Integrand size = 21, antiderivative size = 596 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\frac {c x \left (d-e x^2\right ) \sqrt {d+e x^2}}{8 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )^2}+\frac {c x \sqrt {d+e x^2} \left (d \left (7 c d^2+13 a e^2\right )-6 e \left (c d^2+2 a e^2\right ) x^2\right )}{32 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^4\right )}+\frac {\sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (2 c d^2 e \left (4 c d^2+7 a e^2\right )-\left (21 c^2 d^4+47 a c d^2 e^2+32 a^2 e^4\right ) \left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} \sqrt {c} d \left (c d^2+a e^2\right )^{5/2}}+\frac {\sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (2 c d^2 e \left (4 c d^2+7 a e^2\right )-\left (21 c^2 d^4+47 a c d^2 e^2+32 a^2 e^4\right ) \left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} \sqrt {c} d \left (c d^2+a e^2\right )^{5/2}} \] Output:
1/8*c*x*(-e*x^2+d)*(e*x^2+d)^(1/2)/a/(a*e^2+c*d^2)/(c*x^4+a)^2+1/32*c*x*(e *x^2+d)^(1/2)*(d*(13*a*e^2+7*c*d^2)-6*e*(2*a*e^2+c*d^2)*x^2)/a^2/(a*e^2+c* d^2)^2/(c*x^4+a)+1/128*(a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*(2*c*d^2*e*(7 *a*e^2+4*c*d^2)-(32*a^2*e^4+47*a*c*d^2*e^2+21*c^2*d^4)*(e-(a*e^2+c*d^2)^(1 /2)/a^(1/2)))*arctan(2^(1/2)*a^(1/4)*c^(1/2)*(a^(1/2)*e+(a*e^2+c*d^2)^(1/2 ))^(1/2)*x*(e*x^2+d)^(1/2)/(a^(1/2)*(a^(1/2)*e+(a*e^2+c*d^2)^(1/2))-c*d*x^ 2))*2^(1/2)/a^(9/4)/c^(1/2)/d/(a*e^2+c*d^2)^(5/2)+1/128*(-a^(1/2)*e+(a*e^2 +c*d^2)^(1/2))^(1/2)*(2*c*d^2*e*(7*a*e^2+4*c*d^2)-(32*a^2*e^4+47*a*c*d^2*e ^2+21*c^2*d^4)*(e+(a*e^2+c*d^2)^(1/2)/a^(1/2)))*arctanh(2^(1/2)*a^(1/4)*c^ (1/2)*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*x*(e*x^2+d)^(1/2)/(a^(1/2)*(a ^(1/2)*e-(a*e^2+c*d^2)^(1/2))-c*d*x^2))*2^(1/2)/a^(9/4)/c^(1/2)/d/(a*e^2+c *d^2)^(5/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.55 (sec) , antiderivative size = 1211, normalized size of antiderivative = 2.03 \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:
Integrate[1/(Sqrt[d + e*x^2]*(a + c*x^4)^3),x]
Output:
((c^3*x*Sqrt[d + e*x^2]*(a^2*e^2*(17*d - 16*e*x^2) + c^2*d^2*x^4*(7*d - 6* e*x^2) + a*c*(11*d^3 - 10*d^2*e*x^2 + 13*d*e^2*x^4 - 12*e^3*x^6)))/(a + c* x^4)^2 + (64*a*e^(7/2)*(c*d^2 + a*e^2)*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^ 2*#1^2 + 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (141*c^2*d^4*Log[d + 2*e* x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 656*a*c*d^2*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 512*a^2*e^4*Log[d + 2*e*x^2 - 2*Sqrt [e]*x*Sqrt[d + e*x^2] - #1] - 110*c^2*d^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sq rt[d + e*x^2] - #1]*#1 - 128*a*c*d*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[ d + e*x^2] - #1]*#1 + 29*c^2*d^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e* x^2] - #1]*#1^2 + 32*a*c*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2)/(c*d^3 - 3*c*d^2*#1 - 8*a*e^2*#1 + 3*c*d*#1^2 - c*#1^3) & ])/ d^3 + (e^(3/2)*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^2*#1^2 + 16*a*e^2*#1^2 - 4*c*d*#1^3 + c*#1^4 & , (4*c^4*d^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 9017*a*c^3*d^6*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 51008*a^2*c^2*d^4*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 74752*a^3*c*d^2*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 32768*a^4*e^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^ 2] - #1] + 34*c^4*d^7*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]* #1 + 7120*a*c^3*d^5*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1 ]*#1 + 15296*a^2*c^2*d^3*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x...
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.
\(\displaystyle \int \frac {1}{\left (a+c x^4\right )^3 \sqrt {d+e x^2}} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {1}{\left (a+c x^4\right )^3 \sqrt {d+e x^2}}dx\) |
Input:
Int[1/(Sqrt[d + e*x^2]*(a + c*x^4)^3),x]
Output:
$Aborted
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1842\) vs. \(2(508)=1016\).
Time = 2.40 (sec) , antiderivative size = 1843, normalized size of antiderivative = 3.09
method | result | size |
pseudoelliptic | \(\text {Expression too large to display}\) | \(1843\) |
default | \(\text {Expression too large to display}\) | \(5410\) |
Input:
int(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x,method=_RETURNVERBOSE)
Output:
-1/64/(4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1/2)-2*a*e)^(1/2 )*(-8*(4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1/2)-2*a*e)^(1/2 )*(c*x^4+a)^2*((-a^(17/2)*e^11-129/32*d^2*a^(15/2)*c*e^9-(a*e^2+c*d^2)^(3/ 2)*a^7*e^8-13/2*d^4*a^(13/2)*c^2*e^7-111/32*d^2*(a*e^2+c*d^2)^(3/2)*a^6*c* e^6-85/16*d^6*a^(11/2)*c^3*e^5-147/32*d^4*(a*e^2+c*d^2)^(3/2)*a^5*c^2*e^4- 9/4*d^8*a^(9/2)*c^4*e^3-89/32*d^6*(a*e^2+c*d^2)^(3/2)*a^4*c^3*e^2-13/32*d^ 10*a^(7/2)*c^5*e-21/32*d^8*(a*e^2+c*d^2)^(3/2)*a^3*c^4)*(a*(a*e^2+c*d^2))^ (1/2)+(a^(19/2)*e^11+129/32*d^2*a^(17/2)*c*e^9+(a*e^2+c*d^2)^(3/2)*a^8*e^8 +13/2*d^4*a^(15/2)*c^2*e^7+111/32*d^2*(a*e^2+c*d^2)^(3/2)*a^7*c*e^6+85/16* d^6*a^(13/2)*c^3*e^5+147/32*d^4*(a*e^2+c*d^2)^(3/2)*a^6*c^2*e^4+9/4*d^8*a^ (11/2)*c^4*e^3+89/32*d^6*(a*e^2+c*d^2)^(3/2)*a^5*c^3*e^2+13/32*d^10*a^(9/2 )*c^5*e+21/32*d^8*(a*e^2+c*d^2)^(3/2)*a^4*c^4)*e)*(2*(a*(a*e^2+c*d^2))^(1/ 2)+2*a*e)^(1/2)*ln((a^(1/2)*(e*x^2+d)-(e*x^2+d)^(1/2)*(2*(a*(a*e^2+c*d^2)) ^(1/2)+2*a*e)^(1/2)*x+(a*e^2+c*d^2)^(1/2)*x^2)/x^2)+8*(4*(a*e^2+c*d^2)^(1/ 2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1/2)-2*a*e)^(1/2)*(c*x^4+a)^2*((-a^(17/2)* e^11-129/32*d^2*a^(15/2)*c*e^9-(a*e^2+c*d^2)^(3/2)*a^7*e^8-13/2*d^4*a^(13/ 2)*c^2*e^7-111/32*d^2*(a*e^2+c*d^2)^(3/2)*a^6*c*e^6-85/16*d^6*a^(11/2)*c^3 *e^5-147/32*d^4*(a*e^2+c*d^2)^(3/2)*a^5*c^2*e^4-9/4*d^8*a^(9/2)*c^4*e^3-89 /32*d^6*(a*e^2+c*d^2)^(3/2)*a^4*c^3*e^2-13/32*d^10*a^(7/2)*c^5*e-21/32*d^8 *(a*e^2+c*d^2)^(3/2)*a^3*c^4)*(a*(a*e^2+c*d^2))^(1/2)+(a^(19/2)*e^11+12...
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x**2+d)**(1/2)/(c*x**4+a)**3,x)
Output:
Timed out
\[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\int { \frac {1}{{\left (c x^{4} + a\right )}^{3} \sqrt {e x^{2} + d}} \,d x } \] Input:
integrate(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x, algorithm="maxima")
Output:
integrate(1/((c*x^4 + a)^3*sqrt(e*x^2 + d)), x)
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\int \frac {1}{{\left (c\,x^4+a\right )}^3\,\sqrt {e\,x^2+d}} \,d x \] Input:
int(1/((a + c*x^4)^3*(d + e*x^2)^(1/2)),x)
Output:
int(1/((a + c*x^4)^3*(d + e*x^2)^(1/2)), x)
\[ \int \frac {1}{\sqrt {d+e x^2} \left (a+c x^4\right )^3} \, dx=\int \frac {1}{\sqrt {e \,x^{2}+d}\, \left (c \,x^{4}+a \right )^{3}}d x \] Input:
int(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x)
Output:
int(1/(e*x^2+d)^(1/2)/(c*x^4+a)^3,x)