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

Optimal. Leaf size=115 \[ -\frac {x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{4 d \left (d+e x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1157, 385, 205} \[ -\frac {x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{4 d \left (d+e x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 385

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 110, normalized size = 0.96 \[ \frac {x \left (e \left (a e \left (5 d+3 e x^2\right )+b d \left (e x^2-d\right )\right )-c d^2 \left (3 d+5 e x^2\right )\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 391, normalized size = 3.40 \[ \left [-\frac {2 \, {\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} + {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{16 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac {{\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} - {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{8 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 101, normalized size = 0.88 \[ \frac {{\left (3 \, c d^{2} + b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{8 \, d^{\frac {5}{2}}} - \frac {{\left (5 \, c d^{2} x^{3} e - b d x^{3} e^{2} + 3 \, c d^{3} x - 3 \, a x^{3} e^{3} + b d^{2} x e - 5 \, a d x e^{2}\right )} e^{\left (-2\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 131, normalized size = 1.14 \[ \frac {3 a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d^{2}}+\frac {b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d e}+\frac {3 c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{2}}+\frac {\frac {\left (3 a \,e^{2}+b d e -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-b d e -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.25, size = 121, normalized size = 1.05 \[ -\frac {{\left (5 \, c d^{2} e - b d e^{2} - 3 \, a e^{3}\right )} x^{3} + {\left (3 \, c d^{3} + b d^{2} e - 5 \, a d e^{2}\right )} x}{8 \, {\left (d^{2} e^{4} x^{4} + 2 \, d^{3} e^{3} x^{2} + d^{4} e^{2}\right )}} + \frac {{\left (3 \, c d^{2} + b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.85, size = 112, normalized size = 0.97 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}}-\frac {\frac {x\,\left (3\,c\,d^2+b\,d\,e-5\,a\,e^2\right )}{8\,d\,e^2}-\frac {x^3\,\left (-5\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^2\,e}}{d^2+2\,d\,e\,x^2+e^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.27, size = 196, normalized size = 1.70 \[ - \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (- d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {x^{3} \left (3 a e^{3} + b d e^{2} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - b d^{2} e - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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