Integrand size = 15, antiderivative size = 189 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\frac {x (d+e x)}{4 a \left (a+c x^4\right )}+\frac {e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {c}}-\frac {3 d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}}+\frac {3 d \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{8 \sqrt {2} a^{7/4} \sqrt [4]{c}} \] Output:
1/4*x*(e*x+d)/a/(c*x^4+a)+1/4*e*arctan(c^(1/2)*x^2/a^(1/2))/a^(3/2)/c^(1/2 )+3/16*d*arctan(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/c^(1/4)+3/16 *d*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/c^(1/4)+3/16*d*arct anh(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^(1/ 4)
Time = 0.22 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.19 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\frac {\frac {8 a^{3/4} x (d+e x)}{a+c x^4}-\frac {2 \left (3 \sqrt {2} \sqrt [4]{c} d+4 \sqrt [4]{a} e\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {c}}+\frac {2 \left (3 \sqrt {2} \sqrt [4]{c} d-4 \sqrt [4]{a} e\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {c}}-\frac {3 \sqrt {2} d \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{\sqrt [4]{c}}+\frac {3 \sqrt {2} d \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{\sqrt [4]{c}}}{32 a^{7/4}} \] Input:
Integrate[(d + e*x)/(a + c*x^4)^2,x]
Output:
((8*a^(3/4)*x*(d + e*x))/(a + c*x^4) - (2*(3*Sqrt[2]*c^(1/4)*d + 4*a^(1/4) *e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/Sqrt[c] + (2*(3*Sqrt[2]*c^(1/ 4)*d - 4*a^(1/4)*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/Sqrt[c] - (3* Sqrt[2]*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4) + (3*Sqrt[2]*d*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^( 1/4))/(32*a^(7/4))
Time = 0.64 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2394, 25, 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.
| \(\displaystyle \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {x (d+e x)}{4 a \left (a+c x^4\right )}-\frac {\int -\frac {3 d+2 e x}{c x^4+a}dx}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 d+2 e x}{c x^4+a}dx}{4 a}+\frac {x (d+e x)}{4 a \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 2415 |
| \(\displaystyle \frac {\int \left (\frac {3 d}{c x^4+a}+\frac {2 e x}{c x^4+a}\right )dx}{4 a}+\frac {x (d+e x)}{4 a \left (a+c x^4\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3 d \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {3 d \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{c}}-\frac {3 d \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {3 d \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{c}}+\frac {e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}}{4 a}+\frac {x (d+e x)}{4 a \left (a+c x^4\right )}\) |
Input:
Int[(d + e*x)/(a + c*x^4)^2,x]
Output:
(x*(d + e*x))/(4*a*(a + c*x^4)) + ((e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(Sqrt [a]*Sqrt[c]) - (3*d*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^ (3/4)*c^(1/4)) + (3*d*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]* a^(3/4)*c^(1/4)) - (3*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]* x^2])/(4*Sqrt[2]*a^(3/4)*c^(1/4)) + (3*d*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^( 1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*c^(1/4)))/(4*a)
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) Int[ExpandToSum[n *(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x ] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 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 = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {\frac {e \,x^{2}}{4 a}+\frac {d x}{4 a}}{c \,x^{4}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (2 e \textit {\_R} +3 d \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{16 c a}\) | \(66\) |
| default | \(d \left (\frac {x}{4 a \left (c \,x^{4}+a \right )}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2}}\right )+e \left (\frac {x^{2}}{4 a \left (c \,x^{4}+a \right )}+\frac {\arctan \left (\sqrt {\frac {c}{a}}\, x^{2}\right )}{4 a \sqrt {a c}}\right )\) | \(163\) |
Input:
int((e*x+d)/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
(1/4*e/a*x^2+1/4*d/a*x)/(c*x^4+a)+1/16/c/a*sum((2*_R*e+3*d)/_R^3*ln(x-_R), _R=RootOf(_Z^4*c+a))
Result contains complex when optimal does not.
Time = 2.16 (sec) , antiderivative size = 43065, normalized size of antiderivative = 227.86 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)/(c*x^4+a)^2,x, algorithm="fricas")
Output:
Too large to include
Time = 0.90 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\operatorname {RootSum} {\left (65536 t^{4} a^{7} c^{2} + 2048 t^{2} a^{4} c e^{2} - 1152 t a^{2} c d^{2} e + 16 a e^{4} + 81 c d^{4}, \left ( t \mapsto t \log {\left (x + \frac {- 32768 t^{3} a^{6} c e^{2} - 4608 t^{2} a^{4} c d^{2} e - 512 t a^{3} e^{4} - 1296 t a^{2} c d^{4} + 360 a d^{2} e^{3}}{192 a d e^{4} - 243 c d^{5}} \right )} \right )\right )} + \frac {d x + e x^{2}}{4 a^{2} + 4 a c x^{4}} \] Input:
integrate((e*x+d)/(c*x**4+a)**2,x)
Output:
RootSum(65536*_t**4*a**7*c**2 + 2048*_t**2*a**4*c*e**2 - 1152*_t*a**2*c*d* *2*e + 16*a*e**4 + 81*c*d**4, Lambda(_t, _t*log(x + (-32768*_t**3*a**6*c*e **2 - 4608*_t**2*a**4*c*d**2*e - 512*_t*a**3*e**4 - 1296*_t*a**2*c*d**4 + 360*a*d**2*e**3)/(192*a*d*e**4 - 243*c*d**5)))) + (d*x + e*x**2)/(4*a**2 + 4*a*c*x**4)
Time = 0.11 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.26 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\frac {e x^{2} + d x}{4 \, {\left (a c x^{4} + a^{2}\right )}} + \frac {\frac {3 \, \sqrt {2} d \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {3 \, \sqrt {2} d \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {1}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} d - 4 \, \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {1}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} d + 4 \, \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {c}} c^{\frac {1}{4}}}}{32 \, a} \] Input:
integrate((e*x+d)/(c*x^4+a)^2,x, algorithm="maxima")
Output:
1/4*(e*x^2 + d*x)/(a*c*x^4 + a^2) + 1/32*(3*sqrt(2)*d*log(sqrt(c)*x^2 + sq rt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(1/4)) - 3*sqrt(2)*d*log(sqr t(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^(3/4)*c^(1/4)) + 2*(3*s qrt(2)*a^(1/4)*c^(1/4)*d - 4*sqrt(a)*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^(1/4)) + 2*(3*sqrt(2)*a^(1/4)*c^(1/4)*d + 4*sqrt(a)*e)*arctan(1/2*s qrt(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^(1/4)))/a
Time = 0.12 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.26 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} d \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} - \frac {3 \, \sqrt {2} \left (a c^{3}\right )^{\frac {1}{4}} d \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c} + \frac {e x^{2} + d x}{4 \, {\left (c x^{4} + a\right )} a} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c e + 3 \, \left (a c^{3}\right )^{\frac {1}{4}} c d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{2}} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a c} c e + 3 \, \left (a c^{3}\right )^{\frac {1}{4}} c d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{2}} \] Input:
integrate((e*x+d)/(c*x^4+a)^2,x, algorithm="giac")
Output:
3/32*sqrt(2)*(a*c^3)^(1/4)*d*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/ (a^2*c) - 3/32*sqrt(2)*(a*c^3)^(1/4)*d*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + s qrt(a/c))/(a^2*c) + 1/4*(e*x^2 + d*x)/((c*x^4 + a)*a) + 1/16*sqrt(2)*(2*sq rt(2)*sqrt(a*c)*c*e + 3*(a*c^3)^(1/4)*c*d)*arctan(1/2*sqrt(2)*(2*x + sqrt( 2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^2*c^2) + 1/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c) *c*e + 3*(a*c^3)^(1/4)*c*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4)) /(a/c)^(1/4))/(a^2*c^2)
Time = 0.28 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.49 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx=\left (\sum _{k=1}^4\ln \left (\frac {c^2\,\left (3\,d\,e^2+2\,e^3\,x-{\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )}^2\,a^3\,c\,d\,192+{\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )}^2\,a^3\,c\,e\,x\,128-\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )\,a\,c\,d^2\,x\,36\right )}{a^3\,16}\right )\,\mathrm {root}\left (65536\,a^7\,c^2\,z^4+2048\,a^4\,c\,e^2\,z^2-1152\,a^2\,c\,d^2\,e\,z+81\,c\,d^4+16\,a\,e^4,z,k\right )\right )+\frac {\frac {e\,x^2}{4\,a}+\frac {d\,x}{4\,a}}{c\,x^4+a} \] Input:
int((d + e*x)/(a + c*x^4)^2,x)
Output:
symsum(log((c^2*(3*d*e^2 + 2*e^3*x - 192*root(65536*a^7*c^2*z^4 + 2048*a^4 *c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81*c*d^4 + 16*a*e^4, z, k)^2*a^3*c*d + 1 28*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81*c *d^4 + 16*a*e^4, z, k)^2*a^3*c*e*x - 36*root(65536*a^7*c^2*z^4 + 2048*a^4* c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81*c*d^4 + 16*a*e^4, z, k)*a*c*d^2*x))/(1 6*a^3))*root(65536*a^7*c^2*z^4 + 2048*a^4*c*e^2*z^2 - 1152*a^2*c*d^2*e*z + 81*c*d^4 + 16*a*e^4, z, k), k, 1, 4) + ((e*x^2)/(4*a) + (d*x)/(4*a))/(a + c*x^4)
Time = 0.16 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.48 \[ \int \frac {d+e x}{\left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:
int((e*x+d)/(c*x^4+a)^2,x)
Output:
( - 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 - 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* x**4 - 8*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*e - 8*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*e*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 + 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*x**4 - 8*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*e - 8*sqrt(c)*sqrt(a)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*s qrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c*e*x**4 - 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)*a*d - 3*c **(3/4)*a**(1/4)*sqrt(2)*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sq rt(c)*x**2)*c*d*x**4 + 3*c**(3/4)*a**(1/4)*sqrt(2)*log(c**(1/4)*a**(1/4)*s qrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*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)*c*d*x**4 + 8*a*c*d*x + 8*a*c*e*x**2)/(32*a**2*c*(a + c*x**4))