3.286 \(\int \frac {a+b x^2+c x^4}{x^6 (d+e x^2)^2} \, dx\)

Optimal. Leaf size=136 \[ -\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 c d^2-e (5 b d-7 a e)\right )}{2 d^{9/2}}-\frac {e x \left (a e^2-b d e+c d^2\right )}{2 d^4 \left (d+e x^2\right )}-\frac {c d^2-e (2 b d-3 a e)}{d^4 x}-\frac {b d-2 a e}{3 d^3 x^3}-\frac {a}{5 d^2 x^5} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 136, 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, 1802, 205} \[ -\frac {e x \left (a e^2-b d e+c d^2\right )}{2 d^4 \left (d+e x^2\right )}-\frac {c d^2-e (2 b d-3 a e)}{d^4 x}-\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 c d^2-e (5 b d-7 a e)\right )}{2 d^{9/2}}-\frac {b d-2 a e}{3 d^3 x^3}-\frac {a}{5 d^2 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(x^6*(d + e*x^2)^2),x]

[Out]

-a/(5*d^2*x^5) - (b*d - 2*a*e)/(3*d^3*x^3) - (c*d^2 - e*(2*b*d - 3*a*e))/(d^4*x) - (e*(c*d^2 - b*d*e + a*e^2)*
x)/(2*d^4*(d + e*x^2)) - (Sqrt[e]*(3*c*d^2 - e*(5*b*d - 7*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(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 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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 135, normalized size = 0.99 \[ -\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (7 a e^2-5 b d e+3 c d^2\right )}{2 d^{9/2}}-\frac {e x \left (a e^2-b d e+c d^2\right )}{2 d^4 \left (d+e x^2\right )}+\frac {-3 a e^2+2 b d e-c d^2}{d^4 x}+\frac {2 a e-b d}{3 d^3 x^3}-\frac {a}{5 d^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(x^6*(d + e*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.89, size = 360, normalized size = 2.65 \[ \left [-\frac {30 \, {\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} x^{6} + 20 \, {\left (3 \, c d^{3} - 5 \, b d^{2} e + 7 \, a d e^{2}\right )} x^{4} + 12 \, a d^{3} + 4 \, {\left (5 \, b d^{3} - 7 \, a d^{2} e\right )} x^{2} - 15 \, {\left ({\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} x^{7} + {\left (3 \, c d^{3} - 5 \, b d^{2} e + 7 \, a d e^{2}\right )} x^{5}\right )} \sqrt {-\frac {e}{d}} \log \left (\frac {e x^{2} - 2 \, d x \sqrt {-\frac {e}{d}} - d}{e x^{2} + d}\right )}{60 \, {\left (d^{4} e x^{7} + d^{5} x^{5}\right )}}, -\frac {15 \, {\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} x^{6} + 10 \, {\left (3 \, c d^{3} - 5 \, b d^{2} e + 7 \, a d e^{2}\right )} x^{4} + 6 \, a d^{3} + 2 \, {\left (5 \, b d^{3} - 7 \, a d^{2} e\right )} x^{2} + 15 \, {\left ({\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} x^{7} + {\left (3 \, c d^{3} - 5 \, b d^{2} e + 7 \, a d e^{2}\right )} x^{5}\right )} \sqrt {\frac {e}{d}} \arctan \left (x \sqrt {\frac {e}{d}}\right )}{30 \, {\left (d^{4} e x^{7} + d^{5} x^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^6/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 131, normalized size = 0.96 \[ -\frac {{\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{2 \, d^{\frac {9}{2}}} - \frac {c d^{2} x e - b d x e^{2} + a x e^{3}}{2 \, {\left (x^{2} e + d\right )} d^{4}} - \frac {15 \, c d^{2} x^{4} - 30 \, b d x^{4} e + 45 \, a x^{4} e^{2} + 5 \, b d^{2} x^{2} - 10 \, a d x^{2} e + 3 \, a d^{2}}{15 \, d^{4} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^6/(e*x^2+d)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 183, normalized size = 1.35 \[ -\frac {a \,e^{3} x}{2 \left (e \,x^{2}+d \right ) d^{4}}-\frac {7 a \,e^{3} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{4}}+\frac {b \,e^{2} x}{2 \left (e \,x^{2}+d \right ) d^{3}}+\frac {5 b \,e^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{3}}-\frac {c e x}{2 \left (e \,x^{2}+d \right ) d^{2}}-\frac {3 c e \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d^{2}}-\frac {3 a \,e^{2}}{d^{4} x}+\frac {2 b e}{d^{3} x}-\frac {c}{d^{2} x}+\frac {2 a e}{3 d^{3} x^{3}}-\frac {b}{3 d^{2} x^{3}}-\frac {a}{5 d^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/x^6/(e*x^2+d)^2,x)

[Out]

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

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maxima [A]  time = 2.46, size = 139, normalized size = 1.02 \[ -\frac {15 \, {\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} x^{6} + 10 \, {\left (3 \, c d^{3} - 5 \, b d^{2} e + 7 \, a d e^{2}\right )} x^{4} + 6 \, a d^{3} + 2 \, {\left (5 \, b d^{3} - 7 \, a d^{2} e\right )} x^{2}}{30 \, {\left (d^{4} e x^{7} + d^{5} x^{5}\right )}} - \frac {{\left (3 \, c d^{2} e - 5 \, b d e^{2} + 7 \, a e^{3}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^6/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.38, size = 128, normalized size = 0.94 \[ -\frac {\frac {a}{5\,d}-\frac {x^2\,\left (7\,a\,e-5\,b\,d\right )}{15\,d^2}+\frac {x^4\,\left (3\,c\,d^2-5\,b\,d\,e+7\,a\,e^2\right )}{3\,d^3}+\frac {e\,x^6\,\left (3\,c\,d^2-5\,b\,d\,e+7\,a\,e^2\right )}{2\,d^4}}{e\,x^7+d\,x^5}-\frac {\sqrt {e}\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2-5\,b\,d\,e+7\,a\,e^2\right )}{2\,d^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)/(x^6*(d + e*x^2)^2),x)

[Out]

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

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sympy [B]  time = 2.13, size = 284, normalized size = 2.09 \[ \frac {\sqrt {- \frac {e}{d^{9}}} \left (7 a e^{2} - 5 b d e + 3 c d^{2}\right ) \log {\left (- \frac {d^{5} \sqrt {- \frac {e}{d^{9}}} \left (7 a e^{2} - 5 b d e + 3 c d^{2}\right )}{7 a e^{3} - 5 b d e^{2} + 3 c d^{2} e} + x \right )}}{4} - \frac {\sqrt {- \frac {e}{d^{9}}} \left (7 a e^{2} - 5 b d e + 3 c d^{2}\right ) \log {\left (\frac {d^{5} \sqrt {- \frac {e}{d^{9}}} \left (7 a e^{2} - 5 b d e + 3 c d^{2}\right )}{7 a e^{3} - 5 b d e^{2} + 3 c d^{2} e} + x \right )}}{4} + \frac {- 6 a d^{3} + x^{6} \left (- 105 a e^{3} + 75 b d e^{2} - 45 c d^{2} e\right ) + x^{4} \left (- 70 a d e^{2} + 50 b d^{2} e - 30 c d^{3}\right ) + x^{2} \left (14 a d^{2} e - 10 b d^{3}\right )}{30 d^{5} x^{5} + 30 d^{4} e x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/x**6/(e*x**2+d)**2,x)

[Out]

sqrt(-e/d**9)*(7*a*e**2 - 5*b*d*e + 3*c*d**2)*log(-d**5*sqrt(-e/d**9)*(7*a*e**2 - 5*b*d*e + 3*c*d**2)/(7*a*e**
3 - 5*b*d*e**2 + 3*c*d**2*e) + x)/4 - sqrt(-e/d**9)*(7*a*e**2 - 5*b*d*e + 3*c*d**2)*log(d**5*sqrt(-e/d**9)*(7*
a*e**2 - 5*b*d*e + 3*c*d**2)/(7*a*e**3 - 5*b*d*e**2 + 3*c*d**2*e) + x)/4 + (-6*a*d**3 + x**6*(-105*a*e**3 + 75
*b*d*e**2 - 45*c*d**2*e) + x**4*(-70*a*d*e**2 + 50*b*d**2*e - 30*c*d**3) + x**2*(14*a*d**2*e - 10*b*d**3))/(30
*d**5*x**5 + 30*d**4*e*x**7)

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