3.85 \(\int x (d+e x^2) (a+b \text{csch}^{-1}(c x)) \, dx\)

Optimal. Leaf size=146 \[ \frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{b c d^2 x \tan ^{-1}\left (\sqrt{-c^2 x^2-1}\right )}{4 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (2 c^2 d-e\right )}{4 c^3 \sqrt{-c^2 x^2}}-\frac{b e x \left (-c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt{-c^2 x^2}} \]

[Out]

(b*(2*c^2*d - e)*x*Sqrt[-1 - c^2*x^2])/(4*c^3*Sqrt[-(c^2*x^2)]) - (b*e*x*(-1 - c^2*x^2)^(3/2))/(12*c^3*Sqrt[-(
c^2*x^2)]) + ((d + e*x^2)^2*(a + b*ArcCsch[c*x]))/(4*e) - (b*c*d^2*x*ArcTan[Sqrt[-1 - c^2*x^2]])/(4*e*Sqrt[-(c
^2*x^2)])

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Rubi [A]  time = 0.105749, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6300, 446, 88, 63, 205} \[ \frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{b c d^2 x \tan ^{-1}\left (\sqrt{-c^2 x^2-1}\right )}{4 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (2 c^2 d-e\right )}{4 c^3 \sqrt{-c^2 x^2}}-\frac{b e x \left (-c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x^2)*(a + b*ArcCsch[c*x]),x]

[Out]

(b*(2*c^2*d - e)*x*Sqrt[-1 - c^2*x^2])/(4*c^3*Sqrt[-(c^2*x^2)]) - (b*e*x*(-1 - c^2*x^2)^(3/2))/(12*c^3*Sqrt[-(
c^2*x^2)]) + ((d + e*x^2)^2*(a + b*ArcCsch[c*x]))/(4*e) - (b*c*d^2*x*ArcTan[Sqrt[-1 - c^2*x^2]])/(4*e*Sqrt[-(c
^2*x^2)])

Rule 6300

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rubi steps

\begin{align*} \int x \left (d+e x^2\right ) \left (a+b \text{csch}^{-1}(c x)\right ) \, dx &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{(b c x) \int \frac{\left (d+e x^2\right )^2}{x \sqrt{-1-c^2 x^2}} \, dx}{4 e \sqrt{-c^2 x^2}}\\ &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{(b c x) \operatorname{Subst}\left (\int \frac{(d+e x)^2}{x \sqrt{-1-c^2 x}} \, dx,x,x^2\right )}{8 e \sqrt{-c^2 x^2}}\\ &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{(b c x) \operatorname{Subst}\left (\int \left (-\frac{e \left (-2 c^2 d+e\right )}{c^2 \sqrt{-1-c^2 x}}+\frac{d^2}{x \sqrt{-1-c^2 x}}-\frac{e^2 \sqrt{-1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{8 e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (2 c^2 d-e\right ) x \sqrt{-1-c^2 x^2}}{4 c^3 \sqrt{-c^2 x^2}}-\frac{b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt{-c^2 x^2}}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{\left (b c d^2 x\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1-c^2 x}} \, dx,x,x^2\right )}{8 e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (2 c^2 d-e\right ) x \sqrt{-1-c^2 x^2}}{4 c^3 \sqrt{-c^2 x^2}}-\frac{b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt{-c^2 x^2}}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}+\frac{\left (b d^2 x\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{-1-c^2 x^2}\right )}{4 c e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (2 c^2 d-e\right ) x \sqrt{-1-c^2 x^2}}{4 c^3 \sqrt{-c^2 x^2}}-\frac{b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt{-c^2 x^2}}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 e}-\frac{b c d^2 x \tan ^{-1}\left (\sqrt{-1-c^2 x^2}\right )}{4 e \sqrt{-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0953076, size = 77, normalized size = 0.53 \[ \frac{x \left (3 a c^3 x \left (2 d+e x^2\right )+b \sqrt{\frac{1}{c^2 x^2}+1} \left (c^2 \left (6 d+e x^2\right )-2 e\right )+3 b c^3 x \text{csch}^{-1}(c x) \left (2 d+e x^2\right )\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x^2)*(a + b*ArcCsch[c*x]),x]

[Out]

(x*(3*a*c^3*x*(2*d + e*x^2) + b*Sqrt[1 + 1/(c^2*x^2)]*(-2*e + c^2*(6*d + e*x^2)) + 3*b*c^3*x*(2*d + e*x^2)*Arc
Csch[c*x]))/(12*c^3)

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Maple [A]  time = 0.184, size = 115, normalized size = 0.8 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{a}{{c}^{2}} \left ({\frac{{c}^{4}{x}^{4}e}{4}}+{\frac{{x}^{2}{c}^{4}d}{2}} \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\rm arccsch} \left (cx\right ){c}^{4}{x}^{4}e}{4}}+{\frac{{\rm arccsch} \left (cx\right ){c}^{4}{x}^{2}d}{2}}+{\frac{ \left ({c}^{2}{x}^{2}+1 \right ) \left ({c}^{2}{x}^{2}e+6\,{c}^{2}d-2\,e \right ) }{12\,cx}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)*(a+b*arccsch(c*x)),x)

[Out]

1/c^2*(a/c^2*(1/4*c^4*x^4*e+1/2*x^2*c^4*d)+b/c^2*(1/4*arccsch(c*x)*c^4*x^4*e+1/2*arccsch(c*x)*c^4*x^2*d+1/12*(
c^2*x^2+1)*(c^2*e*x^2+6*c^2*d-2*e)/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/x))

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Maxima [A]  time = 1.00803, size = 128, normalized size = 0.88 \begin{align*} \frac{1}{4} \, a e x^{4} + \frac{1}{2} \, a d x^{2} + \frac{1}{2} \,{\left (x^{2} \operatorname{arcsch}\left (c x\right ) + \frac{x \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c}\right )} b d + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arcsch}\left (c x\right ) + \frac{c^{2} x^{3}{\left (\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 3 \, x \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/2*(x^2*arccsch(c*x) + x*sqrt(1/(c^2*x^2) + 1)/c)*b*d + 1/12*(3*x^4*arccsch(c*x)
+ (c^2*x^3*(1/(c^2*x^2) + 1)^(3/2) - 3*x*sqrt(1/(c^2*x^2) + 1))/c^3)*b*e

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Fricas [A]  time = 2.7772, size = 266, normalized size = 1.82 \begin{align*} \frac{3 \, a c^{3} e x^{4} + 6 \, a c^{3} d x^{2} + 3 \,{\left (b c^{3} e x^{4} + 2 \, b c^{3} d x^{2}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) +{\left (b c^{2} e x^{3} + 2 \,{\left (3 \, b c^{2} d - b e\right )} x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{12 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^3*e*x^4 + 6*a*c^3*d*x^2 + 3*(b*c^3*e*x^4 + 2*b*c^3*d*x^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) +
1)/(c*x)) + (b*c^2*e*x^3 + 2*(3*b*c^2*d - b*e)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x**2+d)*(a+b*acsch(c*x)),x)

[Out]

Integral(x*(a + b*acsch(c*x))*(d + e*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arccsch(c*x) + a)*x, x)