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

Optimal. Leaf size=106 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (c d^2-e (3 b d-5 a e)\right )}{2 d^{7/2} \sqrt {e}}+\frac {x \left (a e^2-b d e+c d^2\right )}{2 d^3 \left (d+e x^2\right )}-\frac {b d-2 a e}{d^3 x}-\frac {a}{3 d^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 106, 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 = {1259, 1261, 205} \[ \frac {x \left (a e^2-b d e+c d^2\right )}{2 d^3 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (c d^2-e (3 b d-5 a e)\right )}{2 d^{7/2} \sqrt {e}}-\frac {b d-2 a e}{d^3 x}-\frac {a}{3 d^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1259

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[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(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] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 105, normalized size = 0.99 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a e^2-3 b d e+c d^2\right )}{2 d^{7/2} \sqrt {e}}+\frac {x \left (a e^2-b d e+c d^2\right )}{2 d^3 \left (d+e x^2\right )}+\frac {2 a e-b d}{d^3 x}-\frac {a}{3 d^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/(d^2*x^3) + (-(b*d) + 2*a*e)/(d^3*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(2*d^3*(d + e*x^2)) + ((c*d^2 - 3*b*
d*e + 5*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(7/2)*Sqrt[e])

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fricas [A]  time = 0.93, size = 316, normalized size = 2.98 \[ \left [-\frac {4 \, a d^{3} e - 6 \, {\left (c d^{3} e - 3 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{4} + 4 \, {\left (3 \, b d^{3} e - 5 \, a d^{2} e^{2}\right )} x^{2} + 3 \, {\left ({\left (c d^{2} e - 3 \, b d e^{2} + 5 \, a e^{3}\right )} x^{5} + {\left (c d^{3} - 3 \, b d^{2} e + 5 \, a d e^{2}\right )} x^{3}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right )}{12 \, {\left (d^{4} e^{2} x^{5} + d^{5} e x^{3}\right )}}, -\frac {2 \, a d^{3} e - 3 \, {\left (c d^{3} e - 3 \, b d^{2} e^{2} + 5 \, a d e^{3}\right )} x^{4} + 2 \, {\left (3 \, b d^{3} e - 5 \, a d^{2} e^{2}\right )} x^{2} - 3 \, {\left ({\left (c d^{2} e - 3 \, b d e^{2} + 5 \, a e^{3}\right )} x^{5} + {\left (c d^{3} - 3 \, b d^{2} e + 5 \, a d e^{2}\right )} x^{3}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right )}{6 \, {\left (d^{4} e^{2} x^{5} + d^{5} e x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 94, normalized size = 0.89 \[ \frac {{\left (c d^{2} - 3 \, b d e + 5 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{2 \, d^{\frac {7}{2}}} + \frac {c d^{2} x - b d x e + a x e^{2}}{2 \, {\left (x^{2} e + d\right )} d^{3}} - \frac {3 \, b d x^{2} - 6 \, a x^{2} e + a d}{3 \, d^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*d^2 - 3*b*d*e + 5*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(7/2) + 1/2*(c*d^2*x - b*d*x*e + a*x*e^2)
/((x^2*e + d)*d^3) - 1/3*(3*b*d*x^2 - 6*a*x^2*e + a*d)/(d^3*x^3)

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maple [A]  time = 0.01, size = 146, normalized size = 1.38 \[ \frac {a \,e^{2} x}{2 \left (e \,x^{2}+d \right ) d^{3}}+\frac {5 a \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{3}}-\frac {b e x}{2 \left (e \,x^{2}+d \right ) d^{2}}-\frac {3 b e \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{2}}+\frac {c x}{2 \left (e \,x^{2}+d \right ) d}+\frac {c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d}+\frac {2 a e}{d^{3} x}-\frac {b}{d^{2} x}-\frac {a}{3 d^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.39, size = 103, normalized size = 0.97 \[ \frac {3 \, {\left (c d^{2} - 3 \, b d e + 5 \, a e^{2}\right )} x^{4} - 2 \, a d^{2} - 2 \, {\left (3 \, b d^{2} - 5 \, a d e\right )} x^{2}}{6 \, {\left (d^{3} e x^{5} + d^{4} x^{3}\right )}} + \frac {{\left (c d^{2} - 3 \, b d e + 5 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(c*d^2 - 3*b*d*e + 5*a*e^2)*x^4 - 2*a*d^2 - 2*(3*b*d^2 - 5*a*d*e)*x^2)/(d^3*e*x^5 + d^4*x^3) + 1/2*(c*d
^2 - 3*b*d*e + 5*a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3)

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mupad [B]  time = 0.36, size = 98, normalized size = 0.92 \[ \frac {\frac {x^2\,\left (5\,a\,e-3\,b\,d\right )}{3\,d^2}-\frac {a}{3\,d}+\frac {x^4\,\left (c\,d^2-3\,b\,d\,e+5\,a\,e^2\right )}{2\,d^3}}{e\,x^5+d\,x^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2-3\,b\,d\,e+5\,a\,e^2\right )}{2\,d^{7/2}\,\sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.53, size = 167, normalized size = 1.58 \[ - \frac {\sqrt {- \frac {1}{d^{7} e}} \left (5 a e^{2} - 3 b d e + c d^{2}\right ) \log {\left (- d^{4} \sqrt {- \frac {1}{d^{7} e}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{7} e}} \left (5 a e^{2} - 3 b d e + c d^{2}\right ) \log {\left (d^{4} \sqrt {- \frac {1}{d^{7} e}} + x \right )}}{4} + \frac {- 2 a d^{2} + x^{4} \left (15 a e^{2} - 9 b d e + 3 c d^{2}\right ) + x^{2} \left (10 a d e - 6 b d^{2}\right )}{6 d^{4} x^{3} + 6 d^{3} e x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(d**7*e))*(5*a*e**2 - 3*b*d*e + c*d**2)*log(-d**4*sqrt(-1/(d**7*e)) + x)/4 + sqrt(-1/(d**7*e))*(5*a*e
**2 - 3*b*d*e + c*d**2)*log(d**4*sqrt(-1/(d**7*e)) + x)/4 + (-2*a*d**2 + x**4*(15*a*e**2 - 9*b*d*e + 3*c*d**2)
+ x**2*(10*a*d*e - 6*b*d**2))/(6*d**4*x**3 + 6*d**3*e*x**5)

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