3.266 \(\int \frac{e^{2 \tan ^{-1}(a x)}}{(c+a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=123 \[ \frac{9 (a x+1) e^{2 \tan ^{-1}(a x)}}{80 a c^4 \left (a^2 x^2+1\right )}+\frac{3 (2 a x+1) e^{2 \tan ^{-1}(a x)}}{40 a c^4 \left (a^2 x^2+1\right )^2}+\frac{(3 a x+1) e^{2 \tan ^{-1}(a x)}}{20 a c^4 \left (a^2 x^2+1\right )^3}+\frac{9 e^{2 \tan ^{-1}(a x)}}{160 a c^4} \]

[Out]

(9*E^(2*ArcTan[a*x]))/(160*a*c^4) + (E^(2*ArcTan[a*x])*(1 + 3*a*x))/(20*a*c^4*(1 + a^2*x^2)^3) + (3*E^(2*ArcTa
n[a*x])*(1 + 2*a*x))/(40*a*c^4*(1 + a^2*x^2)^2) + (9*E^(2*ArcTan[a*x])*(1 + a*x))/(80*a*c^4*(1 + a^2*x^2))

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Rubi [A]  time = 0.119764, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {5070, 5071} \[ \frac{9 (a x+1) e^{2 \tan ^{-1}(a x)}}{80 a c^4 \left (a^2 x^2+1\right )}+\frac{3 (2 a x+1) e^{2 \tan ^{-1}(a x)}}{40 a c^4 \left (a^2 x^2+1\right )^2}+\frac{(3 a x+1) e^{2 \tan ^{-1}(a x)}}{20 a c^4 \left (a^2 x^2+1\right )^3}+\frac{9 e^{2 \tan ^{-1}(a x)}}{160 a c^4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^4,x]

[Out]

(9*E^(2*ArcTan[a*x]))/(160*a*c^4) + (E^(2*ArcTan[a*x])*(1 + 3*a*x))/(20*a*c^4*(1 + a^2*x^2)^3) + (3*E^(2*ArcTa
n[a*x])*(1 + 2*a*x))/(40*a*c^4*(1 + a^2*x^2)^2) + (9*E^(2*ArcTan[a*x])*(1 + a*x))/(80*a*c^4*(1 + a^2*x^2))

Rule 5070

Int[E^(ArcTan[(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*ArcTan[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*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&&  !IntegerQ[I*n] && NeQ[n^2 + 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rule 5071

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcTan[a*x])/(a*c*n), x] /; Fre
eQ[{a, c, d, n}, x] && EqQ[d, a^2*c]

Rubi steps

\begin{align*} \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^4} \, dx &=\frac{e^{2 \tan ^{-1}(a x)} (1+3 a x)}{20 a c^4 \left (1+a^2 x^2\right )^3}+\frac{3 \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^3} \, dx}{4 c}\\ &=\frac{e^{2 \tan ^{-1}(a x)} (1+3 a x)}{20 a c^4 \left (1+a^2 x^2\right )^3}+\frac{3 e^{2 \tan ^{-1}(a x)} (1+2 a x)}{40 a c^4 \left (1+a^2 x^2\right )^2}+\frac{9 \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{20 c^2}\\ &=\frac{e^{2 \tan ^{-1}(a x)} (1+3 a x)}{20 a c^4 \left (1+a^2 x^2\right )^3}+\frac{3 e^{2 \tan ^{-1}(a x)} (1+2 a x)}{40 a c^4 \left (1+a^2 x^2\right )^2}+\frac{9 e^{2 \tan ^{-1}(a x)} (1+a x)}{80 a c^4 \left (1+a^2 x^2\right )}+\frac{9 \int \frac{e^{2 \tan ^{-1}(a x)}}{c+a^2 c x^2} \, dx}{80 c^3}\\ &=\frac{9 e^{2 \tan ^{-1}(a x)}}{160 a c^4}+\frac{e^{2 \tan ^{-1}(a x)} (1+3 a x)}{20 a c^4 \left (1+a^2 x^2\right )^3}+\frac{3 e^{2 \tan ^{-1}(a x)} (1+2 a x)}{40 a c^4 \left (1+a^2 x^2\right )^2}+\frac{9 e^{2 \tan ^{-1}(a x)} (1+a x)}{80 a c^4 \left (1+a^2 x^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.24509, size = 122, normalized size = 0.99 \[ \frac{8 c (3 a x+1) e^{2 \tan ^{-1}(a x)}+3 \left (a^2 c x^2+c\right ) \left (4 (2 a x+1) e^{2 \tan ^{-1}(a x)}+3 (1-i a x)^i (1+i a x)^{-i} (a x-i) (a x+i) \left (a^2 x^2+2 a x+3\right )\right )}{160 a c^2 \left (a^2 c x^2+c\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^4,x]

[Out]

(8*c*E^(2*ArcTan[a*x])*(1 + 3*a*x) + 3*(c + a^2*c*x^2)*(4*E^(2*ArcTan[a*x])*(1 + 2*a*x) + (3*(1 - I*a*x)^I*(-I
+ a*x)*(I + a*x)*(3 + 2*a*x + a^2*x^2))/(1 + I*a*x)^I))/(160*a*c^2*(c + a^2*c*x^2)^3)

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Maple [A]  time = 0.039, size = 73, normalized size = 0.6 \begin{align*}{\frac{{{\rm e}^{2\,\arctan \left ( ax \right ) }} \left ( 9\,{a}^{6}{x}^{6}+18\,{a}^{5}{x}^{5}+45\,{a}^{4}{x}^{4}+60\,{a}^{3}{x}^{3}+75\,{a}^{2}{x}^{2}+66\,ax+47 \right ) }{160\, \left ({a}^{2}{x}^{2}+1 \right ) ^{3}a{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*arctan(a*x))/(a^2*c*x^2+c)^4,x)

[Out]

1/160*exp(2*arctan(a*x))*(9*a^6*x^6+18*a^5*x^5+45*a^4*x^4+60*a^3*x^3+75*a^2*x^2+66*a*x+47)/(a^2*x^2+1)^3/a/c^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (2 \, \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^4, x)

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Fricas [A]  time = 2.02926, size = 212, normalized size = 1.72 \begin{align*} \frac{{\left (9 \, a^{6} x^{6} + 18 \, a^{5} x^{5} + 45 \, a^{4} x^{4} + 60 \, a^{3} x^{3} + 75 \, a^{2} x^{2} + 66 \, a x + 47\right )} e^{\left (2 \, \arctan \left (a x\right )\right )}}{160 \,{\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

1/160*(9*a^6*x^6 + 18*a^5*x^5 + 45*a^4*x^4 + 60*a^3*x^3 + 75*a^2*x^2 + 66*a*x + 47)*e^(2*arctan(a*x))/(a^7*c^4
*x^6 + 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 + a*c^4)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*atan(a*x))/(a**2*c*x**2+c)**4,x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (2 \, \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^4, x)