3.247 \(\int \frac{x^3}{(d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=148 \[ -\frac{d-e x^2}{4 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}-\frac{d e^2 \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac{d e^2 \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}-\frac{e \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 \sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.186862, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1252, 823, 801, 635, 205, 260} \[ -\frac{d-e x^2}{4 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}-\frac{d e^2 \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac{d e^2 \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}-\frac{e \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 \sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

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

Rule 1252

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.148467, size = 114, normalized size = 0.77 \[ \frac{\frac{\left (e x^2-d\right ) \left (a e^2+c d^2\right )}{a+c x^4}+\frac{e \left (a e^2-c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{c}}+d e^2 \log \left (a+c x^4\right )-2 d e^2 \log \left (d+e x^2\right )}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

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

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Maple [A]  time = 0.016, size = 247, normalized size = 1.7 \begin{align*}{\frac{{x}^{2}{e}^{3}a}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{e{x}^{2}{d}^{2}c}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{{e}^{2}da}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{c{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{d{e}^{2}\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{{e}^{3}a}{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{e{d}^{2}c}{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{d{e}^{2}\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^3/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

1/4/(a*e^2+c*d^2)^2/(c*x^4+a)*x^2*e^3*a+1/4/(a*e^2+c*d^2)^2/(c*x^4+a)*x^2*e*d^2*c-1/4/(a*e^2+c*d^2)^2/(c*x^4+a
)*e^2*d*a-1/4/(a*e^2+c*d^2)^2/(c*x^4+a)*c*d^3+1/4*d*e^2*ln(c*x^4+a)/(a*e^2+c*d^2)^2+1/4/(a*e^2+c*d^2)^2/(a*c)^
(1/2)*arctan(c*x^2/(a*c)^(1/2))*a*e^3-1/4/(a*e^2+c*d^2)^2*e/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))*c*d^2-1/2*d*
e^2*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^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.4727, size = 992, normalized size = 6.7 \begin{align*} \left [-\frac{2 \, a c^{2} d^{3} + 2 \, a^{2} c d e^{2} - 2 \,{\left (a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} -{\left (a c d^{2} e - a^{2} e^{3} +{\left (c^{2} d^{2} e - a c e^{3}\right )} x^{4}\right )} \sqrt{-a c} \log \left (\frac{c x^{4} - 2 \, \sqrt{-a c} x^{2} - a}{c x^{4} + a}\right ) - 2 \,{\left (a c^{2} d e^{2} x^{4} + a^{2} c d e^{2}\right )} \log \left (c x^{4} + a\right ) + 4 \,{\left (a c^{2} d e^{2} x^{4} + a^{2} c d e^{2}\right )} \log \left (e x^{2} + d\right )}{8 \,{\left (a^{2} c^{3} d^{4} + 2 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4} +{\left (a c^{4} d^{4} + 2 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4}\right )}}, -\frac{a c^{2} d^{3} + a^{2} c d e^{2} -{\left (a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} -{\left (a c d^{2} e - a^{2} e^{3} +{\left (c^{2} d^{2} e - a c e^{3}\right )} x^{4}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c}}{c x^{2}}\right ) -{\left (a c^{2} d e^{2} x^{4} + a^{2} c d e^{2}\right )} \log \left (c x^{4} + a\right ) + 2 \,{\left (a c^{2} d e^{2} x^{4} + a^{2} c d e^{2}\right )} \log \left (e x^{2} + d\right )}{4 \,{\left (a^{2} c^{3} d^{4} + 2 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4} +{\left (a c^{4} d^{4} + 2 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.10727, size = 254, normalized size = 1.72 \begin{align*} \frac{d e^{2} \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{d e^{3} \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 d^{2} e - a e^{3}\right )} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{4 \,{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{a c}} - \frac{c d^{3} -{\left (c d^{2} e + a e^{3}\right )} x^{2} + a d e^{2}}{4 \,{\left (c x^{4} + a\right )}{\left (c d^{2} + a e^{2}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

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