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

Optimal. Leaf size=187 \[ -b c^2 d^2 \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )+c^2 d^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac{c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}+2 c^2 d^2 \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} b c^3 d^2 x \sqrt{c^2 x^2+1}-\frac{b c d^2 \left (c^2 x^2+1\right )^{3/2}}{2 x}+\frac{1}{4} b c^2 d^2 \sinh ^{-1}(c x) \]

[Out]

(b*c^3*d^2*x*Sqrt[1 + c^2*x^2])/4 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/(2*x) + (b*c^2*d^2*ArcSinh[c*x])/4 + c^2*d^2
*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]) - (d^2*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(2*x^2) + (c^2*d^2*(a + b*Arc
Sinh[c*x])^2)/b + 2*c^2*d^2*(a + b*ArcSinh[c*x])*Log[1 - E^(-2*ArcSinh[c*x])] - b*c^2*d^2*PolyLog[2, E^(-2*Arc
Sinh[c*x])]

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Rubi [A]  time = 0.210942, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5728, 277, 195, 215, 5726, 5659, 3716, 2190, 2279, 2391} \[ b c^2 d^2 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+c^2 d^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac{d^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac{c^2 d^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}+2 c^2 d^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} b c^3 d^2 x \sqrt{c^2 x^2+1}-\frac{b c d^2 \left (c^2 x^2+1\right )^{3/2}}{2 x}+\frac{1}{4} b c^2 d^2 \sinh ^{-1}(c x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*c^3*d^2*x*Sqrt[1 + c^2*x^2])/4 - (b*c*d^2*(1 + c^2*x^2)^(3/2))/(2*x) + (b*c^2*d^2*ArcSinh[c*x])/4 + c^2*d^2
*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]) - (d^2*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(2*x^2) - (c^2*d^2*(a + b*Arc
Sinh[c*x])^2)/b + 2*c^2*d^2*(a + b*ArcSinh[c*x])*Log[1 - E^(2*ArcSinh[c*x])] + b*c^2*d^2*PolyLog[2, E^(2*ArcSi
nh[c*x])]

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 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.34301, size = 143, normalized size = 0.76 \[ \frac{1}{4} d^2 \left (4 c^2 \left (b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )\right )+2 c^4 x^2 \left (a+b \sinh ^{-1}(c x)\right )-\frac{4 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{x^2}-\frac{2 b c \sqrt{c^2 x^2+1}}{x}+b c^2 \left (\sinh ^{-1}(c x)-c x \sqrt{c^2 x^2+1}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.244, size = 262, normalized size = 1.4 \begin{align*}{\frac{{c}^{4}{d}^{2}a{x}^{2}}{2}}+2\,{c}^{2}{d}^{2}a\ln \left ( cx \right ) -{\frac{{d}^{2}a}{2\,{x}^{2}}}-{c}^{2}{d}^{2}b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}+{\frac{{c}^{4}{d}^{2}b{\it Arcsinh} \left ( cx \right ){x}^{2}}{2}}-{\frac{b{c}^{3}{d}^{2}x}{4}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{b{c}^{2}{d}^{2}{\it Arcsinh} \left ( cx \right ) }{4}}+{\frac{{d}^{2}b{c}^{2}}{2}}-{\frac{{d}^{2}bc}{2\,x}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{d}^{2}b{\it Arcsinh} \left ( cx \right ) }{2\,{x}^{2}}}+2\,{c}^{2}{d}^{2}b{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,{c}^{2}{d}^{2}b{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,{c}^{2}{d}^{2}b{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,{c}^{2}{d}^{2}b{\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)^2*(a+b*arcsinh(c*x))/x^3,x)

[Out]

1/2*c^4*d^2*a*x^2+2*c^2*d^2*a*ln(c*x)-1/2*d^2*a/x^2-c^2*d^2*b*arcsinh(c*x)^2+1/2*c^4*d^2*b*arcsinh(c*x)*x^2-1/
4*b*c^3*d^2*x*(c^2*x^2+1)^(1/2)+1/4*b*c^2*d^2*arcsinh(c*x)+1/2*d^2*b*c^2-1/2*c*d^2*b/x*(c^2*x^2+1)^(1/2)-1/2*d
^2*b*arcsinh(c*x)/x^2+2*c^2*d^2*b*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+2*c^2*d^2*b*polylog(2,-c*x-(c^2*x^2
+1)^(1/2))+2*c^2*d^2*b*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+2*c^2*d^2*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^{4} d^{2} x^{2} + 2 \, a c^{2} d^{2} \log \left (x\right ) - \frac{1}{2} \, b d^{2}{\left (\frac{\sqrt{c^{2} x^{2} + 1} c}{x} + \frac{\operatorname{arsinh}\left (c x\right )}{x^{2}}\right )} - \frac{a d^{2}}{2 \, x^{2}} + \int b c^{4} d^{2} x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \frac{2 \, b c^{2} d^{2} \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)^2*(a+b*arcsinh(c*x))/x^3,x, algorithm="maxima")

[Out]

1/2*a*c^4*d^2*x^2 + 2*a*c^2*d^2*log(x) - 1/2*b*d^2*(sqrt(c^2*x^2 + 1)*c/x + arcsinh(c*x)/x^2) - 1/2*a*d^2/x^2
+ integrate(b*c^4*d^2*x*log(c*x + sqrt(c^2*x^2 + 1)) + 2*b*c^2*d^2*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^{4} d^{2} x^{4} + 2 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{4} d^{2} x^{4} + 2 \, b c^{2} d^{2} x^{2} + b d^{2}\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)^2*(a+b*arcsinh(c*x))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{a}{x^{3}}\, dx + \int \frac{2 a c^{2}}{x}\, dx + \int a c^{4} x\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{2 b c^{2} \operatorname{asinh}{\left (c x \right )}}{x}\, dx + \int b c^{4} 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)**2*(a+b*asinh(c*x))/x**3,x)

[Out]

d**2*(Integral(a/x**3, x) + Integral(2*a*c**2/x, x) + Integral(a*c**4*x, x) + Integral(b*asinh(c*x)/x**3, x) +
Integral(2*b*c**2*asinh(c*x)/x, x) + Integral(b*c**4*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 )}^{2}{\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)^2*(a+b*arcsinh(c*x))/x^3,x, algorithm="giac")

[Out]

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