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

Optimal. Leaf size=93 \[ \frac{x \left (\frac{3 a}{d^2}-\frac{5 c}{e^2}\right )}{8 \left (d+e x^2\right )}+\frac{x \left (a+\frac{c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}+\frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0674083, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1158, 385, 205} \[ \frac{x \left (\frac{3 a}{d^2}-\frac{5 c}{e^2}\right )}{8 \left (d+e x^2\right )}+\frac{x \left (a+\frac{c d^2}{e^2}\right )}{4 d \left (d+e x^2\right )^2}+\frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)/(d + e*x^2)^3,x]

[Out]

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

Rule 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + 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, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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

Mathematica [A]  time = 0.0628183, size = 92, normalized size = 0.99 \[ \frac{a e^2 x \left (5 d+3 e x^2\right )-c d^2 x \left (3 d+5 e x^2\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 99, normalized size = 1.1 \begin{align*}{\frac{1}{ \left ( e{x}^{2}+d \right ) ^{2}} \left ({\frac{ \left ( 3\,a{e}^{2}-5\,c{d}^{2} \right ){x}^{3}}{8\,{d}^{2}e}}+{\frac{ \left ( 5\,a{e}^{2}-3\,c{d}^{2} \right ) x}{8\,d{e}^{2}}} \right ) }+{\frac{3\,a}{8\,{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{3\,c}{8\,{e}^{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+a)/(e*x^2+d)^3,x)

[Out]

(1/8*(3*a*e^2-5*c*d^2)/d^2/e*x^3+1/8*(5*a*e^2-3*c*d^2)/d/e^2*x)/(e*x^2+d)^2+3/8/d^2/(d*e)^(1/2)*arctan(e*x/(d*
e)^(1/2))*a+3/8/e^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+a)/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99178, size = 635, normalized size = 6.83 \begin{align*} \left [-\frac{2 \,{\left (5 \, c d^{3} e^{2} - 3 \, a d e^{4}\right )} x^{3} + 3 \,{\left (c d^{4} + a d^{2} e^{2} +{\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \,{\left (c d^{3} e + 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 - 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} - 3 \, a d e^{4}\right )} x^{3} - 3 \,{\left (c d^{4} + a d^{2} e^{2} +{\left (c d^{2} e^{2} + a e^{4}\right )} x^{4} + 2 \,{\left (c d^{3} e + 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 - 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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(2*(5*c*d^3*e^2 - 3*a*d*e^4)*x^3 + 3*(c*d^4 + a*d^2*e^2 + (c*d^2*e^2 + a*e^4)*x^4 + 2*(c*d^3*e + 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 - 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 - 3*a*d*e^4)*x^3 - 3*(c*d^4 + a*d^2*e^2 + (c*d^2*e^2 + a*e^4)*
x^4 + 2*(c*d^3*e + a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + (3*c*d^4*e - 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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Sympy [B]  time = 0.856339, size = 219, normalized size = 2.35 \begin{align*} - \frac{3 \sqrt{- \frac{1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log{\left (- \frac{3 d^{3} e^{2} \sqrt{- \frac{1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right ) \log{\left (\frac{3 d^{3} e^{2} \sqrt{- \frac{1}{d^{5} e^{5}}} \left (a e^{2} + c d^{2}\right )}{3 a e^{2} + 3 c d^{2}} + x \right )}}{16} + \frac{x^{3} \left (3 a e^{3} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(d**5*e**5))*(a*e**2 + c*d**2)*log(-3*d**3*e**2*sqrt(-1/(d**5*e**5))*(a*e**2 + c*d**2)/(3*a*e**2 +
3*c*d**2) + x)/16 + 3*sqrt(-1/(d**5*e**5))*(a*e**2 + c*d**2)*log(3*d**3*e**2*sqrt(-1/(d**5*e**5))*(a*e**2 + c*
d**2)/(3*a*e**2 + 3*c*d**2) + x)/16 + (x**3*(3*a*e**3 - 5*c*d**2*e) + x*(5*a*d*e**2 - 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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Giac [A]  time = 1.12197, size = 104, normalized size = 1.12 \begin{align*} \frac{3 \,{\left (c d^{2} + 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 + 3 \, c d^{3} x - 3 \, a x^{3} e^{3} - 5 \, a d x e^{2}\right )} e^{\left (-2\right )}}{8 \,{\left (x^{2} e + d\right )}^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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