\(\int \frac {(a+c x^4)^2}{(d+e x^2)^3} \, dx\) [134]

Optimal result

Integrand size = 19, antiderivative size = 155 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3}+\frac {\left (c d^2+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}+\frac {\left (3 a^2-\frac {13 c^2 d^4}{e^4}-\frac {10 a c d^2}{e^2}\right ) x}{8 d^2 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1172, 1828, 1167, 211} \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\frac {\left (3 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}}+\frac {x \left (3 a^2-\frac {10 a c d^2}{e^2}-\frac {13 c^2 d^4}{e^4}\right )}{8 d^2 \left (d+e x^2\right )}+\frac {x \left (a e^2+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac {3 c^2 d x}{e^4}+\frac {c^2 x^3}{3 e^3} \]

[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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1167

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 1172

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 1828

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\frac {x \left (3 a^2 e^4 \left (5 d+3 e x^2\right )-6 a c d^2 e^2 \left (3 d+5 e x^2\right )-c^2 d^2 \left (105 d^3+175 d^2 e x^2+56 d e^2 x^4-8 e^3 x^6\right )\right )}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac {\left (35 c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}} \]

[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))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {c^{2} \left (-\frac {1}{3} e \,x^{3}+3 d x \right )}{e^{4}}+\frac {\frac {\frac {e \left (3 a^{2} e^{4}-10 a c \,d^{2} e^{2}-13 c^{2} d^{4}\right ) x^{3}}{8 d^{2}}+\frac {\left (5 a^{2} e^{4}-6 a c \,d^{2} e^{2}-11 c^{2} d^{4}\right ) x}{8 d}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (3 a^{2} e^{4}+6 a c \,d^{2} e^{2}+35 c^{2} d^{4}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 d^{2} \sqrt {e d}}}{e^{4}}\)	\(152\)
risch	\(\frac {c^{2} x^{3}}{3 e^{3}}-\frac {3 c^{2} d x}{e^{4}}+\frac {\frac {e \left (3 a^{2} e^{4}-10 a c \,d^{2} e^{2}-13 c^{2} d^{4}\right ) x^{3}}{8 d^{2}}+\frac {\left (5 a^{2} e^{4}-6 a c \,d^{2} e^{2}-11 c^{2} d^{4}\right ) x}{8 d}}{e^{4} \left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a^{2}}{16 \sqrt {-e d}\, d^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a c}{8 e^{2} \sqrt {-e d}}-\frac {35 d^{2} \ln \left (e x +\sqrt {-e d}\right ) c^{2}}{16 e^{4} \sqrt {-e d}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a^{2}}{16 \sqrt {-e d}\, d^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a c}{8 e^{2} \sqrt {-e d}}+\frac {35 d^{2} \ln \left (-e x +\sqrt {-e d}\right ) c^{2}}{16 e^{4} \sqrt {-e d}}\)	\(263\)

[In]

int((c*x^4+a)^2/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.33 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\left [\frac {16 \, c^{2} d^{3} e^{4} x^{7} - 112 \, c^{2} d^{4} e^{3} x^{5} - 2 \, {\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} + {\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d e^{5}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (35 \, c^{2} d^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{48 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} e^{5}\right )}}, \frac {8 \, c^{2} d^{3} e^{4} x^{7} - 56 \, c^{2} d^{4} e^{3} x^{5} - {\left (175 \, c^{2} d^{5} e^{2} + 30 \, a c d^{3} e^{4} - 9 \, a^{2} d e^{6}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{6} + 6 \, a c d^{4} e^{2} + 3 \, a^{2} d^{2} e^{4} + {\left (35 \, c^{2} d^{4} e^{2} + 6 \, a c d^{2} e^{4} + 3 \, a^{2} e^{6}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e + 6 \, a c d^{3} e^{3} + 3 \, a^{2} d e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (35 \, c^{2} d^{6} e + 6 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x}{24 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} e^{5}\right )}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.66 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=- \frac {3 c^{2} d x}{e^{4}} + \frac {c^{2} x^{3}}{3 e^{3}} - \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \cdot \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (- d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \cdot \left (3 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a^{2} e^{5} - 10 a c d^{2} e^{3} - 13 c^{2} d^{4} e\right ) + x \left (5 a^{2} d e^{4} - 6 a c d^{3} e^{2} - 11 c^{2} d^{5}\right )}{8 d^{4} e^{4} + 16 d^{3} e^{5} x^{2} + 8 d^{2} e^{6} x^{4}} \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\frac {{\left (35 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{4}} - \frac {13 \, c^{2} d^{4} e x^{3} + 10 \, a c d^{2} e^{3} x^{3} - 3 \, a^{2} e^{5} x^{3} + 11 \, c^{2} d^{5} x + 6 \, a c d^{3} e^{2} x - 5 \, a^{2} d e^{4} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{4}} + \frac {c^{2} e^{6} x^{3} - 9 \, c^{2} d e^{5} x}{3 \, e^{9}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx=\frac {c^2\,x^3}{3\,e^3}-\frac {\frac {x^3\,\left (-3\,a^2\,e^5+10\,a\,c\,d^2\,e^3+13\,c^2\,d^4\,e\right )}{8\,d^2}+\frac {x\,\left (-5\,a^2\,e^4+6\,a\,c\,d^2\,e^2+11\,c^2\,d^4\right )}{8\,d}}{d^2\,e^4+2\,d\,e^5\,x^2+e^6\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,a^2\,e^4+6\,a\,c\,d^2\,e^2+35\,c^2\,d^4\right )}{8\,d^{5/2}\,e^{9/2}}-\frac {3\,c^2\,d\,x}{e^4} \]

[In]

int((a + c*x^4)^2/(d + e*x^2)^3,x)

[Out]

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