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

Optimal. Leaf size=124 \[ \frac {x \left (9 c d^2-e (5 b d-a e)\right )}{8 d e^3 \left (d+e x^2\right )}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 c d^2-e (a e+3 b d)\right )}{8 d^{3/2} e^{7/2}}+\frac {c x}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1257, 1157, 388, 205} \[ \frac {x \left (9 c d^2-e (5 b d-a e)\right )}{8 d e^3 \left (d+e x^2\right )}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 c d^2-e (a e+3 b d)\right )}{8 d^{3/2} e^{7/2}}+\frac {c x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - e*(5*b*d - a*e))*x)/(8*d*e^3*(d +
e*x^2)) - ((15*c*d^2 - e*(3*b*d + a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 122, normalized size = 0.98 \[ \frac {x \left (a e^2-5 b d e+9 c d^2\right )}{8 d e^3 \left (d+e x^2\right )}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-a e^2-3 b d e+15 c d^2\right )}{8 d^{3/2} e^{7/2}}+\frac {c x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - 5*b*d*e + a*e^2)*x)/(8*d*e^3*(d +
e*x^2)) - ((15*c*d^2 - 3*b*d*e - a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))

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fricas [A]  time = 0.81, size = 421, normalized size = 3.40 \[ \left [\frac {16 \, c d^{2} e^{3} x^{5} + 2 \, {\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} + {\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} + {\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \, {\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{16 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}, \frac {8 \, c d^{2} e^{3} x^{5} + {\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} - {\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} + {\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \, {\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{8 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*c*d^2*e^3*x^5 + 2*(25*c*d^3*e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 + (15*c*d^4 - 3*b*d^3*e - a*d^2*e^2 + (
15*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*
sqrt(-d*e)*x - d)/(e*x^2 + d)) + 2*(15*c*d^4*e - 3*b*d^3*e^2 - a*d^2*e^3)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^
4*e^4), 1/8*(8*c*d^2*e^3*x^5 + (25*c*d^3*e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 - (15*c*d^4 - 3*b*d^3*e - a*d^2*e^2
+ (15*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d
*e)*x/d) + (15*c*d^4*e - 3*b*d^3*e^2 - a*d^2*e^3)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^4*e^4)]

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giac [A]  time = 0.26, size = 107, normalized size = 0.86 \[ c x e^{\left (-3\right )} - \frac {{\left (15 \, c d^{2} - 3 \, b d e - a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {7}{2}\right )}}{8 \, d^{\frac {3}{2}}} + \frac {{\left (9 \, c d^{2} x^{3} e - 5 \, b d x^{3} e^{2} + 7 \, c d^{3} x + a x^{3} e^{3} - 3 \, b d^{2} x e - a d x e^{2}\right )} e^{\left (-3\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 179, normalized size = 1.44 \[ \frac {a \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} d}-\frac {5 b \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e}+\frac {9 c d \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {a x}{8 \left (e \,x^{2}+d \right )^{2} e}-\frac {3 b d x}{8 \left (e \,x^{2}+d \right )^{2} e^{2}}+\frac {7 c \,d^{2} x}{8 \left (e \,x^{2}+d \right )^{2} e^{3}}+\frac {a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d e}+\frac {3 b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{2}}-\frac {15 c d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{3}}+\frac {c x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x/e^3+1/8/(e*x^2+d)^2/d*x^3*a-5/8/e/(e*x^2+d)^2*x^3*b+9/8/e^2/(e*x^2+d)^2*x^3*c*d-1/8/e/(e*x^2+d)^2*a*x-3/8/
e^2/(e*x^2+d)^2*d*b*x+7/8/e^3/(e*x^2+d)^2*c*d^2*x+1/8/e/d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a+3/8/e^2/(d*e
)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*b-15/8/e^3*d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*c

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maxima [A]  time = 2.46, size = 126, normalized size = 1.02 \[ \frac {{\left (9 \, c d^{2} e - 5 \, b d e^{2} + a e^{3}\right )} x^{3} + {\left (7 \, c d^{3} - 3 \, b d^{2} e - a d e^{2}\right )} x}{8 \, {\left (d e^{5} x^{4} + 2 \, d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}} + \frac {c x}{e^{3}} - \frac {{\left (15 \, c d^{2} - 3 \, b d e - a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 118, normalized size = 0.95 \[ \frac {c\,x}{e^3}-\frac {x\,\left (-\frac {7\,c\,d^2}{8}+\frac {3\,b\,d\,e}{8}+\frac {a\,e^2}{8}\right )-\frac {x^3\,\left (9\,c\,d^2\,e-5\,b\,d\,e^2+a\,e^3\right )}{8\,d}}{d^2\,e^3+2\,d\,e^4\,x^2+e^5\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (-15\,c\,d^2+3\,b\,d\,e+a\,e^2\right )}{8\,d^{3/2}\,e^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.62, size = 201, normalized size = 1.62 \[ \frac {c x}{e^{3}} - \frac {\sqrt {- \frac {1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log {\left (- d^{2} e^{3} \sqrt {- \frac {1}{d^{3} e^{7}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log {\left (d^{2} e^{3} \sqrt {- \frac {1}{d^{3} e^{7}}} + x \right )}}{16} + \frac {x^{3} \left (a e^{3} - 5 b d e^{2} + 9 c d^{2} e\right ) + x \left (- a d e^{2} - 3 b d^{2} e + 7 c d^{3}\right )}{8 d^{3} e^{3} + 16 d^{2} e^{4} x^{2} + 8 d e^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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