\(\int \frac {(d+e x^2)^3}{a+c x^4} \, dx\) [312]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [B] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 278 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\frac {3 d e^2 x}{c}+\frac {e^3 x^3}{3 c}-\frac {\left (\sqrt {c} d \left (c d^2-3 a e^2\right )+\sqrt {a} e \left (3 c d^2-a e^2\right )\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (\sqrt {c} d \left (c d^2-3 a e^2\right )+\sqrt {a} e \left (3 c d^2-a e^2\right )\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{7/4}}+\frac {\left (\sqrt {c} d \left (c d^2-3 a e^2\right )-\sqrt {a} e \left (3 c d^2-a e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{2 \sqrt {2} a^{3/4} c^{7/4}} \] Output:

3*d*e^2*x/c+1/3*e^3*x^3/c+1/4*(c^(1/2)*d*(-3*a*e^2+c*d^2)+a^(1/2)*e*(-a*e^ 
2+3*c*d^2))*arctan(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(3/4)/c^(7/4)+1 
/4*(c^(1/2)*d*(-3*a*e^2+c*d^2)+a^(1/2)*e*(-a*e^2+3*c*d^2))*arctan(1+2^(1/2 
)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(3/4)/c^(7/4)+1/4*(c^(1/2)*d*(-3*a*e^2+c*d^ 
2)-a^(1/2)*e*(-a*e^2+3*c*d^2))*arctanh(2^(1/2)*a^(1/4)*c^(1/4)*x/(a^(1/2)+ 
c^(1/2)*x^2))*2^(1/2)/a^(3/4)/c^(7/4)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.29 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\frac {72 a^{3/4} c^{3/4} d e^2 x+8 a^{3/4} c^{3/4} e^3 x^3+6 \sqrt {2} \left (-c^{3/2} d^3-3 \sqrt {a} c d^2 e+3 a \sqrt {c} d e^2+a^{3/2} e^3\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+6 \sqrt {2} \left (c^{3/2} d^3+3 \sqrt {a} c d^2 e-3 a \sqrt {c} d e^2-a^{3/2} e^3\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-3 \sqrt {2} \left (c^{3/2} d^3-3 \sqrt {a} c d^2 e-3 a \sqrt {c} d e^2+a^{3/2} e^3\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+3 \sqrt {2} \left (c^{3/2} d^3-3 \sqrt {a} c d^2 e-3 a \sqrt {c} d e^2+a^{3/2} e^3\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{24 a^{3/4} c^{7/4}} \] Input:

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

Output:

(72*a^(3/4)*c^(3/4)*d*e^2*x + 8*a^(3/4)*c^(3/4)*e^3*x^3 + 6*Sqrt[2]*(-(c^( 
3/2)*d^3) - 3*Sqrt[a]*c*d^2*e + 3*a*Sqrt[c]*d*e^2 + a^(3/2)*e^3)*ArcTan[1 
- (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 6*Sqrt[2]*(c^(3/2)*d^3 + 3*Sqrt[a]*c*d^2* 
e - 3*a*Sqrt[c]*d*e^2 - a^(3/2)*e^3)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4 
)] - 3*Sqrt[2]*(c^(3/2)*d^3 - 3*Sqrt[a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 + a^(3 
/2)*e^3)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + 3*Sqrt[2 
]*(c^(3/2)*d^3 - 3*Sqrt[a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 + a^(3/2)*e^3)*Log[ 
Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(24*a^(3/4)*c^(7/4))
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1485, 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.

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

\(\Big \downarrow \) 1485

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(3*d*e^2*x)/c + (e^3*x^3)/(3*c) - ((Sqrt[c]*d*(c*d^2 - 3*a*e^2) + Sqrt[a]* 
e*(3*c*d^2 - a*e^2))*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a 
^(3/4)*c^(7/4)) + ((Sqrt[c]*d*(c*d^2 - 3*a*e^2) + Sqrt[a]*e*(3*c*d^2 - a*e 
^2))*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*c^(7/4)) 
- ((Sqrt[c]*d*(c*d^2 - 3*a*e^2) - Sqrt[a]*e*(3*c*d^2 - a*e^2))*Log[Sqrt[a] 
 - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*c^(7/4)) + 
 ((Sqrt[c]*d*(c*d^2 - 3*a*e^2) - Sqrt[a]*e*(3*c*d^2 - a*e^2))*Log[Sqrt[a] 
+ Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*c^(7/4))
 

Defintions of rubi rules used

rule 1485
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[Expa 
ndIntegrand[(d + e*x^2)^q/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e}, x] && 
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]
 

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 = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.29

method result size
risch \(\frac {e^{3} x^{3}}{3 c}+\frac {3 d \,e^{2} x}{c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (e \left (-a \,e^{2}+3 c \,d^{2}\right ) \textit {\_R}^{2}-3 d \,e^{2} a +d^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 c^{2}}\) \(80\)
default \(\frac {e^{2} \left (\frac {1}{3} e \,x^{3}+3 d x \right )}{c}+\frac {\frac {\left (-3 d \,e^{2} a +d^{3} c \right ) \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {\left (-a \,e^{3}+3 c \,d^{2} e \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}}{c}\) \(254\)

Input:

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

Output:

1/3*e^3*x^3/c+3*d*e^2*x/c+1/4/c^2*sum((e*(-a*e^2+3*c*d^2)*_R^2-3*d*e^2*a+d 
^3*c)/_R^3*ln(x-_R),_R=RootOf(_Z^4*c+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2133 vs. \(2 (217) = 434\).

Time = 2.04 (sec) , antiderivative size = 2133, normalized size of antiderivative = 7.67 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*(4*e^3*x^3 + 36*d*e^2*x - 3*c*sqrt(-(6*c^2*d^5*e - 20*a*c*d^3*e^3 + 6 
*a^2*d*e^5 + a*c^3*sqrt(-(c^6*d^12 - 30*a*c^5*d^10*e^2 + 255*a^2*c^4*d^8*e 
^4 - 452*a^3*c^3*d^6*e^6 + 255*a^4*c^2*d^4*e^8 - 30*a^5*c*d^2*e^10 + a^6*e 
^12)/(a^3*c^7)))/(a*c^3))*log(-(c^6*d^12 - 12*a*c^5*d^10*e^2 - 27*a^2*c^4* 
d^8*e^4 + 27*a^4*c^2*d^4*e^8 + 12*a^5*c*d^2*e^10 - a^6*e^12)*x + (a*c^6*d^ 
9 - 18*a^2*c^5*d^7*e^2 + 60*a^3*c^4*d^5*e^4 - 46*a^4*c^3*d^3*e^6 + 3*a^5*c 
^2*d*e^8 + (3*a^3*c^6*d^2*e - a^4*c^5*e^3)*sqrt(-(c^6*d^12 - 30*a*c^5*d^10 
*e^2 + 255*a^2*c^4*d^8*e^4 - 452*a^3*c^3*d^6*e^6 + 255*a^4*c^2*d^4*e^8 - 3 
0*a^5*c*d^2*e^10 + a^6*e^12)/(a^3*c^7)))*sqrt(-(6*c^2*d^5*e - 20*a*c*d^3*e 
^3 + 6*a^2*d*e^5 + a*c^3*sqrt(-(c^6*d^12 - 30*a*c^5*d^10*e^2 + 255*a^2*c^4 
*d^8*e^4 - 452*a^3*c^3*d^6*e^6 + 255*a^4*c^2*d^4*e^8 - 30*a^5*c*d^2*e^10 + 
 a^6*e^12)/(a^3*c^7)))/(a*c^3))) + 3*c*sqrt(-(6*c^2*d^5*e - 20*a*c*d^3*e^3 
 + 6*a^2*d*e^5 + a*c^3*sqrt(-(c^6*d^12 - 30*a*c^5*d^10*e^2 + 255*a^2*c^4*d 
^8*e^4 - 452*a^3*c^3*d^6*e^6 + 255*a^4*c^2*d^4*e^8 - 30*a^5*c*d^2*e^10 + a 
^6*e^12)/(a^3*c^7)))/(a*c^3))*log(-(c^6*d^12 - 12*a*c^5*d^10*e^2 - 27*a^2* 
c^4*d^8*e^4 + 27*a^4*c^2*d^4*e^8 + 12*a^5*c*d^2*e^10 - a^6*e^12)*x - (a*c^ 
6*d^9 - 18*a^2*c^5*d^7*e^2 + 60*a^3*c^4*d^5*e^4 - 46*a^4*c^3*d^3*e^6 + 3*a 
^5*c^2*d*e^8 + (3*a^3*c^6*d^2*e - a^4*c^5*e^3)*sqrt(-(c^6*d^12 - 30*a*c^5* 
d^10*e^2 + 255*a^2*c^4*d^8*e^4 - 452*a^3*c^3*d^6*e^6 + 255*a^4*c^2*d^4*e^8 
 - 30*a^5*c*d^2*e^10 + a^6*e^12)/(a^3*c^7)))*sqrt(-(6*c^2*d^5*e - 20*a*...
 

Sympy [A] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.26 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{3} c^{7} + t^{2} \cdot \left (192 a^{4} c^{4} d e^{5} - 640 a^{3} c^{5} d^{3} e^{3} + 192 a^{2} c^{6} d^{5} e\right ) + a^{6} e^{12} + 6 a^{5} c d^{2} e^{10} + 15 a^{4} c^{2} d^{4} e^{8} + 20 a^{3} c^{3} d^{6} e^{6} + 15 a^{2} c^{4} d^{8} e^{4} + 6 a c^{5} d^{10} e^{2} + c^{6} d^{12}, \left ( t \mapsto t \log {\left (x + \frac {- 64 t^{3} a^{4} c^{5} e^{3} + 192 t^{3} a^{3} c^{6} d^{2} e - 36 t a^{5} c^{2} d e^{8} + 336 t a^{4} c^{3} d^{3} e^{6} - 504 t a^{3} c^{4} d^{5} e^{4} + 144 t a^{2} c^{5} d^{7} e^{2} - 4 t a c^{6} d^{9}}{a^{6} e^{12} - 12 a^{5} c d^{2} e^{10} - 27 a^{4} c^{2} d^{4} e^{8} + 27 a^{2} c^{4} d^{8} e^{4} + 12 a c^{5} d^{10} e^{2} - c^{6} d^{12}} \right )} \right )\right )} + \frac {3 d e^{2} x}{c} + \frac {e^{3} x^{3}}{3 c} \] Input:

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

Output:

RootSum(256*_t**4*a**3*c**7 + _t**2*(192*a**4*c**4*d*e**5 - 640*a**3*c**5* 
d**3*e**3 + 192*a**2*c**6*d**5*e) + a**6*e**12 + 6*a**5*c*d**2*e**10 + 15* 
a**4*c**2*d**4*e**8 + 20*a**3*c**3*d**6*e**6 + 15*a**2*c**4*d**8*e**4 + 6* 
a*c**5*d**10*e**2 + c**6*d**12, Lambda(_t, _t*log(x + (-64*_t**3*a**4*c**5 
*e**3 + 192*_t**3*a**3*c**6*d**2*e - 36*_t*a**5*c**2*d*e**8 + 336*_t*a**4* 
c**3*d**3*e**6 - 504*_t*a**3*c**4*d**5*e**4 + 144*_t*a**2*c**5*d**7*e**2 - 
 4*_t*a*c**6*d**9)/(a**6*e**12 - 12*a**5*c*d**2*e**10 - 27*a**4*c**2*d**4* 
e**8 + 27*a**2*c**4*d**8*e**4 + 12*a*c**5*d**10*e**2 - c**6*d**12)))) + 3* 
d*e**2*x/c + e**3*x**3/(3*c)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\frac {e^{3} x^{3} + 9 \, d e^{2} x}{3 \, c} + \frac {\frac {2 \, \sqrt {2} {\left (c^{\frac {3}{2}} d^{3} + 3 \, \sqrt {a} c d^{2} e - 3 \, a \sqrt {c} d e^{2} - a^{\frac {3}{2}} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (c^{\frac {3}{2}} d^{3} + 3 \, \sqrt {a} c d^{2} e - 3 \, a \sqrt {c} d e^{2} - a^{\frac {3}{2}} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {\sqrt {2} {\left (c^{\frac {3}{2}} d^{3} - 3 \, \sqrt {a} c d^{2} e - 3 \, a \sqrt {c} d e^{2} + a^{\frac {3}{2}} e^{3}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (c^{\frac {3}{2}} d^{3} - 3 \, \sqrt {a} c d^{2} e - 3 \, a \sqrt {c} d e^{2} + a^{\frac {3}{2}} e^{3}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}}{8 \, c} \] Input:

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

Output:

1/3*(e^3*x^3 + 9*d*e^2*x)/c + 1/8*(2*sqrt(2)*(c^(3/2)*d^3 + 3*sqrt(a)*c*d^ 
2*e - 3*a*sqrt(c)*d*e^2 - a^(3/2)*e^3)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x + s 
qrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt( 
c))*sqrt(c)) + 2*sqrt(2)*(c^(3/2)*d^3 + 3*sqrt(a)*c*d^2*e - 3*a*sqrt(c)*d* 
e^2 - a^(3/2)*e^3)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/ 
4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + sqrt( 
2)*(c^(3/2)*d^3 - 3*sqrt(a)*c*d^2*e - 3*a*sqrt(c)*d*e^2 + a^(3/2)*e^3)*log 
(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(3/4)) - sq 
rt(2)*(c^(3/2)*d^3 - 3*sqrt(a)*c*d^2*e - 3*a*sqrt(c)*d*e^2 + a^(3/2)*e^3)* 
log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(3/4)))/ 
c
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.48 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\frac {c^{2} e^{3} x^{3} + 9 \, c^{2} d e^{2} x}{3 \, c^{3}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{3} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d e^{2} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{2} e - \left (a c^{3}\right )^{\frac {3}{4}} a e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{4}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{3} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d e^{2} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{2} e - \left (a c^{3}\right )^{\frac {3}{4}} a e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{4}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{3} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d e^{2} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{2} e + \left (a c^{3}\right )^{\frac {3}{4}} a e^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{4}} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{3} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {1}{4}} a c^{2} d e^{2} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} c d^{2} e + \left (a c^{3}\right )^{\frac {3}{4}} a e^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{4}} \] Input:

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

Output:

1/3*(c^2*e^3*x^3 + 9*c^2*d*e^2*x)/c^3 + 1/4*sqrt(2)*((a*c^3)^(1/4)*c^3*d^3 
 - 3*(a*c^3)^(1/4)*a*c^2*d*e^2 + 3*(a*c^3)^(3/4)*c*d^2*e - (a*c^3)^(3/4)*a 
*e^3)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^4) 
+ 1/4*sqrt(2)*((a*c^3)^(1/4)*c^3*d^3 - 3*(a*c^3)^(1/4)*a*c^2*d*e^2 + 3*(a* 
c^3)^(3/4)*c*d^2*e - (a*c^3)^(3/4)*a*e^3)*arctan(1/2*sqrt(2)*(2*x - sqrt(2 
)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^4) + 1/8*sqrt(2)*((a*c^3)^(1/4)*c^3*d^3 - 
 3*(a*c^3)^(1/4)*a*c^2*d*e^2 - 3*(a*c^3)^(3/4)*c*d^2*e + (a*c^3)^(3/4)*a*e 
^3)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a*c^4) - 1/8*sqrt(2)*((a 
*c^3)^(1/4)*c^3*d^3 - 3*(a*c^3)^(1/4)*a*c^2*d*e^2 - 3*(a*c^3)^(3/4)*c*d^2* 
e + (a*c^3)^(3/4)*a*e^3)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a*c 
^4)
 

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 2712, normalized size of antiderivative = 9.76 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

(e^3*x^3)/(3*c) - atan((a^3*e^6*x*((e^6*(-a^3*c^7)^(1/2))/(16*c^7) + (5*d^ 
3*e^3)/(4*c^2) - (3*d^5*e)/(8*a*c) - (3*a*d*e^5)/(8*c^3) - (d^6*(-a^3*c^7) 
^(1/2))/(16*a^3*c^4) - (15*d^2*e^4*(-a^3*c^7)^(1/2))/(16*a*c^6) + (15*d^4* 
e^2*(-a^3*c^7)^(1/2))/(16*a^2*c^5))^(1/2)*8i)/(6*c^2*d^8*e + (2*a^4*e^9)/c 
^2 + 120*a^2*d^4*e^5 - (36*a^3*d^2*e^7)/c - 92*a*c*d^6*e^3 + (2*d^9*(-a^3* 
c^7)^(1/2))/(a^2*c) + (120*d^5*e^4*(-a^3*c^7)^(1/2))/c^3 - (92*a*d^3*e^6*( 
-a^3*c^7)^(1/2))/c^4 + (6*a^2*d*e^8*(-a^3*c^7)^(1/2))/c^5 - (36*d^7*e^2*(- 
a^3*c^7)^(1/2))/(a*c^2)) - (c^3*d^6*x*((e^6*(-a^3*c^7)^(1/2))/(16*c^7) + ( 
5*d^3*e^3)/(4*c^2) - (3*d^5*e)/(8*a*c) - (3*a*d*e^5)/(8*c^3) - (d^6*(-a^3* 
c^7)^(1/2))/(16*a^3*c^4) - (15*d^2*e^4*(-a^3*c^7)^(1/2))/(16*a*c^6) + (15* 
d^4*e^2*(-a^3*c^7)^(1/2))/(16*a^2*c^5))^(1/2)*8i)/(6*c^2*d^8*e + (2*a^4*e^ 
9)/c^2 + 120*a^2*d^4*e^5 - (36*a^3*d^2*e^7)/c - 92*a*c*d^6*e^3 + (2*d^9*(- 
a^3*c^7)^(1/2))/(a^2*c) + (120*d^5*e^4*(-a^3*c^7)^(1/2))/c^3 - (92*a*d^3*e 
^6*(-a^3*c^7)^(1/2))/c^4 + (6*a^2*d*e^8*(-a^3*c^7)^(1/2))/c^5 - (36*d^7*e^ 
2*(-a^3*c^7)^(1/2))/(a*c^2)) + (a*c^2*d^4*e^2*x*((e^6*(-a^3*c^7)^(1/2))/(1 
6*c^7) + (5*d^3*e^3)/(4*c^2) - (3*d^5*e)/(8*a*c) - (3*a*d*e^5)/(8*c^3) - ( 
d^6*(-a^3*c^7)^(1/2))/(16*a^3*c^4) - (15*d^2*e^4*(-a^3*c^7)^(1/2))/(16*a*c 
^6) + (15*d^4*e^2*(-a^3*c^7)^(1/2))/(16*a^2*c^5))^(1/2)*120i)/(6*c^2*d^8*e 
 + (2*a^4*e^9)/c^2 + 120*a^2*d^4*e^5 - (36*a^3*d^2*e^7)/c - 92*a*c*d^6*e^3 
 + (2*d^9*(-a^3*c^7)^(1/2))/(a^2*c) + (120*d^5*e^4*(-a^3*c^7)^(1/2))/c^...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.24 \[ \int \frac {\left (d+e x^2\right )^3}{a+c x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(6*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x 
)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*e**3 - 18*c**(1/4)*a**(3/4)*sqrt(2)*atan( 
(c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c*d 
**2*e + 18*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*s 
qrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*d*e**2 - 6*c**(3/4)*a**(1/4)*sqrt 
(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt 
(2)))*c*d**3 - 6*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) 
 + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*e**3 + 18*c**(1/4)*a**(3/4) 
*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4) 
*sqrt(2)))*c*d**2*e - 18*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4) 
*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*d*e**2 + 6*c**(3/4) 
*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4) 
*a**(1/4)*sqrt(2)))*c*d**3 - 3*c**(1/4)*a**(3/4)*sqrt(2)*log( - c**(1/4)*a 
**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*e**3 + 9*c**(1/4)*a**(3/4)*s 
qrt(2)*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*c*d**2 
*e + 3*c**(1/4)*a**(3/4)*sqrt(2)*log(c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) 
 + sqrt(c)*x**2)*a*e**3 - 9*c**(1/4)*a**(3/4)*sqrt(2)*log(c**(1/4)*a**(1/4 
)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*c*d**2*e + 9*c**(3/4)*a**(1/4)*sqrt( 
2)*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*d*e**2 - 
 3*c**(3/4)*a**(1/4)*sqrt(2)*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(...