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

Optimal. Leaf size=73 \[ -\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^2 d}+\frac {\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d}+\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^2 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5714, 3718, 2190, 2279, 2391} \[ \frac {b \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^2 d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^2 d}+\frac {\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [B]  time = 0.06, size = 167, normalized size = 2.29 \[ \frac {a \log \left (c^2 x^2+1\right )}{2 c^2 d}+\frac {b \text {Li}_2\left (-\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{c^2 d}+\frac {b \text {Li}_2\left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{c^2 d}-\frac {b \sinh ^{-1}(c x)^2}{2 c^2 d}+\frac {b \sinh ^{-1}(c x) \log \left (1-\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{c^2 d}+\frac {b \sinh ^{-1}(c x) \log \left (\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )}{c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \operatorname {arsinh}\left (c x\right ) + a x}{c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x}{c^{2} d x^{2} + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 98, normalized size = 1.34 \[ \frac {a \ln \left (c^{2} x^{2}+1\right )}{2 c^{2} d}-\frac {b \arcsinh \left (c x \right )^{2}}{2 c^{2} d}+\frac {b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{2} d}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{8} \, b {\left (\frac {\log \left (c^{2} x^{2} + 1\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right ) \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d} + 8 \, \int \frac {\log \left (c^{2} x^{2} + 1\right )}{2 \, {\left (c^{4} d x^{3} + c^{2} d x + {\left (c^{3} d x^{2} + c d\right )} \sqrt {c^{2} x^{2} + 1}\right )}}\,{d x}\right )} + \frac {a \log \left (c^{2} d x^{2} + d\right )}{2 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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