\(\int \frac {(a+c x^4)^2}{(d+e x^2)^4} \, dx\) [135]

Optimal result

Integrand size = 19, antiderivative size = 184 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \]

[Out]

c^2*x/e^4+1/6*(a*e^2+c*d^2)^2*x/d/e^4/(e*x^2+d)^3+1/24*(5*a^2-19*c^2*d^4/e^4-14*a*c*d^2/e^2)*x/d^2/(e*x^2+d)^2
+1/16*(5*a^2+29*c^2*d^4/e^4+2*a*c*d^2/e^2)*x/d^3/(e*x^2+d)-1/16*(-5*a^2*e^4-2*a*c*d^2*e^2+35*c^2*d^4)*arctan(x
*e^(1/2)/d^(1/2))/d^(7/2)/e^(9/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1172, 1828, 1171, 396, 211} \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=-\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{24 d^2 \left (d+e x^2\right )^2}+\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{16 d^3 \left (d+e x^2\right )}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}+\frac {c^2 x}{e^4} \]

[In]

Int[(a + c*x^4)^2/(d + e*x^2)^4,x]

[Out]

(c^2*x)/e^4 + ((c*d^2 + a*e^2)^2*x)/(6*d*e^4*(d + e*x^2)^3) + ((5*a^2 - (19*c^2*d^4)/e^4 - (14*a*c*d^2)/e^2)*x
)/(24*d^2*(d + e*x^2)^2) + ((5*a^2 + (29*c^2*d^4)/e^4 + (2*a*c*d^2)/e^2)*x)/(16*d^3*(d + e*x^2)) - ((35*c^2*d^
4 - 2*a*c*d^2*e^2 - 5*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1171

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

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}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\int \frac {-5 a^2+\frac {c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}-\frac {6 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+\frac {6 c^2 d^2 x^4}{e^2}-\frac {6 c^2 d x^6}{e}}{\left (d+e x^2\right )^3} \, dx}{6 d} \\ & = \frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\int \frac {3 \left (5 a^2+\frac {5 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right )-\frac {48 c^2 d^3 x^2}{e^3}+\frac {24 c^2 d^2 x^4}{e^2}}{\left (d+e x^2\right )^2} \, dx}{24 d^2} \\ & = \frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\int \frac {-3 \left (5 a^2-\frac {19 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right )-\frac {48 c^2 d^3 x^2}{e^3}}{d+e x^2} \, dx}{48 d^3} \\ & = \frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \int \frac {1}{d+e x^2} \, dx}{16 d^3 e^4} \\ & = \frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}+\frac {\left (5 a^2-\frac {19 c^2 d^4}{e^4}-\frac {14 a c d^2}{e^2}\right ) x}{24 d^2 \left (d+e x^2\right )^2}+\frac {\left (5 a^2+\frac {29 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}\right ) x}{16 d^3 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {x \left (-2 a c d^2 e^2 \left (3 d^2+8 d e x^2-3 e^2 x^4\right )+a^2 e^4 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )+c^2 d^3 \left (105 d^3+280 d^2 e x^2+231 d e^2 x^4+48 e^3 x^6\right )\right )}{48 d^3 e^4 \left (d+e x^2\right )^3}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \]

[In]

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

[Out]

(x*(-2*a*c*d^2*e^2*(3*d^2 + 8*d*e*x^2 - 3*e^2*x^4) + a^2*e^4*(33*d^2 + 40*d*e*x^2 + 15*e^2*x^4) + c^2*d^3*(105
*d^3 + 280*d^2*e*x^2 + 231*d*e^2*x^4 + 48*e^3*x^6)))/(48*d^3*e^4*(d + e*x^2)^3) - ((35*c^2*d^4 - 2*a*c*d^2*e^2
- 5*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7/2)*e^(9/2))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {c^{2} x}{e^{4}}+\frac {\frac {\frac {e^{2} \left (5 a^{2} e^{4}+2 a c \,d^{2} e^{2}+29 c^{2} d^{4}\right ) x^{5}}{16 d^{3}}+\frac {e \left (5 a^{2} e^{4}-2 a c \,d^{2} e^{2}+17 c^{2} d^{4}\right ) x^{3}}{6 d^{2}}+\frac {\left (11 a^{2} e^{4}-2 a c \,d^{2} e^{2}+19 c^{2} d^{4}\right ) x}{16 d}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a^{2} e^{4}+2 a c \,d^{2} e^{2}-35 c^{2} d^{4}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{16 d^{3} \sqrt {e d}}}{e^{4}}\)	\(179\)
risch	\(\frac {c^{2} x}{e^{4}}+\frac {\frac {e^{2} \left (5 a^{2} e^{4}+2 a c \,d^{2} e^{2}+29 c^{2} d^{4}\right ) x^{5}}{16 d^{3}}+\frac {e \left (5 a^{2} e^{4}-2 a c \,d^{2} e^{2}+17 c^{2} d^{4}\right ) x^{3}}{6 d^{2}}+\frac {\left (11 a^{2} e^{4}-2 a c \,d^{2} e^{2}+19 c^{2} d^{4}\right ) x}{16 d}}{e^{4} \left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) a^{2}}{32 \sqrt {-e d}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) a c}{16 e^{2} \sqrt {-e d}\, d}+\frac {35 d \ln \left (e x +\sqrt {-e d}\right ) c^{2}}{32 e^{4} \sqrt {-e d}}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) a^{2}}{32 \sqrt {-e d}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) a c}{16 e^{2} \sqrt {-e d}\, d}-\frac {35 d \ln \left (-e x +\sqrt {-e d}\right ) c^{2}}{32 e^{4} \sqrt {-e d}}\)	\(290\)

[In]

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

[Out]

c^2*x/e^4+1/e^4*((1/16*e^2*(5*a^2*e^4+2*a*c*d^2*e^2+29*c^2*d^4)/d^3*x^5+1/6*e*(5*a^2*e^4-2*a*c*d^2*e^2+17*c^2*
d^4)/d^2*x^3+1/16*(11*a^2*e^4-2*a*c*d^2*e^2+19*c^2*d^4)/d*x)/(e*x^2+d)^3+1/16*(5*a^2*e^4+2*a*c*d^2*e^2-35*c^2*
d^4)/d^3/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.60 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\left [\frac {96 \, c^{2} d^{4} e^{4} x^{7} + 6 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 16 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 6 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{96 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}, \frac {48 \, c^{2} d^{4} e^{4} x^{7} + 3 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 8 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 3 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{48 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}\right ] \]

[In]

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

[Out]

[1/96*(96*c^2*d^4*e^4*x^7 + 6*(77*c^2*d^5*e^3 + 2*a*c*d^3*e^5 + 5*a^2*d*e^7)*x^5 + 16*(35*c^2*d^6*e^2 - 2*a*c*
d^4*e^4 + 5*a^2*d^2*e^6)*x^3 + 3*(35*c^2*d^7 - 2*a*c*d^5*e^2 - 5*a^2*d^3*e^4 + (35*c^2*d^4*e^3 - 2*a*c*d^2*e^5
- 5*a^2*e^7)*x^6 + 3*(35*c^2*d^5*e^2 - 2*a*c*d^3*e^4 - 5*a^2*d*e^6)*x^4 + 3*(35*c^2*d^6*e - 2*a*c*d^4*e^3 - 5
*a^2*d^2*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^7*e - 2*a*c*d^5*e^3
+ 11*a^2*d^3*e^5)*x)/(d^4*e^8*x^6 + 3*d^5*e^7*x^4 + 3*d^6*e^6*x^2 + d^7*e^5), 1/48*(48*c^2*d^4*e^4*x^7 + 3*(77
*c^2*d^5*e^3 + 2*a*c*d^3*e^5 + 5*a^2*d*e^7)*x^5 + 8*(35*c^2*d^6*e^2 - 2*a*c*d^4*e^4 + 5*a^2*d^2*e^6)*x^3 - 3*(
35*c^2*d^7 - 2*a*c*d^5*e^2 - 5*a^2*d^3*e^4 + (35*c^2*d^4*e^3 - 2*a*c*d^2*e^5 - 5*a^2*e^7)*x^6 + 3*(35*c^2*d^5*
e^2 - 2*a*c*d^3*e^4 - 5*a^2*d*e^6)*x^4 + 3*(35*c^2*d^6*e - 2*a*c*d^4*e^3 - 5*a^2*d^2*e^5)*x^2)*sqrt(d*e)*arcta
n(sqrt(d*e)*x/d) + 3*(35*c^2*d^7*e - 2*a*c*d^5*e^3 + 11*a^2*d^3*e^5)*x)/(d^4*e^8*x^6 + 3*d^5*e^7*x^4 + 3*d^6*e
^6*x^2 + d^7*e^5)]

Sympy [A] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \cdot \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (- d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \cdot \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a^{2} e^{6} + 6 a c d^{2} e^{4} + 87 c^{2} d^{4} e^{2}\right ) + x^{3} \cdot \left (40 a^{2} d e^{5} - 16 a c d^{3} e^{3} + 136 c^{2} d^{5} e\right ) + x \left (33 a^{2} d^{2} e^{4} - 6 a c d^{4} e^{2} + 57 c^{2} d^{6}\right )}{48 d^{6} e^{4} + 144 d^{5} e^{5} x^{2} + 144 d^{4} e^{6} x^{4} + 48 d^{3} e^{7} x^{6}} \]

[In]

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

[Out]

c**2*x/e**4 - sqrt(-1/(d**7*e**9))*(5*a**2*e**4 + 2*a*c*d**2*e**2 - 35*c**2*d**4)*log(-d**4*e**4*sqrt(-1/(d**7
*e**9)) + x)/32 + sqrt(-1/(d**7*e**9))*(5*a**2*e**4 + 2*a*c*d**2*e**2 - 35*c**2*d**4)*log(d**4*e**4*sqrt(-1/(d
**7*e**9)) + x)/32 + (x**5*(15*a**2*e**6 + 6*a*c*d**2*e**4 + 87*c**2*d**4*e**2) + x**3*(40*a**2*d*e**5 - 16*a*
c*d**3*e**3 + 136*c**2*d**5*e) + x*(33*a**2*d**2*e**4 - 6*a*c*d**4*e**2 + 57*c**2*d**6))/(48*d**6*e**4 + 144*d
**5*e**5*x**2 + 144*d**4*e**6*x**4 + 48*d**3*e**7*x**6)

Maxima [F(-2)]

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

[In]

integrate((c*x^4+a)^2/(e*x^2+d)^4,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 = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {{\left (35 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 5 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{4}} + \frac {87 \, c^{2} d^{4} e^{2} x^{5} + 6 \, a c d^{2} e^{4} x^{5} + 15 \, a^{2} e^{6} x^{5} + 136 \, c^{2} d^{5} e x^{3} - 16 \, a c d^{3} e^{3} x^{3} + 40 \, a^{2} d e^{5} x^{3} + 57 \, c^{2} d^{6} x - 6 \, a c d^{4} e^{2} x + 33 \, a^{2} d^{2} e^{4} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{4}} \]

[In]

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

[Out]

c^2*x/e^4 - 1/16*(35*c^2*d^4 - 2*a*c*d^2*e^2 - 5*a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3*e^4) + 1/48*(87
*c^2*d^4*e^2*x^5 + 6*a*c*d^2*e^4*x^5 + 15*a^2*e^6*x^5 + 136*c^2*d^5*e*x^3 - 16*a*c*d^3*e^3*x^3 + 40*a^2*d*e^5*
x^3 + 57*c^2*d^6*x - 6*a*c*d^4*e^2*x + 33*a^2*d^2*e^4*x)/((e*x^2 + d)^3*d^3*e^4)

Mupad [B] (verification not implemented)

Time = 13.25 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (5\,a^2\,e^5-2\,a\,c\,d^2\,e^3+17\,c^2\,d^4\,e\right )}{6\,d^2}+\frac {x\,\left (11\,a^2\,e^4-2\,a\,c\,d^2\,e^2+19\,c^2\,d^4\right )}{16\,d}+\frac {x^5\,\left (5\,a^2\,e^6+2\,a\,c\,d^2\,e^4+29\,c^2\,d^4\,e^2\right )}{16\,d^3}}{d^3\,e^4+3\,d^2\,e^5\,x^2+3\,d\,e^6\,x^4+e^7\,x^6}+\frac {c^2\,x}{e^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (5\,a^2\,e^4+2\,a\,c\,d^2\,e^2-35\,c^2\,d^4\right )}{16\,d^{7/2}\,e^{9/2}} \]

[In]

int((a + c*x^4)^2/(d + e*x^2)^4,x)

[Out]

((x^3*(5*a^2*e^5 + 17*c^2*d^4*e - 2*a*c*d^2*e^3))/(6*d^2) + (x*(11*a^2*e^4 + 19*c^2*d^4 - 2*a*c*d^2*e^2))/(16*
d) + (x^5*(5*a^2*e^6 + 29*c^2*d^4*e^2 + 2*a*c*d^2*e^4))/(16*d^3))/(d^3*e^4 + e^7*x^6 + 3*d*e^6*x^4 + 3*d^2*e^5
*x^2) + (c^2*x)/e^4 + (atan((e^(1/2)*x)/d^(1/2))*(5*a^2*e^4 - 35*c^2*d^4 + 2*a*c*d^2*e^2))/(16*d^(7/2)*e^(9/2)
)