Integrand size = 21, antiderivative size = 632 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=-\frac {3 e^2 \left (2 c d^2+a e^2\right ) x \sqrt {d+e x^2}}{16 a^2 c^2}-\frac {5 d e^2 x \left (d+e x^2\right )^{3/2}}{32 a^2 c}+\frac {e^2 x \left (d+e x^2\right )^{5/2}}{16 a^2 c}+\frac {x \left (d+e x^2\right )^{9/2}}{8 a \left (a+c x^4\right )^2}+\frac {x \left (7 d-2 e x^2\right ) \left (d+e x^2\right )^{7/2}}{32 a^2 \left (a+c x^4\right )}+\frac {3 \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (11 c^2 d^4 e+11 a c d^2 e^3+4 a^2 e^5-\left (7 c^2 d^4+5 a c d^2 e^2+2 a^2 e^4\right ) \left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} c^{5/2} \sqrt {c d^2+a e^2}}+\frac {3 \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (11 c^2 d^4 e+11 a c d^2 e^3+4 a^2 e^5-\left (7 c^2 d^4+5 a c d^2 e^2+2 a^2 e^4\right ) \left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{64 \sqrt {2} a^{9/4} c^{5/2} \sqrt {c d^2+a e^2}} \] Output:
-3/16*e^2*(a*e^2+2*c*d^2)*x*(e*x^2+d)^(1/2)/a^2/c^2-5/32*d*e^2*x*(e*x^2+d) ^(3/2)/a^2/c+1/16*e^2*x*(e*x^2+d)^(5/2)/a^2/c+1/8*x*(e*x^2+d)^(9/2)/a/(c*x ^4+a)^2+1/32*x*(-2*e*x^2+7*d)*(e*x^2+d)^(7/2)/a^2/(c*x^4+a)+3/128*(a^(1/2) *e+(a*e^2+c*d^2)^(1/2))^(1/2)*(11*c^2*d^4*e+11*a*c*d^2*e^3+4*a^2*e^5-(2*a^ 2*e^4+5*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)/c^(5 /2)/(a*e^2+c*d^2)^(1/2)+3/128*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*(11*c ^2*d^4*e+11*a*c*d^2*e^3+4*a^2*e^5-(2*a^2*e^4+5*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)/c^(5/2)/(a*e^2+c*d^2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 3.09 (sec) , antiderivative size = 2216, normalized size of antiderivative = 3.51 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\text {Result too large to show} \] Input:
Integrate[(d + e*x^2)^(9/2)/(a + c*x^4)^3,x]
Output:
((2*c^3*x*Sqrt[d + e*x^2]*(-6*a^3*e^4 + c^3*d^3*x^4*(7*d + 19*e*x^2) - a^2 *c*e^2*(15*d^2 + d*e*x^2 + 10*e^2*x^4) + a*c^2*d*(11*d^3 + 35*d^2*e*x^2 + 9*d*e^2*x^4 + 15*e^3*x^6)))/(a^2*(a + c*x^4)^2) + 32*c^2*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 & , (161*c*d^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 32*a*e^ 2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 18*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) & ] + (32*e^(7/2)*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 & , (1239*c^5*d^10*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 4394*a*c^4*d^8*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 32487*a^2*c^3*d^6*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 4216*a^3*c^2*d^4*e^6*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 28544*a^4*c*d^2*e^8*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 2048*a^5*e^10*Log[d + 2*e*x^2 - 2*S qrt[e]*x*Sqrt[d + e*x^2] - #1] - 840*c^5*d^9*Log[d + 2*e*x^2 - 2*Sqrt[e]*x *Sqrt[d + e*x^2] - #1]*#1 + 6324*a*c^4*d^7*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e] *x*Sqrt[d + e*x^2] - #1]*#1 + 2160*a^2*c^3*d^5*e^4*Log[d + 2*e*x^2 - 2*Sqr t[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 7360*a^3*c^2*d^3*e^6*Log[d + 2*e*x^2 - 2 *Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 512*a^4*c*d*e^8*Log[d + 2*e*x^2 -...
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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3}dx\) |
Input:
Int[(d + e*x^2)^(9/2)/(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. \(1243\) vs. \(2(532)=1064\).
Time = 1.94 (sec) , antiderivative size = 1244, normalized size of antiderivative = 1.97
method | result | size |
pseudoelliptic | \(\text {Expression too large to display}\) | \(1244\) |
default | \(\text {Expression too large to display}\) | \(53338\) |
Input:
int((e*x^2+d)^(9/2)/(c*x^4+a)^3,x,method=_RETURNVERBOSE)
Output:
-3/32*(d^2*c*(c*x^4+a)^2*(a^(9/2)*(a*e^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2)+a^ 4*(a^2*e^4+5/2*a*c*d^2*e^2+7/2*c^2*d^4))*arctan((2*a^(1/2)*(e*x^2+d)^(1/2) +(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*x)/x/(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))+1/4*(c*x^4+a)^2*((-a^(7/2)*(a*e ^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2)+a^3*(a^2*e^4+5/2*a*c*d^2*e^2+7/2*c^2*d^4 ))*(a*(a*e^2+c*d^2))^(1/2)+(a^(9/2)*(a*e^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2)- a^4*(a^2*e^4+5/2*a*c*d^2*e^2+7/2*c^2*d^4))*e)*(2*(a*(a*e^2+c*d^2))^(1/2)+2 *a*e)^(1/2)*(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*(c*x^4+a)^2*((-a^(7/2)*(a* e^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2)+a^3*(a^2*e^4+5/2*a*c*d^2*e^2+7/2*c^2*d^ 4))*(a*(a*e^2+c*d^2))^(1/2)+(a^(9/2)*(a*e^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2) -a^4*(a^2*e^4+5/2*a*c*d^2*e^2+7/2*c^2*d^4))*e)*(2*(a*(a*e^2+c*d^2))^(1/2)+ 2*a*e)^(1/2)*(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)+d*(-arctan(((2*(a*(a*e^2+c*d^ 2))^(1/2)+2*a*e)^(1/2)*x-2*a^(1/2)*(e*x^2+d)^(1/2))/x/(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))*d*a^(9/2)*(c*x^4+a)^2*( a*e^2+2*c*d^2)*e*(a*e^2+c*d^2)^(1/2)+1/3*(-15*a*c^2*d*e^3*x^6-19*c^3*d^3*e *x^6+10*a^2*c*e^4*x^4-9*a*c^2*d^2*e^2*x^4-7*c^3*d^4*x^4+a^2*c*d*e^3*x^2...
Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x^2+d)^(9/2)/(c*x^4+a)^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x**2+d)**(9/2)/(c*x**4+a)**3,x)
Output:
Timed out
\[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {9}{2}}}{{\left (c x^{4} + a\right )}^{3}} \,d x } \] Input:
integrate((e*x^2+d)^(9/2)/(c*x^4+a)^3,x, algorithm="maxima")
Output:
integrate((e*x^2 + d)^(9/2)/(c*x^4 + a)^3, x)
Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x^2+d)^(9/2)/(c*x^4+a)^3,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{9/2}}{{\left (c\,x^4+a\right )}^3} \,d x \] Input:
int((d + e*x^2)^(9/2)/(a + c*x^4)^3,x)
Output:
int((d + e*x^2)^(9/2)/(a + c*x^4)^3, x)
\[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a+c x^4\right )^3} \, dx=\int \frac {\left (e \,x^{2}+d \right )^{\frac {9}{2}}}{\left (c \,x^{4}+a \right )^{3}}d x \] Input:
int((e*x^2+d)^(9/2)/(c*x^4+a)^3,x)
Output:
int((e*x^2+d)^(9/2)/(c*x^4+a)^3,x)