### 3.61 $$\int \frac{e^{\text{csch}^{-1}(c x)} x^4}{1+c^2 x^2} \, dx$$

Optimal. Leaf size=72 $\frac{x^3 \sqrt{\frac{1}{c^2 x^2}+1}}{3 c^2}+\frac{x^2}{2 c^3}-\frac{2 x \sqrt{\frac{1}{c^2 x^2}+1}}{3 c^4}-\frac{\log \left (c^2 x^2+1\right )}{2 c^5}$

[Out]

(-2*Sqrt[1 + 1/(c^2*x^2)]*x)/(3*c^4) + x^2/(2*c^3) + (Sqrt[1 + 1/(c^2*x^2)]*x^3)/(3*c^2) - Log[1 + c^2*x^2]/(2
*c^5)

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Rubi [A]  time = 0.0975695, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {6342, 271, 191, 266, 43} $\frac{x^3 \sqrt{\frac{1}{c^2 x^2}+1}}{3 c^2}+\frac{x^2}{2 c^3}-\frac{2 x \sqrt{\frac{1}{c^2 x^2}+1}}{3 c^4}-\frac{\log \left (c^2 x^2+1\right )}{2 c^5}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^ArcCsch[c*x]*x^4)/(1 + c^2*x^2),x]

[Out]

(-2*Sqrt[1 + 1/(c^2*x^2)]*x)/(3*c^4) + x^2/(2*c^3) + (Sqrt[1 + 1/(c^2*x^2)]*x^3)/(3*c^2) - Log[1 + c^2*x^2]/(2
*c^5)

Rule 6342

Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d^2/(a*c^2), Int[(d*x)
^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d,
m}, x] && EqQ[b - a*c^2, 0]

Rule 271

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{\text{csch}^{-1}(c x)} x^4}{1+c^2 x^2} \, dx &=\frac{\int \frac{x^2}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{c^2}+\frac{\int \frac{x^3}{1+c^2 x^2} \, dx}{c}\\ &=\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^3}{3 c^2}-\frac{2 \int \frac{1}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{3 c^4}+\frac{\operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )}{2 c}\\ &=-\frac{2 \sqrt{1+\frac{1}{c^2 x^2}} x}{3 c^4}+\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^3}{3 c^2}+\frac{\operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{2 c}\\ &=-\frac{2 \sqrt{1+\frac{1}{c^2 x^2}} x}{3 c^4}+\frac{x^2}{2 c^3}+\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^3}{3 c^2}-\frac{\log \left (1+c^2 x^2\right )}{2 c^5}\\ \end{align*}

Mathematica [A]  time = 0.127156, size = 64, normalized size = 0.89 $\frac{c x \left (2 c^2 x^2 \sqrt{\frac{1}{c^2 x^2}+1}-4 \sqrt{\frac{1}{c^2 x^2}+1}+3 c x\right )-3 \log \left (c^2 x^2+1\right )}{6 c^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^ArcCsch[c*x]*x^4)/(1 + c^2*x^2),x]

[Out]

(c*x*(-4*Sqrt[1 + 1/(c^2*x^2)] + 3*c*x + 2*c^2*Sqrt[1 + 1/(c^2*x^2)]*x^2) - 3*Log[1 + c^2*x^2])/(6*c^5)

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Maple [A]  time = 0.143, size = 120, normalized size = 1.7 \begin{align*}{\frac{x}{3\,{c}^{4}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}} \left ( \left ({\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}} \right ) ^{{\frac{3}{2}}}{c}^{2}-3\,\sqrt{-{\frac{ \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }{{c}^{4}}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}+{\frac{{x}^{2}}{2\,{c}^{3}}}-{\frac{\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{5}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*x^4/(c^2*x^2+1),x)

[Out]

1/3*((c^2*x^2+1)/c^2/x^2)^(1/2)*x/c^4*(((c^2*x^2+1)/c^2)^(3/2)*c^2-3*(-(x*c^2+(-c^2)^(1/2))*(-x*c^2+(-c^2)^(1/
2))/c^4)^(1/2))/((c^2*x^2+1)/c^2)^(1/2)+1/2*x^2/c^3-1/2*ln(c^2*x^2+1)/c^5

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Maxima [A]  time = 1.02718, size = 66, normalized size = 0.92 \begin{align*} \frac{x^{2}}{2 \, c^{3}} + \frac{\sqrt{c^{2} x^{2} + 1}{\left (c^{2} x^{2} - 2\right )}}{3 \, c^{5}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{2 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^4/(c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*x^2/c^3 + 1/3*sqrt(c^2*x^2 + 1)*(c^2*x^2 - 2)/c^5 - 1/2*log(c^2*x^2 + 1)/c^5

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Fricas [A]  time = 2.98138, size = 127, normalized size = 1.76 \begin{align*} \frac{3 \, c^{2} x^{2} + 2 \,{\left (c^{3} x^{3} - 2 \, c x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 3 \, \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^4/(c^2*x^2+1),x, algorithm="fricas")

[Out]

1/6*(3*c^2*x^2 + 2*(c^3*x^3 - 2*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 3*log(c^2*x^2 + 1))/c^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac{c x^{4} \sqrt{1 + \frac{1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))*x**4/(c**2*x**2+1),x)

[Out]

(Integral(x**3/(c**2*x**2 + 1), x) + Integral(c*x**4*sqrt(1 + 1/(c**2*x**2))/(c**2*x**2 + 1), x))/c

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Giac [A]  time = 1.14811, size = 101, normalized size = 1.4 \begin{align*} \frac{2 \,{\left | c \right |} \mathrm{sgn}\left (x\right )}{3 \, c^{6}} + \frac{2 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left | c \right |} \mathrm{sgn}\left (x\right ) - 6 \, \sqrt{c^{2} x^{2} + 1}{\left | c \right |} \mathrm{sgn}\left (x\right ) + 3 \,{\left (c^{2} x^{2} + 1\right )} c - 3 \, c \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^4/(c^2*x^2+1),x, algorithm="giac")

[Out]

2/3*abs(c)*sgn(x)/c^6 + 1/6*(2*(c^2*x^2 + 1)^(3/2)*abs(c)*sgn(x) - 6*sqrt(c^2*x^2 + 1)*abs(c)*sgn(x) + 3*(c^2*
x^2 + 1)*c - 3*c*log(c^2*x^2 + 1))/c^6