\(\int \frac {(a+c x^4)^2}{(d+e x^2)^5} \, dx\) [310]

Optimal result

Integrand size = 19, antiderivative size = 226 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\left (c d^2+a e^2\right )^2 x}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\left (25 c d^2-7 a e^2\right ) \left (c d^2+a e^2\right ) x}{48 d^2 e^4 \left (d+e x^2\right )^3}+\frac {\left (163 c^2 d^4+6 a c d^2 e^2+35 a^2 e^4\right ) x}{192 d^3 e^4 \left (d+e x^2\right )^2}-\frac {\left (93 c^2 d^4-6 a c d^2 e^2-35 a^2 e^4\right ) x}{128 d^4 e^4 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4+6 a c d^2 e^2+35 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{128 d^{9/2} e^{9/2}} \] Output:

1/8*(a*e^2+c*d^2)^2*x/d/e^4/(e*x^2+d)^4-1/48*(-7*a*e^2+25*c*d^2)*(a*e^2+c* 
d^2)*x/d^2/e^4/(e*x^2+d)^3+1/192*(35*a^2*e^4+6*a*c*d^2*e^2+163*c^2*d^4)*x/ 
d^3/e^4/(e*x^2+d)^2-1/128*(-35*a^2*e^4-6*a*c*d^2*e^2+93*c^2*d^4)*x/d^4/e^4 
/(e*x^2+d)+1/128*(35*a^2*e^4+6*a*c*d^2*e^2+35*c^2*d^4)*arctan(e^(1/2)*x/d^ 
(1/2))/d^(9/2)/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\frac {\sqrt {d} \sqrt {e} x \left (-6 a c d^2 e^2 \left (3 d^3+11 d^2 e x^2-11 d e^2 x^4-3 e^3 x^6\right )+a^2 e^4 \left (279 d^3+511 d^2 e x^2+385 d e^2 x^4+105 e^3 x^6\right )-c^2 d^4 \left (105 d^3+385 d^2 e x^2+511 d e^2 x^4+279 e^3 x^6\right )\right )}{\left (d+e x^2\right )^4}+3 \left (35 c^2 d^4+6 a c d^2 e^2+35 a^2 e^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{384 d^{9/2} e^{9/2}} \] Input:

Integrate[(a + c*x^4)^2/(d + e*x^2)^5,x]
 

Output:

((Sqrt[d]*Sqrt[e]*x*(-6*a*c*d^2*e^2*(3*d^3 + 11*d^2*e*x^2 - 11*d*e^2*x^4 - 
 3*e^3*x^6) + a^2*e^4*(279*d^3 + 511*d^2*e*x^2 + 385*d*e^2*x^4 + 105*e^3*x 
^6) - c^2*d^4*(105*d^3 + 385*d^2*e*x^2 + 511*d*e^2*x^4 + 279*e^3*x^6)))/(d 
 + e*x^2)^4 + 3*(35*c^2*d^4 + 6*a*c*d^2*e^2 + 35*a^2*e^4)*ArcTan[(Sqrt[e]* 
x)/Sqrt[d]])/(384*d^(9/2)*e^(9/2))
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1472, 25, 2345, 25, 1471, 27, 25, 298, 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.

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

\(\Big \downarrow \) 1472

\(\displaystyle \frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}-\frac {\int -\frac {\frac {8 c^2 d x^6}{e}-\frac {8 c^2 d^2 x^4}{e^2}+\frac {8 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+7 a^2-\frac {2 a c d^2}{e^2}-\frac {c^2 d^4}{e^4}}{\left (e x^2+d\right )^4}dx}{8 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\frac {8 c^2 d x^6}{e}-\frac {8 c^2 d^2 x^4}{e^2}+\frac {8 c d \left (c d^2+2 a e^2\right ) x^2}{e^3}+7 a^2-\frac {2 a c d^2}{e^2}-\frac {c^2 d^4}{e^4}}{\left (e x^2+d\right )^4}dx}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}-\frac {\int -\frac {\frac {19 c^2 d^4}{e^4}-\frac {96 c^2 x^2 d^3}{e^3}+\frac {48 c^2 x^4 d^2}{e^2}+\frac {6 a c d^2}{e^2}+35 a^2}{\left (e x^2+d\right )^3}dx}{6 d}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\frac {19 c^2 d^4}{e^4}-\frac {96 c^2 x^2 d^3}{e^3}+\frac {48 c^2 x^4 d^2}{e^2}+\frac {6 a c d^2}{e^2}+35 a^2}{\left (e x^2+d\right )^3}dx}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {\frac {x \left (35 a^2+\frac {6 a c d^2}{e^2}+\frac {163 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}-\frac {\int -\frac {3 \left (64 c^2 x^2 d^3+\left (-\frac {29 c^2 d^4}{e^4}+\frac {6 a c d^2}{e^2}+35 a^2\right ) e^3\right )}{e^3 \left (e x^2+d\right )^2}dx}{4 d}}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int -\frac {\frac {29 c^2 d^4}{e}-64 c^2 x^2 d^3-6 a c e d^2-35 a^2 e^3}{\left (e x^2+d\right )^2}dx}{4 d e^3}+\frac {x \left (35 a^2+\frac {6 a c d^2}{e^2}+\frac {163 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {x \left (35 a^2+\frac {6 a c d^2}{e^2}+\frac {163 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}-\frac {3 \int \frac {\frac {29 c^2 d^4}{e}-64 c^2 x^2 d^3-6 a c e d^2-35 a^2 e^3}{\left (e x^2+d\right )^2}dx}{4 d e^3}}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {\frac {x \left (35 a^2+\frac {6 a c d^2}{e^2}+\frac {163 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (-35 a^2 e^4-6 a c d^2 e^2+93 c^2 d^4\right )}{2 d e \left (d+e x^2\right )}-\frac {\left (35 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \int \frac {1}{e x^2+d}dx}{2 d e}\right )}{4 d e^3}}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {x \left (35 a^2+\frac {6 a c d^2}{e^2}+\frac {163 c^2 d^4}{e^4}\right )}{4 d \left (d+e x^2\right )^2}-\frac {3 \left (\frac {x \left (-35 a^2 e^4-6 a c d^2 e^2+93 c^2 d^4\right )}{2 d e \left (d+e x^2\right )}-\frac {\left (35 a^2 e^4+6 a c d^2 e^2+35 c^2 d^4\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{3/2}}\right )}{4 d e^3}}{6 d}+\frac {x \left (7 a^2-\frac {18 a c d^2}{e^2}-\frac {25 c^2 d^4}{e^4}\right )}{6 d \left (d+e x^2\right )^3}}{8 d}+\frac {x \left (a e^2+c d^2\right )^2}{8 d e^4 \left (d+e x^2\right )^4}\)

Input:

Int[(a + c*x^4)^2/(d + e*x^2)^5,x]
 

Output:

((c*d^2 + a*e^2)^2*x)/(8*d*e^4*(d + e*x^2)^4) + (((7*a^2 - (25*c^2*d^4)/e^ 
4 - (18*a*c*d^2)/e^2)*x)/(6*d*(d + e*x^2)^3) + (((35*a^2 + (163*c^2*d^4)/e 
^4 + (6*a*c*d^2)/e^2)*x)/(4*d*(d + e*x^2)^2) - (3*(((93*c^2*d^4 - 6*a*c*d^ 
2*e^2 - 35*a^2*e^4)*x)/(2*d*e*(d + e*x^2)) - ((35*c^2*d^4 + 6*a*c*d^2*e^2 
+ 35*a^2*e^4)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(3/2))))/(4*d*e^3) 
)/(6*d))/(8*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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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 = 0.15 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\frac {\left (35 a^{2} e^{4}+6 a c \,d^{2} e^{2}-93 c^{2} d^{4}\right ) x^{7}}{128 d^{4} e}+\frac {\left (385 a^{2} e^{4}+66 a c \,d^{2} e^{2}-511 c^{2} d^{4}\right ) x^{5}}{384 d^{3} e^{2}}+\frac {\left (511 a^{2} e^{4}-66 a c \,d^{2} e^{2}-385 c^{2} d^{4}\right ) x^{3}}{384 d^{2} e^{3}}+\frac {\left (93 a^{2} e^{4}-6 a c \,d^{2} e^{2}-35 c^{2} d^{4}\right ) x}{128 d \,e^{4}}}{\left (e \,x^{2}+d \right )^{4}}+\frac {\left (35 a^{2} e^{4}+6 a c \,d^{2} e^{2}+35 c^{2} d^{4}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{128 d^{4} e^{4} \sqrt {d e}}\)	\(212\)
risch	\(\frac {\frac {\left (35 a^{2} e^{4}+6 a c \,d^{2} e^{2}-93 c^{2} d^{4}\right ) x^{7}}{128 d^{4} e}+\frac {\left (385 a^{2} e^{4}+66 a c \,d^{2} e^{2}-511 c^{2} d^{4}\right ) x^{5}}{384 d^{3} e^{2}}+\frac {\left (511 a^{2} e^{4}-66 a c \,d^{2} e^{2}-385 c^{2} d^{4}\right ) x^{3}}{384 d^{2} e^{3}}+\frac {\left (93 a^{2} e^{4}-6 a c \,d^{2} e^{2}-35 c^{2} d^{4}\right ) x}{128 d \,e^{4}}}{\left (e \,x^{2}+d \right )^{4}}-\frac {35 \ln \left (e x +\sqrt {-d e}\right ) a^{2}}{256 \sqrt {-d e}\, d^{4}}-\frac {3 \ln \left (e x +\sqrt {-d e}\right ) a c}{128 \sqrt {-d e}\, e^{2} d^{2}}-\frac {35 \ln \left (e x +\sqrt {-d e}\right ) c^{2}}{256 \sqrt {-d e}\, e^{4}}+\frac {35 \ln \left (-e x +\sqrt {-d e}\right ) a^{2}}{256 \sqrt {-d e}\, d^{4}}+\frac {3 \ln \left (-e x +\sqrt {-d e}\right ) a c}{128 \sqrt {-d e}\, e^{2} d^{2}}+\frac {35 \ln \left (-e x +\sqrt {-d e}\right ) c^{2}}{256 \sqrt {-d e}\, e^{4}}\)	\(320\)

Input:

int((c*x^4+a)^2/(e*x^2+d)^5,x,method=_RETURNVERBOSE)
 

Output:

(1/128*(35*a^2*e^4+6*a*c*d^2*e^2-93*c^2*d^4)/d^4/e*x^7+1/384*(385*a^2*e^4+ 
66*a*c*d^2*e^2-511*c^2*d^4)/d^3/e^2*x^5+1/384*(511*a^2*e^4-66*a*c*d^2*e^2- 
385*c^2*d^4)/d^2/e^3*x^3+1/128*(93*a^2*e^4-6*a*c*d^2*e^2-35*c^2*d^4)/d/e^4 
*x)/(e*x^2+d)^4+1/128*(35*a^2*e^4+6*a*c*d^2*e^2+35*c^2*d^4)/d^4/e^4/(d*e)^ 
(1/2)*arctan(e*x/(d*e)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 806, normalized size of antiderivative = 3.57 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx =\text {Too large to display} \] Input:

integrate((c*x^4+a)^2/(e*x^2+d)^5,x, algorithm="fricas")
 

Output:

[-1/768*(6*(93*c^2*d^5*e^4 - 6*a*c*d^3*e^6 - 35*a^2*d*e^8)*x^7 + 2*(511*c^ 
2*d^6*e^3 - 66*a*c*d^4*e^5 - 385*a^2*d^2*e^7)*x^5 + 2*(385*c^2*d^7*e^2 + 6 
6*a*c*d^5*e^4 - 511*a^2*d^3*e^6)*x^3 + 3*(35*c^2*d^8 + 6*a*c*d^6*e^2 + 35* 
a^2*d^4*e^4 + (35*c^2*d^4*e^4 + 6*a*c*d^2*e^6 + 35*a^2*e^8)*x^8 + 4*(35*c^ 
2*d^5*e^3 + 6*a*c*d^3*e^5 + 35*a^2*d*e^7)*x^6 + 6*(35*c^2*d^6*e^2 + 6*a*c* 
d^4*e^4 + 35*a^2*d^2*e^6)*x^4 + 4*(35*c^2*d^7*e + 6*a*c*d^5*e^3 + 35*a^2*d 
^3*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^8*e + 6*a*c*d^6*e^3 - 93*a^2*d^4*e^5)*x)/(d^5*e^9*x^8 + 4*d^6*e^ 
8*x^6 + 6*d^7*e^7*x^4 + 4*d^8*e^6*x^2 + d^9*e^5), -1/384*(3*(93*c^2*d^5*e^ 
4 - 6*a*c*d^3*e^6 - 35*a^2*d*e^8)*x^7 + (511*c^2*d^6*e^3 - 66*a*c*d^4*e^5 
- 385*a^2*d^2*e^7)*x^5 + (385*c^2*d^7*e^2 + 66*a*c*d^5*e^4 - 511*a^2*d^3*e 
^6)*x^3 - 3*(35*c^2*d^8 + 6*a*c*d^6*e^2 + 35*a^2*d^4*e^4 + (35*c^2*d^4*e^4 
 + 6*a*c*d^2*e^6 + 35*a^2*e^8)*x^8 + 4*(35*c^2*d^5*e^3 + 6*a*c*d^3*e^5 + 3 
5*a^2*d*e^7)*x^6 + 6*(35*c^2*d^6*e^2 + 6*a*c*d^4*e^4 + 35*a^2*d^2*e^6)*x^4 
 + 4*(35*c^2*d^7*e + 6*a*c*d^5*e^3 + 35*a^2*d^3*e^5)*x^2)*sqrt(d*e)*arctan 
(sqrt(d*e)*x/d) + 3*(35*c^2*d^8*e + 6*a*c*d^6*e^3 - 93*a^2*d^4*e^5)*x)/(d^ 
5*e^9*x^8 + 4*d^6*e^8*x^6 + 6*d^7*e^7*x^4 + 4*d^8*e^6*x^2 + d^9*e^5)]
 

Sympy [A] (verification not implemented)

Time = 2.06 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=- \frac {\sqrt {- \frac {1}{d^{9} e^{9}}} \cdot \left (35 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (- d^{5} e^{4} \sqrt {- \frac {1}{d^{9} e^{9}}} + x \right )}}{256} + \frac {\sqrt {- \frac {1}{d^{9} e^{9}}} \cdot \left (35 a^{2} e^{4} + 6 a c d^{2} e^{2} + 35 c^{2} d^{4}\right ) \log {\left (d^{5} e^{4} \sqrt {- \frac {1}{d^{9} e^{9}}} + x \right )}}{256} + \frac {x^{7} \cdot \left (105 a^{2} e^{7} + 18 a c d^{2} e^{5} - 279 c^{2} d^{4} e^{3}\right ) + x^{5} \cdot \left (385 a^{2} d e^{6} + 66 a c d^{3} e^{4} - 511 c^{2} d^{5} e^{2}\right ) + x^{3} \cdot \left (511 a^{2} d^{2} e^{5} - 66 a c d^{4} e^{3} - 385 c^{2} d^{6} e\right ) + x \left (279 a^{2} d^{3} e^{4} - 18 a c d^{5} e^{2} - 105 c^{2} d^{7}\right )}{384 d^{8} e^{4} + 1536 d^{7} e^{5} x^{2} + 2304 d^{6} e^{6} x^{4} + 1536 d^{5} e^{7} x^{6} + 384 d^{4} e^{8} x^{8}} \] Input:

integrate((c*x**4+a)**2/(e*x**2+d)**5,x)
 

Output:

-sqrt(-1/(d**9*e**9))*(35*a**2*e**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4)*log( 
-d**5*e**4*sqrt(-1/(d**9*e**9)) + x)/256 + sqrt(-1/(d**9*e**9))*(35*a**2*e 
**4 + 6*a*c*d**2*e**2 + 35*c**2*d**4)*log(d**5*e**4*sqrt(-1/(d**9*e**9)) + 
 x)/256 + (x**7*(105*a**2*e**7 + 18*a*c*d**2*e**5 - 279*c**2*d**4*e**3) + 
x**5*(385*a**2*d*e**6 + 66*a*c*d**3*e**4 - 511*c**2*d**5*e**2) + x**3*(511 
*a**2*d**2*e**5 - 66*a*c*d**4*e**3 - 385*c**2*d**6*e) + x*(279*a**2*d**3*e 
**4 - 18*a*c*d**5*e**2 - 105*c**2*d**7))/(384*d**8*e**4 + 1536*d**7*e**5*x 
**2 + 2304*d**6*e**6*x**4 + 1536*d**5*e**7*x**6 + 384*d**4*e**8*x**8)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^4+a)^2/(e*x^2+d)^5,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 = 216, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {{\left (35 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + 35 \, a^{2} e^{4}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{128 \, \sqrt {d e} d^{4} e^{4}} - \frac {279 \, c^{2} d^{4} e^{3} x^{7} - 18 \, a c d^{2} e^{5} x^{7} - 105 \, a^{2} e^{7} x^{7} + 511 \, c^{2} d^{5} e^{2} x^{5} - 66 \, a c d^{3} e^{4} x^{5} - 385 \, a^{2} d e^{6} x^{5} + 385 \, c^{2} d^{6} e x^{3} + 66 \, a c d^{4} e^{3} x^{3} - 511 \, a^{2} d^{2} e^{5} x^{3} + 105 \, c^{2} d^{7} x + 18 \, a c d^{5} e^{2} x - 279 \, a^{2} d^{3} e^{4} x}{384 \, {\left (e x^{2} + d\right )}^{4} d^{4} e^{4}} \] Input:

integrate((c*x^4+a)^2/(e*x^2+d)^5,x, algorithm="giac")
 

Output:

1/128*(35*c^2*d^4 + 6*a*c*d^2*e^2 + 35*a^2*e^4)*arctan(e*x/sqrt(d*e))/(sqr 
t(d*e)*d^4*e^4) - 1/384*(279*c^2*d^4*e^3*x^7 - 18*a*c*d^2*e^5*x^7 - 105*a^ 
2*e^7*x^7 + 511*c^2*d^5*e^2*x^5 - 66*a*c*d^3*e^4*x^5 - 385*a^2*d*e^6*x^5 + 
 385*c^2*d^6*e*x^3 + 66*a*c*d^4*e^3*x^3 - 511*a^2*d^2*e^5*x^3 + 105*c^2*d^ 
7*x + 18*a*c*d^5*e^2*x - 279*a^2*d^3*e^4*x)/((e*x^2 + d)^4*d^4*e^4)
 

Mupad [B] (verification not implemented)

Time = 17.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,a^2\,e^4+6\,a\,c\,d^2\,e^2+35\,c^2\,d^4\right )}{128\,d^{9/2}\,e^{9/2}}-\frac {\frac {x\,\left (-93\,a^2\,e^4+6\,a\,c\,d^2\,e^2+35\,c^2\,d^4\right )}{128\,d\,e^4}-\frac {x^7\,\left (35\,a^2\,e^4+6\,a\,c\,d^2\,e^2-93\,c^2\,d^4\right )}{128\,d^4\,e}+\frac {x^3\,\left (-511\,a^2\,e^4+66\,a\,c\,d^2\,e^2+385\,c^2\,d^4\right )}{384\,d^2\,e^3}-\frac {x^5\,\left (385\,a^2\,e^4+66\,a\,c\,d^2\,e^2-511\,c^2\,d^4\right )}{384\,d^3\,e^2}}{d^4+4\,d^3\,e\,x^2+6\,d^2\,e^2\,x^4+4\,d\,e^3\,x^6+e^4\,x^8} \] Input:

int((a + c*x^4)^2/(d + e*x^2)^5,x)
                                                                                    
                                                                                    
 

Output:

(atan((e^(1/2)*x)/d^(1/2))*(35*a^2*e^4 + 35*c^2*d^4 + 6*a*c*d^2*e^2))/(128 
*d^(9/2)*e^(9/2)) - ((x*(35*c^2*d^4 - 93*a^2*e^4 + 6*a*c*d^2*e^2))/(128*d* 
e^4) - (x^7*(35*a^2*e^4 - 93*c^2*d^4 + 6*a*c*d^2*e^2))/(128*d^4*e) + (x^3* 
(385*c^2*d^4 - 511*a^2*e^4 + 66*a*c*d^2*e^2))/(384*d^2*e^3) - (x^5*(385*a^ 
2*e^4 - 511*c^2*d^4 + 66*a*c*d^2*e^2))/(384*d^3*e^2))/(d^4 + e^4*x^8 + 4*d 
^3*e*x^2 + 4*d*e^3*x^6 + 6*d^2*e^2*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 631, normalized size of antiderivative = 2.79 \[ \int \frac {\left (a+c x^4\right )^2}{\left (d+e x^2\right )^5} \, dx =\text {Too large to display} \] Input:

int((c*x^4+a)^2/(e*x^2+d)^5,x)
 

Output:

(105*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d**4*e**4 + 420*sq 
rt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d**3*e**5*x**2 + 630*sqrt 
(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d**2*e**6*x**4 + 420*sqrt(e 
)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*d*e**7*x**6 + 105*sqrt(e)*sqr 
t(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a**2*e**8*x**8 + 18*sqrt(e)*sqrt(d)*ata 
n((e*x)/(sqrt(e)*sqrt(d)))*a*c*d**6*e**2 + 72*sqrt(e)*sqrt(d)*atan((e*x)/( 
sqrt(e)*sqrt(d)))*a*c*d**5*e**3*x**2 + 108*sqrt(e)*sqrt(d)*atan((e*x)/(sqr 
t(e)*sqrt(d)))*a*c*d**4*e**4*x**4 + 72*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e) 
*sqrt(d)))*a*c*d**3*e**5*x**6 + 18*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqr 
t(d)))*a*c*d**2*e**6*x**8 + 105*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d 
)))*c**2*d**8 + 420*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d** 
7*e*x**2 + 630*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**6*e** 
2*x**4 + 420*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**5*e**3* 
x**6 + 105*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*c**2*d**4*e**4*x* 
*8 + 279*a**2*d**4*e**5*x + 511*a**2*d**3*e**6*x**3 + 385*a**2*d**2*e**7*x 
**5 + 105*a**2*d*e**8*x**7 - 18*a*c*d**6*e**3*x - 66*a*c*d**5*e**4*x**3 + 
66*a*c*d**4*e**5*x**5 + 18*a*c*d**3*e**6*x**7 - 105*c**2*d**8*e*x - 385*c* 
*2*d**7*e**2*x**3 - 511*c**2*d**6*e**3*x**5 - 279*c**2*d**5*e**4*x**7)/(38 
4*d**5*e**5*(d**4 + 4*d**3*e*x**2 + 6*d**2*e**2*x**4 + 4*d*e**3*x**6 + e** 
4*x**8))