3.288 \(\int \frac{x^6 (a+b x^2+c x^4)}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=173 \[ \frac{d x \left (17 c d^2-e (13 b d-9 a e)\right )}{8 e^5 \left (d+e x^2\right )}-\frac{d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}+\frac{x \left (6 c d^2-e (3 b d-a e)\right )}{e^5}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (15 a e^2-35 b d e+63 c d^2\right )}{8 e^{11/2}}-\frac{x^3 (3 c d-b e)}{3 e^4}+\frac{c x^5}{5 e^3} \]

[Out]

((6*c*d^2 - e*(3*b*d - a*e))*x)/e^5 - ((3*c*d - b*e)*x^3)/(3*e^4) + (c*x^5)/(5*e^3) - (d^2*(c*d^2 - b*d*e + a*
e^2)*x)/(4*e^5*(d + e*x^2)^2) + (d*(17*c*d^2 - e*(13*b*d - 9*a*e))*x)/(8*e^5*(d + e*x^2)) - (Sqrt[d]*(63*c*d^2
- 35*b*d*e + 15*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*e^(11/2))

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Rubi [A]  time = 0.321216, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1257, 1814, 1810, 205} \[ \frac{d x \left (17 c d^2-e (13 b d-9 a e)\right )}{8 e^5 \left (d+e x^2\right )}-\frac{d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}+\frac{x \left (6 c d^2-e (3 b d-a e)\right )}{e^5}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (15 a e^2-35 b d e+63 c d^2\right )}{8 e^{11/2}}-\frac{x^3 (3 c d-b e)}{3 e^4}+\frac{c x^5}{5 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*x^2 + c*x^4))/(d + e*x^2)^3,x]

[Out]

((6*c*d^2 - e*(3*b*d - a*e))*x)/e^5 - ((3*c*d - b*e)*x^3)/(3*e^4) + (c*x^5)/(5*e^3) - (d^2*(c*d^2 - b*d*e + a*
e^2)*x)/(4*e^5*(d + e*x^2)^2) + (d*(17*c*d^2 - e*(13*b*d - 9*a*e))*x)/(8*e^5*(d + e*x^2)) - (Sqrt[d]*(63*c*d^2
- 35*b*d*e + 15*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*e^(11/2))

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

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]

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]

Rubi steps

\begin{align*} \int \frac{x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac{d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}-\frac{\int \frac{-d^2 \left (c d^2-b d e+a e^2\right )+4 d e \left (c d^2-b d e+a e^2\right ) x^2-4 e^2 \left (c d^2-b d e+a e^2\right ) x^4+4 e^3 (c d-b e) x^6-4 c e^4 x^8}{\left (d+e x^2\right )^2} \, dx}{4 e^5}\\ &=-\frac{d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac{d \left (17 c d^2-e (13 b d-9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}+\frac{\int \frac{-d^2 \left (15 c d^2-e (11 b d-7 a e)\right )+8 d e \left (3 c d^2-e (2 b d-a e)\right ) x^2-8 d e^2 (2 c d-b e) x^4+8 c d e^3 x^6}{d+e x^2} \, dx}{8 d e^5}\\ &=-\frac{d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac{d \left (17 c d^2-e (13 b d-9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}+\frac{\int \left (8 d \left (6 c d^2-e (3 b d-a e)\right )-8 d e (3 c d-b e) x^2+8 c d e^2 x^4+\frac{-63 c d^4+35 b d^3 e-15 a d^2 e^2}{d+e x^2}\right ) \, dx}{8 d e^5}\\ &=\frac{\left (6 c d^2-e (3 b d-a e)\right ) x}{e^5}-\frac{(3 c d-b e) x^3}{3 e^4}+\frac{c x^5}{5 e^3}-\frac{d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac{d \left (17 c d^2-e (13 b d-9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}+\frac{\left (-63 c d^4+35 b d^3 e-15 a d^2 e^2\right ) \int \frac{1}{d+e x^2} \, dx}{8 d e^5}\\ &=\frac{\left (6 c d^2-e (3 b d-a e)\right ) x}{e^5}-\frac{(3 c d-b e) x^3}{3 e^4}+\frac{c x^5}{5 e^3}-\frac{d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac{d \left (17 c d^2-e (13 b d-9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}-\frac{\sqrt{d} \left (63 c d^2-5 e (7 b d-3 a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 e^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.111269, size = 170, normalized size = 0.98 \[ \frac{x \left (d e (9 a e-13 b d)+17 c d^3\right )}{8 e^5 \left (d+e x^2\right )}-\frac{x \left (d^2 e (a e-b d)+c d^4\right )}{4 e^5 \left (d+e x^2\right )^2}+\frac{x \left (e (a e-3 b d)+6 c d^2\right )}{e^5}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (5 e (3 a e-7 b d)+63 c d^2\right )}{8 e^{11/2}}+\frac{x^3 (b e-3 c d)}{3 e^4}+\frac{c x^5}{5 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*x^2 + c*x^4))/(d + e*x^2)^3,x]

[Out]

((6*c*d^2 + e*(-3*b*d + a*e))*x)/e^5 + ((-3*c*d + b*e)*x^3)/(3*e^4) + (c*x^5)/(5*e^3) - ((c*d^4 + d^2*e*(-(b*d
) + a*e))*x)/(4*e^5*(d + e*x^2)^2) + ((17*c*d^3 + d*e*(-13*b*d + 9*a*e))*x)/(8*e^5*(d + e*x^2)) - (Sqrt[d]*(63
*c*d^2 + 5*e*(-7*b*d + 3*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*e^(11/2))

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Maple [A]  time = 0.011, size = 239, normalized size = 1.4 \begin{align*}{\frac{c{x}^{5}}{5\,{e}^{3}}}+{\frac{{x}^{3}b}{3\,{e}^{3}}}-{\frac{{x}^{3}cd}{{e}^{4}}}+{\frac{ax}{{e}^{3}}}-3\,{\frac{bdx}{{e}^{4}}}+6\,{\frac{c{d}^{2}x}{{e}^{5}}}+{\frac{9\,d{x}^{3}a}{8\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{13\,{d}^{2}{x}^{3}b}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{17\,{d}^{3}{x}^{3}c}{8\,{e}^{4} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{7\,a{d}^{2}x}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{11\,b{d}^{3}x}{8\,{e}^{4} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{15\,c{d}^{4}x}{8\,{e}^{5} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{15\,ad}{8\,{e}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{35\,{d}^{2}b}{8\,{e}^{4}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{63\,c{d}^{3}}{8\,{e}^{5}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

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)^3,x)

[Out]

1/5*c*x^5/e^3+1/3/e^3*x^3*b-1/e^4*x^3*c*d+1/e^3*a*x-3/e^4*d*b*x+6/e^5*c*d^2*x+9/8*d/e^2/(e*x^2+d)^2*x^3*a-13/8
*d^2/e^3/(e*x^2+d)^2*x^3*b+17/8*d^3/e^4/(e*x^2+d)^2*x^3*c+7/8*d^2/e^3/(e*x^2+d)^2*a*x-11/8*d^3/e^4/(e*x^2+d)^2
*b*x+15/8*d^4/e^5/(e*x^2+d)^2*c*x-15/8*d/e^3/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*a+35/8*d^2/e^4/(d*e)^(1/2)*ar
ctan(e*x/(d*e)^(1/2))*b-63/8*d^3/e^5/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*c

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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^6*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86095, size = 1125, normalized size = 6.5 \begin{align*} \left [\frac{48 \, c e^{4} x^{9} - 16 \,{\left (9 \, c d e^{3} - 5 \, b e^{4}\right )} x^{7} + 16 \,{\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{5} + 50 \,{\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{3} + 15 \,{\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2} +{\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{4} + 2 \,{\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, 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 ) + 30 \,{\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2}\right )} x}{240 \,{\left (e^{7} x^{4} + 2 \, d e^{6} x^{2} + d^{2} e^{5}\right )}}, \frac{24 \, c e^{4} x^{9} - 8 \,{\left (9 \, c d e^{3} - 5 \, b e^{4}\right )} x^{7} + 8 \,{\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{5} + 25 \,{\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{3} - 15 \,{\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2} +{\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{4} + 2 \,{\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{2}\right )} \sqrt{\frac{d}{e}} \arctan \left (\frac{e x \sqrt{\frac{d}{e}}}{d}\right ) + 15 \,{\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2}\right )} x}{120 \,{\left (e^{7} x^{4} + 2 \, d e^{6} x^{2} + d^{2} e^{5}\right )}}\right ] \end{align*}

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)^3,x, algorithm="fricas")

[Out]

[1/240*(48*c*e^4*x^9 - 16*(9*c*d*e^3 - 5*b*e^4)*x^7 + 16*(63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e^4)*x^5 + 50*(63*c
*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^3 + 15*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2 + (63*c*d^2*e^2 - 35*b*d*e^
3 + 15*a*e^4)*x^4 + 2*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^2)*sqrt(-d/e)*log((e*x^2 - 2*e*x*sqrt(-d/e) -
d)/(e*x^2 + d)) + 30*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2)*x)/(e^7*x^4 + 2*d*e^6*x^2 + d^2*e^5), 1/120*(24*c
*e^4*x^9 - 8*(9*c*d*e^3 - 5*b*e^4)*x^7 + 8*(63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e^4)*x^5 + 25*(63*c*d^3*e - 35*b*
d^2*e^2 + 15*a*d*e^3)*x^3 - 15*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2 + (63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e^4)*
x^4 + 2*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^2)*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 15*(63*c*d^4 - 35*b*
d^3*e + 15*a*d^2*e^2)*x)/(e^7*x^4 + 2*d*e^6*x^2 + d^2*e^5)]

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Sympy [A]  time = 4.37259, size = 233, normalized size = 1.35 \begin{align*} \frac{c x^{5}}{5 e^{3}} + \frac{\sqrt{- \frac{d}{e^{11}}} \left (15 a e^{2} - 35 b d e + 63 c d^{2}\right ) \log{\left (- e^{5} \sqrt{- \frac{d}{e^{11}}} + x \right )}}{16} - \frac{\sqrt{- \frac{d}{e^{11}}} \left (15 a e^{2} - 35 b d e + 63 c d^{2}\right ) \log{\left (e^{5} \sqrt{- \frac{d}{e^{11}}} + x \right )}}{16} + \frac{x^{3} \left (9 a d e^{3} - 13 b d^{2} e^{2} + 17 c d^{3} e\right ) + x \left (7 a d^{2} e^{2} - 11 b d^{3} e + 15 c d^{4}\right )}{8 d^{2} e^{5} + 16 d e^{6} x^{2} + 8 e^{7} x^{4}} + \frac{x^{3} \left (b e - 3 c d\right )}{3 e^{4}} + \frac{x \left (a e^{2} - 3 b d e + 6 c d^{2}\right )}{e^{5}} \end{align*}

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)**3,x)

[Out]

c*x**5/(5*e**3) + sqrt(-d/e**11)*(15*a*e**2 - 35*b*d*e + 63*c*d**2)*log(-e**5*sqrt(-d/e**11) + x)/16 - sqrt(-d
/e**11)*(15*a*e**2 - 35*b*d*e + 63*c*d**2)*log(e**5*sqrt(-d/e**11) + x)/16 + (x**3*(9*a*d*e**3 - 13*b*d**2*e**
2 + 17*c*d**3*e) + x*(7*a*d**2*e**2 - 11*b*d**3*e + 15*c*d**4))/(8*d**2*e**5 + 16*d*e**6*x**2 + 8*e**7*x**4) +
x**3*(b*e - 3*c*d)/(3*e**4) + x*(a*e**2 - 3*b*d*e + 6*c*d**2)/e**5

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Giac [A]  time = 1.09991, size = 216, normalized size = 1.25 \begin{align*} -\frac{{\left (63 \, c d^{3} - 35 \, b d^{2} e + 15 \, a d e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{11}{2}\right )}}{8 \, \sqrt{d}} + \frac{1}{15} \,{\left (3 \, c x^{5} e^{12} - 15 \, c d x^{3} e^{11} + 5 \, b x^{3} e^{12} + 90 \, c d^{2} x e^{10} - 45 \, b d x e^{11} + 15 \, a x e^{12}\right )} e^{\left (-15\right )} + \frac{{\left (17 \, c d^{3} x^{3} e - 13 \, b d^{2} x^{3} e^{2} + 15 \, c d^{4} x + 9 \, a d x^{3} e^{3} - 11 \, b d^{3} x e + 7 \, a d^{2} x e^{2}\right )} e^{\left (-5\right )}}{8 \,{\left (x^{2} e + d\right )}^{2}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

-1/8*(63*c*d^3 - 35*b*d^2*e + 15*a*d*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-11/2)/sqrt(d) + 1/15*(3*c*x^5*e^12 - 1
5*c*d*x^3*e^11 + 5*b*x^3*e^12 + 90*c*d^2*x*e^10 - 45*b*d*x*e^11 + 15*a*x*e^12)*e^(-15) + 1/8*(17*c*d^3*x^3*e -
13*b*d^2*x^3*e^2 + 15*c*d^4*x + 9*a*d*x^3*e^3 - 11*b*d^3*x*e + 7*a*d^2*x*e^2)*e^(-5)/(x^2*e + d)^2