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

Optimal. Leaf size=178 \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c d^3}+\frac{3 b}{8 c d^3 \sqrt{c^2 x^2+1}}+\frac{b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.140758, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5690, 5693, 4180, 2279, 2391, 261} \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{8 c d^3}+\frac{3 x \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c d^3}+\frac{3 b}{8 c d^3 \sqrt{c^2 x^2+1}}+\frac{b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

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

Mathematica [A]  time = 0.152428, size = 341, normalized size = 1.92 \[ \frac{-9 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+9 i b \left (c^2 x^2+1\right )^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+9 a c^3 x^3+9 a c^4 x^4 \tan ^{-1}(c x)+18 a c^2 x^2 \tan ^{-1}(c x)+15 a c x+9 a \tan ^{-1}(c x)+9 b c^2 x^2 \sqrt{c^2 x^2+1}+11 b \sqrt{c^2 x^2+1}+9 b c^3 x^3 \sinh ^{-1}(c x)+9 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-9 i b c^4 x^4 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+18 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-18 i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+15 b c x \sinh ^{-1}(c x)+9 i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-9 i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{24 c d^3 \left (c^2 x^2+1\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(15*a*c*x + 9*a*c^3*x^3 + 11*b*Sqrt[1 + c^2*x^2] + 9*b*c^2*x^2*Sqrt[1 + c^2*x^2] + 15*b*c*x*ArcSinh[c*x] + 9*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*ArcSinh
[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*Ar
cSinh[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*ArcSi
nh[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*d^3*(1 +
c^2*x^2)^2)

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Maple [A]  time = 0.01, size = 295, normalized size = 1.7 \begin{align*}{\frac{ax}{4\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,ax}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,a\arctan \left ( cx \right ) }{8\,c{d}^{3}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) x}{4\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,b{\it Arcsinh} \left ( cx \right ) x}{8\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,b{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{8\,c{d}^{3}}}+{\frac{3\,bc{x}^{2}}{8\,{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{11\,b}{24\,c{d}^{3}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{3\,b\arctan \left ( cx \right ) }{8\,c{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{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{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{d}^{3}}{\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((a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x)

[Out]

1/4*a/d^3*x/(c^2*x^2+1)^2+3/8*a/d^3*x/(c^2*x^2+1)+3/8/c*a/d^3*arctan(c*x)+1/4*b/d^3*arcsinh(c*x)*x/(c^2*x^2+1)
^2+3/8*b/d^3*arcsinh(c*x)*x/(c^2*x^2+1)+3/8/c*b/d^3*arcsinh(c*x)*arctan(c*x)+3/8*c*b/d^3*x^2/(c^2*x^2+1)^(3/2)
+11/24*b/c/d^3/(c^2*x^2+1)^(3/2)+3/8/c*b/d^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8/c*b/d^3*arcta
n(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8*I/c*b/d^3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/8*I/c*b/d^3*
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}{8} \, a{\left (\frac{3 \, c^{2} x^{3} + 5 \, x}{c^{4} d^{3} x^{4} + 2 \, c^{2} d^{3} x^{2} + d^{3}} + \frac{3 \, \arctan \left (c x\right )}{c d^{3}}\right )} + b \int \frac{\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((a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

1/8*a*((3*c^2*x^3 + 5*x)/(c^4*d^3*x^4 + 2*c^2*d^3*x^2 + d^3) + 3*arctan(c*x)/(c*d^3)) + b*integrate(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 \operatorname{arsinh}\left (c x\right ) + a}{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((a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(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}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{b \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((a+b*asinh(c*x))/(c**2*d*x**2+d)**3,x)

[Out]

(Integral(a/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*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{b \operatorname{arsinh}\left (c x\right ) + a}{{\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((a+b*arcsinh(c*x))/(c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

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