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

Optimal. Leaf size=135 \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (7 c d^2-e (5 b d-3 a e)\right )}{2 e^{9/2}}+\frac {x \left (3 c d^2-e (2 b d-a e)\right )}{e^4}+\frac {d x \left (a e^2-b d e+c d^2\right )}{2 e^4 \left (d+e x^2\right )}-\frac {x^3 (2 c d-b e)}{3 e^3}+\frac {c x^5}{5 e^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 135, 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} \[ \frac {d x \left (a e^2-b d e+c d^2\right )}{2 e^4 \left (d+e x^2\right )}+\frac {x \left (3 c d^2-e (2 b d-a e)\right )}{e^4}-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (7 c d^2-e (5 b d-3 a e)\right )}{2 e^{9/2}}-\frac {x^3 (2 c d-b e)}{3 e^3}+\frac {c x^5}{5 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 133, normalized size = 0.99 \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 a e^2-5 b d e+7 c d^2\right )}{2 e^{9/2}}+\frac {x \left (a e^2-2 b d e+3 c d^2\right )}{e^4}+\frac {x \left (a d e^2-b d^2 e+c d^3\right )}{2 e^4 \left (d+e x^2\right )}+\frac {x^3 (b e-2 c d)}{3 e^3}+\frac {c x^5}{5 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 350, normalized size = 2.59 \[ \left [\frac {12 \, c e^{3} x^{7} - 4 \, {\left (7 \, c d e^{2} - 5 \, b e^{3}\right )} x^{5} + 20 \, {\left (7 \, c d^{2} e - 5 \, b d e^{2} + 3 \, a e^{3}\right )} x^{3} + 15 \, {\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2} + {\left (7 \, c d^{2} e - 5 \, b d e^{2} + 3 \, a e^{3}\right )} x^{2}\right )} \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 30 \, {\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} x}{60 \, {\left (e^{5} x^{2} + d e^{4}\right )}}, \frac {6 \, c e^{3} x^{7} - 2 \, {\left (7 \, c d e^{2} - 5 \, b e^{3}\right )} x^{5} + 10 \, {\left (7 \, c d^{2} e - 5 \, b d e^{2} + 3 \, a e^{3}\right )} x^{3} - 15 \, {\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2} + {\left (7 \, c d^{2} e - 5 \, b d e^{2} + 3 \, a e^{3}\right )} x^{2}\right )} \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 15 \, {\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} x}{30 \, {\left (e^{5} x^{2} + d e^{4}\right )}}\right ] \]

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)^2,x, algorithm="fricas")

[Out]

[1/60*(12*c*e^3*x^7 - 4*(7*c*d*e^2 - 5*b*e^3)*x^5 + 20*(7*c*d^2*e - 5*b*d*e^2 + 3*a*e^3)*x^3 + 15*(7*c*d^3 - 5
*b*d^2*e + 3*a*d*e^2 + (7*c*d^2*e - 5*b*d*e^2 + 3*a*e^3)*x^2)*sqrt(-d/e)*log((e*x^2 - 2*e*x*sqrt(-d/e) - d)/(e
*x^2 + d)) + 30*(7*c*d^3 - 5*b*d^2*e + 3*a*d*e^2)*x)/(e^5*x^2 + d*e^4), 1/30*(6*c*e^3*x^7 - 2*(7*c*d*e^2 - 5*b
*e^3)*x^5 + 10*(7*c*d^2*e - 5*b*d*e^2 + 3*a*e^3)*x^3 - 15*(7*c*d^3 - 5*b*d^2*e + 3*a*d*e^2 + (7*c*d^2*e - 5*b*
d*e^2 + 3*a*e^3)*x^2)*sqrt(d/e)*arctan(e*x*sqrt(d/e)/d) + 15*(7*c*d^3 - 5*b*d^2*e + 3*a*d*e^2)*x)/(e^5*x^2 + d
*e^4)]

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giac [A]  time = 0.28, size = 125, normalized size = 0.93 \[ -\frac {{\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{2 \, \sqrt {d}} + \frac {1}{15} \, {\left (3 \, c x^{5} e^{8} - 10 \, c d x^{3} e^{7} + 5 \, b x^{3} e^{8} + 45 \, c d^{2} x e^{6} - 30 \, b d x e^{7} + 15 \, a x e^{8}\right )} e^{\left (-10\right )} + \frac {{\left (c d^{3} x - b d^{2} x e + a d x e^{2}\right )} e^{\left (-4\right )}}{2 \, {\left (x^{2} e + d\right )}} \]

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)^2,x, algorithm="giac")

[Out]

-1/2*(7*c*d^3 - 5*b*d^2*e + 3*a*d*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/sqrt(d) + 1/15*(3*c*x^5*e^8 - 10*c*d
*x^3*e^7 + 5*b*x^3*e^8 + 45*c*d^2*x*e^6 - 30*b*d*x*e^7 + 15*a*x*e^8)*e^(-10) + 1/2*(c*d^3*x - b*d^2*x*e + a*d*
x*e^2)*e^(-4)/(x^2*e + d)

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maple [A]  time = 0.01, size = 176, normalized size = 1.30 \[ \frac {c \,x^{5}}{5 e^{2}}+\frac {b \,x^{3}}{3 e^{2}}-\frac {2 c d \,x^{3}}{3 e^{3}}+\frac {a d x}{2 \left (e \,x^{2}+d \right ) e^{2}}-\frac {3 a d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{2}}-\frac {b \,d^{2} x}{2 \left (e \,x^{2}+d \right ) e^{3}}+\frac {5 b \,d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{3}}+\frac {c \,d^{3} x}{2 \left (e \,x^{2}+d \right ) e^{4}}-\frac {7 c \,d^{3} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{4}}+\frac {a x}{e^{2}}-\frac {2 b d x}{e^{3}}+\frac {3 c \,d^{2} x}{e^{4}} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 2.45, size = 130, normalized size = 0.96 \[ \frac {{\left (c d^{3} - b d^{2} e + a d e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + d e^{4}\right )}} - \frac {{\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} e^{4}} + \frac {3 \, c e^{2} x^{5} - 5 \, {\left (2 \, c d e - b e^{2}\right )} x^{3} + 15 \, {\left (3 \, c d^{2} - 2 \, b d e + a e^{2}\right )} x}{15 \, e^{4}} \]

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)^2,x, algorithm="maxima")

[Out]

1/2*(c*d^3 - b*d^2*e + a*d*e^2)*x/(e^5*x^2 + d*e^4) - 1/2*(7*c*d^3 - 5*b*d^2*e + 3*a*d*e^2)*arctan(e*x/sqrt(d*
e))/(sqrt(d*e)*e^4) + 1/15*(3*c*e^2*x^5 - 5*(2*c*d*e - b*e^2)*x^3 + 15*(3*c*d^2 - 2*b*d*e + a*e^2)*x)/e^4

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mupad [B]  time = 0.32, size = 179, normalized size = 1.33 \[ x^3\,\left (\frac {b}{3\,e^2}-\frac {2\,c\,d}{3\,e^3}\right )-x\,\left (\frac {c\,d^2}{e^4}-\frac {a}{e^2}+\frac {2\,d\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )}{e}\right )+\frac {c\,x^5}{5\,e^2}+\frac {x\,\left (\frac {c\,d^3}{2}-\frac {b\,d^2\,e}{2}+\frac {a\,d\,e^2}{2}\right )}{e^5\,x^2+d\,e^4}-\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,x\,\left (7\,c\,d^2-5\,b\,d\,e+3\,a\,e^2\right )}{7\,c\,d^3-5\,b\,d^2\,e+3\,a\,d\,e^2}\right )\,\left (7\,c\,d^2-5\,b\,d\,e+3\,a\,e^2\right )}{2\,e^{9/2}} \]

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)^2,x)

[Out]

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

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sympy [A]  time = 1.08, size = 189, normalized size = 1.40 \[ \frac {c x^{5}}{5 e^{2}} + x^{3} \left (\frac {b}{3 e^{2}} - \frac {2 c d}{3 e^{3}}\right ) + x \left (\frac {a}{e^{2}} - \frac {2 b d}{e^{3}} + \frac {3 c d^{2}}{e^{4}}\right ) + \frac {x \left (a d e^{2} - b d^{2} e + c d^{3}\right )}{2 d e^{4} + 2 e^{5} x^{2}} + \frac {\sqrt {- \frac {d}{e^{9}}} \left (3 a e^{2} - 5 b d e + 7 c d^{2}\right ) \log {\left (- e^{4} \sqrt {- \frac {d}{e^{9}}} + x \right )}}{4} - \frac {\sqrt {- \frac {d}{e^{9}}} \left (3 a e^{2} - 5 b d e + 7 c d^{2}\right ) \log {\left (e^{4} \sqrt {- \frac {d}{e^{9}}} + x \right )}}{4} \]

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)**2,x)

[Out]

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

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