\(\int \frac {(a+c x^4)^2}{(d+e x^2)^4} \, dx\) [309]

Optimal result

Integrand size = 19, antiderivative size = 187 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^2 x}{e^4}+\frac {\left (c d^2+a e^2\right )^2 x}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\left (19 c d^2-5 a e^2\right ) \left (c d^2+a e^2\right ) x}{24 d^2 e^4 \left (d+e x^2\right )^2}+\frac {\left (29 c^2 d^4+2 a c d^2 e^2+5 a^2 e^4\right ) x}{16 d^3 e^4 \left (d+e x^2\right )}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {x \left (-2 a c d^2 e^2 \left (3 d^2+8 d e x^2-3 e^2 x^4\right )+a^2 e^4 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )+c^2 d^3 \left (105 d^3+280 d^2 e x^2+231 d e^2 x^4+48 e^3 x^6\right )\right )}{48 d^3 e^4 \left (d+e x^2\right )^3}-\frac {\left (35 c^2 d^4-2 a c d^2 e^2-5 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{9/2}} \] Input:

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

Output:

(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))
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1472, 25, 2345, 27, 1471, 25, 25, 27, 299, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1472

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}-\frac {\int -\frac {\frac {6 c^2 d x^6}{e}-\frac {6 c^2 d^2 x^4}{e^2}+\frac {6 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+5 a^2-\frac {2 a c d^2}{e^2}-\frac {c^2 d^4}{e^4}}{\left (e x^2+d\right )^3}dx}{6 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\frac {6 c^2 d x^6}{e}-\frac {6 c^2 d^2 x^4}{e^2}+\frac {6 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+5 a^2-\frac {2 a c d^2}{e^2}-\frac {c^2 d^4}{e^4}}{\left (e x^2+d\right )^3}dx}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\int -\frac {3 \left (\frac {8 c^2 d^2 x^4}{e^2}-\frac {16 c^2 d^3 x^2}{e^3}+\frac {5 c^2 d^4+2 a c e^2 d^2+5 a^2 e^4}{e^4}\right )}{\left (e x^2+d\right )^2}dx}{4 d}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {\frac {5 c^2 d^4}{e^4}-\frac {16 c^2 x^2 d^3}{e^3}+\frac {8 c^2 x^4 d^2}{e^2}+\frac {2 a c d^2}{e^2}+5 a^2}{\left (e x^2+d\right )^2}dx}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {3 \left (\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}-\frac {\int -\frac {16 c^2 x^2 d^3+\left (-\frac {19 c^2 d^4}{e^4}+\frac {2 a c d^2}{e^2}+5 a^2\right ) e^3}{e^3 \left (e x^2+d\right )}dx}{2 d}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {\int -\frac {\frac {19 c^2 d^4}{e}-16 c^2 x^2 d^3-2 a c e d^2-5 a^2 e^3}{e^3 \left (e x^2+d\right )}dx}{2 d}+\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}-\frac {\int \frac {\frac {19 c^2 d^4}{e}-16 c^2 x^2 d^3-2 a c e d^2-5 a^2 e^3}{e^3 \left (e x^2+d\right )}dx}{2 d}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \left (\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}-\frac {\int \frac {\frac {19 c^2 d^4}{e}-16 c^2 x^2 d^3-2 a c e d^2-5 a^2 e^3}{e x^2+d}dx}{2 d e^3}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {3 \left (\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}-\frac {\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \int \frac {1}{e x^2+d}dx}{e}-\frac {16 c^2 d^3 x}{e}}{2 d e^3}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {3 \left (\frac {x \left (5 a^2+\frac {2 a c d^2}{e^2}+\frac {29 c^2 d^4}{e^4}\right )}{2 d \left (d+e x^2\right )}-\frac {\frac {\left (-5 a^2 e^4-2 a c d^2 e^2+35 c^2 d^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} e^{3/2}}-\frac {16 c^2 d^3 x}{e}}{2 d e^3}\right )}{4 d}+\frac {x \left (5 a^2-\frac {14 a c d^2}{e^2}-\frac {19 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (a e^2+c d^2\right )^2}{6 d e^4 \left (d+e x^2\right )^3}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(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] + Simp[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 1472

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi 
th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom 
ialRemainder[(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] + Simp[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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[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] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.96

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

Input:

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

Output:

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/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.54 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\left [\frac {96 \, c^{2} d^{4} e^{4} x^{7} + 6 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 16 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 6 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{96 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}, \frac {48 \, c^{2} d^{4} e^{4} x^{7} + 3 \, {\left (77 \, c^{2} d^{5} e^{3} + 2 \, a c d^{3} e^{5} + 5 \, a^{2} d e^{7}\right )} x^{5} + 8 \, {\left (35 \, c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + 5 \, a^{2} d^{2} e^{6}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{7} - 2 \, a c d^{5} e^{2} - 5 \, a^{2} d^{3} e^{4} + {\left (35 \, c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} - 5 \, a^{2} e^{7}\right )} x^{6} + 3 \, {\left (35 \, c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} - 5 \, a^{2} d e^{6}\right )} x^{4} + 3 \, {\left (35 \, c^{2} d^{6} e - 2 \, a c d^{4} e^{3} - 5 \, a^{2} d^{2} e^{5}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 3 \, {\left (35 \, c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + 11 \, a^{2} d^{3} e^{5}\right )} x}{48 \, {\left (d^{4} e^{8} x^{6} + 3 \, d^{5} e^{7} x^{4} + 3 \, d^{6} e^{6} x^{2} + d^{7} e^{5}\right )}}\right ] \] Input:

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

Output:

[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)*arctan(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 = 1.20 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \cdot \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (- d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{9}}} \cdot \left (5 a^{2} e^{4} + 2 a c d^{2} e^{2} - 35 c^{2} d^{4}\right ) \log {\left (d^{4} e^{4} \sqrt {- \frac {1}{d^{7} e^{9}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a^{2} e^{6} + 6 a c d^{2} e^{4} + 87 c^{2} d^{4} e^{2}\right ) + x^{3} \cdot \left (40 a^{2} d e^{5} - 16 a c d^{3} e^{3} + 136 c^{2} d^{5} e\right ) + x \left (33 a^{2} d^{2} e^{4} - 6 a c d^{4} e^{2} + 57 c^{2} d^{6}\right )}{48 d^{6} e^{4} + 144 d^{5} e^{5} x^{2} + 144 d^{4} e^{6} x^{4} + 48 d^{3} e^{7} x^{6}} \] Input:

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

Output:

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {{\left (35 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - 5 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{4}} + \frac {87 \, c^{2} d^{4} e^{2} x^{5} + 6 \, a c d^{2} e^{4} x^{5} + 15 \, a^{2} e^{6} x^{5} + 136 \, c^{2} d^{5} e x^{3} - 16 \, a c d^{3} e^{3} x^{3} + 40 \, a^{2} d e^{5} x^{3} + 57 \, c^{2} d^{6} x - 6 \, a c d^{4} e^{2} x + 33 \, a^{2} d^{2} e^{4} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{4}} \] Input:

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

Output:

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 = 17.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (5\,a^2\,e^5-2\,a\,c\,d^2\,e^3+17\,c^2\,d^4\,e\right )}{6\,d^2}+\frac {x\,\left (11\,a^2\,e^4-2\,a\,c\,d^2\,e^2+19\,c^2\,d^4\right )}{16\,d}+\frac {x^5\,\left (5\,a^2\,e^6+2\,a\,c\,d^2\,e^4+29\,c^2\,d^4\,e^2\right )}{16\,d^3}}{d^3\,e^4+3\,d^2\,e^5\,x^2+3\,d\,e^6\,x^4+e^7\,x^6}+\frac {c^2\,x}{e^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (5\,a^2\,e^4+2\,a\,c\,d^2\,e^2-35\,c^2\,d^4\right )}{16\,d^{7/2}\,e^{9/2}} \] Input:

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

Output:

((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))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.70 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^4} \, dx=\frac {15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} d^{3} e^{4}+45 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} d^{2} e^{5} x^{2}+45 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} d \,e^{6} x^{4}+15 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} e^{7} x^{6}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a c \,d^{5} e^{2}+18 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a c \,d^{4} e^{3} x^{2}+18 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a c \,d^{3} e^{4} x^{4}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a c \,d^{2} e^{5} x^{6}-105 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c^{2} d^{7}-315 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c^{2} d^{6} e \,x^{2}-315 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c^{2} d^{5} e^{2} x^{4}-105 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) c^{2} d^{4} e^{3} x^{6}+33 a^{2} d^{3} e^{5} x +40 a^{2} d^{2} e^{6} x^{3}+15 a^{2} d \,e^{7} x^{5}-6 a c \,d^{5} e^{3} x -16 a c \,d^{4} e^{4} x^{3}+6 a c \,d^{3} e^{5} x^{5}+105 c^{2} d^{7} e x +280 c^{2} d^{6} e^{2} x^{3}+231 c^{2} d^{5} e^{3} x^{5}+48 c^{2} d^{4} e^{4} x^{7}}{48 d^{4} e^{5} \left (e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}\right )} \] Input:

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

Output:

(15*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d**3*e**4 + 45*sqrt 
(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d**2*e**5*x**2 + 45*sqrt(e) 
*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d*e**6*x**4 + 15*sqrt(e)*sqrt( 
d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*e**7*x**6 + 6*sqrt(e)*sqrt(d)*atan(( 
e*x)/(sqrt(e)*sqrt(d)))*a*c*d**5*e**2 + 18*sqrt(e)*sqrt(d)*atan((e*x)/(sqr 
t(e)*sqrt(d)))*a*c*d**4*e**3*x**2 + 18*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e) 
*sqrt(d)))*a*c*d**3*e**4*x**4 + 6*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt 
(d)))*a*c*d**2*e**5*x**6 - 105*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d) 
))*c**2*d**7 - 315*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**6 
*e*x**2 - 315*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**5*e**2 
*x**4 - 105*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**4*e**3*x 
**6 + 33*a**2*d**3*e**5*x + 40*a**2*d**2*e**6*x**3 + 15*a**2*d*e**7*x**5 - 
 6*a*c*d**5*e**3*x - 16*a*c*d**4*e**4*x**3 + 6*a*c*d**3*e**5*x**5 + 105*c* 
*2*d**7*e*x + 280*c**2*d**6*e**2*x**3 + 231*c**2*d**5*e**3*x**5 + 48*c**2* 
d**4*e**4*x**7)/(48*d**4*e**5*(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3 
*x**6))