(4*c^5*d^3) - (((3*I)/8)*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^5*d^3) + (((3*I)/8)*b*PolyLog[2,
I*E^ArcSinh[c*x]])/(c^5*d^3)

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 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 )}{\left (d+c^2 d x^2\right )^3} \, dx &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{b \int \frac{x^3}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}+\frac{3 \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac{(3 b) \int \frac{x}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 c^3 d^3}+\frac{b \operatorname{Subst}\left (\int \frac{x}{\left (1+c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{8 c d^3}+\frac{3 \int \frac{a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx}{8 c^4 d^2}\\ &=-\frac{3 b}{8 c^5 d^3 \sqrt{1+c^2 x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac{3 \operatorname{Subst}\left (\int (a+b x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5 d^3}+\frac{b \operatorname{Subst}\left (\int \left (-\frac{1}{c^2 \left (1+c^2 x\right )^{5/2}}+\frac{1}{c^2 \left (1+c^2 x\right )^{3/2}}\right ) \, dx,x,x^2\right )}{8 c d^3}\\ &=\frac{b}{12 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{5 b}{8 c^5 d^3 \sqrt{1+c^2 x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac{3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac{(3 i b) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5 d^3}+\frac{(3 i b) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5 d^3}\\ &=\frac{b}{12 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{5 b}{8 c^5 d^3 \sqrt{1+c^2 x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac{3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac{(3 i b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{(3 i b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}\\ &=\frac{b}{12 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{5 b}{8 c^5 d^3 \sqrt{1+c^2 x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac{3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac{3 i b \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}+\frac{3 i b \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{8 c^5 d^3}\\ \end{align*}

Mathematica [A]  time = 0.64223, size = 341, normalized size = 1.83 \[ -\frac{9 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-9 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+15 a c^3 x^3-9 a c^4 x^4 \tan ^{-1}(c x)-18 a c^2 x^2 \tan ^{-1}(c x)+9 a c x-9 a \tan ^{-1}(c x)+15 b c^2 x^2 \sqrt{c^2 x^2+1}+13 b \sqrt{c^2 x^2+1}+15 b c^3 x^3 \sinh ^{-1}(c x)-9 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+9 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-18 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+18 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+9 b c x \sinh ^{-1}(c x)-9 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+9 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{24 c^5 d^3 \left (c^2 x^2+1\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^3,x]

[Out]

-(9*a*c*x + 15*a*c^3*x^3 + 13*b*Sqrt[1 + c^2*x^2] + 15*b*c^2*x^2*Sqrt[1 + c^2*x^2] + 9*b*c*x*ArcSinh[c*x] + 15
*b*c^3*x^3*ArcSinh[c*x] - 9*a*ArcTan[c*x] - 18*a*c^2*x^2*ArcTan[c*x] - 9*a*c^4*x^4*ArcTan[c*x] - (9*I)*b*ArcSi
nh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - (18*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - (9*I)*b*c^4*x^4*
ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + (9*I)*b*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + (18*I)*b*c^2*x^2*Arc
Sinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + (9*I)*b*c^4*x^4*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + (9*I)*b*(1 + c^
2*x^2)^2*PolyLog[2, (-I)*E^ArcSinh[c*x]] - (9*I)*b*(1 + c^2*x^2)^2*PolyLog[2, I*E^ArcSinh[c*x]])/(24*c^5*d^3*(
1 + c^2*x^2)^2)

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Maple [A]  time = 0.016, size = 313, normalized size = 1.7 \begin{align*} -{\frac{5\,a{x}^{3}}{8\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{3\,ax}{8\,{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,a\arctan \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}}-{\frac{5\,b{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{3\,b{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{4}{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}}+{\frac{3\,b\arctan \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{3\,b\arctan \left ( cx \right ) }{8\,{c}^{5}{d}^{3}}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{3\,i}{8}}b}{{c}^{5}{d}^{3}}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{3\,i}{8}}b}{{c}^{5}{d}^{3}}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{5\,b{x}^{2}}{8\,{c}^{3}{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{13\,b}{24\,{c}^{5}{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \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)^3,x)

[Out]

-5/8/c^2*a/d^3*x^3/(c^2*x^2+1)^2-3/8/c^4*a/d^3*x/(c^2*x^2+1)^2+3/8/c^5*a/d^3*arctan(c*x)-5/8/c^2*b/d^3*arcsinh
(c*x)*x^3/(c^2*x^2+1)^2-3/8/c^4*b/d^3*arcsinh(c*x)*x/(c^2*x^2+1)^2+3/8/c^5*b/d^3*arcsinh(c*x)*arctan(c*x)+3/8/
c^5*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8/c^5*b/d^3*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+
1)^(1/2))-3/8*I/c^5*b/d^3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/8*I/c^5*b/d^3*dilog(1-I*(1+I*c*x)/(c^2*x^2+
1)^(1/2))-5/8/c^3*b/d^3*x^2/(c^2*x^2+1)^(3/2)-13/24*b/c^5/d^3/(c^2*x^2+1)^(3/2)

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

[Out]

-1/8*a*((5*c^2*x^3 + 3*x)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3) - 3*arctan(c*x)/(c^5*d^3)) + b*integrate(x^4
*log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3), 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^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, 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)^3,x, algorithm="fricas")

[Out]

integral((b*x^4*arcsinh(c*x) + a*x^4)/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{4}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{b x^{4} \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a*x**4/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*x**4*asinh(c*x)/(c**6*x**6 + 3*c
**4*x**4 + 3*c**2*x**2 + 1), x))/d**3

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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}}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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