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

Optimal. Leaf size=61 \[ -\frac {1}{3} c^2 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}+c^2 \sqrt {\frac {1}{c^2 x^2}+1}+c^2 \tan ^{-1}(c x)-\frac {1}{3 c x^3}+\frac {c}{x} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6342, 266, 43, 325, 203} \[ -\frac {1}{3} c^2 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}+c^2 \sqrt {\frac {1}{c^2 x^2}+1}+c^2 \tan ^{-1}(c x)-\frac {1}{3 c x^3}+\frac {c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 \left (1+c^2 x^2\right )} \, dx &=\frac {\int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5} \, dx}{c^2}+\frac {\int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx}{c}\\ &=-\frac {1}{3 c x^3}-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{2 c^2}-c \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {1}{3 c x^3}+\frac {c}{x}-\frac {\operatorname {Subst}\left (\int \left (-\frac {c^2}{\sqrt {1+\frac {x}{c^2}}}+c^2 \sqrt {1+\frac {x}{c^2}}\right ) \, dx,x,\frac {1}{x^2}\right )}{2 c^2}+c^3 \int \frac {1}{1+c^2 x^2} \, dx\\ &=c^2 \sqrt {1+\frac {1}{c^2 x^2}}-\frac {1}{3} c^2 \left (1+\frac {1}{c^2 x^2}\right )^{3/2}-\frac {1}{3 c x^3}+\frac {c}{x}+c^2 \tan ^{-1}(c x)\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 54, normalized size = 0.89 \[ \frac {\sqrt {\frac {1}{c^2 x^2}+1} \left (2 c^2 x^2-1\right )}{3 x^2}+c^2 \tan ^{-1}(c x)-\frac {1}{3 c x^3}+\frac {c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 70, normalized size = 1.15 \[ \frac {3 \, c^{3} x^{3} \arctan \left (c x\right ) + 2 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + {\left (2 \, c^{3} x^{3} - c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1}{3 \, c x^{3}} \]

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]

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

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giac [A]  time = 0.16, size = 82, normalized size = 1.34 \[ c^{2} \arctan \left (c x\right ) + \frac {4 \, {\left (3 \, {\left (x {\left | c \right |} - \sqrt {c^{2} x^{2} + 1}\right )}^{2} - 1\right )} c^{2} \mathrm {sgn}\relax (x)}{3 \, {\left ({\left (x {\left | c \right |} - \sqrt {c^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} + \frac {3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \]

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]

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

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maple [B]  time = 0.07, size = 193, normalized size = 3.16 \[ \frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{2} \left (3 \left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} x^{2} c^{2}-3 \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, x^{4} c^{2}-3 \ln \left (x +\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\right ) x^{3}+3 \ln \left (x +\sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\right ) x^{3}-\left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}}\right )}{3 x^{2} \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}}-\frac {1}{3 c \,x^{3}}+\frac {c}{x}+c^{2} \arctan \left (c x \right ) \]

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]

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

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maxima [A]  time = 0.43, size = 56, normalized size = 0.92 \[ c^{2} \arctan \left (c x\right ) + \frac {{\left (2 \, c^{2} x^{2} - 1\right )} \sqrt {c^{2} x^{2} + 1}}{3 \, c x^{3}} + \frac {3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \]

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]

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

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mupad [B]  time = 2.22, size = 57, normalized size = 0.93 \[ \frac {c+\frac {2\,c^2\,x\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3}}{x}-\frac {\frac {x\,\sqrt {\frac {1}{c^2\,x^2}+1}}{3}+\frac {1}{3\,c}}{x^3}+c^2\,\mathrm {atan}\left (c\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + (2*c^2*x*(1/(c^2*x^2) + 1)^(1/2))/3)/x - ((x*(1/(c^2*x^2) + 1)^(1/2))/3 + 1/(3*c))/x^3 + c^2*atan(c*x)

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sympy [A]  time = 5.23, size = 75, normalized size = 1.23 \[ - 2 c^{5} \left (\frac {\left (1 + \frac {1}{c^{2} x^{2}}\right )^{\frac {3}{2}}}{6 c^{3}} - \frac {\sqrt {1 + \frac {1}{c^{2} x^{2}}}}{2 c^{3}}\right ) - \frac {c^{3} \operatorname {atan}{\left (\frac {1}{x \sqrt {c^{2}}} \right )}}{\sqrt {c^{2}}} + \frac {c}{x} - \frac {1}{3 c x^{3}} \]

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]

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

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