3.248 \(\int \frac{x}{(d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=151 \[ \frac{\sqrt{c} d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \left (a e^2+c d^2\right )^2}+\frac{a e+c d x^2}{4 a \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{e^3 \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{e^3 \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2} \]

[Out]

(a*e + c*d*x^2)/(4*a*(c*d^2 + a*e^2)*(a + c*x^4)) + (Sqrt[c]*d*(c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]]
)/(4*a^(3/2)*(c*d^2 + a*e^2)^2) + (e^3*Log[d + e*x^2])/(2*(c*d^2 + a*e^2)^2) - (e^3*Log[a + c*x^4])/(4*(c*d^2
+ a*e^2)^2)

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Rubi [A]  time = 0.180638, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1248, 741, 801, 635, 205, 260} \[ \frac{\sqrt{c} d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \left (a e^2+c d^2\right )^2}+\frac{a e+c d x^2}{4 a \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{e^3 \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{e^3 \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(a*e + c*d*x^2)/(4*a*(c*d^2 + a*e^2)*(a + c*x^4)) + (Sqrt[c]*d*(c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]]
)/(4*a^(3/2)*(c*d^2 + a*e^2)^2) + (e^3*Log[d + e*x^2])/(2*(c*d^2 + a*e^2)^2) - (e^3*Log[a + c*x^4])/(4*(c*d^2
+ a*e^2)^2)

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.138577, size = 117, normalized size = 0.77 \[ \frac{\frac{\sqrt{c} d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{a^{3/2}}+\frac{\left (a e^2+c d^2\right ) \left (a e+c d x^2\right )}{a \left (a+c x^4\right )}-e^3 \log \left (a+c x^4\right )+2 e^3 \log \left (d+e x^2\right )}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(((c*d^2 + a*e^2)*(a*e + c*d*x^2))/(a*(a + c*x^4)) + (Sqrt[c]*d*(c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]
])/a^(3/2) + 2*e^3*Log[d + e*x^2] - e^3*Log[a + c*x^4])/(4*(c*d^2 + a*e^2)^2)

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Maple [A]  time = 0.016, size = 255, normalized size = 1.7 \begin{align*}{\frac{{x}^{2}cd{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{{c}^{2}{d}^{3}{x}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) a}}+{\frac{{e}^{3}a}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{e{d}^{2}c}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{{e}^{3}\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{3\,cd{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{c}^{2}{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{e}^{3}\ln \left ( e{x}^{2}+d \right ) }{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 39.4054, size = 925, normalized size = 6.13 \begin{align*} \left [\frac{2 \, a c d^{2} e + 2 \, a^{2} e^{3} + 2 \,{\left (c^{2} d^{3} + a c d e^{2}\right )} x^{2} +{\left (a c d^{3} + 3 \, a^{2} d e^{2} +{\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} x^{4}\right )} \sqrt{-\frac{c}{a}} \log \left (\frac{c x^{4} + 2 \, a x^{2} \sqrt{-\frac{c}{a}} - a}{c x^{4} + a}\right ) - 2 \,{\left (a c e^{3} x^{4} + a^{2} e^{3}\right )} \log \left (c x^{4} + a\right ) + 4 \,{\left (a c e^{3} x^{4} + a^{2} e^{3}\right )} \log \left (e x^{2} + d\right )}{8 \,{\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} +{\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{4}\right )}}, \frac{a c d^{2} e + a^{2} e^{3} +{\left (c^{2} d^{3} + a c d e^{2}\right )} x^{2} -{\left (a c d^{3} + 3 \, a^{2} d e^{2} +{\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} x^{4}\right )} \sqrt{\frac{c}{a}} \arctan \left (\frac{a \sqrt{\frac{c}{a}}}{c x^{2}}\right ) -{\left (a c e^{3} x^{4} + a^{2} e^{3}\right )} \log \left (c x^{4} + a\right ) + 2 \,{\left (a c e^{3} x^{4} + a^{2} e^{3}\right )} \log \left (e x^{2} + d\right )}{4 \,{\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} +{\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(2*a*c*d^2*e + 2*a^2*e^3 + 2*(c^2*d^3 + a*c*d*e^2)*x^2 + (a*c*d^3 + 3*a^2*d*e^2 + (c^2*d^3 + 3*a*c*d*e^2)
*x^4)*sqrt(-c/a)*log((c*x^4 + 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) - 2*(a*c*e^3*x^4 + a^2*e^3)*log(c*x^4 + a)
+ 4*(a*c*e^3*x^4 + a^2*e^3)*log(e*x^2 + d))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (a*c^3*d^4 + 2*a^2*c^2*
d^2*e^2 + a^3*c*e^4)*x^4), 1/4*(a*c*d^2*e + a^2*e^3 + (c^2*d^3 + a*c*d*e^2)*x^2 - (a*c*d^3 + 3*a^2*d*e^2 + (c^
2*d^3 + 3*a*c*d*e^2)*x^4)*sqrt(c/a)*arctan(a*sqrt(c/a)/(c*x^2)) - (a*c*e^3*x^4 + a^2*e^3)*log(c*x^4 + a) + 2*(
a*c*e^3*x^4 + a^2*e^3)*log(e*x^2 + d))/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4 + (a*c^3*d^4 + 2*a^2*c^2*d^2*e
^2 + a^3*c*e^4)*x^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.09404, size = 269, normalized size = 1.78 \begin{align*} -\frac{e^{3} \log \left (c x^{4} + a\right )}{4 \,{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac{e^{4} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )}} + \frac{{\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{4 \,{\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt{a c}} + \frac{a c d^{2} e +{\left (c^{2} d^{3} + a c d e^{2}\right )} x^{2} + a^{2} e^{3}}{4 \,{\left (c x^{4} + a\right )}{\left (c d^{2} + a e^{2}\right )}^{2} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*e^3*log(c*x^4 + a)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) + 1/2*e^4*log(abs(x^2*e + d))/(c^2*d^4*e + 2*a*c*d
^2*e^3 + a^2*e^5) + 1/4*(c^2*d^3 + 3*a*c*d*e^2)*arctan(c*x^2/sqrt(a*c))/((a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^
4)*sqrt(a*c)) + 1/4*(a*c*d^2*e + (c^2*d^3 + a*c*d*e^2)*x^2 + a^2*e^3)/((c*x^4 + a)*(c*d^2 + a*e^2)^2*a)