3.6 \(\int \frac {e^{\cot ^{-1}(x)}}{(a+a x^2)^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ -\frac {3 (1-x) e^{\cot ^{-1}(x)}}{10 a^2 \sqrt {a x^2+a}}-\frac {(1-3 x) e^{\cot ^{-1}(x)}}{10 a \left (a x^2+a\right )^{3/2}} \]

[Out]

-1/10*exp(arccot(x))*(1-3*x)/a/(a*x^2+a)^(3/2)-3/10*exp(arccot(x))*(1-x)/a^2/(a*x^2+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5115, 5114} \[ -\frac {3 (1-x) e^{\cot ^{-1}(x)}}{10 a^2 \sqrt {a x^2+a}}-\frac {(1-3 x) e^{\cot ^{-1}(x)}}{10 a \left (a x^2+a\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCot[x]/(a + a*x^2)^(5/2),x]

[Out]

-(E^ArcCot[x]*(1 - 3*x))/(10*a*(a + a*x^2)^(3/2)) - (3*E^ArcCot[x]*(1 - x))/(10*a^2*Sqrt[a + a*x^2])

Rule 5114

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> -Simp[((n - a*x)*E^(n*ArcCot[a*x]))
/(a*c*(n^2 + 1)*Sqrt[c + d*x^2]), x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] &&  !IntegerQ[(I*n - 1)/2]

Rule 5115

Int[E^(ArcCot[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((n + 2*a*(p + 1)*x)*(c + d*x^
2)^(p + 1)*E^(n*ArcCot[a*x]))/(a*c*(n^2 + 4*(p + 1)^2)), x] + Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 + 4*(p + 1)^2
)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCot[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&& NeQ[p, -3/2] && NeQ[n^2 + 4*(p + 1)^2, 0] &&  !(IntegerQ[p] && IntegerQ[(I*n)/2]) &&  !( !IntegerQ[p] && I
ntegerQ[(I*n - 1)/2])

Rubi steps

\begin {align*} \int \frac {e^{\cot ^{-1}(x)}}{\left (a+a x^2\right )^{5/2}} \, dx &=-\frac {e^{\cot ^{-1}(x)} (1-3 x)}{10 a \left (a+a x^2\right )^{3/2}}+\frac {3 \int \frac {e^{\cot ^{-1}(x)}}{\left (a+a x^2\right )^{3/2}} \, dx}{5 a}\\ &=-\frac {e^{\cot ^{-1}(x)} (1-3 x)}{10 a \left (a+a x^2\right )^{3/2}}-\frac {3 e^{\cot ^{-1}(x)} (1-x)}{10 a^2 \sqrt {a+a x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 51, normalized size = 0.93 \[ \frac {e^{\cot ^{-1}(x)} \left (-3 \sqrt {\frac {1}{x^2}+1} x \cos \left (3 \cot ^{-1}(x)\right )+15 x+2 \cos \left (2 \cot ^{-1}(x)\right )-14\right )}{40 a^2 \sqrt {a \left (x^2+1\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCot[x]/(a + a*x^2)^(5/2),x]

[Out]

(E^ArcCot[x]*(-14 + 15*x + 2*Cos[2*ArcCot[x]] - 3*Sqrt[1 + x^(-2)]*x*Cos[3*ArcCot[x]]))/(40*a^2*Sqrt[a*(1 + x^
2)])

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fricas [A]  time = 0.75, size = 50, normalized size = 0.91 \[ \frac {\sqrt {a x^{2} + a} {\left (3 \, x^{3} - 3 \, x^{2} + 6 \, x - 4\right )} e^{\operatorname {arccot}\relax (x)}}{10 \, {\left (a^{3} x^{4} + 2 \, a^{3} x^{2} + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccot(x))/(a*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/10*sqrt(a*x^2 + a)*(3*x^3 - 3*x^2 + 6*x - 4)*e^arccot(x)/(a^3*x^4 + 2*a^3*x^2 + a^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\operatorname {arccot}\relax (x)}}{{\left (a x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccot(x))/(a*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate(e^arccot(x)/(a*x^2 + a)^(5/2), x)

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maple [A]  time = 0.04, size = 35, normalized size = 0.64 \[ \frac {\left (x^{2}+1\right ) \left (3 x^{3}-3 x^{2}+6 x -4\right ) {\mathrm e}^{\mathrm {arccot}\relax (x )}}{10 \left (a \,x^{2}+a \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arccot(x))/(a*x^2+a)^(5/2),x)

[Out]

1/10*(x^2+1)*(3*x^3-3*x^2+6*x-4)*exp(arccot(x))/(a*x^2+a)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\operatorname {arccot}\relax (x)}}{{\left (a x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arccot(x))/(a*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(e^arccot(x)/(a*x^2 + a)^(5/2), x)

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mupad [B]  time = 0.14, size = 42, normalized size = 0.76 \[ -\frac {4\,{\mathrm {e}}^{\mathrm {acot}\relax (x)}-6\,x\,{\mathrm {e}}^{\mathrm {acot}\relax (x)}+3\,x^2\,{\mathrm {e}}^{\mathrm {acot}\relax (x)}-3\,x^3\,{\mathrm {e}}^{\mathrm {acot}\relax (x)}}{10\,a\,{\left (a\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(acot(x))/(a + a*x^2)^(5/2),x)

[Out]

-(4*exp(acot(x)) - 6*x*exp(acot(x)) + 3*x^2*exp(acot(x)) - 3*x^3*exp(acot(x)))/(10*a*(a + a*x^2)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\operatorname {acot}{\relax (x )}}}{\left (a \left (x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(acot(x))/(a*x**2+a)**(5/2),x)

[Out]

Integral(exp(acot(x))/(a*(x**2 + 1))**(5/2), x)

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