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

Optimal. Leaf size=143 \[ -\frac{x \left (13 c d^2-e (9 b d-5 a e)\right )}{8 e^4 \left (d+e x^2\right )}+\frac{d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (35 c d^2-3 e (5 b d-a e)\right )}{8 \sqrt{d} e^{9/2}}-\frac{x (3 c d-b e)}{e^4}+\frac{c x^3}{3 e^3} \]

[Out]

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

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Rubi [A]  time = 0.2102, antiderivative size = 143, 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, 1153, 205} \[ -\frac{x \left (13 c d^2-e (9 b d-5 a e)\right )}{8 e^4 \left (d+e x^2\right )}+\frac{d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (35 c d^2-3 e (5 b d-a e)\right )}{8 \sqrt{d} e^{9/2}}-\frac{x (3 c d-b e)}{e^4}+\frac{c x^3}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx &=\frac{d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac{\int \frac{d \left (c d^2-b d e+a e^2\right )-4 e \left (c d^2-b d e+a e^2\right ) x^2+4 e^2 (c d-b e) x^4-4 c e^3 x^6}{\left (d+e x^2\right )^2} \, dx}{4 e^4}\\ &=\frac{d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac{\left (13 c d^2-e (9 b d-5 a e)\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac{\int \frac{d \left (11 c d^2-e (7 b d-3 a e)\right )-8 d e (2 c d-b e) x^2+8 c d e^2 x^4}{d+e x^2} \, dx}{8 d e^4}\\ &=\frac{d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac{\left (13 c d^2-e (9 b d-5 a e)\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac{\int \left (-8 d (3 c d-b e)+8 c d e x^2+\frac{35 c d^3-15 b d^2 e+3 a d e^2}{d+e x^2}\right ) \, dx}{8 d e^4}\\ &=-\frac{(3 c d-b e) x}{e^4}+\frac{c x^3}{3 e^3}+\frac{d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac{\left (13 c d^2-e (9 b d-5 a e)\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac{\left (35 c d^2-3 e (5 b d-a e)\right ) \int \frac{1}{d+e x^2} \, dx}{8 e^4}\\ &=-\frac{(3 c d-b e) x}{e^4}+\frac{c x^3}{3 e^3}+\frac{d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac{\left (13 c d^2-e (9 b d-5 a e)\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac{\left (35 c d^2-3 e (5 b d-a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 \sqrt{d} e^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0873107, size = 141, normalized size = 0.99 \[ -\frac{x \left (5 a e^2-9 b d e+13 c d^2\right )}{8 e^4 \left (d+e x^2\right )}+\frac{x \left (a d e^2-b d^2 e+c d^3\right )}{4 e^4 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (3 a e^2-15 b d e+35 c d^2\right )}{8 \sqrt{d} e^{9/2}}+\frac{x (b e-3 c d)}{e^4}+\frac{c x^3}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 202, normalized size = 1.4 \begin{align*}{\frac{c{x}^{3}}{3\,{e}^{3}}}+{\frac{bx}{{e}^{3}}}-3\,{\frac{cdx}{{e}^{4}}}-{\frac{5\,{x}^{3}a}{8\,e \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{9\,{x}^{3}bd}{8\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{13\,{x}^{3}c{d}^{2}}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{3\,adx}{8\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{7\,{d}^{2}bx}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{11\,c{d}^{3}x}{8\,{e}^{4} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{3\,a}{8\,{e}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{15\,bd}{8\,{e}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{35\,c{d}^{2}}{8\,{e}^{4}}\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^4*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x)

[Out]

1/3*c*x^3/e^3+1/e^3*b*x-3/e^4*c*d*x-5/8/e/(e*x^2+d)^2*x^3*a+9/8/e^2/(e*x^2+d)^2*x^3*b*d-13/8/e^3/(e*x^2+d)^2*x
^3*c*d^2-3/8/e^2/(e*x^2+d)^2*d*a*x+7/8/e^3/(e*x^2+d)^2*d^2*b*x-11/8/e^4/(e*x^2+d)^2*c*d^3*x+3/8/e^2/(d*e)^(1/2
)*arctan(e*x/(d*e)^(1/2))*a-15/8/e^3/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*d*b+35/8/e^4/(d*e)^(1/2)*arctan(e*x/(
d*e)^(1/2))*c*d^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^4*(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 = 1.85582, size = 1010, normalized size = 7.06 \begin{align*} \left [\frac{16 \, c d e^{4} x^{7} - 16 \,{\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 10 \,{\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} - 3 \,{\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} +{\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \,{\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt{-d e} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) - 6 \,{\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{48 \,{\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac{8 \, c d e^{4} x^{7} - 8 \,{\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 5 \,{\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} + 3 \,{\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} +{\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \,{\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) - 3 \,{\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{24 \,{\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(16*c*d*e^4*x^7 - 16*(7*c*d^2*e^3 - 3*b*d*e^4)*x^5 - 10*(35*c*d^3*e^2 - 15*b*d^2*e^3 + 3*a*d*e^4)*x^3 -
3*(35*c*d^4 - 15*b*d^3*e + 3*a*d^2*e^2 + (35*c*d^2*e^2 - 15*b*d*e^3 + 3*a*e^4)*x^4 + 2*(35*c*d^3*e - 15*b*d^2*
e^2 + 3*a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(35*c*d^4*e - 15*b*d^3*e^2
+ 3*a*d^2*e^3)*x)/(d*e^7*x^4 + 2*d^2*e^6*x^2 + d^3*e^5), 1/24*(8*c*d*e^4*x^7 - 8*(7*c*d^2*e^3 - 3*b*d*e^4)*x^5
- 5*(35*c*d^3*e^2 - 15*b*d^2*e^3 + 3*a*d*e^4)*x^3 + 3*(35*c*d^4 - 15*b*d^3*e + 3*a*d^2*e^2 + (35*c*d^2*e^2 -
15*b*d*e^3 + 3*a*e^4)*x^4 + 2*(35*c*d^3*e - 15*b*d^2*e^2 + 3*a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3
*(35*c*d^4*e - 15*b*d^3*e^2 + 3*a*d^2*e^3)*x)/(d*e^7*x^4 + 2*d^2*e^6*x^2 + d^3*e^5)]

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Sympy [A]  time = 3.93966, size = 211, normalized size = 1.48 \begin{align*} \frac{c x^{3}}{3 e^{3}} - \frac{\sqrt{- \frac{1}{d e^{9}}} \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log{\left (- d e^{4} \sqrt{- \frac{1}{d e^{9}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{d e^{9}}} \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log{\left (d e^{4} \sqrt{- \frac{1}{d e^{9}}} + x \right )}}{16} - \frac{x^{3} \left (5 a e^{3} - 9 b d e^{2} + 13 c d^{2} e\right ) + x \left (3 a d e^{2} - 7 b d^{2} e + 11 c d^{3}\right )}{8 d^{2} e^{4} + 16 d e^{5} x^{2} + 8 e^{6} x^{4}} + \frac{x \left (b e - 3 c d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(c*x**4+b*x**2+a)/(e*x**2+d)**3,x)

[Out]

c*x**3/(3*e**3) - sqrt(-1/(d*e**9))*(3*a*e**2 - 15*b*d*e + 35*c*d**2)*log(-d*e**4*sqrt(-1/(d*e**9)) + x)/16 +
sqrt(-1/(d*e**9))*(3*a*e**2 - 15*b*d*e + 35*c*d**2)*log(d*e**4*sqrt(-1/(d*e**9)) + x)/16 - (x**3*(5*a*e**3 - 9
*b*d*e**2 + 13*c*d**2*e) + x*(3*a*d*e**2 - 7*b*d**2*e + 11*c*d**3))/(8*d**2*e**4 + 16*d*e**5*x**2 + 8*e**6*x**
4) + x*(b*e - 3*c*d)/e**4

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Giac [A]  time = 1.07332, size = 169, normalized size = 1.18 \begin{align*} \frac{{\left (35 \, c d^{2} - 15 \, b d e + 3 \, a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{9}{2}\right )}}{8 \, \sqrt{d}} + \frac{1}{3} \,{\left (c x^{3} e^{6} - 9 \, c d x e^{5} + 3 \, b x e^{6}\right )} e^{\left (-9\right )} - \frac{{\left (13 \, c d^{2} x^{3} e - 9 \, b d x^{3} e^{2} + 11 \, c d^{3} x + 5 \, a x^{3} e^{3} - 7 \, b d^{2} x e + 3 \, a d x e^{2}\right )} e^{\left (-4\right )}}{8 \,{\left (x^{2} e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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