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

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

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.09, size = 141, normalized size = 0.99 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 a e^2-15 b d e+35 c d^2\right )}{8 \sqrt {d} e^{9/2}}-\frac {x \left (5 a e^2-9 b d e+13 c d^2\right )}{8 e^4 \left (d+e x^2\right )}+\frac {x \left (a d e^2-b d^2 e+c d^3\right )}{4 e^4 \left (d+e x^2\right )^2}+\frac {x (b e-3 c d)}{e^4}+\frac {c x^3}{3 e^3} \]

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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fricas [A]  time = 0.68, size = 462, normalized size = 3.23 \[ \left [\frac {16 \, c d e^{4} x^{7} - 16 \, {\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 10 \, {\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} - 3 \, {\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} + {\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{48 \, {\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac {8 \, c d e^{4} x^{7} - 8 \, {\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 5 \, {\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} + 3 \, {\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} + {\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{24 \, {\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \]

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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giac [A]  time = 0.47, size = 125, normalized size = 0.87 \[ \frac {{\left (35 \, c d^{2} - 15 \, b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{8 \, \sqrt {d}} + \frac {1}{3} \, {\left (c x^{3} e^{6} - 9 \, c d x e^{5} + 3 \, b x e^{6}\right )} e^{\left (-9\right )} - \frac {{\left (13 \, c d^{2} x^{3} e - 9 \, b d x^{3} e^{2} + 11 \, c d^{3} x + 5 \, a x^{3} e^{3} - 7 \, b d^{2} x e + 3 \, a d x e^{2}\right )} e^{\left (-4\right )}}{8 \, {\left (x^{2} e + d\right )}^{2}} \]

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

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maple [A]  time = 0.01, size = 202, normalized size = 1.41 \[ -\frac {5 a \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e}+\frac {9 b d \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {13 c \,d^{2} x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e^{3}}-\frac {3 a d x}{8 \left (e \,x^{2}+d \right )^{2} e^{2}}+\frac {7 b \,d^{2} x}{8 \left (e \,x^{2}+d \right )^{2} e^{3}}-\frac {11 c \,d^{3} x}{8 \left (e \,x^{2}+d \right )^{2} e^{4}}+\frac {c \,x^{3}}{3 e^{3}}+\frac {3 a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{2}}-\frac {15 b d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{3}}+\frac {35 c \,d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{4}}+\frac {b x}{e^{3}}-\frac {3 c d x}{e^{4}} \]

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*a*d*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(1/(d*e)^(1/2)*e*x)*a-15/8/e^3/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*d*b+35/8/e^4/(d*e)^(1/2)*arctan(1
/(d*e)^(1/2)*e*x)*c*d^2

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maxima [A]  time = 2.51, size = 139, normalized size = 0.97 \[ -\frac {{\left (13 \, c d^{2} e - 9 \, b d e^{2} + 5 \, a e^{3}\right )} x^{3} + {\left (11 \, c d^{3} - 7 \, b d^{2} e + 3 \, a d e^{2}\right )} x}{8 \, {\left (e^{6} x^{4} + 2 \, d e^{5} x^{2} + d^{2} e^{4}\right )}} + \frac {{\left (35 \, c d^{2} - 15 \, b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} e^{4}} + \frac {c e x^{3} - 3 \, {\left (3 \, c d - b e\right )} x}{3 \, e^{4}} \]

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]

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

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mupad [B]  time = 0.34, size = 137, normalized size = 0.96 \[ x\,\left (\frac {b}{e^3}-\frac {3\,c\,d}{e^4}\right )-\frac {\left (\frac {13\,c\,d^2\,e}{8}-\frac {9\,b\,d\,e^2}{8}+\frac {5\,a\,e^3}{8}\right )\,x^3+\left (\frac {11\,c\,d^3}{8}-\frac {7\,b\,d^2\,e}{8}+\frac {3\,a\,d\,e^2}{8}\right )\,x}{d^2\,e^4+2\,d\,e^5\,x^2+e^6\,x^4}+\frac {c\,x^3}{3\,e^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,c\,d^2-15\,b\,d\,e+3\,a\,e^2\right )}{8\,\sqrt {d}\,e^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.37, size = 212, normalized size = 1.48 \[ \frac {c x^{3}}{3 e^{3}} + x \left (\frac {b}{e^{3}} - \frac {3 c d}{e^{4}}\right ) - \frac {\sqrt {- \frac {1}{d e^{9}}} \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log {\left (- d e^{4} \sqrt {- \frac {1}{d e^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d e^{9}}} \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log {\left (d e^{4} \sqrt {- \frac {1}{d e^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 5 a e^{3} + 9 b d e^{2} - 13 c d^{2} e\right ) + x \left (- 3 a d e^{2} + 7 b d^{2} e - 11 c d^{3}\right )}{8 d^{2} e^{4} + 16 d e^{5} x^{2} + 8 e^{6} x^{4}} \]

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) + x*(b/e**3 - 3*c*d/e**4) - sqrt(-1/(d*e**9))*(3*a*e**2 - 15*b*d*e + 35*c*d**2)*log(-d*e**4*sq
rt(-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)

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