\(\int \frac {x^2 (a+b x^4)^3}{(c+d x^4)^2} \, dx\) [214]

Optimal result

Integrand size = 22, antiderivative size = 255 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=-\frac {b^2 (2 b c-3 a d) x^3}{3 d^3}+\frac {b^3 x^7}{7 d^2}-\frac {(b c-a d)^3 x^3}{4 c d^3 \left (c+d x^4\right )}-\frac {(b c-a d)^2 (11 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{5/4} d^{15/4}}+\frac {(b c-a d)^2 (11 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{5/4} d^{15/4}}-\frac {(b c-a d)^2 (11 b c+a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{8 \sqrt {2} c^{5/4} d^{15/4}} \] Output:

-1/3*b^2*(-3*a*d+2*b*c)*x^3/d^3+1/7*b^3*x^7/d^2-1/4*(-a*d+b*c)^3*x^3/c/d^3 
/(d*x^4+c)+1/16*(-a*d+b*c)^2*(a*d+11*b*c)*arctan(-1+2^(1/2)*d^(1/4)*x/c^(1 
/4))*2^(1/2)/c^(5/4)/d^(15/4)+1/16*(-a*d+b*c)^2*(a*d+11*b*c)*arctan(1+2^(1 
/2)*d^(1/4)*x/c^(1/4))*2^(1/2)/c^(5/4)/d^(15/4)-1/16*(-a*d+b*c)^2*(a*d+11* 
b*c)*arctanh(2^(1/2)*c^(1/4)*d^(1/4)*x/(c^(1/2)+d^(1/2)*x^2))*2^(1/2)/c^(5 
/4)/d^(15/4)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\frac {-224 b^2 d^{3/4} (2 b c-3 a d) x^3+96 b^3 d^{7/4} x^7+\frac {168 d^{3/4} (-b c+a d)^3 x^3}{c \left (c+d x^4\right )}-\frac {42 \sqrt {2} (b c-a d)^2 (11 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{5/4}}+\frac {42 \sqrt {2} (b c-a d)^2 (11 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{5/4}}+\frac {21 \sqrt {2} (b c-a d)^2 (11 b c+a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{5/4}}-\frac {21 \sqrt {2} (b c-a d)^2 (11 b c+a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{5/4}}}{672 d^{15/4}} \] Input:

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

Output:

(-224*b^2*d^(3/4)*(2*b*c - 3*a*d)*x^3 + 96*b^3*d^(7/4)*x^7 + (168*d^(3/4)* 
(-(b*c) + a*d)^3*x^3)/(c*(c + d*x^4)) - (42*Sqrt[2]*(b*c - a*d)^2*(11*b*c 
+ a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(5/4) + (42*Sqrt[2]*(b*c 
 - a*d)^2*(11*b*c + a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(5/4) 
+ (21*Sqrt[2]*(b*c - a*d)^2*(11*b*c + a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d 
^(1/4)*x + Sqrt[d]*x^2])/c^(5/4) - (21*Sqrt[2]*(b*c - a*d)^2*(11*b*c + a*d 
)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/c^(5/4))/(672*d^ 
(15/4))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {968, 25, 1040, 2009}

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.

\(\displaystyle \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 968

\(\displaystyle -\frac {\int -\frac {x^2 \left (b x^4+a\right ) \left (b (11 b c-7 a d) x^4+a (3 b c+a d)\right )}{d x^4+c}dx}{4 c d}-\frac {x^3 \left (a+b x^4\right )^2 (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {x^2 \left (b x^4+a\right ) \left (b (11 b c-7 a d) x^4+a (3 b c+a d)\right )}{d x^4+c}dx}{4 c d}-\frac {x^3 \left (a+b x^4\right )^2 (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 1040

\(\displaystyle \frac {\int \left (\frac {b^2 (11 b c-7 a d) x^6}{d}-\frac {b \left (11 b^2 c^2-21 a b d c+6 a^2 d^2\right ) x^2}{d^2}+\frac {\left (11 b^3 c^3-21 a b^2 d c^2+9 a^2 b d^2 c+a^3 d^3\right ) x^2}{d^2 \left (d x^4+c\right )}\right )dx}{4 c d}-\frac {x^3 \left (a+b x^4\right )^2 (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {b x^3 \left (6 a^2 d^2-21 a b c d+11 b^2 c^2\right )}{3 d^2}-\frac {(b c-a d)^2 (a d+11 b c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {(b c-a d)^2 (a d+11 b c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} \sqrt [4]{c} d^{11/4}}+\frac {b^2 x^7 (11 b c-7 a d)}{7 d}+\frac {(b c-a d)^2 (a d+11 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} d^{11/4}}-\frac {(b c-a d)^2 (a d+11 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} d^{11/4}}}{4 c d}-\frac {x^3 \left (a+b x^4\right )^2 (b c-a d)}{4 c d \left (c+d x^4\right )}\)

Input:

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

Output:

-1/4*((b*c - a*d)*x^3*(a + b*x^4)^2)/(c*d*(c + d*x^4)) + (-1/3*(b*(11*b^2* 
c^2 - 21*a*b*c*d + 6*a^2*d^2)*x^3)/d^2 + (b^2*(11*b*c - 7*a*d)*x^7)/(7*d) 
- ((b*c - a*d)^2*(11*b*c + a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/( 
2*Sqrt[2]*c^(1/4)*d^(11/4)) + ((b*c - a*d)^2*(11*b*c + a*d)*ArcTan[1 + (Sq 
rt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*d^(11/4)) + ((b*c - a*d)^2*( 
11*b*c + a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*S 
qrt[2]*c^(1/4)*d^(11/4)) - ((b*c - a*d)^2*(11*b*c + a*d)*Log[Sqrt[c] + Sqr 
t[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*Sqrt[2]*c^(1/4)*d^(11/4)))/(4*c* 
d)
 

Defintions of rubi rules used

rule 25

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

rule 968

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) 
*((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1))   Int 
[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c 
*b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ 
n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, 
 x]
 

rule 1040

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ 
(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {b^{3} x^{7}}{7 d^{2}}+\frac {b^{2} a \,x^{3}}{d^{2}}-\frac {2 b^{3} c \,x^{3}}{3 d^{3}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x^{3}}{4 c \,d^{3} \left (d \,x^{4}+c \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{4}+c \right )}{\sum }\frac {\left (a^{3} d^{3}+9 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +11 b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{16 d^{4} c}\)	\(157\)
default	\(\frac {b^{2} \left (\frac {b d \,x^{7}}{7}+\frac {\left (3 a d -2 c b \right ) x^{3}}{3}\right )}{d^{3}}+\frac {\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x^{3}}{4 c \left (d \,x^{4}+c \right )}+\frac {\left (a^{3} d^{3}+9 a^{2} b c \,d^{2}-21 a \,b^{2} c^{2} d +11 b^{3} c^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c d \left (\frac {c}{d}\right )^{\frac {1}{4}}}}{d^{3}}\)	\(229\)

Input:

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

Output:

1/7*b^3*x^7/d^2+b^2/d^2*a*x^3-2/3*b^3/d^3*c*x^3+1/4*(a^3*d^3-3*a^2*b*c*d^2 
+3*a*b^2*c^2*d-b^3*c^3)/c*x^3/d^3/(d*x^4+c)+1/16/d^4/c*sum((a^3*d^3+9*a^2* 
b*c*d^2-21*a*b^2*c^2*d+11*b^3*c^3)/_R*ln(x-_R),_R=RootOf(_Z^4*d+c))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 2079, normalized size of antiderivative = 8.15 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/336*(48*b^3*c*d^2*x^11 - 16*(11*b^3*c^2*d - 21*a*b^2*c*d^2)*x^7 - 28*(11 
*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 3*a^3*d^3)*x^3 + 21*(c*d^4*x^4 
 + c^2*d^3)*(-(14641*b^12*c^12 - 111804*a*b^11*c^11*d + 368082*a^2*b^10*c^ 
10*d^2 - 676588*a^3*b^9*c^9*d^3 + 746703*a^4*b^8*c^8*d^4 - 486648*a^5*b^7* 
c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 5688*a^7*b^5*c^5*d^7 - 10017*a^8*b^4*c^ 
4*d^8 + 692*a^9*b^3*c^3*d^9 + 402*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a 
^12*d^12)/(c^5*d^15))^(1/4)*log(c^4*d^11*(-(14641*b^12*c^12 - 111804*a*b^1 
1*c^11*d + 368082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9*c^9*d^3 + 746703*a^4* 
b^8*c^8*d^4 - 486648*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 5688*a^7*b 
^5*c^5*d^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9*b^3*c^3*d^9 + 402*a^10*b^2*c^ 
2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)/(c^5*d^15))^(3/4) + (1331*b^9*c^9 - 
 7623*a*b^8*c^8*d + 17820*a^2*b^7*c^7*d^2 - 21372*a^3*b^6*c^6*d^3 + 13194* 
a^4*b^5*c^5*d^4 - 3186*a^5*b^4*c^4*d^5 - 372*a^6*b^3*c^3*d^6 + 180*a^7*b^2 
*c^2*d^7 + 27*a^8*b*c*d^8 + a^9*d^9)*x) - 21*(I*c*d^4*x^4 + I*c^2*d^3)*(-( 
14641*b^12*c^12 - 111804*a*b^11*c^11*d + 368082*a^2*b^10*c^10*d^2 - 676588 
*a^3*b^9*c^9*d^3 + 746703*a^4*b^8*c^8*d^4 - 486648*a^5*b^7*c^7*d^5 + 16018 
8*a^6*b^6*c^6*d^6 - 5688*a^7*b^5*c^5*d^7 - 10017*a^8*b^4*c^4*d^8 + 692*a^9 
*b^3*c^3*d^9 + 402*a^10*b^2*c^2*d^10 + 36*a^11*b*c*d^11 + a^12*d^12)/(c^5* 
d^15))^(1/4)*log(I*c^4*d^11*(-(14641*b^12*c^12 - 111804*a*b^11*c^11*d + 36 
8082*a^2*b^10*c^10*d^2 - 676588*a^3*b^9*c^9*d^3 + 746703*a^4*b^8*c^8*d^...
 

Sympy [A] (verification not implemented)

Time = 8.90 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.69 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\frac {b^{3} x^{7}}{7 d^{2}} + x^{3} \left (\frac {a b^{2}}{d^{2}} - \frac {2 b^{3} c}{3 d^{3}}\right ) + \frac {x^{3} \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{4 c^{2} d^{3} + 4 c d^{4} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} c^{5} d^{15} + a^{12} d^{12} + 36 a^{11} b c d^{11} + 402 a^{10} b^{2} c^{2} d^{10} + 692 a^{9} b^{3} c^{3} d^{9} - 10017 a^{8} b^{4} c^{4} d^{8} - 5688 a^{7} b^{5} c^{5} d^{7} + 160188 a^{6} b^{6} c^{6} d^{6} - 486648 a^{5} b^{7} c^{7} d^{5} + 746703 a^{4} b^{8} c^{8} d^{4} - 676588 a^{3} b^{9} c^{9} d^{3} + 368082 a^{2} b^{10} c^{10} d^{2} - 111804 a b^{11} c^{11} d + 14641 b^{12} c^{12}, \left ( t \mapsto t \log {\left (\frac {4096 t^{3} c^{4} d^{11}}{a^{9} d^{9} + 27 a^{8} b c d^{8} + 180 a^{7} b^{2} c^{2} d^{7} - 372 a^{6} b^{3} c^{3} d^{6} - 3186 a^{5} b^{4} c^{4} d^{5} + 13194 a^{4} b^{5} c^{5} d^{4} - 21372 a^{3} b^{6} c^{6} d^{3} + 17820 a^{2} b^{7} c^{7} d^{2} - 7623 a b^{8} c^{8} d + 1331 b^{9} c^{9}} + x \right )} \right )\right )} \] Input:

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

Output:

b**3*x**7/(7*d**2) + x**3*(a*b**2/d**2 - 2*b**3*c/(3*d**3)) + x**3*(a**3*d 
**3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(4*c**2*d**3 + 4*c*d* 
*4*x**4) + RootSum(65536*_t**4*c**5*d**15 + a**12*d**12 + 36*a**11*b*c*d** 
11 + 402*a**10*b**2*c**2*d**10 + 692*a**9*b**3*c**3*d**9 - 10017*a**8*b**4 
*c**4*d**8 - 5688*a**7*b**5*c**5*d**7 + 160188*a**6*b**6*c**6*d**6 - 48664 
8*a**5*b**7*c**7*d**5 + 746703*a**4*b**8*c**8*d**4 - 676588*a**3*b**9*c**9 
*d**3 + 368082*a**2*b**10*c**10*d**2 - 111804*a*b**11*c**11*d + 14641*b**1 
2*c**12, Lambda(_t, _t*log(4096*_t**3*c**4*d**11/(a**9*d**9 + 27*a**8*b*c* 
d**8 + 180*a**7*b**2*c**2*d**7 - 372*a**6*b**3*c**3*d**6 - 3186*a**5*b**4* 
c**4*d**5 + 13194*a**4*b**5*c**5*d**4 - 21372*a**3*b**6*c**6*d**3 + 17820* 
a**2*b**7*c**7*d**2 - 7623*a*b**8*c**8*d + 1331*b**9*c**9) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=-\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{3}}{4 \, {\left (c d^{4} x^{4} + c^{2} d^{3}\right )}} + \frac {3 \, b^{3} d x^{7} - 7 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{3}}{21 \, d^{3}} + \frac {{\left (11 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{32 \, c d^{3}} \] Input:

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

Output:

-1/4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^3/(c*d^4*x^4 + 
c^2*d^3) + 1/21*(3*b^3*d*x^7 - 7*(2*b^3*c - 3*a*b^2*d)*x^3)/d^3 + 1/32*(11 
*b^3*c^3 - 21*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*(2*sqrt(2)*arctan(1/2 
*sqrt(2)*(2*sqrt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(s 
qrt(sqrt(c)*sqrt(d))*sqrt(d)) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x 
- sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*s 
qrt(d)) - sqrt(2)*log(sqrt(d)*x^2 + sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/( 
c^(1/4)*d^(3/4)) + sqrt(2)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^(1/4)*x + s 
qrt(c))/(c^(1/4)*d^(3/4)))/(c*d^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (202) = 404\).

Time = 0.13 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.01 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=-\frac {b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{2} d x^{3} + 3 \, a^{2} b c d^{2} x^{3} - a^{3} d^{3} x^{3}}{4 \, {\left (d x^{4} + c\right )} c d^{3}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 21 \, \left (c d^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 9 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{6}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 21 \, \left (c d^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 9 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{6}} - \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 21 \, \left (c d^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 9 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{6}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 21 \, \left (c d^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 9 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{6}} + \frac {3 \, b^{3} d^{12} x^{7} - 14 \, b^{3} c d^{11} x^{3} + 21 \, a b^{2} d^{12} x^{3}}{21 \, d^{14}} \] Input:

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

Output:

-1/4*(b^3*c^3*x^3 - 3*a*b^2*c^2*d*x^3 + 3*a^2*b*c*d^2*x^3 - a^3*d^3*x^3)/( 
(d*x^4 + c)*c*d^3) + 1/16*sqrt(2)*(11*(c*d^3)^(3/4)*b^3*c^3 - 21*(c*d^3)^( 
3/4)*a*b^2*c^2*d + 9*(c*d^3)^(3/4)*a^2*b*c*d^2 + (c*d^3)^(3/4)*a^3*d^3)*ar 
ctan(1/2*sqrt(2)*(2*x + sqrt(2)*(c/d)^(1/4))/(c/d)^(1/4))/(c^2*d^6) + 1/16 
*sqrt(2)*(11*(c*d^3)^(3/4)*b^3*c^3 - 21*(c*d^3)^(3/4)*a*b^2*c^2*d + 9*(c*d 
^3)^(3/4)*a^2*b*c*d^2 + (c*d^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x - s 
qrt(2)*(c/d)^(1/4))/(c/d)^(1/4))/(c^2*d^6) - 1/32*sqrt(2)*(11*(c*d^3)^(3/4 
)*b^3*c^3 - 21*(c*d^3)^(3/4)*a*b^2*c^2*d + 9*(c*d^3)^(3/4)*a^2*b*c*d^2 + ( 
c*d^3)^(3/4)*a^3*d^3)*log(x^2 + sqrt(2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^ 
6) + 1/32*sqrt(2)*(11*(c*d^3)^(3/4)*b^3*c^3 - 21*(c*d^3)^(3/4)*a*b^2*c^2*d 
 + 9*(c*d^3)^(3/4)*a^2*b*c*d^2 + (c*d^3)^(3/4)*a^3*d^3)*log(x^2 - sqrt(2)* 
x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^6) + 1/21*(3*b^3*d^12*x^7 - 14*b^3*c*d^1 
1*x^3 + 21*a*b^2*d^12*x^3)/d^14
 

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 610, normalized size of antiderivative = 2.39 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=x^3\,\left (\frac {a\,b^2}{d^2}-\frac {2\,b^3\,c}{3\,d^3}\right )+\frac {b^3\,x^7}{7\,d^2}+\frac {x^3\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{4\,c\,\left (d^4\,x^4+c\,d^3\right )}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,x\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+11\,b\,c\right )\,\left (a^6\,d^6+18\,a^5\,b\,c\,d^5+39\,a^4\,b^2\,c^2\,d^4-356\,a^3\,b^3\,c^3\,d^3+639\,a^2\,b^4\,c^4\,d^2-462\,a\,b^5\,c^5\,d+121\,b^6\,c^6\right )}{{\left (-c\right )}^{1/4}\,\left (a^9\,d^9+27\,a^8\,b\,c\,d^8+180\,a^7\,b^2\,c^2\,d^7-372\,a^6\,b^3\,c^3\,d^6-3186\,a^5\,b^4\,c^4\,d^5+13194\,a^4\,b^5\,c^5\,d^4-21372\,a^3\,b^6\,c^6\,d^3+17820\,a^2\,b^7\,c^7\,d^2-7623\,a\,b^8\,c^8\,d+1331\,b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+11\,b\,c\right )}{8\,{\left (-c\right )}^{5/4}\,d^{15/4}}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,x\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+11\,b\,c\right )\,\left (a^6\,d^6+18\,a^5\,b\,c\,d^5+39\,a^4\,b^2\,c^2\,d^4-356\,a^3\,b^3\,c^3\,d^3+639\,a^2\,b^4\,c^4\,d^2-462\,a\,b^5\,c^5\,d+121\,b^6\,c^6\right )\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}\,\left (a^9\,d^9+27\,a^8\,b\,c\,d^8+180\,a^7\,b^2\,c^2\,d^7-372\,a^6\,b^3\,c^3\,d^6-3186\,a^5\,b^4\,c^4\,d^5+13194\,a^4\,b^5\,c^5\,d^4-21372\,a^3\,b^6\,c^6\,d^3+17820\,a^2\,b^7\,c^7\,d^2-7623\,a\,b^8\,c^8\,d+1331\,b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+11\,b\,c\right )\,1{}\mathrm {i}}{8\,{\left (-c\right )}^{5/4}\,d^{15/4}} \] Input:

int((x^2*(a + b*x^4)^3)/(c + d*x^4)^2,x)
 

Output:

x^3*((a*b^2)/d^2 - (2*b^3*c)/(3*d^3)) + (b^3*x^7)/(7*d^2) + (x^3*(a^3*d^3 
- b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(4*c*(c*d^3 + d^4*x^4)) - (ata 
n((d^(1/4)*x*(a*d - b*c)^2*(a*d + 11*b*c)*(a^6*d^6 + 121*b^6*c^6 + 639*a^2 
*b^4*c^4*d^2 - 356*a^3*b^3*c^3*d^3 + 39*a^4*b^2*c^2*d^4 - 462*a*b^5*c^5*d 
+ 18*a^5*b*c*d^5))/((-c)^(1/4)*(a^9*d^9 + 1331*b^9*c^9 + 17820*a^2*b^7*c^7 
*d^2 - 21372*a^3*b^6*c^6*d^3 + 13194*a^4*b^5*c^5*d^4 - 3186*a^5*b^4*c^4*d^ 
5 - 372*a^6*b^3*c^3*d^6 + 180*a^7*b^2*c^2*d^7 - 7623*a*b^8*c^8*d + 27*a^8* 
b*c*d^8)))*(a*d - b*c)^2*(a*d + 11*b*c))/(8*(-c)^(5/4)*d^(15/4)) - (atan(( 
d^(1/4)*x*(a*d - b*c)^2*(a*d + 11*b*c)*(a^6*d^6 + 121*b^6*c^6 + 639*a^2*b^ 
4*c^4*d^2 - 356*a^3*b^3*c^3*d^3 + 39*a^4*b^2*c^2*d^4 - 462*a*b^5*c^5*d + 1 
8*a^5*b*c*d^5)*1i)/((-c)^(1/4)*(a^9*d^9 + 1331*b^9*c^9 + 17820*a^2*b^7*c^7 
*d^2 - 21372*a^3*b^6*c^6*d^3 + 13194*a^4*b^5*c^5*d^4 - 3186*a^5*b^4*c^4*d^ 
5 - 372*a^6*b^3*c^3*d^6 + 180*a^7*b^2*c^2*d^7 - 7623*a*b^8*c^8*d + 27*a^8* 
b*c*d^8)))*(a*d - b*c)^2*(a*d + 11*b*c)*1i)/(8*(-c)^(5/4)*d^(15/4))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1355, normalized size of antiderivative = 5.31 \[ \int \frac {x^2 \left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^2*(b*x^4+a)^3/(d*x^4+c)^2,x)
 

Output:

( - 42*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt( 
d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*c*d**3 - 42*d**(1/4)*c**(3/4)*sqrt 
(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt 
(2)))*a**3*d**4*x**4 - 378*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/ 
4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c**2*d**2 - 
378*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)* 
x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c*d**3*x**4 + 882*d**(1/4)*c**(3/4) 
*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4) 
*sqrt(2)))*a*b**2*c**3*d + 882*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c* 
*(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a*b**2*c**2*d** 
2*x**4 - 462*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2 
*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*b**3*c**4 - 462*d**(1/4)*c**(3/4) 
*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4) 
*sqrt(2)))*b**3*c**3*d*x**4 + 42*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)* 
c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*c*d**3 + 
 42*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)* 
x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*d**4*x**4 + 378*d**(1/4)*c**(3/4)*sqr 
t(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqr 
t(2)))*a**2*b*c**2*d**2 + 378*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c** 
(1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c*d**3...