3.1.39 \(\int \frac {x^2 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^2} \, dx\) [39]

Optimal. Leaf size=127 \[ -\frac {b}{2 c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c^3 d^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5810, 5789, 4265, 2317, 2438, 267} \begin {gather*} \frac {\text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 c^3 d^2}-\frac {b}{2 c^3 d^2 \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

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 2317

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 2438

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 4265

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 5789

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 5810

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/(2*c*(p + 1)))*Simp[
(d + e*x^2)^p/(1 + c^2*x^2)^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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 221, normalized size = 1.74 \begin {gather*} -\frac {a c x+b \sqrt {1+c^2 x^2}+b c x \sinh ^{-1}(c x)-a \text {ArcTan}(c x)-a c^2 x^2 \text {ArcTan}(c x)-i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c^3 d^2 \left (1+c^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 3.12, size = 219, normalized size = 1.72

method	result	size
derivativedivides	\(\frac {-\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {a \arctan \left (c x \right )}{2 d^{2}}-\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}}{c^{3}}\)	\(219\)
default	\(\frac {-\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {a \arctan \left (c x \right )}{2 d^{2}}-\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}}{c^{3}}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^3*(-1/2*a/d^2*c*x/(c^2*x^2+1)+1/2*a/d^2*arctan(c*x)-1/2*b/d^2*arcsinh(c*x)*c*x/(c^2*x^2+1)+1/2*b/d^2*arcsi
nh(c*x)*arctan(c*x)+1/2*b/d^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*b/d^2*arctan(c*x)*ln(1-I*(1+
I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*b/d^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*b/d^2*dilog(1-I*(1+I*c*x)/(
c^2*x^2+1)^(1/2))-1/2*b/d^2/(c^2*x^2+1)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{2}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{2} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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