Integrand size = 17, antiderivative size = 248 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\frac {3 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {c}}-\frac {d \left (\sqrt {c} d^2+3 \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2+3 \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {e^3 \log \left (a+c x^4\right )}{4 c} \] Output:
3/2*d^2*e*arctan(c^(1/2)*x^2/a^(1/2))/a^(1/2)/c^(1/2)+1/4*d*(c^(1/2)*d^2+3 *a^(1/2)*e^2)*arctan(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(3/4)/c^(3/4) +1/4*d*(c^(1/2)*d^2+3*a^(1/2)*e^2)*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^( 1/2)/a^(3/4)/c^(3/4)+1/4*d*(c^(1/2)*d^2-3*a^(1/2)*e^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^(3/4)+1/4*e^3*ln(c *x^4+a)/c
Time = 0.25 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.30 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\frac {-2 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2+6 \sqrt [4]{a} \sqrt [4]{c} d e+3 \sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2-6 \sqrt [4]{a} \sqrt [4]{c} d e+3 \sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-\sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-3 a^{3/4} d e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+\sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-3 a^{3/4} d e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+2 a e^3 \log \left (a+c x^4\right )}{8 a c} \] Input:
Integrate[(d + e*x)^3/(a + c*x^4),x]
Output:
(-2*a^(1/4)*c^(1/4)*d*(Sqrt[2]*Sqrt[c]*d^2 + 6*a^(1/4)*c^(1/4)*d*e + 3*Sqr t[2]*Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 2*a^(1/4)*c^(1 /4)*d*(Sqrt[2]*Sqrt[c]*d^2 - 6*a^(1/4)*c^(1/4)*d*e + 3*Sqrt[2]*Sqrt[a]*e^2 )*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[ c]*d^3 - 3*a^(3/4)*d*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c ]*x^2] + Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[c]*d^3 - 3*a^(3/4)*d*e^2)*Log[Sqrt[ a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + 2*a*e^3*Log[a + c*x^4])/(8 *a*c)
Time = 0.77 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2415, 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.
| \(\displaystyle \int \frac {(d+e x)^3}{a+c x^4} \, dx\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \int \left (\frac {d^3+3 d e^2 x^2}{a+c x^4}+\frac {x \left (3 d^2 e+e^3 x^2\right )}{a+c x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (3 \sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2-3 \sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {3 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {c}}+\frac {e^3 \log \left (a+c x^4\right )}{4 c}\) |
Input:
Int[(d + e*x)^3/(a + c*x^4),x]
Output:
(3*d^2*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*Sqrt[a]*Sqrt[c]) - (d*(Sqrt[c]* d^2 + 3*Sqrt[a]*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a ^(3/4)*c^(3/4)) + (d*(Sqrt[c]*d^2 + 3*Sqrt[a]*e^2)*ArcTan[1 + (Sqrt[2]*c^( 1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*c^(3/4)) - (d*(Sqrt[c]*d^2 - 3*Sqrt[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^(3/4)) + (d*(Sqrt[c]*d^2 - 3*Sqrt[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^(3/4)) + (e^3*Log[a + c*x^4])/(4*c)
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff [Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1 }]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.22
| method | result | size |
| risch | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3} e^{3}+3 \textit {\_R}^{2} d \,e^{2}+3 \textit {\_R} \,d^{2} e +d^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 c}\) | \(54\) |
| default | \(\frac {d^{3} \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {3 d^{2} e \arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{2 \sqrt {a c}}+\frac {3 d \,e^{2} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}+\frac {e^{3} \ln \left (c \,x^{4}+a \right )}{4 c}\) | \(250\) |
Input:
int((e*x+d)^3/(c*x^4+a),x,method=_RETURNVERBOSE)
Output:
1/4/c*sum((_R^3*e^3+3*_R^2*d*e^2+3*_R*d^2*e+d^3)/_R^3*ln(x-_R),_R=RootOf(_ Z^4*c+a))
Result contains complex when optimal does not.
Time = 15.66 (sec) , antiderivative size = 141845, normalized size of antiderivative = 571.96 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)^3/(c*x^4+a),x, algorithm="fricas")
Output:
Too large to include
Time = 3.87 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{3} c^{4} - 256 t^{3} a^{3} c^{3} e^{3} + t^{2} \cdot \left (96 a^{3} c^{2} e^{6} + 480 a^{2} c^{3} d^{4} e^{2}\right ) + t \left (- 16 a^{3} c e^{9} + 192 a^{2} c^{2} d^{4} e^{5} - 48 a c^{3} d^{8} e\right ) + a^{3} e^{12} + 3 a^{2} c d^{4} e^{8} + 3 a c^{2} d^{8} e^{4} + c^{3} d^{12}, \left ( t \mapsto t \log {\left (x + \frac {1728 t^{3} a^{4} c^{3} e^{6} + 960 t^{3} a^{3} c^{4} d^{4} e^{2} - 1296 t^{2} a^{4} c^{2} e^{9} - 2016 t^{2} a^{3} c^{3} d^{4} e^{5} + 48 t^{2} a^{2} c^{4} d^{8} e + 324 t a^{4} c e^{12} + 4716 t a^{3} c^{2} d^{4} e^{8} + 1452 t a^{2} c^{3} d^{8} e^{4} + 4 t a c^{4} d^{12} - 27 a^{4} e^{15} + 1119 a^{3} c d^{4} e^{11} - 609 a^{2} c^{2} d^{8} e^{7} - 91 a c^{3} d^{12} e^{3}}{729 a^{3} c d^{3} e^{12} - 1053 a^{2} c^{2} d^{7} e^{8} - 117 a c^{3} d^{11} e^{4} + c^{4} d^{15}} \right )} \right )\right )} \] Input:
integrate((e*x+d)**3/(c*x**4+a),x)
Output:
RootSum(256*_t**4*a**3*c**4 - 256*_t**3*a**3*c**3*e**3 + _t**2*(96*a**3*c* *2*e**6 + 480*a**2*c**3*d**4*e**2) + _t*(-16*a**3*c*e**9 + 192*a**2*c**2*d **4*e**5 - 48*a*c**3*d**8*e) + a**3*e**12 + 3*a**2*c*d**4*e**8 + 3*a*c**2* d**8*e**4 + c**3*d**12, Lambda(_t, _t*log(x + (1728*_t**3*a**4*c**3*e**6 + 960*_t**3*a**3*c**4*d**4*e**2 - 1296*_t**2*a**4*c**2*e**9 - 2016*_t**2*a* *3*c**3*d**4*e**5 + 48*_t**2*a**2*c**4*d**8*e + 324*_t*a**4*c*e**12 + 4716 *_t*a**3*c**2*d**4*e**8 + 1452*_t*a**2*c**3*d**8*e**4 + 4*_t*a*c**4*d**12 - 27*a**4*e**15 + 1119*a**3*c*d**4*e**11 - 609*a**2*c**2*d**8*e**7 - 91*a* c**3*d**12*e**3)/(729*a**3*c*d**3*e**12 - 1053*a**2*c**2*d**7*e**8 - 117*a *c**3*d**11*e**4 + c**4*d**15))))
Time = 0.11 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.25 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} + c d^{3} - 3 \, \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{3} - c d^{3} + 3 \, \sqrt {a} \sqrt {c} d e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{8 \, a^{\frac {3}{4}} c^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + 3 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} - 6 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} + \frac {{\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {5}{4}} d^{3} + 3 \, \sqrt {2} a^{\frac {3}{4}} c^{\frac {3}{4}} d e^{2} + 6 \, \sqrt {a} c d^{2} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{4 \, a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {5}{4}}} \] Input:
integrate((e*x+d)^3/(c*x^4+a),x, algorithm="maxima")
Output:
1/8*sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*e^3 + c*d^3 - 3*sqrt(a)*sqrt(c)*d*e^2 )*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(5/4)) + 1/8*sqrt(2)*(sqrt(2)*a^(3/4)*c^(1/4)*e^3 - c*d^3 + 3*sqrt(a)*sqrt(c)*d* e^2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(5/ 4)) + 1/4*(sqrt(2)*a^(1/4)*c^(5/4)*d^3 + 3*sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 - 6*sqrt(a)*c*d^2*e)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1 /4))/sqrt(sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(5/4)) + 1/4* (sqrt(2)*a^(1/4)*c^(5/4)*d^3 + 3*sqrt(2)*a^(3/4)*c^(3/4)*d*e^2 + 6*sqrt(a) *c*d^2*e)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/4))/sqrt( sqrt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(5/4))
Time = 0.12 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\frac {e^{3} \log \left ({\left | c x^{4} + a \right |}\right )}{4 \, c} + \frac {\sqrt {2} {\left (3 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (3 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{4 \, a c^{3}} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - 3 \, \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{8 \, a c^{3}} \] Input:
integrate((e*x+d)^3/(c*x^4+a),x, algorithm="giac")
Output:
1/4*e^3*log(abs(c*x^4 + a))/c + 1/4*sqrt(2)*(3*sqrt(2)*sqrt(a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + 3*(a*c^3)^(3/4)*d*e^2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^3) + 1/4*sqrt(2)*(3*sqrt(2)*sqrt( a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^3 + 3*(a*c^3)^(3/4)*d*e^2)*arctan(1/2 *sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a*c^3) + 1/8*sqrt(2)*(( a*c^3)^(1/4)*c^2*d^3 - 3*(a*c^3)^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1 /4) + sqrt(a/c))/(a*c^3) - 1/8*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - 3*(a*c^3)^ (3/4)*d*e^2)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a*c^3)
Time = 0.65 (sec) , antiderivative size = 894, normalized size of antiderivative = 3.60 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\sum _{k=1}^4\ln \left (-c\,d^2\,\left (-3\,c\,d^5\,e^2+5\,a\,d\,e^6+3\,a\,e^7\,x+{\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )}^2\,a\,c^2\,d\,8+\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,c^2\,d^4\,x\,2-5\,c\,d^4\,e^3\,x-{\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )}^2\,a\,c^2\,e\,x\,24+\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,a\,c\,d\,e^3\,32-\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right )\,a\,c\,e^4\,x\,6\right )\,2\right )\,\mathrm {root}\left (256\,a^3\,c^4\,z^4-256\,a^3\,c^3\,e^3\,z^3+480\,a^2\,c^3\,d^4\,e^2\,z^2+96\,a^3\,c^2\,e^6\,z^2+192\,a^2\,c^2\,d^4\,e^5\,z-48\,a\,c^3\,d^8\,e\,z-16\,a^3\,c\,e^9\,z+3\,a^2\,c\,d^4\,e^8+3\,a\,c^2\,d^8\,e^4+c^3\,d^{12}+a^3\,e^{12},z,k\right ) \] Input:
int((d + e*x)^3/(a + c*x^4),x)
Output:
symsum(log(-2*c*d^2*(5*a*d*e^6 - 3*c*d^5*e^2 + 3*a*e^7*x + 8*root(256*a^3* c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z ^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d ^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)^2*a*c^2*d + 2*root(2 56*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^ 2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3* a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)*c^2*d^4*x - 5 *c*d^4*e^3*x - 24*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 480*a^2*c^3 *d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e *z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e ^12, z, k)^2*a*c^2*e*x + 32*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3*z^3 + 4 80*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a *c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^ 12 + a^3*e^12, z, k)*a*c*d*e^3 - 6*root(256*a^3*c^4*z^4 - 256*a^3*c^3*e^3* z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4*e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8*e^4 + c^3*d^12 + a^3*e^12, z, k)*a*c*e^4*x))*root(256*a^3*c^4*z^4 - 256*a^3*c^3 *e^3*z^3 + 480*a^2*c^3*d^4*e^2*z^2 + 96*a^3*c^2*e^6*z^2 + 192*a^2*c^2*d^4* e^5*z - 48*a*c^3*d^8*e*z - 16*a^3*c*e^9*z + 3*a^2*c*d^4*e^8 + 3*a*c^2*d^8* e^4 + c^3*d^12 + a^3*e^12, z, k), k, 1, 4)
Time = 0.14 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.75 \[ \int \frac {(d+e x)^3}{a+c x^4} \, dx=\frac {-6 c^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d \,e^{2}-2 c^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d^{3}-12 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d^{2} e +6 c^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d \,e^{2}+2 c^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d^{3}-12 \sqrt {c}\, \sqrt {a}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) d^{2} e +3 c^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) d \,e^{2}-3 c^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) d \,e^{2}-c^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) d^{3}+c^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) d^{3}+2 \,\mathrm {log}\left (-c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a \,e^{3}+2 \,\mathrm {log}\left (c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a \,e^{3}}{8 a c} \] Input:
int((e*x+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)))*d*e**2 - 2*c**(3/4)*a**(1/4)*sqrt(2)*ata n((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*d **3 - 12*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c **(1/4)*a**(1/4)*sqrt(2)))*d**2*e + 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)))*d*e**2 + 2*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)))*d**3 - 12*sqrt(c)*sqrt(a)*atan((c**(1/4)*a* *(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*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)*d*e**2 - 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)*d*e**2 - c**(3/4)*a**(1/4)*sqrt(2)*log( - c* *(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*d**3 + c**(3/4)*a**(1/ 4)*sqrt(2)*log(c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*d**3 + 2*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*e**3 + 2*log(c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*e**3)/(8*a*c )