Integrand size = 24, antiderivative size = 411 \[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\frac {x \left (d+e x^2\right )^{3/2}}{6 a \left (a-c x^4\right )^{3/2}}+\frac {x \sqrt {d+e x^2} \left (5 d+2 e x^2\right )}{12 a^2 \sqrt {a-c x^4}}+\frac {e \sqrt {d+e x^2} \sqrt {a-c x^4}}{6 a^2 c x}+\frac {e \left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{6 a^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}+\frac {\left (5 c d^2-2 a e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{12 a^{5/2} \sqrt {c} \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
1/6*x*(e*x^2+d)^(3/2)/a/(-c*x^4+a)^(3/2)+1/12*x*(e*x^2+d)^(1/2)*(2*e*x^2+5 *d)/a^2/(-c*x^4+a)^(1/2)+1/6*e*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/a^2/c/x+1/ 6*e*(d+a^(1/2)*e/c^(1/2))*(1-a/c/x^4)^(1/2)*x^3*(a^(1/2)*(e*x^2+d)/(c^(1/2 )*d+a^(1/2)*e)/x^2)^(1/2)*EllipticE(1/2*(1-a^(1/2)/c^(1/2)/x^2)^(1/2)*2^(1 /2),2^(1/2)*(d/(d+a^(1/2)*e/c^(1/2)))^(1/2))/a^2/(e*x^2+d)^(1/2)/(-c*x^4+a )^(1/2)+1/12*(-2*a*e^2+5*c*d^2)*(1-a/c/x^4)^(1/2)*x^3*(a^(1/2)*(e*x^2+d)/( c^(1/2)*d+a^(1/2)*e)/x^2)^(1/2)*EllipticF(1/2*(1-a^(1/2)/c^(1/2)/x^2)^(1/2 )*2^(1/2),2^(1/2)*(d/(d+a^(1/2)*e/c^(1/2)))^(1/2))/a^(5/2)/c^(1/2)/(e*x^2+ d)^(1/2)/(-c*x^4+a)^(1/2)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx \] Input:
Integrate[(d + e*x^2)^(3/2)/(a - c*x^4)^(5/2),x]
Output:
Integrate[(d + e*x^2)^(3/2)/(a - c*x^4)^(5/2), 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1571 |
| \(\displaystyle \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}}dx\) |
Input:
Int[(d + e*x^2)^(3/2)/(a - c*x^4)^(5/2),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]
\[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{\left (-c \,x^{4}+a \right )^{\frac {5}{2}}}d x\]
Input:
int((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x)
Output:
int((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}}}{{\left (-c x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x, algorithm="fricas")
Output:
integral(-sqrt(-c*x^4 + a)*(e*x^2 + d)^(3/2)/(c^3*x^12 - 3*a*c^2*x^8 + 3*a ^2*c*x^4 - a^3), x)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int \frac {\left (d + e x^{2}\right )^{\frac {3}{2}}}{\left (a - c x^{4}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((e*x**2+d)**(3/2)/(-c*x**4+a)**(5/2),x)
Output:
Integral((d + e*x**2)**(3/2)/(a - c*x**4)**(5/2), x)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}}}{{\left (-c x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x, algorithm="maxima")
Output:
integrate((e*x^2 + d)^(3/2)/(-c*x^4 + a)^(5/2), x)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}}}{{\left (-c x^{4} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x, algorithm="giac")
Output:
integrate((e*x^2 + d)^(3/2)/(-c*x^4 + a)^(5/2), x)
Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{3/2}}{{\left (a-c\,x^4\right )}^{5/2}} \,d x \] Input:
int((d + e*x^2)^(3/2)/(a - c*x^4)^(5/2),x)
Output:
int((d + e*x^2)^(3/2)/(a - c*x^4)^(5/2), x)
\[ \int \frac {\left (d+e x^2\right )^{3/2}}{\left (a-c x^4\right )^{5/2}} \, dx=\text {too large to display} \] Input:
int((e*x^2+d)^(3/2)/(-c*x^4+a)^(5/2),x)
Output:
(12*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*d*e**2*x - 3*sqrt(d + e*x**2)*sqrt (a - c*x**4)*c*d**3*x + 4*sqrt(d + e*x**2)*sqrt(a - c*x**4)*c*d**2*e*x**3 - 8*sqrt(d + e*x**2)*sqrt(a - c*x**4)*c*d*e**2*x**5 + 48*int((sqrt(d + e*x **2)*sqrt(a - c*x**4)*x**4)/(4*a**4*d*e**2 + 4*a**4*e**3*x**2 + a**3*c*d** 3 + a**3*c*d**2*e*x**2 - 12*a**3*c*d*e**2*x**4 - 12*a**3*c*e**3*x**6 - 3*a **2*c**2*d**3*x**4 - 3*a**2*c**2*d**2*e*x**6 + 12*a**2*c**2*d*e**2*x**8 + 12*a**2*c**2*e**3*x**10 + 3*a*c**3*d**3*x**8 + 3*a*c**3*d**2*e*x**10 - 4*a *c**3*d*e**2*x**12 - 4*a*c**3*e**3*x**14 - c**4*d**3*x**12 - c**4*d**2*e*x **14),x)*a**5*e**6 - 120*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a **4*d*e**2 + 4*a**4*e**3*x**2 + a**3*c*d**3 + a**3*c*d**2*e*x**2 - 12*a**3 *c*d*e**2*x**4 - 12*a**3*c*e**3*x**6 - 3*a**2*c**2*d**3*x**4 - 3*a**2*c**2 *d**2*e*x**6 + 12*a**2*c**2*d*e**2*x**8 + 12*a**2*c**2*e**3*x**10 + 3*a*c* *3*d**3*x**8 + 3*a*c**3*d**2*e*x**10 - 4*a*c**3*d*e**2*x**12 - 4*a*c**3*e* *3*x**14 - c**4*d**3*x**12 - c**4*d**2*e*x**14),x)*a**4*c*d**2*e**4 - 96*i nt((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a**4*d*e**2 + 4*a**4*e**3*x **2 + a**3*c*d**3 + a**3*c*d**2*e*x**2 - 12*a**3*c*d*e**2*x**4 - 12*a**3*c *e**3*x**6 - 3*a**2*c**2*d**3*x**4 - 3*a**2*c**2*d**2*e*x**6 + 12*a**2*c** 2*d*e**2*x**8 + 12*a**2*c**2*e**3*x**10 + 3*a*c**3*d**3*x**8 + 3*a*c**3*d* *2*e*x**10 - 4*a*c**3*d*e**2*x**12 - 4*a*c**3*e**3*x**14 - c**4*d**3*x**12 - c**4*d**2*e*x**14),x)*a**4*c*e**6*x**4 + 27*int((sqrt(d + e*x**2)*sq...