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

Optimal. Leaf size=155 \[ \frac{x \left (3 a^2-\frac{10 a c d^2}{e^2}-\frac{13 c^2 d^4}{e^4}\right )}{8 d^2 \left (d+e x^2\right )}+\frac{\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{9/2}}+\frac{x \left (a e^2+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac{3 c^2 d x}{e^4}+\frac{c^2 x^3}{3 e^3} \]

[Out]

(-3*c^2*d*x)/e^4 + (c^2*x^3)/(3*e^3) + ((c*d^2 + a*e^2)^2*x)/(4*d*e^4*(d + e*x^2)^2) + ((3*a^2 - (13*c^2*d^4)/
e^4 - (10*a*c*d^2)/e^2)*x)/(8*d^2*(d + e*x^2)) + ((35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTan[(Sqrt[e]*x)/
Sqrt[d]])/(8*d^(5/2)*e^(9/2))

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Rubi [A]  time = 0.252468, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1158, 1814, 1153, 205} \[ \frac{x \left (3 a^2-\frac{10 a c d^2}{e^2}-\frac{13 c^2 d^4}{e^4}\right )}{8 d^2 \left (d+e x^2\right )}+\frac{\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{9/2}}+\frac{x \left (a e^2+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac{3 c^2 d x}{e^4}+\frac{c^2 x^3}{3 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

(-3*c^2*d*x)/e^4 + (c^2*x^3)/(3*e^3) + ((c*d^2 + a*e^2)^2*x)/(4*d*e^4*(d + e*x^2)^2) + ((3*a^2 - (13*c^2*d^4)/
e^4 - (10*a*c*d^2)/e^2)*x)/(8*d^2*(d + e*x^2)) + ((35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTan[(Sqrt[e]*x)/
Sqrt[d]])/(8*d^(5/2)*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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 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{\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx &=\frac{\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac{\int \frac{-3 a^2+\frac{c^2 d^4}{e^4}+\frac{2 a c d^2}{e^2}-\frac{4 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac{4 c^2 d^2 x^4}{e^2}-\frac{4 c^2 d x^6}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d}\\ &=\frac{\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac{\left (3 a^2-\frac{13 c^2 d^4}{e^4}-\frac{10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac{\int \frac{3 a^2+\frac{11 c^2 d^4}{e^4}+\frac{6 a c d^2}{e^2}-\frac{16 c^2 d^3 x^2}{e^3}+\frac{8 c^2 d^2 x^4}{e^2}}{d+e x^2} \, dx}{8 d^2}\\ &=\frac{\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac{\left (3 a^2-\frac{13 c^2 d^4}{e^4}-\frac{10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac{\int \left (-\frac{24 c^2 d^3}{e^4}+\frac{8 c^2 d^2 x^2}{e^3}+\frac{35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{8 d^2}\\ &=-\frac{3 c^2 d x}{e^4}+\frac{c^2 x^3}{3 e^3}+\frac{\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac{\left (3 a^2-\frac{13 c^2 d^4}{e^4}-\frac{10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac{\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \int \frac{1}{d+e x^2} \, dx}{8 d^2 e^4}\\ &=-\frac{3 c^2 d x}{e^4}+\frac{c^2 x^3}{3 e^3}+\frac{\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac{\left (3 a^2-\frac{13 c^2 d^4}{e^4}-\frac{10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac{\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.111127, size = 154, normalized size = 0.99 \[ \frac{x \left (3 a^2 e^4 \left (5 d+3 e x^2\right )-6 a c d^2 e^2 \left (3 d+5 e x^2\right )-c^2 d^2 \left (175 d^2 e x^2+105 d^3+56 d e^2 x^4-8 e^3 x^6\right )\right )}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac{\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*a^2*e^4*(5*d + 3*e*x^2) - 6*a*c*d^2*e^2*(3*d + 5*e*x^2) - c^2*d^2*(105*d^3 + 175*d^2*e*x^2 + 56*d*e^2*x^
4 - 8*e^3*x^6)))/(24*d^2*e^4*(d + e*x^2)^2) + ((35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqr
t[d]])/(8*d^(5/2)*e^(9/2))

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Maple [A]  time = 0.053, size = 211, normalized size = 1.4 \begin{align*}{\frac{{c}^{2}{x}^{3}}{3\,{e}^{3}}}-3\,{\frac{{c}^{2}dx}{{e}^{4}}}+{\frac{3\,{a}^{2}e{x}^{3}}{8\, \left ( e{x}^{2}+d \right ) ^{2}{d}^{2}}}-{\frac{5\,a{x}^{3}c}{4\,e \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{13\,{d}^{2}{x}^{3}{c}^{2}}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{5\,{a}^{2}x}{8\, \left ( e{x}^{2}+d \right ) ^{2}d}}-{\frac{3\,adxc}{4\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{11\,{d}^{3}x{c}^{2}}{8\,{e}^{4} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{3\,{a}^{2}}{8\,{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{3\,ac}{4\,{e}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{35\,{c}^{2}{d}^{2}}{8\,{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)^3,x)

[Out]

1/3*c^2*x^3/e^3-3*c^2*d*x/e^4+3/8*e/(e*x^2+d)^2/d^2*x^3*a^2-5/4/e/(e*x^2+d)^2*x^3*a*c-13/8/e^3/(e*x^2+d)^2*d^2
*x^3*c^2+5/8/(e*x^2+d)^2/d*x*a^2-3/4/e^2/(e*x^2+d)^2*d*x*a*c-11/8/e^4/(e*x^2+d)^2*d^3*x*c^2+3/8/d^2/(d*e)^(1/2
)*arctan(e*x/(d*e)^(1/2))*a^2+3/4/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*a*c+35/8/e^4*d^2/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9364, size = 1071, normalized size = 6.91 \begin{align*} \left [\frac{16 \, c^{2} d^{3} e^{4} x^{7} - 112 \, c^{2} d^{4} e^{3} x^{5} - 2 \,{\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} - 3 \,{\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} +{\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \,{\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d 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 ) - 6 \,{\left (35 \, c^{2} d^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{48 \,{\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} e^{5}\right )}}, \frac{8 \, c^{2} d^{3} e^{4} x^{7} - 56 \, c^{2} d^{4} e^{3} x^{5} -{\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} + 3 \,{\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} +{\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \,{\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d e^{5}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) - 3 \,{\left (35 \, c^{2} d^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{24 \,{\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} 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)^3,x, algorithm="fricas")

[Out]

[1/48*(16*c^2*d^3*e^4*x^7 - 112*c^2*d^4*e^3*x^5 - 2*(175*c^2*d^5*e^2 + 30*a*c*d^3*e^4 - 9*a^2*d*e^6)*x^3 - 3*(
35*c^2*d^6 + 6*a*c*d^4*e^2 + 3*a^2*d^2*e^4 + (35*c^2*d^4*e^2 + 6*a*c*d^2*e^4 + 3*a^2*e^6)*x^4 + 2*(35*c^2*d^5*
e + 6*a*c*d^3*e^3 + 3*a^2*d*e^5)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(35*c^2*d^6
*e + 6*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x)/(d^3*e^7*x^4 + 2*d^4*e^6*x^2 + d^5*e^5), 1/24*(8*c^2*d^3*e^4*x^7 - 56*c
^2*d^4*e^3*x^5 - (175*c^2*d^5*e^2 + 30*a*c*d^3*e^4 - 9*a^2*d*e^6)*x^3 + 3*(35*c^2*d^6 + 6*a*c*d^4*e^2 + 3*a^2*
d^2*e^4 + (35*c^2*d^4*e^2 + 6*a*c*d^2*e^4 + 3*a^2*e^6)*x^4 + 2*(35*c^2*d^5*e + 6*a*c*d^3*e^3 + 3*a^2*d*e^5)*x^
2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(35*c^2*d^6*e + 6*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x)/(d^3*e^7*x^4 + 2*d^4*
e^6*x^2 + d^5*e^5)]

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Sympy [A]  time = 1.85728, size = 257, normalized size = 1.66 \begin{align*} - \frac{3 c^{2} d x}{e^{4}} + \frac{c^{2} x^{3}}{3 e^{3}} - \frac{\sqrt{- \frac{1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log{\left (- d^{3} e^{4} \sqrt{- \frac{1}{d^{5} e^{9}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log{\left (d^{3} e^{4} \sqrt{- \frac{1}{d^{5} e^{9}}} + x \right )}}{16} + \frac{x^{3} \left (3 a^{2} e^{5} - 10 a c d^{2} e^{3} - 13 c^{2} d^{4} e\right ) + x \left (5 a^{2} d e^{4} - 6 a c d^{3} e^{2} - 11 c^{2} d^{5}\right )}{8 d^{4} e^{4} + 16 d^{3} e^{5} x^{2} + 8 d^{2} e^{6} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*c**2*d*x/e**4 + c**2*x**3/(3*e**3) - sqrt(-1/(d**5*e**9))*(3*a**2*e**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4)*lo
g(-d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + sqrt(-1/(d**5*e**9))*(3*a**2*e**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4
)*log(d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + (x**3*(3*a**2*e**5 - 10*a*c*d**2*e**3 - 13*c**2*d**4*e) + x*(5*
a**2*d*e**4 - 6*a*c*d**3*e**2 - 11*c**2*d**5))/(8*d**4*e**4 + 16*d**3*e**5*x**2 + 8*d**2*e**6*x**4)

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Giac [A]  time = 1.19466, size = 196, normalized size = 1.26 \begin{align*} \frac{1}{3} \,{\left (c^{2} x^{3} e^{6} - 9 \, c^{2} d x e^{5}\right )} e^{\left (-9\right )} + \frac{{\left (35 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{9}{2}\right )}}{8 \, d^{\frac{5}{2}}} - \frac{{\left (13 \, c^{2} d^{4} x^{3} e + 11 \, c^{2} d^{5} x + 10 \, a c d^{2} x^{3} e^{3} + 6 \, a c d^{3} x e^{2} - 3 \, a^{2} x^{3} e^{5} - 5 \, a^{2} d x e^{4}\right )} e^{\left (-4\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)^2/(e*x^2+d)^3,x, algorithm="giac")

[Out]

1/3*(c^2*x^3*e^6 - 9*c^2*d*x*e^5)*e^(-9) + 1/8*(35*c^2*d^4 + 6*a*c*d^2*e^2 + 3*a^2*e^4)*arctan(x*e^(1/2)/sqrt(
d))*e^(-9/2)/d^(5/2) - 1/8*(13*c^2*d^4*x^3*e + 11*c^2*d^5*x + 10*a*c*d^2*x^3*e^3 + 6*a*c*d^3*x*e^2 - 3*a^2*x^3
*e^5 - 5*a^2*d*x*e^4)*e^(-4)/((x^2*e + d)^2*d^2)