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

Optimal. Leaf size=111 \[ -\frac{1}{2} b d \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{4} b c d x \sqrt{c^2 x^2+1}-\frac{1}{4} b d \sinh ^{-1}(c x) \]

[Out]

-(b*c*d*x*Sqrt[1 + c^2*x^2])/4 - (b*d*ArcSinh[c*x])/4 + (d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/2 + (d*(a + b*A
rcSinh[c*x])^2)/(2*b) + d*(a + b*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - (b*d*PolyLog[2, E^(-2*ArcSinh[c*
x])])/2

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Rubi [A]  time = 0.122861, antiderivative size = 111, 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 = {5726, 5659, 3716, 2190, 2279, 2391, 195, 215} \[ \frac{1}{2} b d \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{4} b c d x \sqrt{c^2 x^2+1}-\frac{1}{4} b d \sinh ^{-1}(c x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(b*c*d*x*Sqrt[1 + c^2*x^2])/4 - (b*d*ArcSinh[c*x])/4 + (d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/2 - (d*(a + b*A
rcSinh[c*x])^2)/(2*b) + d*(a + b*ArcSinh[c*x])*Log[1 - E^(2*ArcSinh[c*x])] + (b*d*PolyLog[2, E^(2*ArcSinh[c*x]
)])/2

Rule 5726

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

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]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+d \int \frac{a+b \sinh ^{-1}(c x)}{x} \, dx-\frac{1}{2} (b c d) \int \sqrt{1+c^2 x^2} \, dx\\ &=-\frac{1}{4} b c d x \sqrt{1+c^2 x^2}+\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )+d \operatorname{Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}(c x)\right )-\frac{1}{4} (b c d) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx\\ &=-\frac{1}{4} b c d x \sqrt{1+c^2 x^2}-\frac{1}{4} b d \sinh ^{-1}(c x)+\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}-(2 d) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{1}{4} b c d x \sqrt{1+c^2 x^2}-\frac{1}{4} b d \sinh ^{-1}(c x)+\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-(b d) \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{1}{4} b c d x \sqrt{1+c^2 x^2}-\frac{1}{4} b d \sinh ^{-1}(c x)+\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac{1}{2} (b d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )\\ &=-\frac{1}{4} b c d x \sqrt{1+c^2 x^2}-\frac{1}{4} b d \sinh ^{-1}(c x)+\frac{1}{2} d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} b d \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.0606308, size = 113, normalized size = 1.02 \[ \frac{1}{2} b d \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} a c^2 d x^2+a d \log (x)-\frac{1}{4} b c d x \sqrt{c^2 x^2+1}+\frac{1}{2} b c^2 d x^2 \sinh ^{-1}(c x)-\frac{1}{2} b d \sinh ^{-1}(c x)^2+\frac{1}{4} b d \sinh ^{-1}(c x)+b d \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*c^2*d*x^2)/2 - (b*c*d*x*Sqrt[1 + c^2*x^2])/4 + (b*d*ArcSinh[c*x])/4 + (b*c^2*d*x^2*ArcSinh[c*x])/2 - (b*d*A
rcSinh[c*x]^2)/2 + b*d*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + a*d*Log[x] + (b*d*PolyLog[2, E^(2*ArcSinh[c*
x])])/2

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Maple [A]  time = 0.086, size = 162, normalized size = 1.5 \begin{align*}{\frac{da{c}^{2}{x}^{2}}{2}}+da\ln \left ( cx \right ) -{\frac{db \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2}}+{\frac{db{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{2}}-{\frac{dbcx}{4}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{bd{\it Arcsinh} \left ( cx \right ) }{4}}+db{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) +db{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +db{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +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,x)

[Out]

1/2*d*a*c^2*x^2+d*a*ln(c*x)-1/2*d*b*arcsinh(c*x)^2+1/2*d*b*arcsinh(c*x)*c^2*x^2-1/4*b*c*d*x*(c^2*x^2+1)^(1/2)+
1/4*b*d*arcsinh(c*x)+d*b*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+d*b*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+d*b*ar
csinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+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*} \frac{1}{2} \, a c^{2} d x^{2} + a d \log \left (x\right ) + \int b c^{2} d x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \frac{b d \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{x}\,{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,x, algorithm="maxima")

[Out]

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

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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}, 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,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, x)

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

[Out]

d*(Integral(a/x, x) + Integral(a*c**2*x, x) + Integral(b*asinh(c*x)/x, x) + Integral(b*c**2*x*asinh(c*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}\,{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,x, algorithm="giac")

[Out]

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