3.3.44 \(\int \frac {x^9}{(d+e x^2) (a+c x^4)^2} \, dx\) [244]

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

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 = {1266, 1661, 1643, 649, 211, 266} \begin {gather*} -\frac {\sqrt {a} d \text {ArcTan}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right ) \left (a e^2+3 c d^2\right )}{4 c^{3/2} \left (a e^2+c d^2\right )^2}+\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 {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 {d^4 \log \left (d+e x^2\right )}{2 e \left (a e^2+c d^2\right )^2} \end {gather*}

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 211

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

Rule 266

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]

Rule 649

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 1266

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 1643

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 1661

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]

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} \text {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 {\text {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 {\text {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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.14, size = 135, normalized size = 0.80 \begin {gather*} \frac {\frac {a \left (c d^2+a e^2\right ) \left (a e+c d x^2\right )}{c^2 \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 )}{c^{3/2}}+\frac {2 d^4 \log \left (d+e x^2\right )}{e}+\frac {a e \left (2 c d^2+a e^2\right ) \log \left (a+c x^4\right )}{c^2}}{4 \left (c d^2+a e^2\right )^2} \end {gather*}

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 = 0.23, size = 160, normalized size = 0.95

method	result	size
default	\(\frac {d^{4} \ln \left (e \,x^{2}+d \right )}{2 e \left (a \,e^{2}+c \,d^{2}\right )^{2}}-\frac {a \left (\frac {-\frac {d \left (a \,e^{2}+c \,d^{2}\right ) x^{2}}{2 c}-\frac {a e \left (a \,e^{2}+c \,d^{2}\right )}{2 c^{2}}}{c \,x^{4}+a}+\frac {\frac {\left (-2 a \,e^{3}-4 c \,d^{2} e \right ) \ln \left (c \,x^{4}+a \right )}{2 c}+\frac {\left (d \,e^{2} a +3 c \,d^{3}\right ) \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 c}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2}}\)	\(160\)
risch	\(\text {Expression too large to display}\)	\(2233\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 214, normalized size = 1.27 \begin {gather*} \frac {d^{4} \log \left (x^{2} e + d\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 {a c d x^{2} + a^{2} e}{4 \, {\left (a c^{3} d^{2} + {\left (c^{4} d^{2} + a c^{3} e^{2}\right )} x^{4} + a^{2} c^{2} e^{2}\right )}} \end {gather*}

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]

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

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Fricas [A]
time = 9.53, size = 548, normalized size = 3.24 \begin {gather*} \left [\frac {2 \, a c^{2} d^{3} x^{2} e + 2 \, a^{2} c d x^{2} e^{3} + 2 \, a^{2} c d^{2} e^{2} + 2 \, a^{3} e^{4} + {\left ({\left (a c^{2} d x^{4} + a^{2} c d\right )} e^{3} + 3 \, {\left (c^{3} d^{3} x^{4} + a c^{2} d^{3}\right )} e\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 ({\left (a^{2} c x^{4} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{4} + a^{2} c d^{2}\right )} e^{2}\right )} \log \left (c x^{4} + a\right ) + 4 \, {\left (c^{3} d^{4} x^{4} + a c^{2} d^{4}\right )} \log \left (x^{2} e + d\right )}{8 \, {\left ({\left (a^{2} c^{3} x^{4} + a^{3} c^{2}\right )} e^{5} + 2 \, {\left (a c^{4} d^{2} x^{4} + a^{2} c^{3} d^{2}\right )} e^{3} + {\left (c^{5} d^{4} x^{4} + a c^{4} d^{4}\right )} e\right )}}, \frac {a c^{2} d^{3} x^{2} e + a^{2} c d x^{2} e^{3} + a^{2} c d^{2} e^{2} + a^{3} e^{4} - {\left ({\left (a c^{2} d x^{4} + a^{2} c d\right )} e^{3} + 3 \, {\left (c^{3} d^{3} x^{4} + a c^{2} d^{3}\right )} e\right )} \sqrt {\frac {a}{c}} \arctan \left (\frac {c x^{2} \sqrt {\frac {a}{c}}}{a}\right ) + {\left ({\left (a^{2} c x^{4} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{4} + a^{2} c d^{2}\right )} e^{2}\right )} \log \left (c x^{4} + a\right ) + 2 \, {\left (c^{3} d^{4} x^{4} + a c^{2} d^{4}\right )} \log \left (x^{2} e + d\right )}{4 \, {\left ({\left (a^{2} c^{3} x^{4} + a^{3} c^{2}\right )} e^{5} + 2 \, {\left (a c^{4} d^{2} x^{4} + a^{2} c^{3} d^{2}\right )} e^{3} + {\left (c^{5} d^{4} x^{4} + a c^{4} d^{4}\right )} e\right )}}\right ] \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 3.62, size = 251, normalized size = 1.49 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 1.30, size = 305, normalized size = 1.80 \begin {gather*} \frac {\frac {a^2\,e}{4\,c^2\,\left (c\,d^2+a\,e^2\right )}+\frac {a\,d\,x^2}{4\,c\,\left (c\,d^2+a\,e^2\right )}}{c\,x^4+a}-\frac {\ln \left (\sqrt {-a\,c^5}+c^3\,x^2\right )\,\left (3\,c\,d^3\,\sqrt {-a\,c^5}-2\,a^2\,c^2\,e^3-4\,a\,c^3\,d^2\,e+a\,d\,e^2\,\sqrt {-a\,c^5}\right )}{8\,\left (a^2\,c^4\,e^4+2\,a\,c^5\,d^2\,e^2+c^6\,d^4\right )}+\frac {\ln \left (\sqrt {-a\,c^5}-c^3\,x^2\right )\,\left (3\,c\,d^3\,\sqrt {-a\,c^5}+2\,a^2\,c^2\,e^3+4\,a\,c^3\,d^2\,e+a\,d\,e^2\,\sqrt {-a\,c^5}\right )}{8\,\left (a^2\,c^4\,e^4+2\,a\,c^5\,d^2\,e^2+c^6\,d^4\right )}+\frac {d^4\,\ln \left (e\,x^2+d\right )}{2\,a^2\,e^5+4\,a\,c\,d^2\,e^3+2\,c^2\,d^4\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*e)/(4*c^2*(a*e^2 + c*d^2)) + (a*d*x^2)/(4*c*(a*e^2 + c*d^2)))/(a + c*x^4) - (log((-a*c^5)^(1/2) + c^3*x^
2)*(3*c*d^3*(-a*c^5)^(1/2) - 2*a^2*c^2*e^3 - 4*a*c^3*d^2*e + a*d*e^2*(-a*c^5)^(1/2)))/(8*(c^6*d^4 + a^2*c^4*e^
4 + 2*a*c^5*d^2*e^2)) + (log((-a*c^5)^(1/2) - c^3*x^2)*(3*c*d^3*(-a*c^5)^(1/2) + 2*a^2*c^2*e^3 + 4*a*c^3*d^2*e
+ a*d*e^2*(-a*c^5)^(1/2)))/(8*(c^6*d^4 + a^2*c^4*e^4 + 2*a*c^5*d^2*e^2)) + (d^4*log(d + e*x^2))/(2*a^2*e^5 +
2*c^2*d^4*e + 4*a*c*d^2*e^3)

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