3.60 \(\int \frac {e^{\text {csch}^{-1}(c x)} x^5}{1+c^2 x^2} \, dx\)

Optimal. Leaf size=92 \[ \frac {\tan ^{-1}(c x)}{c^6}-\frac {x}{c^5}+\frac {x^3}{3 c^3}+\frac {x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{4 c^2}+\frac {3 \tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{8 c^6}-\frac {3 x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{8 c^4} \]

[Out]

-x/c^5+1/3*x^3/c^3+arctan(c*x)/c^6+3/8*arctanh((1+1/c^2/x^2)^(1/2))/c^6-3/8*x^2*(1+1/c^2/x^2)^(1/2)/c^4+1/4*x^
4*(1+1/c^2/x^2)^(1/2)/c^2

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Rubi [A]  time = 0.11, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 9, 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, 302, 203} \[ \frac {x^4 \sqrt {\frac {1}{c^2 x^2}+1}}{4 c^2}+\frac {x^3}{3 c^3}-\frac {3 x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{8 c^4}+\frac {3 \tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{8 c^6}-\frac {x}{c^5}+\frac {\tan ^{-1}(c x)}{c^6} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCsch[c*x]*x^5)/(1 + c^2*x^2),x]

[Out]

-(x/c^5) - (3*Sqrt[1 + 1/(c^2*x^2)]*x^2)/(8*c^4) + x^3/(3*c^3) + (Sqrt[1 + 1/(c^2*x^2)]*x^4)/(4*c^2) + ArcTan[
c*x]/c^6 + (3*ArcTanh[Sqrt[1 + 1/(c^2*x^2)]])/(8*c^6)

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^5}{1+c^2 x^2} \, dx &=\frac {\int \frac {x^3}{\sqrt {1+\frac {1}{c^2 x^2}}} \, dx}{c^2}+\frac {\int \frac {x^4}{1+c^2 x^2} \, dx}{c}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{2 c^2}+\frac {\int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{c}\\ &=-\frac {x}{c^5}+\frac {x^3}{3 c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^4}{4 c^2}+\frac {\int \frac {1}{1+c^2 x^2} \, dx}{c^5}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{8 c^4}\\ &=-\frac {x}{c^5}-\frac {3 \sqrt {1+\frac {1}{c^2 x^2}} x^2}{8 c^4}+\frac {x^3}{3 c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^4}{4 c^2}+\frac {\tan ^{-1}(c x)}{c^6}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{16 c^6}\\ &=-\frac {x}{c^5}-\frac {3 \sqrt {1+\frac {1}{c^2 x^2}} x^2}{8 c^4}+\frac {x^3}{3 c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^4}{4 c^2}+\frac {\tan ^{-1}(c x)}{c^6}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-c^2+c^2 x^2} \, dx,x,\sqrt {1+\frac {1}{c^2 x^2}}\right )}{8 c^4}\\ &=-\frac {x}{c^5}-\frac {3 \sqrt {1+\frac {1}{c^2 x^2}} x^2}{8 c^4}+\frac {x^3}{3 c^3}+\frac {\sqrt {1+\frac {1}{c^2 x^2}} x^4}{4 c^2}+\frac {\tan ^{-1}(c x)}{c^6}+\frac {3 \tanh ^{-1}\left (\sqrt {1+\frac {1}{c^2 x^2}}\right )}{8 c^6}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 85, normalized size = 0.92 \[ \frac {9 \log \left (x \left (\sqrt {\frac {1}{c^2 x^2}+1}+1\right )\right )+c x \left (8 c^2 x^2-9 c x \sqrt {\frac {1}{c^2 x^2}+1}+6 c^3 x^3 \sqrt {\frac {1}{c^2 x^2}+1}-24\right )+24 \tan ^{-1}(c x)}{24 c^6} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcCsch[c*x]*x^5)/(1 + c^2*x^2),x]

[Out]

(c*x*(-24 - 9*c*Sqrt[1 + 1/(c^2*x^2)]*x + 8*c^2*x^2 + 6*c^3*Sqrt[1 + 1/(c^2*x^2)]*x^3) + 24*ArcTan[c*x] + 9*Lo
g[(1 + Sqrt[1 + 1/(c^2*x^2)])*x])/(24*c^6)

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fricas [A]  time = 1.64, size = 90, normalized size = 0.98 \[ \frac {8 \, c^{3} x^{3} - 24 \, c x + 3 \, {\left (2 \, c^{4} x^{4} - 3 \, c^{2} x^{2}\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 24 \, \arctan \left (c x\right ) - 9 \, \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{24 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^5/(c^2*x^2+1),x, algorithm="fricas")

[Out]

1/24*(8*c^3*x^3 - 24*c*x + 3*(2*c^4*x^4 - 3*c^2*x^2)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 24*arctan(c*x) - 9*log(c*
x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x))/c^6

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giac [A]  time = 0.15, size = 89, normalized size = 0.97 \[ \frac {1}{8} \, \sqrt {c^{2} x^{2} + 1} x {\left (\frac {2 \, x^{2} {\left | c \right |} \mathrm {sgn}\relax (x)}{c^{4}} - \frac {3 \, {\left | c \right |} \mathrm {sgn}\relax (x)}{c^{6}}\right )} - \frac {3 \, \log \left (-x {\left | c \right |} + \sqrt {c^{2} x^{2} + 1}\right ) \mathrm {sgn}\relax (x)}{8 \, c^{6}} + \frac {\arctan \left (c x\right )}{c^{6}} + \frac {c^{6} x^{3} - 3 \, c^{4} x}{3 \, c^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^5/(c^2*x^2+1),x, algorithm="giac")

[Out]

1/8*sqrt(c^2*x^2 + 1)*x*(2*x^2*abs(c)*sgn(x)/c^4 - 3*abs(c)*sgn(x)/c^6) - 3/8*log(-x*abs(c) + sqrt(c^2*x^2 + 1
))*sgn(x)/c^6 + arctan(c*x)/c^6 + 1/3*(c^6*x^3 - 3*c^4*x)/c^9

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maple [B]  time = 0.12, size = 165, normalized size = 1.79 \[ \frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, x \left (2 x \left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} c^{4}-5 x \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2}-5 \ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right )+8 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right )\right )}{8 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{6}}+\frac {x^{3}}{3 c^{3}}-\frac {x}{c^{5}}+\frac {\arctan \left (c x \right )}{c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*x^5/(c^2*x^2+1),x)

[Out]

1/8*((c^2*x^2+1)/c^2/x^2)^(1/2)*x*(2*x*((c^2*x^2+1)/c^2)^(3/2)*c^4-5*x*((c^2*x^2+1)/c^2)^(1/2)*c^2-5*ln(x+((c^
2*x^2+1)/c^2)^(1/2))+8*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^6+1/3*x^3/c^3-x/c^5+arctan(c*x)/c^6

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maxima [B]  time = 0.40, size = 162, normalized size = 1.76 \[ \frac {c^{2} x^{3} - 3 \, x}{3 \, c^{5}} - \frac {\frac {2 \, {\left (\frac {5 \, \sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - \frac {3 \, \left (\frac {c^{2} x^{2} + 1}{x^{2}}\right )^{\frac {3}{2}}}{c^{3}}\right )}}{\frac {2 \, {\left (c^{2} x^{2} + 1\right )}}{c^{2} x^{2}} - \frac {{\left (c^{2} x^{2} + 1\right )}^{2}}{c^{4} x^{4}} - 1} - 3 \, \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} + 1\right ) + 3 \, \log \left (\frac {\sqrt {\frac {c^{2} x^{2} + 1}{x^{2}}}}{c} - 1\right )}{16 \, c^{6}} + \frac {\arctan \left (c x\right )}{c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x^5/(c^2*x^2+1),x, algorithm="maxima")

[Out]

1/3*(c^2*x^3 - 3*x)/c^5 - 1/16*(2*(5*sqrt((c^2*x^2 + 1)/x^2)/c - 3*((c^2*x^2 + 1)/x^2)^(3/2)/c^3)/(2*(c^2*x^2
+ 1)/(c^2*x^2) - (c^2*x^2 + 1)^2/(c^4*x^4) - 1) - 3*log(sqrt((c^2*x^2 + 1)/x^2)/c + 1) + 3*log(sqrt((c^2*x^2 +
1)/x^2)/c - 1))/c^6 + arctan(c*x)/c^6

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mupad [B]  time = 2.43, size = 79, normalized size = 0.86 \[ \frac {3\,\mathrm {atanh}\left (\sqrt {\frac {1}{c^2\,x^2}+1}\right )}{8\,c^6}+\frac {3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3}{3\,c^6}+\frac {x^4\,\sqrt {\frac {1}{c^2\,x^2}+1}}{4\,c^2}-\frac {3\,x^2\,\sqrt {\frac {1}{c^2\,x^2}+1}}{8\,c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*((1/(c^2*x^2) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 + 1),x)

[Out]

(3*atanh((1/(c^2*x^2) + 1)^(1/2)))/(8*c^6) + (3*atan(c*x) - 3*c*x + c^3*x^3)/(3*c^6) + (x^4*(1/(c^2*x^2) + 1)^
(1/2))/(4*c^2) - (3*x^2*(1/(c^2*x^2) + 1)^(1/2))/(8*c^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{4}}{c^{2} x^{2} + 1}\, dx + \int \frac {c x^{5} \sqrt {1 + \frac {1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))*x**5/(c**2*x**2+1),x)

[Out]

(Integral(x**4/(c**2*x**2 + 1), x) + Integral(c*x**5*sqrt(1 + 1/(c**2*x**2))/(c**2*x**2 + 1), x))/c

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