3.1.61 \(\int \frac {e^{\text {csch}^{-1}(c x)} x^4}{1+c^2 x^2} \, dx\) [61]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 = {6477, 277, 197, 272, 45} \begin {gather*} \frac {x^2}{2 c^3}+\frac {x^3 \sqrt {\frac {1}{c^2 x^2}+1}}{3 c^2}-\frac {\log \left (c^2 x^2+1\right )}{2 c^5}-\frac {2 x \sqrt {\frac {1}{c^2 x^2}+1}}{3 c^4} \end {gather*}

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 45

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])

Rule 197

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 272

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 277

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 6477

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]

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 {\text {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 {\text {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*}

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Mathematica [A]
time = 0.08, size = 64, normalized size = 0.89 \begin {gather*} \frac {c x \left (-4 \sqrt {1+\frac {1}{c^2 x^2}}+3 c x+2 c^2 \sqrt {1+\frac {1}{c^2 x^2}} x^2\right )-3 \log \left (1+c^2 x^2\right )}{6 c^5} \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(124\) vs. \(2(60)=120\).
time = 0.83, size = 125, normalized size = 1.74

method	result	size
default	\(\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \left (\left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} c^{2}-3 \sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right )}{3 c^{4} \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}}+\frac {\frac {x^{2}}{2 c^{2}}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2 c^{4}}}{c}\)	\(125\)

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,method=_RETURNVERBOSE)

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

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Maxima [A]
time = 0.28, size = 49, normalized size = 0.68 \begin {gather*} \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 {gather*}

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 = 0.41, size = 58, normalized size = 0.81 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.40, size = 85, normalized size = 1.18 \begin {gather*} -\frac {\log \left (c^{2} x^{2} + 1\right )}{2 \, c^{5}} + \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}} c^{12} {\left | c \right |} \mathrm {sgn}\left (x\right ) - 6 \, \sqrt {c^{2} x^{2} + 1} c^{12} {\left | c \right |} \mathrm {sgn}\left (x\right ) + 3 \, {\left (c^{2} x^{2} + 1\right )} c^{13}}{6 \, c^{18}} \end {gather*}

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]

-1/2*log(c^2*x^2 + 1)/c^5 + 2/3*abs(c)*sgn(x)/c^6 + 1/6*(2*(c^2*x^2 + 1)^(3/2)*c^12*abs(c)*sgn(x) - 6*sqrt(c^2
*x^2 + 1)*c^12*abs(c)*sgn(x) + 3*(c^2*x^2 + 1)*c^13)/c^18

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Mupad [B]
time = 2.35, size = 61, normalized size = 0.85 \begin {gather*} \frac {x^3\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3\,c^2}-\frac {2\,x\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3\,c^4}-\frac {\ln \left (c^2\,x^2+1\right )-c^2\,x^2}{2\,c^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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