3.280 \(\int \frac {x^6 (a+b x^2+c x^4)}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=168 \[ -\frac {d x \left (4 c d^2-e (3 b d-2 a e)\right )}{e^5}+\frac {x^3 \left (3 c d^2-e (2 b d-a e)\right )}{3 e^4}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{2 e^5 \left (d+e x^2\right )}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (9 c d^2-e (7 b d-5 a e)\right )}{2 e^{11/2}}-\frac {x^5 (2 c d-b e)}{5 e^3}+\frac {c x^7}{7 e^2} \]

[Out]

-d*(4*c*d^2-e*(-2*a*e+3*b*d))*x/e^5+1/3*(3*c*d^2-e*(-a*e+2*b*d))*x^3/e^4-1/5*(-b*e+2*c*d)*x^5/e^3+1/7*c*x^7/e^
2-1/2*d^2*(a*e^2-b*d*e+c*d^2)*x/e^5/(e*x^2+d)+1/2*d^(3/2)*(9*c*d^2-e*(-5*a*e+7*b*d))*arctan(x*e^(1/2)/d^(1/2))
/e^(11/2)

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Rubi [A]  time = 0.23, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1257, 1810, 205} \[ \frac {x^3 \left (3 c d^2-e (2 b d-a e)\right )}{3 e^4}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{2 e^5 \left (d+e x^2\right )}-\frac {d x \left (4 c d^2-e (3 b d-2 a e)\right )}{e^5}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (9 c d^2-e (7 b d-5 a e)\right )}{2 e^{11/2}}-\frac {x^5 (2 c d-b e)}{5 e^3}+\frac {c x^7}{7 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*x^2 + c*x^4))/(d + e*x^2)^2,x]

[Out]

-((d*(4*c*d^2 - e*(3*b*d - 2*a*e))*x)/e^5) + ((3*c*d^2 - e*(2*b*d - a*e))*x^3)/(3*e^4) - ((2*c*d - b*e)*x^5)/(
5*e^3) + (c*x^7)/(7*e^2) - (d^2*(c*d^2 - b*d*e + a*e^2)*x)/(2*e^5*(d + e*x^2)) + (d^(3/2)*(9*c*d^2 - e*(7*b*d
- 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*e^(11/2))

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 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^2} \, dx &=-\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{2 e^5 \left (d+e x^2\right )}-\frac {\int \frac {-d^2 \left (c d^2-b d e+a e^2\right )+2 d e \left (c d^2-b d e+a e^2\right ) x^2-2 e^2 \left (c d^2-b d e+a e^2\right ) x^4+2 e^3 (c d-b e) x^6-2 c e^4 x^8}{d+e x^2} \, dx}{2 e^5}\\ &=-\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{2 e^5 \left (d+e x^2\right )}-\frac {\int \left (2 d \left (4 c d^2-e (3 b d-2 a e)\right )-2 e \left (3 c d^2-e (2 b d-a e)\right ) x^2+2 e^2 (2 c d-b e) x^4-2 c e^3 x^6+\frac {-9 c d^4+7 b d^3 e-5 a d^2 e^2}{d+e x^2}\right ) \, dx}{2 e^5}\\ &=-\frac {d \left (4 c d^2-e (3 b d-2 a e)\right ) x}{e^5}+\frac {\left (3 c d^2-e (2 b d-a e)\right ) x^3}{3 e^4}-\frac {(2 c d-b e) x^5}{5 e^3}+\frac {c x^7}{7 e^2}-\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{2 e^5 \left (d+e x^2\right )}+\frac {\left (d^2 \left (9 c d^2-e (7 b d-5 a e)\right )\right ) \int \frac {1}{d+e x^2} \, dx}{2 e^5}\\ &=-\frac {d \left (4 c d^2-e (3 b d-2 a e)\right ) x}{e^5}+\frac {\left (3 c d^2-e (2 b d-a e)\right ) x^3}{3 e^4}-\frac {(2 c d-b e) x^5}{5 e^3}+\frac {c x^7}{7 e^2}-\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{2 e^5 \left (d+e x^2\right )}+\frac {d^{3/2} \left (9 c d^2-e (7 b d-5 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 165, normalized size = 0.98 \[ -\frac {d x \left (2 a e^2-3 b d e+4 c d^2\right )}{e^5}+\frac {x^3 \left (a e^2-2 b d e+3 c d^2\right )}{3 e^4}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a e^2-7 b d e+9 c d^2\right )}{2 e^{11/2}}-\frac {x \left (a d^2 e^2-b d^3 e+c d^4\right )}{2 e^5 \left (d+e x^2\right )}+\frac {x^5 (b e-2 c d)}{5 e^3}+\frac {c x^7}{7 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*x^2 + c*x^4))/(d + e*x^2)^2,x]

[Out]

-((d*(4*c*d^2 - 3*b*d*e + 2*a*e^2)*x)/e^5) + ((3*c*d^2 - 2*b*d*e + a*e^2)*x^3)/(3*e^4) + ((-2*c*d + b*e)*x^5)/
(5*e^3) + (c*x^7)/(7*e^2) - ((c*d^4 - b*d^3*e + a*d^2*e^2)*x)/(2*e^5*(d + e*x^2)) + (d^(3/2)*(9*c*d^2 - 7*b*d*
e + 5*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*e^(11/2))

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fricas [A]  time = 0.94, size = 426, normalized size = 2.54 \[ \left [\frac {60 \, c e^{4} x^{9} - 12 \, {\left (9 \, c d e^{3} - 7 \, b e^{4}\right )} x^{7} + 28 \, {\left (9 \, c d^{2} e^{2} - 7 \, b d e^{3} + 5 \, a e^{4}\right )} x^{5} - 140 \, {\left (9 \, c d^{3} e - 7 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{3} + 105 \, {\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2} + {\left (9 \, c d^{3} e - 7 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) - 210 \, {\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2}\right )} x}{420 \, {\left (e^{6} x^{2} + d e^{5}\right )}}, \frac {30 \, c e^{4} x^{9} - 6 \, {\left (9 \, c d e^{3} - 7 \, b e^{4}\right )} x^{7} + 14 \, {\left (9 \, c d^{2} e^{2} - 7 \, b d e^{3} + 5 \, a e^{4}\right )} x^{5} - 70 \, {\left (9 \, c d^{3} e - 7 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{3} + 105 \, {\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2} + {\left (9 \, c d^{3} e - 7 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{2}\right )} \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) - 105 \, {\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2}\right )} x}{210 \, {\left (e^{6} x^{2} + d e^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

[1/420*(60*c*e^4*x^9 - 12*(9*c*d*e^3 - 7*b*e^4)*x^7 + 28*(9*c*d^2*e^2 - 7*b*d*e^3 + 5*a*e^4)*x^5 - 140*(9*c*d^
3*e - 7*b*d^2*e^2 + 5*a*d*e^3)*x^3 + 105*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2 + (9*c*d^3*e - 7*b*d^2*e^2 + 5*a*d
*e^3)*x^2)*sqrt(-d/e)*log((e*x^2 + 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) - 210*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2
)*x)/(e^6*x^2 + d*e^5), 1/210*(30*c*e^4*x^9 - 6*(9*c*d*e^3 - 7*b*e^4)*x^7 + 14*(9*c*d^2*e^2 - 7*b*d*e^3 + 5*a*
e^4)*x^5 - 70*(9*c*d^3*e - 7*b*d^2*e^2 + 5*a*d*e^3)*x^3 + 105*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2 + (9*c*d^3*e
- 7*b*d^2*e^2 + 5*a*d*e^3)*x^2)*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) - 105*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2)*x)
/(e^6*x^2 + d*e^5)]

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giac [A]  time = 0.32, size = 160, normalized size = 0.95 \[ \frac {{\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {11}{2}\right )}}{2 \, \sqrt {d}} + \frac {1}{105} \, {\left (15 \, c x^{7} e^{12} - 42 \, c d x^{5} e^{11} + 21 \, b x^{5} e^{12} + 105 \, c d^{2} x^{3} e^{10} - 70 \, b d x^{3} e^{11} - 420 \, c d^{3} x e^{9} + 35 \, a x^{3} e^{12} + 315 \, b d^{2} x e^{10} - 210 \, a d x e^{11}\right )} e^{\left (-14\right )} - \frac {{\left (c d^{4} x - b d^{3} x e + a d^{2} x e^{2}\right )} e^{\left (-5\right )}}{2 \, {\left (x^{2} e + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x, algorithm="giac")

[Out]

1/2*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-11/2)/sqrt(d) + 1/105*(15*c*x^7*e^12 - 4
2*c*d*x^5*e^11 + 21*b*x^5*e^12 + 105*c*d^2*x^3*e^10 - 70*b*d*x^3*e^11 - 420*c*d^3*x*e^9 + 35*a*x^3*e^12 + 315*
b*d^2*x*e^10 - 210*a*d*x*e^11)*e^(-14) - 1/2*(c*d^4*x - b*d^3*x*e + a*d^2*x*e^2)*e^(-5)/(x^2*e + d)

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maple [A]  time = 0.01, size = 214, normalized size = 1.27 \[ \frac {c \,x^{7}}{7 e^{2}}+\frac {b \,x^{5}}{5 e^{2}}-\frac {2 c d \,x^{5}}{5 e^{3}}+\frac {a \,x^{3}}{3 e^{2}}-\frac {2 b d \,x^{3}}{3 e^{3}}+\frac {c \,d^{2} x^{3}}{e^{4}}-\frac {a \,d^{2} x}{2 \left (e \,x^{2}+d \right ) e^{3}}+\frac {5 a \,d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{3}}+\frac {b \,d^{3} x}{2 \left (e \,x^{2}+d \right ) e^{4}}-\frac {7 b \,d^{3} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{4}}-\frac {c \,d^{4} x}{2 \left (e \,x^{2}+d \right ) e^{5}}+\frac {9 c \,d^{4} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{5}}-\frac {2 a d x}{e^{3}}+\frac {3 b \,d^{2} x}{e^{4}}-\frac {4 c \,d^{3} x}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x)

[Out]

1/7*c*x^7/e^2+1/5/e^2*x^5*b-2/5/e^3*x^5*c*d+1/3/e^2*x^3*a-2/3/e^3*x^3*b*d+1/e^4*x^3*c*d^2-2/e^3*a*d*x+3/e^4*d^
2*b*x-4/e^5*c*d^3*x-1/2*d^2/e^3*x/(e*x^2+d)*a+1/2*d^3/e^4*x/(e*x^2+d)*b-1/2*d^4/e^5*x/(e*x^2+d)*c+5/2*d^2/e^3/
(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a-7/2*d^3/e^4/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*b+9/2*d^4/e^5/(d*e)^
(1/2)*arctan(1/(d*e)^(1/2)*e*x)*c

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maxima [A]  time = 2.49, size = 165, normalized size = 0.98 \[ -\frac {{\left (c d^{4} - b d^{3} e + a d^{2} e^{2}\right )} x}{2 \, {\left (e^{6} x^{2} + d e^{5}\right )}} + \frac {{\left (9 \, c d^{4} - 7 \, b d^{3} e + 5 \, a d^{2} e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} e^{5}} + \frac {15 \, c e^{3} x^{7} - 21 \, {\left (2 \, c d e^{2} - b e^{3}\right )} x^{5} + 35 \, {\left (3 \, c d^{2} e - 2 \, b d e^{2} + a e^{3}\right )} x^{3} - 105 \, {\left (4 \, c d^{3} - 3 \, b d^{2} e + 2 \, a d e^{2}\right )} x}{105 \, e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/2*(c*d^4 - b*d^3*e + a*d^2*e^2)*x/(e^6*x^2 + d*e^5) + 1/2*(9*c*d^4 - 7*b*d^3*e + 5*a*d^2*e^2)*arctan(e*x/sq
rt(d*e))/(sqrt(d*e)*e^5) + 1/105*(15*c*e^3*x^7 - 21*(2*c*d*e^2 - b*e^3)*x^5 + 35*(3*c*d^2*e - 2*b*d*e^2 + a*e^
3)*x^3 - 105*(4*c*d^3 - 3*b*d^2*e + 2*a*d*e^2)*x)/e^5

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mupad [B]  time = 0.33, size = 251, normalized size = 1.49 \[ x^5\,\left (\frac {b}{5\,e^2}-\frac {2\,c\,d}{5\,e^3}\right )-x^3\,\left (\frac {c\,d^2}{3\,e^4}-\frac {a}{3\,e^2}+\frac {2\,d\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )}{3\,e}\right )+x\,\left (\frac {2\,d\,\left (\frac {c\,d^2}{e^4}-\frac {a}{e^2}+\frac {2\,d\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )}{e}\right )}{e}-\frac {d^2\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )}{e^2}\right )-\frac {x\,\left (\frac {c\,d^4}{2}-\frac {b\,d^3\,e}{2}+\frac {a\,d^2\,e^2}{2}\right )}{e^6\,x^2+d\,e^5}+\frac {c\,x^7}{7\,e^2}+\frac {d^{3/2}\,\mathrm {atan}\left (\frac {d^{3/2}\,\sqrt {e}\,x\,\left (9\,c\,d^2-7\,b\,d\,e+5\,a\,e^2\right )}{9\,c\,d^4-7\,b\,d^3\,e+5\,a\,d^2\,e^2}\right )\,\left (9\,c\,d^2-7\,b\,d\,e+5\,a\,e^2\right )}{2\,e^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(a + b*x^2 + c*x^4))/(d + e*x^2)^2,x)

[Out]

x^5*(b/(5*e^2) - (2*c*d)/(5*e^3)) - x^3*((c*d^2)/(3*e^4) - a/(3*e^2) + (2*d*(b/e^2 - (2*c*d)/e^3))/(3*e)) + x*
((2*d*((c*d^2)/e^4 - a/e^2 + (2*d*(b/e^2 - (2*c*d)/e^3))/e))/e - (d^2*(b/e^2 - (2*c*d)/e^3))/e^2) - (x*((c*d^4
)/2 + (a*d^2*e^2)/2 - (b*d^3*e)/2))/(d*e^5 + e^6*x^2) + (c*x^7)/(7*e^2) + (d^(3/2)*atan((d^(3/2)*e^(1/2)*x*(5*
a*e^2 + 9*c*d^2 - 7*b*d*e))/(9*c*d^4 + 5*a*d^2*e^2 - 7*b*d^3*e))*(5*a*e^2 + 9*c*d^2 - 7*b*d*e))/(2*e^(11/2))

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sympy [B]  time = 1.18, size = 320, normalized size = 1.90 \[ \frac {c x^{7}}{7 e^{2}} + x^{5} \left (\frac {b}{5 e^{2}} - \frac {2 c d}{5 e^{3}}\right ) + x^{3} \left (\frac {a}{3 e^{2}} - \frac {2 b d}{3 e^{3}} + \frac {c d^{2}}{e^{4}}\right ) + x \left (- \frac {2 a d}{e^{3}} + \frac {3 b d^{2}}{e^{4}} - \frac {4 c d^{3}}{e^{5}}\right ) + \frac {x \left (- a d^{2} e^{2} + b d^{3} e - c d^{4}\right )}{2 d e^{5} + 2 e^{6} x^{2}} - \frac {\sqrt {- \frac {d^{3}}{e^{11}}} \left (5 a e^{2} - 7 b d e + 9 c d^{2}\right ) \log {\left (- \frac {e^{5} \sqrt {- \frac {d^{3}}{e^{11}}} \left (5 a e^{2} - 7 b d e + 9 c d^{2}\right )}{5 a d e^{2} - 7 b d^{2} e + 9 c d^{3}} + x \right )}}{4} + \frac {\sqrt {- \frac {d^{3}}{e^{11}}} \left (5 a e^{2} - 7 b d e + 9 c d^{2}\right ) \log {\left (\frac {e^{5} \sqrt {- \frac {d^{3}}{e^{11}}} \left (5 a e^{2} - 7 b d e + 9 c d^{2}\right )}{5 a d e^{2} - 7 b d^{2} e + 9 c d^{3}} + x \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(c*x**4+b*x**2+a)/(e*x**2+d)**2,x)

[Out]

c*x**7/(7*e**2) + x**5*(b/(5*e**2) - 2*c*d/(5*e**3)) + x**3*(a/(3*e**2) - 2*b*d/(3*e**3) + c*d**2/e**4) + x*(-
2*a*d/e**3 + 3*b*d**2/e**4 - 4*c*d**3/e**5) + x*(-a*d**2*e**2 + b*d**3*e - c*d**4)/(2*d*e**5 + 2*e**6*x**2) -
sqrt(-d**3/e**11)*(5*a*e**2 - 7*b*d*e + 9*c*d**2)*log(-e**5*sqrt(-d**3/e**11)*(5*a*e**2 - 7*b*d*e + 9*c*d**2)/
(5*a*d*e**2 - 7*b*d**2*e + 9*c*d**3) + x)/4 + sqrt(-d**3/e**11)*(5*a*e**2 - 7*b*d*e + 9*c*d**2)*log(e**5*sqrt(
-d**3/e**11)*(5*a*e**2 - 7*b*d*e + 9*c*d**2)/(5*a*d*e**2 - 7*b*d**2*e + 9*c*d**3) + x)/4

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