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

Optimal. Leaf size=131 \[ \frac{x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac{c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac{\left (7 c d^2-a e^2\right ) \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} e^{9/2}}-\frac{2 c^2 d x^3}{3 e^3}+\frac{c^2 x^5}{5 e^2} \]

[Out]

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

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Rubi [A]  time = 0.187989, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1158, 1810, 205} \[ \frac{x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac{c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac{\left (7 c d^2-a e^2\right ) \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} e^{9/2}}-\frac{2 c^2 d x^3}{3 e^3}+\frac{c^2 x^5}{5 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^2,x]

[Out]

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

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 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -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{\left (a+c x^4\right )^2}{\left (d+e x^2\right )^2} \, dx &=\frac{\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac{\int \frac{-a^2+\frac{c^2 d^4}{e^4}+\frac{2 a c d^2}{e^2}-\frac{2 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac{2 c^2 d^2 x^4}{e^2}-\frac{2 c^2 d x^6}{e}}{d+e x^2} \, dx}{2 d}\\ &=\frac{\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac{\int \left (-\frac{2 c d \left (3 c d^2+2 a e^2\right )}{e^4}+\frac{4 c^2 d^2 x^2}{e^3}-\frac{2 c^2 d x^4}{e^2}+\frac{7 c^2 d^4+6 a c d^2 e^2-a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{2 d}\\ &=\frac{c \left (3 c d^2+2 a e^2\right ) x}{e^4}-\frac{2 c^2 d x^3}{3 e^3}+\frac{c^2 x^5}{5 e^2}+\frac{\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac{\left (\left (7 c d^2-a e^2\right ) \left (c d^2+a e^2\right )\right ) \int \frac{1}{d+e x^2} \, dx}{2 d e^4}\\ &=\frac{c \left (3 c d^2+2 a e^2\right ) x}{e^4}-\frac{2 c^2 d x^3}{3 e^3}+\frac{c^2 x^5}{5 e^2}+\frac{\left (c d^2+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac{\left (7 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} e^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.109365, size = 134, normalized size = 1.02 \[ -\frac{\left (-a^2 e^4+6 a c d^2 e^2+7 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 d^{3/2} e^{9/2}}+\frac{x \left (a e^2+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac{c x \left (2 a e^2+3 c d^2\right )}{e^4}-\frac{2 c^2 d x^3}{3 e^3}+\frac{c^2 x^5}{5 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(3*c*d^2 + 2*a*e^2)*x)/e^4 - (2*c^2*d*x^3)/(3*e^3) + (c^2*x^5)/(5*e^2) + ((c*d^2 + a*e^2)^2*x)/(2*d*e^4*(d
+ e*x^2)) - ((7*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(9/2))

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Maple [A]  time = 0.053, size = 170, normalized size = 1.3 \begin{align*}{\frac{{c}^{2}{x}^{5}}{5\,{e}^{2}}}-{\frac{2\,{c}^{2}d{x}^{3}}{3\,{e}^{3}}}+2\,{\frac{acx}{{e}^{2}}}+3\,{\frac{{c}^{2}{d}^{2}x}{{e}^{4}}}+{\frac{{a}^{2}x}{2\,d \left ( e{x}^{2}+d \right ) }}+{\frac{adxc}{{e}^{2} \left ( e{x}^{2}+d \right ) }}+{\frac{{d}^{3}x{c}^{2}}{2\,{e}^{4} \left ( e{x}^{2}+d \right ) }}+{\frac{{a}^{2}}{2\,d}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-3\,{\frac{acd}{{e}^{2}\sqrt{de}}\arctan \left ({\frac{ex}{\sqrt{de}}} \right ) }-{\frac{7\,{c}^{2}{d}^{3}}{2\,{e}^{4}}\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)^2/(e*x^2+d)^2,x)

[Out]

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

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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)^2/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86485, size = 807, normalized size = 6.16 \begin{align*} \left [\frac{12 \, c^{2} d^{2} e^{4} x^{7} - 28 \, c^{2} d^{3} e^{3} x^{5} + 20 \,{\left (7 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4}\right )} x^{3} + 15 \,{\left (7 \, c^{2} d^{5} + 6 \, a c d^{3} e^{2} - a^{2} d e^{4} +{\left (7 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} - a^{2} e^{5}\right )} x^{2}\right )} \sqrt{-d e} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) + 30 \,{\left (7 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x}{60 \,{\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac{6 \, c^{2} d^{2} e^{4} x^{7} - 14 \, c^{2} d^{3} e^{3} x^{5} + 10 \,{\left (7 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4}\right )} x^{3} - 15 \,{\left (7 \, c^{2} d^{5} + 6 \, a c d^{3} e^{2} - a^{2} d e^{4} +{\left (7 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} - a^{2} e^{5}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) + 15 \,{\left (7 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x}{30 \,{\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*c^2*d^2*e^4*x^7 - 28*c^2*d^3*e^3*x^5 + 20*(7*c^2*d^4*e^2 + 6*a*c*d^2*e^4)*x^3 + 15*(7*c^2*d^5 + 6*a*
c*d^3*e^2 - a^2*d*e^4 + (7*c^2*d^4*e + 6*a*c*d^2*e^3 - a^2*e^5)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x -
d)/(e*x^2 + d)) + 30*(7*c^2*d^5*e + 6*a*c*d^3*e^3 + a^2*d*e^5)*x)/(d^2*e^6*x^2 + d^3*e^5), 1/30*(6*c^2*d^2*e^4
*x^7 - 14*c^2*d^3*e^3*x^5 + 10*(7*c^2*d^4*e^2 + 6*a*c*d^2*e^4)*x^3 - 15*(7*c^2*d^5 + 6*a*c*d^3*e^2 - a^2*d*e^4
+ (7*c^2*d^4*e + 6*a*c*d^2*e^3 - a^2*e^5)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + 15*(7*c^2*d^5*e + 6*a*c*d^3*
e^3 + a^2*d*e^5)*x)/(d^2*e^6*x^2 + d^3*e^5)]

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Sympy [B]  time = 1.02083, size = 314, normalized size = 2.4 \begin{align*} - \frac{2 c^{2} d x^{3}}{3 e^{3}} + \frac{c^{2} x^{5}}{5 e^{2}} + \frac{x \left (a^{2} e^{4} + 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{2 d^{2} e^{4} + 2 d e^{5} x^{2}} - \frac{\sqrt{- \frac{1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right ) \log{\left (- \frac{d^{2} e^{4} \sqrt{- \frac{1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right )}{a^{2} e^{4} - 6 a c d^{2} e^{2} - 7 c^{2} d^{4}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right ) \log{\left (\frac{d^{2} e^{4} \sqrt{- \frac{1}{d^{3} e^{9}}} \left (a e^{2} - 7 c d^{2}\right ) \left (a e^{2} + c d^{2}\right )}{a^{2} e^{4} - 6 a c d^{2} e^{2} - 7 c^{2} d^{4}} + x \right )}}{4} + \frac{x \left (2 a c e^{2} + 3 c^{2} d^{2}\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15645, size = 173, normalized size = 1.32 \begin{align*} \frac{1}{15} \,{\left (3 \, c^{2} x^{5} e^{8} - 10 \, c^{2} d x^{3} e^{7} + 45 \, c^{2} d^{2} x e^{6} + 30 \, a c x e^{8}\right )} e^{\left (-10\right )} - \frac{{\left (7 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{9}{2}\right )}}{2 \, d^{\frac{3}{2}}} + \frac{{\left (c^{2} d^{4} x + 2 \, a c d^{2} x e^{2} + a^{2} x e^{4}\right )} e^{\left (-4\right )}}{2 \,{\left (x^{2} e + d\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*c^2*x^5*e^8 - 10*c^2*d*x^3*e^7 + 45*c^2*d^2*x*e^6 + 30*a*c*x*e^8)*e^(-10) - 1/2*(7*c^2*d^4 + 6*a*c*d^2
*e^2 - a^2*e^4)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/d^(3/2) + 1/2*(c^2*d^4*x + 2*a*c*d^2*x*e^2 + a^2*x*e^4)*e^(
-4)/((x^2*e + d)*d)