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

Optimal. Leaf size=128 \[ -\frac{1}{2} b c^2 d \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )-\frac{d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+c^2 d \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{b c d \sqrt{c^2 x^2+1}}{2 x}+\frac{1}{2} b c^2 d \sinh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.126375, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {5728, 277, 215, 5659, 3716, 2190, 2279, 2391} \[ \frac{1}{2} b c^2 d \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-\frac{d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+c^2 d \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{b c d \sqrt{c^2 x^2+1}}{2 x}+\frac{1}{2} b c^2 d \sinh ^{-1}(c x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 5728

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((f*x
)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSinh[c*x]))/(f*(m + 1)), x] + (-Dist[(b*c*d^p)/(f*(m + 1)), Int[(f*x)^(m + 1
)*(1 + c^2*x^2)^(p - 1/2), x], x] - Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*A
rcSinh[c*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] && ILtQ[(m + 1)/2, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.0599271, size = 111, normalized size = 0.87 \[ \frac{1}{2} b c^2 d \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+a c^2 d \log (x)-\frac{a d}{2 x^2}-\frac{b c d \sqrt{c^2 x^2+1}}{2 x}-\frac{1}{2} b c^2 d \sinh ^{-1}(c x)^2+b c^2 d \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac{b d \sinh ^{-1}(c x)}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.167, size = 175, normalized size = 1.4 \begin{align*}{c}^{2}da\ln \left ( cx \right ) -{\frac{da}{2\,{x}^{2}}}-{\frac{{c}^{2}db \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2}}-{\frac{bcd}{2\,x}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{{c}^{2}db}{2}}-{\frac{db{\it Arcsinh} \left ( cx \right ) }{2\,{x}^{2}}}+{c}^{2}db{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) +{c}^{2}db{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +{c}^{2}db{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +{c}^{2}db{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c^2*d*integrate(log(c*x + sqrt(c^2*x^2 + 1))/x, x) + a*c^2*d*log(x) - 1/2*b*d*(sqrt(c^2*x^2 + 1)*c/x + arcsi
nh(c*x)/x^2) - 1/2*a*d/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a c^{2} d x^{2} + a d +{\left (b c^{2} d x^{2} + b d\right )} \operatorname{arsinh}\left (c x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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