3.62 \(\int \frac {e^{\text {csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx\)

Optimal. Leaf size=59 \[ -\frac {\tan ^{-1}(c x)}{c^4}+\frac {x}{c^3}+\frac {x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}-\frac {\tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{2 c^4} \]

[Out]

x/c^3-arctan(c*x)/c^4-1/2*arctanh((1+1/c^2/x^2)^(1/2))/c^4+1/2*x^2*(1+1/c^2/x^2)^(1/2)/c^2

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Rubi [A]  time = 0.09, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6342, 266, 51, 63, 208, 321, 203} \[ \frac {x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{2 c^2}-\frac {\tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{2 c^4}+\frac {x}{c^3}-\frac {\tan ^{-1}(c x)}{c^4} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCsch[c*x]*x^3)/(1 + c^2*x^2),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 208

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

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 321

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 54, normalized size = 0.92 \[ -\frac {-c x \left (c x \sqrt {\frac {1}{c^2 x^2}+1}+2\right )+\log \left (x \left (\sqrt {\frac {1}{c^2 x^2}+1}+1\right )\right )+2 \tan ^{-1}(c x)}{2 c^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcCsch[c*x]*x^3)/(1 + c^2*x^2),x]

[Out]

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

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fricas [A]  time = 0.89, size = 68, normalized size = 1.15 \[ \frac {c^{2} x^{2} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, c x - 2 \, \arctan \left (c x\right ) + \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{2 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="fricas")

[Out]

1/2*(c^2*x^2*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 2*c*x - 2*arctan(c*x) + log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c
*x))/c^4

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giac [A]  time = 0.14, size = 61, normalized size = 1.03 \[ \frac {\sqrt {c^{2} x^{2} + 1} x {\left | c \right |} \mathrm {sgn}\relax (x)}{2 \, c^{4}} + \frac {x}{c^{3}} + \frac {\log \left (-x {\left | c \right |} + \sqrt {c^{2} x^{2} + 1}\right ) \mathrm {sgn}\relax (x)}{2 \, c^{4}} - \frac {\arctan \left (c x\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="giac")

[Out]

1/2*sqrt(c^2*x^2 + 1)*x*abs(c)*sgn(x)/c^4 + x/c^3 + 1/2*log(-x*abs(c) + sqrt(c^2*x^2 + 1))*sgn(x)/c^4 - arctan
(c*x)/c^4

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maple [B]  time = 0.06, size = 133, normalized size = 2.25 \[ \frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \left (x \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2}+\ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right )-2 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right )\right )}{2 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{4}}+\frac {x}{c^{3}}-\frac {\arctan \left (c x \right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x)

[Out]

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

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maxima [B]  time = 0.49, size = 107, normalized size = 1.81 \[ \frac {x}{c^{3}} + \frac {\frac {2 \, \sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c {\left (\frac {c^{2} x^{2} + 1}{c^{2} x^{2}} - 1\right )}} - \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} + 1\right ) + \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - 1\right )}{4 \, c^{4}} - \frac {\arctan \left (c x\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^3/(c^2*x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.34, size = 51, normalized size = 0.86 \[ \frac {x^2\,\sqrt {\frac {1}{c^2\,x^2}+1}}{2\,c^2}-\frac {\mathrm {atan}\left (c\,x\right )-c\,x}{c^4}-\frac {\mathrm {atanh}\left (\sqrt {\frac {1}{c^2\,x^2}+1}\right )}{2\,c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*((1/(c^2*x^2) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 + 1),x)

[Out]

(x^2*(1/(c^2*x^2) + 1)^(1/2))/(2*c^2) - (atan(c*x) - c*x)/c^4 - atanh((1/(c^2*x^2) + 1)^(1/2))/(2*c^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {c x^{3} \sqrt {1 + \frac {1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))*x**3/(c**2*x**2+1),x)

[Out]

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

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