3.244 \(\int \frac{x^9}{(d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=169 \[ \frac{a \left (a e+c d x^2\right )}{4 c^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{a e \left (a e^2+2 c d^2\right ) \log \left (a+c x^4\right )}{4 c^2 \left (a e^2+c d^2\right )^2}-\frac{\sqrt{a} d \left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 c^{3/2} \left (a e^2+c d^2\right )^2}+\frac{d^4 \log \left (d+e x^2\right )}{2 e \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.366727, antiderivative size = 169, 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, 1629, 635, 205, 260} \[ \frac{a \left (a e+c d x^2\right )}{4 c^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{a e \left (a e^2+2 c d^2\right ) \log \left (a+c x^4\right )}{4 c^2 \left (a e^2+c d^2\right )^2}-\frac{\sqrt{a} d \left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 c^{3/2} \left (a e^2+c d^2\right )^2}+\frac{d^4 \log \left (d+e x^2\right )}{2 e \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^9/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(a*(a*e + c*d*x^2))/(4*c^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (Sqrt[a]*d*(3*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sq
rt[a]])/(4*c^(3/2)*(c*d^2 + a*e^2)^2) + (d^4*Log[d + e*x^2])/(2*e*(c*d^2 + a*e^2)^2) + (a*e*(2*c*d^2 + a*e^2)*
Log[a + c*x^4])/(4*c^2*(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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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^9}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{(d+e x) \left (a+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{a \left (a e+c d x^2\right )}{4 c^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 d^2}{c d^2+a e^2}-\frac{a^2 d e x}{c d^2+a e^2}-2 a x^2}{(d+e x) \left (a+c x^2\right )} \, dx,x,x^2\right )}{4 a c}\\ &=\frac{a \left (a e+c d x^2\right )}{4 c^2 \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^4}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac{a^2 \left (d \left (3 c d^2+a e^2\right )-2 e \left (2 c d^2+a e^2\right ) 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 \left (a e+c d x^2\right )}{4 c^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac{d^4 \log \left (d+e x^2\right )}{2 e \left (c d^2+a e^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \frac{d \left (3 c d^2+a e^2\right )-2 e \left (2 c d^2+a e^2\right ) x}{a+c x^2} \, dx,x,x^2\right )}{4 c \left (c d^2+a e^2\right )^2}\\ &=\frac{a \left (a e+c d x^2\right )}{4 c^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac{d^4 \log \left (d+e x^2\right )}{2 e \left (c d^2+a e^2\right )^2}+\frac{\left (a e \left (2 c d^2+a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{a+c x^2} \, dx,x,x^2\right )}{2 c \left (c d^2+a e^2\right )^2}-\frac{\left (a d \left (3 c d^2+a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+c x^2} \, dx,x,x^2\right )}{4 c \left (c d^2+a e^2\right )^2}\\ &=\frac{a \left (a e+c d x^2\right )}{4 c^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac{\sqrt{a} d \left (3 c d^2+a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 c^{3/2} \left (c d^2+a e^2\right )^2}+\frac{d^4 \log \left (d+e x^2\right )}{2 e \left (c d^2+a e^2\right )^2}+\frac{a e \left (2 c d^2+a e^2\right ) \log \left (a+c x^4\right )}{4 c^2 \left (c d^2+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.21659, size = 135, normalized size = 0.8 \[ \frac{\frac{a \left (a e^2+c d^2\right ) \left (a e+c d x^2\right )}{c^2 \left (a+c x^4\right )}+\frac{a e \left (a e^2+2 c d^2\right ) \log \left (a+c x^4\right )}{c^2}-\frac{\sqrt{a} d \left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{c^{3/2}}+\frac{2 d^4 \log \left (d+e x^2\right )}{e}}{4 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/((d + e*x^2)*(a + c*x^4)^2),x]

[Out]

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

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Maple [A]  time = 0.033, size = 305, normalized size = 1.8 \begin{align*}{\frac{d{a}^{2}{x}^{2}{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) c}}+{\frac{a{x}^{2}{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}+{\frac{{a}^{3}{e}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ){c}^{2}}}+{\frac{{a}^{2}e{d}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) c}}+{\frac{{a}^{2}\ln \left ( c{x}^{4}+a \right ){e}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}{c}^{2}}}+{\frac{a\ln \left ( c{x}^{4}+a \right ) e{d}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}c}}-{\frac{d{a}^{2}{e}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}c}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{3\,a{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{{d}^{4}\ln \left ( e{x}^{2}+d \right ) }{2\,e \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

1/4*a^2/(a*e^2+c*d^2)^2/(c*x^4+a)*d/c*x^2*e^2+1/4*a/(a*e^2+c*d^2)^2/(c*x^4+a)*x^2*d^3+1/4*a^3/(a*e^2+c*d^2)^2/
(c*x^4+a)*e^3/c^2+1/4*a^2/(a*e^2+c*d^2)^2/(c*x^4+a)*e/c*d^2+1/4*a^2/(a*e^2+c*d^2)^2/c^2*ln(c*x^4+a)*e^3+1/2*a/
(a*e^2+c*d^2)^2/c*ln(c*x^4+a)*e*d^2-1/4*a^2/(a*e^2+c*d^2)^2/c/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))*d*e^2-3/4*
a/(a*e^2+c*d^2)^2/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))*d^3+1/2*d^4*ln(e*x^2+d)/e/(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^9/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 81.2695, size = 1106, normalized size = 6.54 \begin{align*} \left [\frac{2 \, a^{2} c d^{2} e^{2} + 2 \, a^{3} e^{4} + 2 \,{\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{2} +{\left (3 \, a c^{2} d^{3} e + a^{2} c d e^{3} +{\left (3 \, c^{3} d^{3} e + a c^{2} d e^{3}\right )} x^{4}\right )} \sqrt{-\frac{a}{c}} \log \left (\frac{c x^{4} - 2 \, c x^{2} \sqrt{-\frac{a}{c}} - a}{c x^{4} + a}\right ) + 2 \,{\left (2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} +{\left (2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{4}\right )} \log \left (c x^{4} + a\right ) + 4 \,{\left (c^{3} d^{4} x^{4} + a c^{2} d^{4}\right )} \log \left (e x^{2} + d\right )}{8 \,{\left (a c^{4} d^{4} e + 2 \, a^{2} c^{3} d^{2} e^{3} + a^{3} c^{2} e^{5} +{\left (c^{5} d^{4} e + 2 \, a c^{4} d^{2} e^{3} + a^{2} c^{3} e^{5}\right )} x^{4}\right )}}, \frac{a^{2} c d^{2} e^{2} + a^{3} e^{4} +{\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{2} -{\left (3 \, a c^{2} d^{3} e + a^{2} c d e^{3} +{\left (3 \, c^{3} d^{3} e + a c^{2} d e^{3}\right )} x^{4}\right )} \sqrt{\frac{a}{c}} \arctan \left (\frac{c x^{2} \sqrt{\frac{a}{c}}}{a}\right ) +{\left (2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4} +{\left (2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} x^{4}\right )} \log \left (c x^{4} + a\right ) + 2 \,{\left (c^{3} d^{4} x^{4} + a c^{2} d^{4}\right )} \log \left (e x^{2} + d\right )}{4 \,{\left (a c^{4} d^{4} e + 2 \, a^{2} c^{3} d^{2} e^{3} + a^{3} c^{2} e^{5} +{\left (c^{5} d^{4} e + 2 \, a c^{4} d^{2} e^{3} + a^{2} c^{3} e^{5}\right )} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

[1/8*(2*a^2*c*d^2*e^2 + 2*a^3*e^4 + 2*(a*c^2*d^3*e + a^2*c*d*e^3)*x^2 + (3*a*c^2*d^3*e + a^2*c*d*e^3 + (3*c^3*
d^3*e + a*c^2*d*e^3)*x^4)*sqrt(-a/c)*log((c*x^4 - 2*c*x^2*sqrt(-a/c) - a)/(c*x^4 + a)) + 2*(2*a^2*c*d^2*e^2 +
a^3*e^4 + (2*a*c^2*d^2*e^2 + a^2*c*e^4)*x^4)*log(c*x^4 + a) + 4*(c^3*d^4*x^4 + a*c^2*d^4)*log(e*x^2 + d))/(a*c
^4*d^4*e + 2*a^2*c^3*d^2*e^3 + a^3*c^2*e^5 + (c^5*d^4*e + 2*a*c^4*d^2*e^3 + a^2*c^3*e^5)*x^4), 1/4*(a^2*c*d^2*
e^2 + a^3*e^4 + (a*c^2*d^3*e + a^2*c*d*e^3)*x^2 - (3*a*c^2*d^3*e + a^2*c*d*e^3 + (3*c^3*d^3*e + a*c^2*d*e^3)*x
^4)*sqrt(a/c)*arctan(c*x^2*sqrt(a/c)/a) + (2*a^2*c*d^2*e^2 + a^3*e^4 + (2*a*c^2*d^2*e^2 + a^2*c*e^4)*x^4)*log(
c*x^4 + a) + 2*(c^3*d^4*x^4 + a*c^2*d^4)*log(e*x^2 + d))/(a*c^4*d^4*e + 2*a^2*c^3*d^2*e^3 + a^3*c^2*e^5 + (c^5
*d^4*e + 2*a*c^4*d^2*e^3 + a^2*c^3*e^5)*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**9/(e*x**2+d)/(c*x**4+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.09794, size = 339, normalized size = 2.01 \begin{align*} \frac{d^{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 (2 \, a c d^{2} e + a^{2} e^{3}\right )} \log \left (c x^{4} + a\right )}{4 \,{\left (c^{4} d^{4} + 2 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )}} - \frac{{\left (3 \, a c d^{3} + a^{2} d e^{2}\right )} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{4 \,{\left (c^{3} d^{4} + 2 \, a c^{2} d^{2} e^{2} + a^{2} c e^{4}\right )} \sqrt{a c}} - \frac{2 \, a c d^{2} x^{4} e - a c d^{3} x^{2} + a^{2} x^{4} e^{3} - a^{2} d x^{2} e^{2} + a^{2} d^{2} e}{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^9/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

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