3.246 \(\int \frac{x^5}{(d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=153 \[ -\frac{a e+c d x^2}{4 c \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{d^2 e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{d^2 e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}+\frac{d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 \sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.247646, antiderivative size = 153, 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, 1647, 801, 635, 205, 260} \[ -\frac{a e+c d x^2}{4 c \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{d^2 e \log \left (d+e x^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{d^2 e \log \left (a+c x^4\right )}{4 \left (a e^2+c d^2\right )^2}+\frac{d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 \sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

-(a*e + c*d*x^2)/(4*c*(c*d^2 + a*e^2)*(a + c*x^4)) + (d*(c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(4*Sqrt
[a]*Sqrt[c]*(c*d^2 + a*e^2)^2) + (d^2*e*Log[d + e*x^2])/(2*(c*d^2 + a*e^2)^2) - (d^2*e*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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

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^5}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{a e+c d x^2}{4 c \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{a c d^2}{c d^2+a e^2}+\frac{a c d e x}{c d^2+a e^2}}{(d+e x) \left (a+c x^2\right )} \, dx,x,x^2\right )}{4 a c}\\ &=-\frac{a e+c d x^2}{4 c \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^2 e^2}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac{a c d \left (-c d^2+a e^2+2 c d e x\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}\right ) \, dx,x,x^2\right )}{4 a c}\\ &=-\frac{a e+c d x^2}{4 c \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac{d^2 e \log \left (d+e x^2\right )}{2 \left (c d^2+a e^2\right )^2}-\frac{d \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{a e+c d x^2}{4 c \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac{d^2 e \log \left (d+e x^2\right )}{2 \left (c d^2+a e^2\right )^2}-\frac{\left (c d^2 e\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 (d \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{a e+c d x^2}{4 c \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac{d \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^2 e \log \left (d+e x^2\right )}{2 \left (c d^2+a e^2\right )^2}-\frac{d^2 e \log \left (a+c x^4\right )}{4 \left (c d^2+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.157612, size = 120, normalized size = 0.78 \[ \frac{-\frac{\left (a e^2+c d^2\right ) \left (a e+c d x^2\right )}{c \left (a+c x^4\right )}+\frac{d \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{c}}-d^2 e \log \left (a+c x^4\right )+2 d^2 e \log \left (d+e x^2\right )}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

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

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Maple [A]  time = 0.016, size = 252, normalized size = 1.7 \begin{align*} -{\frac{{x}^{2}{e}^{2}da}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{c{x}^{2}{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{{a}^{2}{e}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) c}}-{\frac{ae{d}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{e{d}^{2}\ln \left ( c{x}^{4}+a \right ) }{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{{e}^{2}da}{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{d}^{3}}{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}\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^5/(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^2*d*a-1/4/(a*e^2+c*d^2)^2/(c*x^4+a)*x^2*c*d^3-1/4/(a*e^2+c*d^2)^2/(c*x^4+
a)*a^2*e^3/c-1/4/(a*e^2+c*d^2)^2/(c*x^4+a)*a*e*d^2-1/4*d^2*e*ln(c*x^4+a)/(a*e^2+c*d^2)^2-1/4/(a*e^2+c*d^2)^2*d
/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))*a*e^2+1/4/(a*e^2+c*d^2)^2/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))*c*d^3+1
/2*d^2*e*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^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.3171, size = 987, normalized size = 6.45 \begin{align*} \left [-\frac{2 \, a^{2} c d^{2} e + 2 \, a^{3} e^{3} + 2 \,{\left (a c^{2} d^{3} + a^{2} c d e^{2}\right )} x^{2} -{\left (a c d^{3} - a^{2} d e^{2} +{\left (c^{2} d^{3} - a c d e^{2}\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^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (c x^{4} + a\right ) - 4 \,{\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\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^{2} c d^{2} e + a^{3} e^{3} +{\left (a c^{2} d^{3} + a^{2} c d e^{2}\right )} x^{2} +{\left (a c d^{3} - a^{2} d e^{2} +{\left (c^{2} d^{3} - a c d e^{2}\right )} x^{4}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c}}{c x^{2}}\right ) +{\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\right )} \log \left (c x^{4} + a\right ) - 2 \,{\left (a c^{2} d^{2} e x^{4} + a^{2} c d^{2} e\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^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

[-1/8*(2*a^2*c*d^2*e + 2*a^3*e^3 + 2*(a*c^2*d^3 + a^2*c*d*e^2)*x^2 - (a*c*d^3 - a^2*d*e^2 + (c^2*d^3 - a*c*d*e
^2)*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^2*e*x^4 + a^2*c*d^2*e)*log(c*
x^4 + a) - 4*(a*c^2*d^2*e*x^4 + a^2*c*d^2*e)*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^2*c*d^2*e + a^3*e^3 + (a*c^2*d^3 + a^2*c*d*e^2)*x^2
+ (a*c*d^3 - a^2*d*e^2 + (c^2*d^3 - a*c*d*e^2)*x^4)*sqrt(a*c)*arctan(sqrt(a*c)/(c*x^2)) + (a*c^2*d^2*e*x^4 + a
^2*c*d^2*e)*log(c*x^4 + a) - 2*(a*c^2*d^2*e*x^4 + a^2*c*d^2*e)*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**5/(e*x**2+d)/(c*x**4+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.08849, size = 297, normalized size = 1.94 \begin{align*} -\frac{d^{2} e \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^{2} e^{2} \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^{3} - a d e^{2}\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^{2} d^{2} x^{4} e - c^{2} d^{3} x^{2} - a c d x^{2} e^{2} - a^{2} e^{3}}{4 \,{\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )}{\left (c x^{4} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

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