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

Optimal. Leaf size=142 \[ \frac{x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (35 a e^2-15 b d e+3 c d^2\right )}{8 d^{9/2} \sqrt{e}}+\frac{x \left (3 c d^2-e (7 b d-11 a e)\right )}{8 d^4 \left (d+e x^2\right )}-\frac{b d-3 a e}{d^4 x}-\frac{a}{3 d^3 x^3} \]

[Out]

-a/(3*d^3*x^3) - (b*d - 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((3*c*d^2 - e*(7*
b*d - 11*a*e))*x)/(8*d^4*(d + e*x^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(9/
2)*Sqrt[e])

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Rubi [A]  time = 0.217412, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1259, 1261, 205} \[ \frac{x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (35 a e^2-15 b d e+3 c d^2\right )}{8 d^{9/2} \sqrt{e}}+\frac{x \left (3 c d^2-e (7 b d-11 a e)\right )}{8 d^4 \left (d+e x^2\right )}-\frac{b d-3 a e}{d^4 x}-\frac{a}{3 d^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(3*d^3*x^3) - (b*d - 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((3*c*d^2 - e*(7*
b*d - 11*a*e))*x)/(8*d^4*(d + e*x^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(9/
2)*Sqrt[e])

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]

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]

Rubi steps

\begin{align*} \int \frac{a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx &=\frac{\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac{\int \frac{4 a d^2 e^2+4 d e^2 (b d-a e) x^2+3 e^2 \left (c d^2-b d e+a e^2\right ) x^4}{x^4 \left (d+e x^2\right )^2} \, dx}{4 d^3 e^2}\\ &=\frac{\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac{\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac{\int \frac{8 a d^4 e^4+8 d^3 e^4 (b d-2 a e) x^2+d^2 e^4 \left (3 c d^2-e (7 b d-11 a e)\right ) x^4}{x^4 \left (d+e x^2\right )} \, dx}{8 d^6 e^4}\\ &=\frac{\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac{\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac{\int \left (\frac{8 a d^3 e^4}{x^4}+\frac{8 d^2 e^4 (b d-3 a e)}{x^2}+\frac{d^2 e^4 \left (3 c d^2-15 b d e+35 a e^2\right )}{d+e x^2}\right ) \, dx}{8 d^6 e^4}\\ &=-\frac{a}{3 d^3 x^3}-\frac{b d-3 a e}{d^4 x}+\frac{\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac{\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac{\left (3 c d^2-15 b d e+35 a e^2\right ) \int \frac{1}{d+e x^2} \, dx}{8 d^4}\\ &=-\frac{a}{3 d^3 x^3}-\frac{b d-3 a e}{d^4 x}+\frac{\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac{\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac{\left (3 c d^2-15 b d e+35 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{9/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.0858429, size = 141, normalized size = 0.99 \[ \frac{x \left (11 a e^2-7 b d e+3 c d^2\right )}{8 d^4 \left (d+e x^2\right )}+\frac{x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (35 a e^2-15 b d e+3 c d^2\right )}{8 d^{9/2} \sqrt{e}}+\frac{3 a e-b d}{d^4 x}-\frac{a}{3 d^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(3*d^3*x^3) + (-(b*d) + 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((3*c*d^2 - 7*
b*d*e + 11*a*e^2)*x)/(8*d^4*(d + e*x^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^
(9/2)*Sqrt[e])

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Maple [A]  time = 0.014, size = 207, normalized size = 1.5 \begin{align*} -{\frac{a}{3\,{d}^{3}{x}^{3}}}+3\,{\frac{ae}{{d}^{4}x}}-{\frac{b}{{d}^{3}x}}+{\frac{11\,{x}^{3}a{e}^{3}}{8\,{d}^{4} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{7\,{x}^{3}b{e}^{2}}{8\,{d}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{3\,{x}^{3}ce}{8\,{d}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{13\,a{e}^{2}x}{8\,{d}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{9\,bex}{8\,{d}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{5\,cx}{8\,d \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{35\,a{e}^{2}}{8\,{d}^{4}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{15\,be}{8\,{d}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{3\,c}{8\,{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

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)^3,x)

[Out]

-1/3*a/d^3/x^3+3/d^4/x*a*e-1/d^3/x*b+11/8/d^4/(e*x^2+d)^2*x^3*a*e^3-7/8/d^3/(e*x^2+d)^2*x^3*b*e^2+3/8/d^2/(e*x
^2+d)^2*x^3*c*e+13/8/d^3/(e*x^2+d)^2*e^2*a*x-9/8/d^2/(e*x^2+d)^2*b*e*x+5/8/d/(e*x^2+d)^2*c*x+35/8/d^4/(d*e)^(1
/2)*arctan(e*x/(d*e)^(1/2))*a*e^2-15/8/d^3/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*e*b+3/8/d^2/(d*e)^(1/2)*arctan(
e*x/(d*e)^(1/2))*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91339, size = 1031, normalized size = 7.26 \begin{align*} \left [\frac{6 \,{\left (3 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 35 \, a d e^{4}\right )} x^{6} - 16 \, a d^{4} e + 10 \,{\left (3 \, c d^{4} e - 15 \, b d^{3} e^{2} + 35 \, a d^{2} e^{3}\right )} x^{4} - 16 \,{\left (3 \, b d^{4} e - 7 \, a d^{3} e^{2}\right )} x^{2} - 3 \,{\left ({\left (3 \, c d^{2} e^{2} - 15 \, b d e^{3} + 35 \, a e^{4}\right )} x^{7} + 2 \,{\left (3 \, c d^{3} e - 15 \, b d^{2} e^{2} + 35 \, a d e^{3}\right )} x^{5} +{\left (3 \, c d^{4} - 15 \, b d^{3} e + 35 \, a d^{2} 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 )}{48 \,{\left (d^{5} e^{3} x^{7} + 2 \, d^{6} e^{2} x^{5} + d^{7} e x^{3}\right )}}, \frac{3 \,{\left (3 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 35 \, a d e^{4}\right )} x^{6} - 8 \, a d^{4} e + 5 \,{\left (3 \, c d^{4} e - 15 \, b d^{3} e^{2} + 35 \, a d^{2} e^{3}\right )} x^{4} - 8 \,{\left (3 \, b d^{4} e - 7 \, a d^{3} e^{2}\right )} x^{2} + 3 \,{\left ({\left (3 \, c d^{2} e^{2} - 15 \, b d e^{3} + 35 \, a e^{4}\right )} x^{7} + 2 \,{\left (3 \, c d^{3} e - 15 \, b d^{2} e^{2} + 35 \, a d e^{3}\right )} x^{5} +{\left (3 \, c d^{4} - 15 \, b d^{3} e + 35 \, a d^{2} e^{2}\right )} x^{3}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right )}{24 \,{\left (d^{5} e^{3} x^{7} + 2 \, d^{6} e^{2} x^{5} + d^{7} e x^{3}\right )}}\right ] \end{align*}

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)^3,x, algorithm="fricas")

[Out]

[1/48*(6*(3*c*d^3*e^2 - 15*b*d^2*e^3 + 35*a*d*e^4)*x^6 - 16*a*d^4*e + 10*(3*c*d^4*e - 15*b*d^3*e^2 + 35*a*d^2*
e^3)*x^4 - 16*(3*b*d^4*e - 7*a*d^3*e^2)*x^2 - 3*((3*c*d^2*e^2 - 15*b*d*e^3 + 35*a*e^4)*x^7 + 2*(3*c*d^3*e - 15
*b*d^2*e^2 + 35*a*d*e^3)*x^5 + (3*c*d^4 - 15*b*d^3*e + 35*a*d^2*e^2)*x^3)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)
*x - d)/(e*x^2 + d)))/(d^5*e^3*x^7 + 2*d^6*e^2*x^5 + d^7*e*x^3), 1/24*(3*(3*c*d^3*e^2 - 15*b*d^2*e^3 + 35*a*d*
e^4)*x^6 - 8*a*d^4*e + 5*(3*c*d^4*e - 15*b*d^3*e^2 + 35*a*d^2*e^3)*x^4 - 8*(3*b*d^4*e - 7*a*d^3*e^2)*x^2 + 3*(
(3*c*d^2*e^2 - 15*b*d*e^3 + 35*a*e^4)*x^7 + 2*(3*c*d^3*e - 15*b*d^2*e^2 + 35*a*d*e^3)*x^5 + (3*c*d^4 - 15*b*d^
3*e + 35*a*d^2*e^2)*x^3)*sqrt(d*e)*arctan(sqrt(d*e)*x/d))/(d^5*e^3*x^7 + 2*d^6*e^2*x^5 + d^7*e*x^3)]

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Sympy [A]  time = 3.40683, size = 214, normalized size = 1.51 \begin{align*} - \frac{\sqrt{- \frac{1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log{\left (- d^{5} \sqrt{- \frac{1}{d^{9} e}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log{\left (d^{5} \sqrt{- \frac{1}{d^{9} e}} + x \right )}}{16} + \frac{- 8 a d^{3} + x^{6} \left (105 a e^{3} - 45 b d e^{2} + 9 c d^{2} e\right ) + x^{4} \left (175 a d e^{2} - 75 b d^{2} e + 15 c d^{3}\right ) + x^{2} \left (56 a d^{2} e - 24 b d^{3}\right )}{24 d^{6} x^{3} + 48 d^{5} e x^{5} + 24 d^{4} e^{2} x^{7}} \end{align*}

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)**3,x)

[Out]

-sqrt(-1/(d**9*e))*(35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(-d**5*sqrt(-1/(d**9*e)) + x)/16 + sqrt(-1/(d**9*e))*(
35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(d**5*sqrt(-1/(d**9*e)) + x)/16 + (-8*a*d**3 + x**6*(105*a*e**3 - 45*b*d*e
**2 + 9*c*d**2*e) + x**4*(175*a*d*e**2 - 75*b*d**2*e + 15*c*d**3) + x**2*(56*a*d**2*e - 24*b*d**3))/(24*d**6*x
**3 + 48*d**5*e*x**5 + 24*d**4*e**2*x**7)

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Giac [A]  time = 1.11525, size = 173, normalized size = 1.22 \begin{align*} \frac{{\left (3 \, c d^{2} - 15 \, b d e + 35 \, a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{1}{2}\right )}}{8 \, d^{\frac{9}{2}}} + \frac{3 \, c d^{2} x^{3} e - 7 \, b d x^{3} e^{2} + 5 \, c d^{3} x + 11 \, a x^{3} e^{3} - 9 \, b d^{2} x e + 13 \, a d x e^{2}}{8 \,{\left (x^{2} e + d\right )}^{2} d^{4}} - \frac{3 \, b d x^{2} - 9 \, a x^{2} e + a d}{3 \, d^{4} x^{3}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

1/8*(3*c*d^2 - 15*b*d*e + 35*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(9/2) + 1/8*(3*c*d^2*x^3*e - 7*b*d*x^
3*e^2 + 5*c*d^3*x + 11*a*x^3*e^3 - 9*b*d^2*x*e + 13*a*d*x*e^2)/((x^2*e + d)^2*d^4) - 1/3*(3*b*d*x^2 - 9*a*x^2*
e + a*d)/(d^4*x^3)