Integrand size = 19, antiderivative size = 371 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {c x \left (d-e x^2\right )}{4 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \left (c d^2+a e^2\right )^2}+\frac {\sqrt [4]{c} \left (\sqrt {a} e \left (c d^2+5 a e^2\right )-\sqrt {c} d \left (3 c d^2+7 a e^2\right )\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}-\frac {\sqrt [4]{c} \left (\sqrt {a} e \left (c d^2+5 a e^2\right )-\sqrt {c} d \left (3 c d^2+7 a e^2\right )\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}+\frac {\sqrt [4]{c} \left (\sqrt {a} e \left (c d^2+5 a e^2\right )+\sqrt {c} d \left (3 c d^2+7 a e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2} \] Output:
1/4*c*x*(-e*x^2+d)/a/(a*e^2+c*d^2)/(c*x^4+a)+e^(7/2)*arctan(e^(1/2)*x/d^(1 /2))/d^(1/2)/(a*e^2+c*d^2)^2-1/16*c^(1/4)*(a^(1/2)*e*(5*a*e^2+c*d^2)-c^(1/ 2)*d*(7*a*e^2+3*c*d^2))*arctan(-1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/ 4)/(a*e^2+c*d^2)^2-1/16*c^(1/4)*(a^(1/2)*e*(5*a*e^2+c*d^2)-c^(1/2)*d*(7*a* e^2+3*c*d^2))*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/(a*e^2+c *d^2)^2+1/16*c^(1/4)*(a^(1/2)*e*(5*a*e^2+c*d^2)+c^(1/2)*d*(7*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^(7/4 )/(a*e^2+c*d^2)^2
Time = 0.20 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {\frac {8 c \left (c d^2+a e^2\right ) x \left (d-e x^2\right )}{a \left (a+c x^4\right )}+\frac {32 e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {2 \sqrt {2} \sqrt [4]{c} \left (-3 c^{3/2} d^3+\sqrt {a} c d^2 e-7 a \sqrt {c} d e^2+5 a^{3/2} e^3\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{7/4}}-\frac {2 \sqrt {2} \sqrt [4]{c} \left (-3 c^{3/2} d^3+\sqrt {a} c d^2 e-7 a \sqrt {c} d e^2+5 a^{3/2} e^3\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{7/4}}-\frac {\sqrt {2} \sqrt [4]{c} \left (3 c^{3/2} d^3+\sqrt {a} c d^2 e+7 a \sqrt {c} d e^2+5 a^{3/2} e^3\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{7/4}}+\frac {\sqrt {2} \sqrt [4]{c} \left (3 c^{3/2} d^3+\sqrt {a} c d^2 e+7 a \sqrt {c} d e^2+5 a^{3/2} e^3\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{7/4}}}{32 \left (c d^2+a e^2\right )^2} \] Input:
Integrate[1/((d + e*x^2)*(a + c*x^4)^2),x]
Output:
((8*c*(c*d^2 + a*e^2)*x*(d - e*x^2))/(a*(a + c*x^4)) + (32*e^(7/2)*ArcTan[ (Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] + (2*Sqrt[2]*c^(1/4)*(-3*c^(3/2)*d^3 + Sqrt[ a]*c*d^2*e - 7*a*Sqrt[c]*d*e^2 + 5*a^(3/2)*e^3)*ArcTan[1 - (Sqrt[2]*c^(1/4 )*x)/a^(1/4)])/a^(7/4) - (2*Sqrt[2]*c^(1/4)*(-3*c^(3/2)*d^3 + Sqrt[a]*c*d^ 2*e - 7*a*Sqrt[c]*d*e^2 + 5*a^(3/2)*e^3)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^ (1/4)])/a^(7/4) - (Sqrt[2]*c^(1/4)*(3*c^(3/2)*d^3 + Sqrt[a]*c*d^2*e + 7*a* Sqrt[c]*d*e^2 + 5*a^(3/2)*e^3)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + S qrt[c]*x^2])/a^(7/4) + (Sqrt[2]*c^(1/4)*(3*c^(3/2)*d^3 + Sqrt[a]*c*d^2*e + 7*a*Sqrt[c]*d*e^2 + 5*a^(3/2)*e^3)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)* x + Sqrt[c]*x^2])/a^(7/4))/(32*(c*d^2 + a*e^2)^2)
Time = 1.38 (sec) , antiderivative size = 689, normalized size of antiderivative = 1.86, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1568, 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.
| \(\displaystyle \int \frac {1}{\left (a+c x^4\right )^2 \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1568 |
| \(\displaystyle \int \left (-\frac {c e^2 \left (e x^2-d\right )}{\left (a+c x^4\right ) \left (a e^2+c d^2\right )^2}+\frac {c \left (d-e x^2\right )}{\left (a+c x^4\right )^2 \left (a e^2+c d^2\right )}+\frac {e^4}{\left (d+e x^2\right ) \left (a e^2+c d^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{c} e^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (\sqrt {c} d-\sqrt {a} e\right )}{2 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )^2}+\frac {\sqrt [4]{c} e^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {c} d-\sqrt {a} e\right )}{2 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )^2}-\frac {\sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) \left (3 \sqrt {c} d-\sqrt {a} e\right )}{8 \sqrt {2} a^{7/4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) \left (3 \sqrt {c} d-\sqrt {a} e\right )}{8 \sqrt {2} a^{7/4} \left (a e^2+c d^2\right )}-\frac {\sqrt [4]{c} e^2 \left (\sqrt {a} e+\sqrt {c} d\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )^2}+\frac {\sqrt [4]{c} e^2 \left (\sqrt {a} e+\sqrt {c} d\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{4 \sqrt {2} a^{3/4} \left (a e^2+c d^2\right )^2}-\frac {\sqrt [4]{c} \left (\sqrt {a} e+3 \sqrt {c} d\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \left (a e^2+c d^2\right )}+\frac {\sqrt [4]{c} \left (\sqrt {a} e+3 \sqrt {c} d\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} \left (a e^2+c d^2\right )}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \left (a e^2+c d^2\right )^2}+\frac {c x \left (d-e x^2\right )}{4 a \left (a+c x^4\right ) \left (a e^2+c d^2\right )}\) |
Input:
Int[1/((d + e*x^2)*(a + c*x^4)^2),x]
Output:
(c*x*(d - e*x^2))/(4*a*(c*d^2 + a*e^2)*(a + c*x^4)) + (e^(7/2)*ArcTan[(Sqr t[e]*x)/Sqrt[d]])/(Sqrt[d]*(c*d^2 + a*e^2)^2) - (c^(1/4)*e^2*(Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*(c* d^2 + a*e^2)^2) - (c^(1/4)*(3*Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 - (Sqrt[2]*c ^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)) + (c^(1/4)*e^2*(Sq rt[c]*d - Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a ^(3/4)*(c*d^2 + a*e^2)^2) + (c^(1/4)*(3*Sqrt[c]*d - Sqrt[a]*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)) - (c^(1/ 4)*e^2*(Sqrt[c]*d + Sqrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + S qrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)^2) - (c^(1/4)*(3*Sqrt[c]*d + Sqrt[a]*e)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16* Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)) + (c^(1/4)*e^2*(Sqrt[c]*d + Sqrt[a]*e)*Lo g[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*( c*d^2 + a*e^2)^2) + (c^(1/4)*(3*Sqrt[c]*d + Sqrt[a]*e)*Log[Sqrt[a] + Sqrt[ 2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2))
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int [ExpandIntegrand[(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, p, q}, x] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0])
Time = 0.20 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.90
| method | result | size |
| default | \(\frac {e^{4} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {d e}}+\frac {c \left (\frac {-\frac {e \left (a \,e^{2}+c \,d^{2}\right ) x^{3}}{4 a}+\frac {d \left (a \,e^{2}+c \,d^{2}\right ) x}{4 a}}{c \,x^{4}+a}+\frac {\frac {\left (7 d \,e^{2} a +3 d^{3} c \right ) \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {\left (-5 a \,e^{3}-c \,d^{2} e \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}}{4 a}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2}}\) | \(334\) |
| risch | \(\text {Expression too large to display}\) | \(1470\) |
Input:
int(1/(e*x^2+d)/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
e^4/(a*e^2+c*d^2)^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+c/(a*e^2+c*d^2)^2* ((-1/4*e*(a*e^2+c*d^2)/a*x^3+1/4*d*(a*e^2+c*d^2)/a*x)/(c*x^4+a)+1/4/a*(1/8 *(7*a*d*e^2+3*c*d^3)*(1/c*a)^(1/4)/a*2^(1/2)*(ln((x^2+(1/c*a)^(1/4)*x*2^(1 /2)+(1/c*a)^(1/2))/(x^2-(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))+2*arctan(2 ^(1/2)/(1/c*a)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1))+1/8*(-5*a*e ^3-c*d^2*e)/c/(1/c*a)^(1/4)*2^(1/2)*(ln((x^2-(1/c*a)^(1/4)*x*2^(1/2)+(1/c* a)^(1/2))/(x^2+(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))+2*arctan(2^(1/2)/(1 /c*a)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1))))
Leaf count of result is larger than twice the leaf count of optimal. 4934 vs. \(2 (303) = 606\).
Time = 11.08 (sec) , antiderivative size = 9892, normalized size of antiderivative = 26.66 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x**2+d)/(c*x**4+a)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (303) = 606\).
Time = 0.13 (sec) , antiderivative size = 621, normalized size of antiderivative = 1.67 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(1/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")
Output:
e^4*arctan(e*x/sqrt(d*e))/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(d*e)) + 1/8*(3*(a*c^3)^(1/4)*c^3*d^3 + 7*(a*c^3)^(1/4)*a*c^2*d*e^2 - (a*c^3)^(3/ 4)*c*d^2*e - 5*(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))/(sqrt(2)*a^2*c^4*d^4 + 2*sqrt(2)*a^3*c^3*d^2*e^2 + s qrt(2)*a^4*c^2*e^4) + 1/8*(3*(a*c^3)^(1/4)*c^3*d^3 + 7*(a*c^3)^(1/4)*a*c^2 *d*e^2 - (a*c^3)^(3/4)*c*d^2*e - 5*(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))/(sqrt(2)*a^2*c^4*d^4 + 2*sqrt(2) *a^3*c^3*d^2*e^2 + sqrt(2)*a^4*c^2*e^4) + 1/16*(3*(a*c^3)^(1/4)*c^3*d^3 + 7*(a*c^3)^(1/4)*a*c^2*d*e^2 + (a*c^3)^(3/4)*c*d^2*e + 5*(a*c^3)^(3/4)*a*e^ 3)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^4*d^4 + 2*s qrt(2)*a^3*c^3*d^2*e^2 + sqrt(2)*a^4*c^2*e^4) - 1/16*(3*(a*c^3)^(1/4)*c^3* d^3 + 7*(a*c^3)^(1/4)*a*c^2*d*e^2 + (a*c^3)^(3/4)*c*d^2*e + 5*(a*c^3)^(3/4 )*a*e^3)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^4*d^4 + 2*sqrt(2)*a^3*c^3*d^2*e^2 + sqrt(2)*a^4*c^2*e^4) - 1/4*(c*e*x^3 - c*d*x )/((c*x^4 + a)*(a*c*d^2 + a^2*e^2))
Time = 19.52 (sec) , antiderivative size = 17945, normalized size of antiderivative = 48.37 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
int(1/((a + c*x^4)^2*(d + e*x^2)),x)
Output:
((c*d*x)/(4*a*(a*e^2 + c*d^2)) - (c*e*x^3)/(4*a*(a*e^2 + c*d^2)))/(a + c*x ^4) - atan(((((((65536*a^11*c^4*e^16 - 12288*a^4*c^11*d^14*e^2 - 57344*a^5 *c^10*d^12*e^4 - 36864*a^6*c^9*d^10*e^6 + 245760*a^7*c^8*d^8*e^8 + 634880* a^8*c^7*d^6*e^10 + 663552*a^9*c^6*d^4*e^12 + 331776*a^10*c^5*d^2*e^14)/(25 6*(a^8*e^8 + a^4*c^4*d^8 + 4*a^7*c*d^2*e^6 + 4*a^5*c^3*d^6*e^2 + 6*a^6*c^2 *d^4*e^4)) - (x*((9*c^3*d^6*(-a^7*c)^(1/2) - 25*a^3*e^6*(-a^7*c)^(1/2) + 6 *a^4*c^3*d^5*e + 44*a^5*c^2*d^3*e^3 + 70*a^6*c*d*e^5 + 41*a*c^2*d^4*e^2*(- a^7*c)^(1/2) + 39*a^2*c*d^2*e^4*(-a^7*c)^(1/2))/(256*(a^11*e^8 + a^7*c^4*d ^8 + 4*a^10*c*d^2*e^6 + 4*a^8*c^3*d^6*e^2 + 6*a^9*c^2*d^4*e^4)))^(1/2)*(65 536*a^13*c^4*e^17 - 65536*a^6*c^11*d^14*e^3 - 327680*a^7*c^10*d^12*e^5 - 5 89824*a^8*c^9*d^10*e^7 - 327680*a^9*c^8*d^8*e^9 + 327680*a^10*c^7*d^6*e^11 + 589824*a^11*c^6*d^4*e^13 + 327680*a^12*c^5*d^2*e^15))/(128*(a^8*e^8 + a ^4*c^4*d^8 + 4*a^7*c*d^2*e^6 + 4*a^5*c^3*d^6*e^2 + 6*a^6*c^2*d^4*e^4)))*(( 9*c^3*d^6*(-a^7*c)^(1/2) - 25*a^3*e^6*(-a^7*c)^(1/2) + 6*a^4*c^3*d^5*e + 4 4*a^5*c^2*d^3*e^3 + 70*a^6*c*d*e^5 + 41*a*c^2*d^4*e^2*(-a^7*c)^(1/2) + 39* a^2*c*d^2*e^4*(-a^7*c)^(1/2))/(256*(a^11*e^8 + a^7*c^4*d^8 + 4*a^10*c*d^2* e^6 + 4*a^8*c^3*d^6*e^2 + 6*a^9*c^2*d^4*e^4)))^(1/2) - (x*(1152*a^2*c^11*d ^13*e^2 - 49024*a^8*c^5*d*e^14 + 7936*a^3*c^10*d^11*e^4 + 20352*a^4*c^9*d^ 9*e^6 + 8704*a^5*c^8*d^7*e^8 - 66688*a^6*c^7*d^5*e^10 - 110848*a^7*c^6*d^3 *e^12))/(128*(a^8*e^8 + a^4*c^4*d^8 + 4*a^7*c*d^2*e^6 + 4*a^5*c^3*d^6*e...
Time = 0.21 (sec) , antiderivative size = 1438, normalized size of antiderivative = 3.88 \[ \int \frac {1}{\left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/(e*x^2+d)/(c*x^4+a)^2,x)
Output:
(10*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**2*d*e**3 + 2*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*c*d**3*e + 10*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*c*d*e**3*x**4 + 2*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**2*d**3*e*x**4 - 14*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**2 *d**2*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)))*a*c*d**4 - 14*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*c*d**2*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)))*c**2*d**4*x* *4 - 10*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**2*d*e**3 - 2*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*c*d**3*e - 10*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sq rt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c*d*e**3*x**4 - 2*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**2*d**3*e*x**4 + 14*c**(3/4)*a**(1/4)*sqrt(2)...