Integrand size = 21, antiderivative size = 668 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=-\frac {e^2 \left (11 c^2 d^4+39 a c d^2 e^2-32 a^2 e^4\right ) x}{32 a^2 d \left (c d^2+a e^2\right )^3 \sqrt {d+e x^2}}+\frac {c x \left (d-e x^2\right )}{8 a \left (c d^2+a e^2\right ) \sqrt {d+e x^2} \left (a+c x^4\right )^2}+\frac {c x \left (d \left (7 c d^2+17 a e^2\right )-2 e \left (2 c d^2+7 a e^2\right ) x^2\right )}{32 a^2 \left (c d^2+a e^2\right )^2 \sqrt {d+e x^2} \left (a+c x^4\right )}+\frac {3 \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (e \left (c d^2-4 a e^2\right ) \left (c d^2+5 a e^2\right )-\left (7 c^2 d^4+25 a c d^2 e^2+38 a^2 e^4\right ) \left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^{7/2}}+\frac {3 \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (e \left (c d^2-4 a e^2\right ) \left (c d^2+5 a e^2\right )-\left (7 c^2 d^4+25 a c d^2 e^2+38 a^2 e^4\right ) \left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} \left (c d^2+a e^2\right )^{7/2}} \] Output:
-1/32*e^2*(-32*a^2*e^4+39*a*c*d^2*e^2+11*c^2*d^4)*x/a^2/d/(a*e^2+c*d^2)^3/ (e*x^2+d)^(1/2)+1/8*c*x*(-e*x^2+d)/a/(a*e^2+c*d^2)/(e*x^2+d)^(1/2)/(c*x^4+ a)^2+1/32*c*x*(d*(17*a*e^2+7*c*d^2)-2*e*(7*a*e^2+2*c*d^2)*x^2)/a^2/(a*e^2+ c*d^2)^2/(e*x^2+d)^(1/2)/(c*x^4+a)+3/128*c^(1/2)*(a^(1/2)*e+(a*e^2+c*d^2)^ (1/2))^(1/2)*(e*(-4*a*e^2+c*d^2)*(5*a*e^2+c*d^2)-(38*a^2*e^4+25*a*c*d^2*e^ 2+7*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)/(a*e^2+c*d^2)^(7/2)+3/ 128*c^(1/2)*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*(e*(-4*a*e^2+c*d^2)*(5* a*e^2+c*d^2)-(38*a^2*e^4+25*a*c*d^2*e^2+7*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)/(a*e^2+c*d^2)^(7/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.42 (sec) , antiderivative size = 1993, normalized size of antiderivative = 2.98 \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:
Integrate[1/((d + e*x^2)^(3/2)*(a + c*x^4)^3),x]
Output:
((2*x*(32*a^4*e^6 + c^4*d^4*x^4*(7*d^2 - 4*d*e*x^2 - 11*e^2*x^4) + 2*a^3*c *e^4*(-9*d^2 - 9*d*e*x^2 + 32*e^2*x^4) + a*c^3*d^2*(11*d^4 - 8*d^3*e*x^2 + 2*d^2*e^2*x^4 - 18*d*e^3*x^6 - 39*e^4*x^8) + a^2*c^2*e^2*(21*d^4 - 26*d^3 *e*x^2 - 61*d^2*e^2*x^4 - 14*d*e^3*x^6 + 32*e^4*x^8)))/(a^2*d*Sqrt[d + e*x ^2]*(a + c*x^4)^2) - 32*e^(11/2)*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 & , (17*c*d^2*Log[d + 2*e*x^2 - 2*S qrt[e]*x*Sqrt[d + e*x^2] - #1] + 16*a*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sq rt[d + e*x^2] - #1] - 6*c*d*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + c*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) & ] + (16*e^(7/2)*Roo tSum[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 & , (860*c^4*d^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 6091*a*c^3*d^6*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 1 3680*a^2*c^2*d^4*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 12544*a^3*c*d^2*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 4096*a^4*e^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 706*c^ 4*d^7*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 2364*a*c^3* d^5*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 2688*a^2* c^2*d^3*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 1024* a^3*c*d*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 18...
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.
\(\displaystyle \int \frac {1}{\left (a+c x^4\right )^3 \left (d+e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {1}{\left (a+c x^4\right )^3 \left (d+e x^2\right )^{3/2}}dx\) |
Input:
Int[1/((d + e*x^2)^(3/2)*(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. \(1597\) vs. \(2(576)=1152\).
Time = 2.96 (sec) , antiderivative size = 1598, normalized size of antiderivative = 2.39
method | result | size |
pseudoelliptic | \(\text {Expression too large to display}\) | \(1598\) |
default | \(\text {Expression too large to display}\) | \(11178\) |
Input:
int(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x,method=_RETURNVERBOSE)
Output:
-57/32*(-1/4*(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*(e*x^2+d)^(1/2)*((-(c *x^4+a)^2*(a^2*e^4+25/38*a*c*d^2*e^2+7/38*c^2*d^4)*(a*e^2+c*d^2)^(1/2)-12/ 19*((29/6*e^2*x^4+d^2)*c*e^2*a^(7/2)+1/4*c^2*(29/3*e^4*x^8+8*d^2*e^2*x^4+d ^4)*a^(5/2)+29/12*a^(9/2)*e^4+1/4*x^4*d^2*((4*e^2*x^4+2*d^2)*a^(3/2)+c*d^2 *x^4*a^(1/2))*c^3)*e)*(a*(a*e^2+c*d^2))^(1/2)+(a*(c*x^4+a)^2*(a^2*e^4+25/3 8*a*c*d^2*e^2+7/38*c^2*d^4)*(a*e^2+c*d^2)^(1/2)+3/19*(c^2*(29/3*e^4*x^8+8* d^2*e^2*x^4+d^4)*a^(7/2)+2*c^3*d^2*x^4*(2*e^2*x^4+d^2)*a^(5/2)+4*(29/6*e^2 *x^4+d^2)*c*e^2*a^(9/2)+a^(3/2)*c^4*d^4*x^8+29/3*a^(11/2)*e^4)*e)*e)*(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)*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)+1/4*(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*(e* x^2+d)^(1/2)*((-(c*x^4+a)^2*(a^2*e^4+25/38*a*c*d^2*e^2+7/38*c^2*d^4)*(a*e^ 2+c*d^2)^(1/2)-12/19*((29/6*e^2*x^4+d^2)*c*e^2*a^(7/2)+1/4*c^2*(29/3*e^4*x ^8+8*d^2*e^2*x^4+d^4)*a^(5/2)+29/12*a^(9/2)*e^4+1/4*x^4*d^2*((4*e^2*x^4+2* d^2)*a^(3/2)+c*d^2*x^4*a^(1/2))*c^3)*e)*(a*(a*e^2+c*d^2))^(1/2)+(a*(c*x^4+ a)^2*(a^2*e^4+25/38*a*c*d^2*e^2+7/38*c^2*d^4)*(a*e^2+c*d^2)^(1/2)+3/19*(c^ 2*(29/3*e^4*x^8+8*d^2*e^2*x^4+d^4)*a^(7/2)+2*c^3*d^2*x^4*(2*e^2*x^4+d^2)*a ^(5/2)+4*(29/6*e^2*x^4+d^2)*c*e^2*a^(9/2)+a^(3/2)*c^4*d^4*x^8+29/3*a^(11/2 )*e^4)*e)*e)*(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)*ln((a^(1/2)*(e*x^2+d)+(e*x^2+d)^(1/2)*(2*(a*(a*e^2+c*d^2))^(1/...
Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x**2+d)**(3/2)/(c*x**4+a)**3,x)
Output:
Timed out
\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\int { \frac {1}{{\left (c x^{4} + a\right )}^{3} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x, algorithm="maxima")
Output:
integrate(1/((c*x^4 + a)^3*(e*x^2 + d)^(3/2)), x)
Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\int \frac {1}{{\left (c\,x^4+a\right )}^3\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \] Input:
int(1/((a + c*x^4)^3*(d + e*x^2)^(3/2)),x)
Output:
int(1/((a + c*x^4)^3*(d + e*x^2)^(3/2)), x)
\[ \int \frac {1}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^3} \, dx=\int \frac {1}{\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (c \,x^{4}+a \right )^{3}}d x \] Input:
int(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x)
Output:
int(1/(e*x^2+d)^(3/2)/(c*x^4+a)^3,x)