Integrand size = 17, antiderivative size = 284 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=-\frac {e^3}{4 c \left (a+c x^4\right )}+\frac {x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a \left (a+c x^4\right )}+\frac {3 d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {3 d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}} \] Output:
-1/4*e^3/c/(c*x^4+a)+1/4*x*(3*d*e^2*x^2+3*d^2*e*x+d^3)/a/(c*x^4+a)+3/4*d^2 *e*arctan(c^(1/2)*x^2/a^(1/2))/a^(3/2)/c^(1/2)+3/16*d*(c^(1/2)*d^2+a^(1/2) *e^2)*arctan(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/c^(3/4)+3/16*d* (c^(1/2)*d^2+a^(1/2)*e^2)*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7 /4)/c^(3/4)+3/16*d*(c^(1/2)*d^2-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^(7/4)/c^(3/4)
Time = 0.39 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.22 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {-\frac {8 a \left (a e^3-c d x \left (d^2+3 d e x+3 e^2 x^2\right )\right )}{a+c x^4}-6 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2+4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+6 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {2} \sqrt {c} d^2-4 \sqrt [4]{a} \sqrt [4]{c} d e+\sqrt {2} \sqrt {a} e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+3 \sqrt {2} \sqrt [4]{c} \left (-\sqrt [4]{a} \sqrt {c} d^3+a^{3/4} d e^2\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+3 \sqrt {2} \sqrt [4]{c} \left (\sqrt [4]{a} \sqrt {c} d^3-a^{3/4} d e^2\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{32 a^2 c} \] Input:
Integrate[(d + e*x)^3/(a + c*x^4)^2,x]
Output:
((-8*a*(a*e^3 - c*d*x*(d^2 + 3*d*e*x + 3*e^2*x^2)))/(a + c*x^4) - 6*a^(1/4 )*c^(1/4)*d*(Sqrt[2]*Sqrt[c]*d^2 + 4*a^(1/4)*c^(1/4)*d*e + Sqrt[2]*Sqrt[a] *e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 6*a^(1/4)*c^(1/4)*d*(Sqrt[ 2]*Sqrt[c]*d^2 - 4*a^(1/4)*c^(1/4)*d*e + Sqrt[2]*Sqrt[a]*e^2)*ArcTan[1 + ( Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 3*Sqrt[2]*c^(1/4)*(-(a^(1/4)*Sqrt[c]*d^3) + a^(3/4)*d*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2] + 3* Sqrt[2]*c^(1/4)*(a^(1/4)*Sqrt[c]*d^3 - a^(3/4)*d*e^2)*Log[Sqrt[a] + Sqrt[2 ]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(32*a^2*c)
Time = 0.89 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.25, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2393, 27, 2006, 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.
| \(\displaystyle \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 2393 |
| \(\displaystyle -\frac {\int -\frac {3 \left (d^3+2 e x d^2+e^2 x^2 d\right )}{c x^4+a}dx}{4 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \int \frac {d^3+2 e x d^2+e^2 x^2 d}{c x^4+a}dx}{4 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 2006 |
| \(\displaystyle \frac {3 \int \frac {\left (d^{3/2}+e x \sqrt {d}\right )^2}{c x^4+a}dx}{4 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \frac {3 \int \left (\frac {2 e x d^2}{c x^4+a}+\frac {d^3+e^2 x^2 d}{c x^4+a}\right )dx}{4 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \left (-\frac {d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}{2 \sqrt {2} a^{3/4} c^{3/4}}-\frac {d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} c^{3/4}}+\frac {d^2 e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}\right )}{4 a}-\frac {a e^3-c x \left (d^3+3 d^2 e x+3 d e^2 x^2\right )}{4 a c \left (a+c x^4\right )}\) |
Input:
Int[(d + e*x)^3/(a + c*x^4)^2,x]
Output:
-1/4*(a*e^3 - c*x*(d^3 + 3*d^2*e*x + 3*d*e^2*x^2))/(a*c*(a + c*x^4)) + (3* ((d^2*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(Sqrt[a]*Sqrt[c]) - (d*(Sqrt[c]*d^2 + 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 + 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 - Sqrt[a]*e^2)*Lo g[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 - 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))))/(4*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^Expon[Px , x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; PolyQ[Px, x] && GtQ[Expon[P x, x], 1] && NeQ[Coeff[Px, x, 0], 0] && !MatchQ[Px, (a_.)*(v_)^Expon[Px, x ] /; FreeQ[a, x] && LinearQ[v, x]]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q , x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) In t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n , 0] && LtQ[p, -1]
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 = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {\frac {3 d \,e^{2} x^{3}}{4 a}+\frac {3 d^{2} e \,x^{2}}{4 a}+\frac {d^{3} x}{4 a}-\frac {e^{3}}{4 c}}{c \,x^{4}+a}+\frac {3 d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (e^{2} \textit {\_R}^{2}+2 d e \textit {\_R} +d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16 a c}\) | \(100\) |
| default | \(d^{3} \left (\frac {x}{4 a \left (c \,x^{4}+a \right )}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2}}\right )+3 d^{2} e \left (\frac {x^{2}}{4 a \left (c \,x^{4}+a \right )}+\frac {\arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{4 a \sqrt {a c}}\right )+3 d \,e^{2} \left (\frac {x^{3}}{4 a \left (c \,x^{4}+a \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )+\frac {e^{3} x^{4}}{4 a \left (c \,x^{4}+a \right )}\) | \(317\) |
Input:
int((e*x+d)^3/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
(3/4*d*e^2/a*x^3+3/4*d^2*e/a*x^2+1/4*d^3/a*x-1/4*e^3/c)/(c*x^4+a)+3/16*d/a /c*sum((_R^2*e^2+2*_R*d*e+d^2)/_R^3*ln(x-_R),_R=RootOf(_Z^4*c+a))
Result contains complex when optimal does not.
Time = 15.25 (sec) , antiderivative size = 91191, normalized size of antiderivative = 321.10 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="fricas")
Output:
Too large to include
Time = 9.11 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\operatorname {RootSum} {\left (65536 t^{4} a^{7} c^{3} + 27648 t^{2} a^{4} c^{2} d^{4} e^{2} + t \left (3456 a^{3} c d^{4} e^{5} - 3456 a^{2} c^{2} d^{8} e\right ) + 81 a^{2} d^{4} e^{8} + 162 a c d^{8} e^{4} + 81 c^{2} d^{12}, \left ( t \mapsto t \log {\left (x + \frac {4096 t^{3} a^{7} c^{2} e^{6} + 28672 t^{3} a^{6} c^{3} d^{4} e^{2} - 7680 t^{2} a^{5} c^{2} d^{4} e^{5} + 1536 t^{2} a^{4} c^{3} d^{8} e + 2160 t a^{4} c d^{4} e^{8} + 9216 t a^{3} c^{2} d^{8} e^{4} + 144 t a^{2} c^{3} d^{12} + 162 a^{3} d^{4} e^{11} - 648 a^{2} c d^{8} e^{7} - 810 a c^{2} d^{12} e^{3}}{27 a^{3} d^{3} e^{12} - 891 a^{2} c d^{7} e^{8} - 891 a c^{2} d^{11} e^{4} + 27 c^{3} d^{15}} \right )} \right )\right )} + \frac {- a e^{3} + c d^{3} x + 3 c d^{2} e x^{2} + 3 c d e^{2} x^{3}}{4 a^{2} c + 4 a c^{2} x^{4}} \] Input:
integrate((e*x+d)**3/(c*x**4+a)**2,x)
Output:
RootSum(65536*_t**4*a**7*c**3 + 27648*_t**2*a**4*c**2*d**4*e**2 + _t*(3456 *a**3*c*d**4*e**5 - 3456*a**2*c**2*d**8*e) + 81*a**2*d**4*e**8 + 162*a*c*d **8*e**4 + 81*c**2*d**12, Lambda(_t, _t*log(x + (4096*_t**3*a**7*c**2*e**6 + 28672*_t**3*a**6*c**3*d**4*e**2 - 7680*_t**2*a**5*c**2*d**4*e**5 + 1536 *_t**2*a**4*c**3*d**8*e + 2160*_t*a**4*c*d**4*e**8 + 9216*_t*a**3*c**2*d** 8*e**4 + 144*_t*a**2*c**3*d**12 + 162*a**3*d**4*e**11 - 648*a**2*c*d**8*e* *7 - 810*a*c**2*d**12*e**3)/(27*a**3*d**3*e**12 - 891*a**2*c*d**7*e**8 - 8 91*a*c**2*d**11*e**4 + 27*c**3*d**15)))) + (-a*e**3 + c*d**3*x + 3*c*d**2* e*x**2 + 3*c*d*e**2*x**3)/(4*a**2*c + 4*a*c**2*x**4)
Time = 0.13 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, d {\left (\frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (\sqrt {c} d^{2} - \sqrt {a} e^{2}\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} - 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}} + \frac {2 \, {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {3}{4}} d^{2} + \sqrt {2} a^{\frac {3}{4}} c^{\frac {1}{4}} e^{2} + 4 \, \sqrt {a} \sqrt {c} d e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {3}{4}}}\right )}}{32 \, a} + \frac {3 \, c d e^{2} x^{3} + 3 \, c d^{2} e x^{2} + c d^{3} x - a e^{3}}{4 \, {\left (a c^{2} x^{4} + a^{2} c\right )}} \] Input:
integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="maxima")
Output:
3/32*d*(sqrt(2)*(sqrt(c)*d^2 - sqrt(a)*e^2)*log(sqrt(c)*x^2 + sqrt(2)*a^(1 /4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(3/4)) - sqrt(2)*(sqrt(c)*d^2 - sqrt(a )*e^2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^( 3/4)) + 2*(sqrt(2)*a^(1/4)*c^(3/4)*d^2 + sqrt(2)*a^(3/4)*c^(1/4)*e^2 - 4*s qrt(a)*sqrt(c)*d*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^(3/4)) + 2*(s qrt(2)*a^(1/4)*c^(3/4)*d^2 + sqrt(2)*a^(3/4)*c^(1/4)*e^2 + 4*sqrt(a)*sqrt( c)*d*e)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sq rt(a)*sqrt(c)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(c))*c^(3/4)))/a + 1/4*(3*c*d*e^ 2*x^3 + 3*c*d^2*e*x^2 + c*d^3*x - a*e^3)/(a*c^2*x^4 + a^2*c)
Time = 0.12 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, c d e^{2} x^{3} + 3 \, c d^{2} e x^{2} + c d^{3} x - a e^{3}}{4 \, {\left (c x^{4} + a\right )} a c} + \frac {3 \, \sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {3 \, \sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c^{2} d^{2} e + \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} + \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {3 \, \sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} - \frac {3 \, \sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} c^{2} d^{3} - \left (a c^{3}\right )^{\frac {3}{4}} d e^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} \] Input:
integrate((e*x+d)^3/(c*x^4+a)^2,x, algorithm="giac")
Output:
1/4*(3*c*d*e^2*x^3 + 3*c*d^2*e*x^2 + c*d^3*x - a*e^3)/((c*x^4 + a)*a*c) + 3/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c)*c^2*d^2*e + (a*c^3)^(1/4)*c^2*d^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^2*c^3) + 3/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c)*c^2*d^2*e + (a*c^3)^(1/4) *c^2*d^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^2*c^3) + 3/32*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - (a*c^3 )^(3/4)*d*e^2)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3) - 3/ 32*sqrt(2)*((a*c^3)^(1/4)*c^2*d^3 - (a*c^3)^(3/4)*d*e^2)*log(x^2 - sqrt(2) *x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c^3)
Time = 22.35 (sec) , antiderivative size = 670, normalized size of antiderivative = 2.36 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx=\left (\sum _{k=1}^4\ln \left (\frac {c\,d^2\,\left (27\,c\,d^5\,e^2-9\,a\,d\,e^6+36\,c\,d^4\,e^3\,x-{\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )}^2\,a^3\,c^2\,d\,256-\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a\,c^2\,d^4\,x\,48+\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a^2\,c\,e^4\,x\,48+{\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )}^2\,a^3\,c^2\,e\,x\,512-\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\,a^2\,c\,d\,e^3\,192\right )\,3}{a^3\,64}\right )\,\mathrm {root}\left (65536\,a^7\,c^3\,z^4+27648\,a^4\,c^2\,d^4\,e^2\,z^2+3456\,a^3\,c\,d^4\,e^5\,z-3456\,a^2\,c^2\,d^8\,e\,z+162\,a\,c\,d^8\,e^4+81\,a^2\,d^4\,e^8+81\,c^2\,d^{12},z,k\right )\right )+\frac {\frac {d^3\,x}{4\,a}-\frac {e^3}{4\,c}+\frac {3\,d^2\,e\,x^2}{4\,a}+\frac {3\,d\,e^2\,x^3}{4\,a}}{c\,x^4+a} \] Input:
int((d + e*x)^3/(a + c*x^4)^2,x)
Output:
symsum(log((3*c*d^2*(27*c*d^5*e^2 - 9*a*d*e^6 + 36*c*d^4*e^3*x - 256*root( 65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 345 6*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)^ 2*a^3*c^2*d - 48*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456 *a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)*a*c^2*d^4*x + 48*root(65536*a^7*c^3*z^4 + 27648*a^4* c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^ 8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)*a^2*c*e^4*x + 512*root(65536*a ^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c ^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k)^2*a^3*c ^2*e*x - 192*root(65536*a^7*c^3*z^4 + 27648*a^4*c^2*d^4*e^2*z^2 + 3456*a^3 *c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c*d^8*e^4 + 81*a^2*d^4*e^8 + 8 1*c^2*d^12, z, k)*a^2*c*d*e^3))/(64*a^3))*root(65536*a^7*c^3*z^4 + 27648*a ^4*c^2*d^4*e^2*z^2 + 3456*a^3*c*d^4*e^5*z - 3456*a^2*c^2*d^8*e*z + 162*a*c *d^8*e^4 + 81*a^2*d^4*e^8 + 81*c^2*d^12, z, k), k, 1, 4) + ((d^3*x)/(4*a) - e^3/(4*c) + (3*d^2*e*x^2)/(4*a) + (3*d*e^2*x^3)/(4*a))/(a + c*x^4)
Time = 0.15 (sec) , antiderivative size = 836, normalized size of antiderivative = 2.94 \[ \int \frac {(d+e x)^3}{\left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:
int((e*x+d)^3/(c*x^4+a)^2,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*d*e**2 - 6*c**(1/4)*a**(3/4)*sqrt(2)*a tan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2))) *c*d*e**2*x**4 - 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)))*a*d**3 - 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*x**4 - 24*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)))*a*d**2*e - 24*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)))*c*d**2*e*x**4 + 6*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*s qrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*d*e**2 + 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)))*c*d*e**2*x**4 + 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)))*a*d**3 + 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*x**4 - 24*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)))*a*d**2*e - 24*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)))*c*d**2*e*x**4 + 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*d*e**2 + 3*c**(1...