3.3.29 \(\int \frac {x^9}{(d+e x^2) (a+c x^4)} \, dx\) [229]

Optimal. Leaf size=134 \[ -\frac {d x^2}{2 c e^2}+\frac {x^4}{4 c e}+\frac {a^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{3/2} \left (c d^2+a e^2\right )}+\frac {d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2+a e^2\right )}-\frac {a^2 e \log \left (a+c x^4\right )}{4 c^2 \left (c d^2+a e^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1266, 1643, 649, 211, 266} \begin {gather*} \frac {a^{3/2} d \text {ArcTan}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{3/2} \left (a e^2+c d^2\right )}-\frac {a^2 e \log \left (a+c x^4\right )}{4 c^2 \left (a e^2+c d^2\right )}+\frac {d^4 \log \left (d+e x^2\right )}{2 e^3 \left (a e^2+c d^2\right )}-\frac {d x^2}{2 c e^2}+\frac {x^4}{4 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/((d + e*x^2)*(a + c*x^4)),x]

[Out]

-1/2*(d*x^2)/(c*e^2) + x^4/(4*c*e) + (a^(3/2)*d*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*c^(3/2)*(c*d^2 + a*e^2)) + (
d^4*Log[d + e*x^2])/(2*e^3*(c*d^2 + a*e^2)) - (a^2*e*Log[a + c*x^4])/(4*c^2*(c*d^2 + a*e^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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 134, normalized size = 1.00 \begin {gather*} -\frac {d x^2}{2 c e^2}+\frac {x^4}{4 c e}+\frac {a^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{3/2} \left (c d^2+a e^2\right )}+\frac {d^4 \log \left (d+e x^2\right )}{2 e^3 \left (c d^2+a e^2\right )}-\frac {a^2 e \log \left (a+c x^4\right )}{4 c^2 \left (c d^2+a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d*x^2)/(c*e^2) + x^4/(4*c*e) + (a^(3/2)*d*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*c^(3/2)*(c*d^2 + a*e^2)) + (
d^4*Log[d + e*x^2])/(2*e^3*(c*d^2 + a*e^2)) - (a^2*e*Log[a + c*x^4])/(4*c^2*(c*d^2 + a*e^2))

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Maple [A]
time = 0.18, size = 103, normalized size = 0.77

method	result	size
default	\(\frac {\left (-e \,x^{2}+d \right )^{2}}{4 c \,e^{3}}+\frac {d^{4} \ln \left (e \,x^{2}+d \right )}{2 e^{3} \left (a \,e^{2}+c \,d^{2}\right )}+\frac {a^{2} \left (-\frac {e \ln \left (c \,x^{4}+a \right )}{2 c}+\frac {d \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) c}\)	\(103\)
risch	\(\frac {x^{4}}{4 c e}-\frac {d \,x^{2}}{2 c \,e^{2}}+\frac {d^{2}}{4 c \,e^{3}}+\frac {d^{4} \ln \left (e \,x^{2}+d \right )}{2 e^{3} \left (a \,e^{2}+c \,d^{2}\right )}+\frac {\munderset {\textit {\_R} =\RootOf \left (\left (a \,c^{2} e^{2}+c^{3} d^{2}\right ) \textit {\_Z}^{2}+2 a^{2} c \,e^{3} \textit {\_Z} +e^{4} a^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (2 a \,c^{2} e^{4}-2 c^{3} d^{2} e^{2}\right ) \textit {\_R}^{2}+\left (4 a^{2} c \,e^{5}-5 a \,c^{2} d^{2} e^{3}+4 c^{3} d^{4} e \right ) \textit {\_R} +2 e^{6} a^{3}-2 a^{2} c \,d^{2} e^{4}+2 a \,c^{2} d^{4} e^{2}\right ) x^{2}+\left (3 a \,c^{2} d \,e^{3}-c^{3} d^{3} e \right ) \textit {\_R}^{2}+\left (5 a^{2} c d \,e^{4}-4 a \,c^{2} d^{3} e^{2}+2 c^{3} d^{5}\right ) \textit {\_R} +2 a^{3} d \,e^{5}-2 a^{2} c \,d^{3} e^{3}\right )}{4 e^{2} c}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 118, normalized size = 0.88 \begin {gather*} \frac {d^{4} \log \left (x^{2} e + d\right )}{2 \, {\left (c d^{2} e^{3} + a e^{5}\right )}} - \frac {a^{2} e \log \left (c x^{4} + a\right )}{4 \, {\left (c^{3} d^{2} + a c^{2} e^{2}\right )}} + \frac {a^{2} d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c^{2} d^{2} + a c e^{2}\right )} \sqrt {a c}} + \frac {{\left (x^{4} e - 2 \, d x^{2}\right )} e^{\left (-2\right )}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.27, size = 273, normalized size = 2.04 \begin {gather*} \left [\frac {c^{2} d^{2} x^{4} e^{2} - 2 \, c^{2} d^{3} x^{2} e + a c x^{4} e^{4} + 2 \, c^{2} d^{4} \log \left (x^{2} e + d\right ) - 2 \, a c d x^{2} e^{3} + a c d \sqrt {-\frac {a}{c}} e^{3} \log \left (\frac {c x^{4} + 2 \, c x^{2} \sqrt {-\frac {a}{c}} - a}{c x^{4} + a}\right ) - a^{2} e^{4} \log \left (c x^{4} + a\right )}{4 \, {\left (c^{3} d^{2} e^{3} + a c^{2} e^{5}\right )}}, \frac {c^{2} d^{2} x^{4} e^{2} - 2 \, c^{2} d^{3} x^{2} e + a c x^{4} e^{4} + 2 \, c^{2} d^{4} \log \left (x^{2} e + d\right ) - 2 \, a c d x^{2} e^{3} + 2 \, a c d \sqrt {\frac {a}{c}} \arctan \left (\frac {c x^{2} \sqrt {\frac {a}{c}}}{a}\right ) e^{3} - a^{2} e^{4} \log \left (c x^{4} + a\right )}{4 \, {\left (c^{3} d^{2} e^{3} + a c^{2} e^{5}\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),x, algorithm="fricas")

[Out]

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

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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),x)

[Out]

Timed out

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Giac [A]
time = 5.69, size = 121, normalized size = 0.90 \begin {gather*} \frac {d^{4} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c d^{2} e^{3} + a e^{5}\right )}} - \frac {a^{2} e \log \left (c x^{4} + a\right )}{4 \, {\left (c^{3} d^{2} + a c^{2} e^{2}\right )}} + \frac {a^{2} d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c^{2} d^{2} + a c e^{2}\right )} \sqrt {a c}} + \frac {{\left (c x^{4} e - 2 \, c d x^{2}\right )} e^{\left (-2\right )}}{4 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.87, size = 181, normalized size = 1.35 \begin {gather*} \frac {\ln \left (\sqrt {-a^3\,c^5}+a\,c^3\,x^2\right )\,\left (d\,\sqrt {-a^3\,c^5}-a^2\,c^2\,e\right )}{4\,c^5\,d^2+4\,a\,c^4\,e^2}-\frac {\ln \left (\sqrt {-a^3\,c^5}-a\,c^3\,x^2\right )\,\left (d\,\sqrt {-a^3\,c^5}+a^2\,c^2\,e\right )}{4\,\left (c^5\,d^2+a\,c^4\,e^2\right )}+\frac {d^4\,\ln \left (e\,x^2+d\right )}{2\,c\,d^2\,e^3+2\,a\,e^5}+\frac {x^4}{4\,c\,e}-\frac {d\,x^2}{2\,c\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/((a + c*x^4)*(d + e*x^2)),x)

[Out]

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

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