Integrand size = 24, antiderivative size = 598 \[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {\left (a-\frac {c d^2}{e^2}\right ) x \sqrt {a-c x^4}}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 \left (\frac {a}{d^2}+\frac {3 c}{e^2}\right ) x \sqrt {a-c x^4}}{3 \sqrt {d+e x^2}}-\frac {\left (15 c d^2+4 a e^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{6 d^2 e^3 x}-\frac {\sqrt {c} \left (\sqrt {c} d+\sqrt {a} e\right ) \left (15 c d^2+4 a e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{6 d^2 e^3 \sqrt {d+e x^2} \sqrt {a-c x^4}}+\frac {\sqrt {a} \sqrt {c} \left (5 c d^2+4 a e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{6 d^2 e^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {5 c^2 d \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{2 e^3 \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
1/3*(a-c*d^2/e^2)*x*(-c*x^4+a)^(1/2)/d/(e*x^2+d)^(3/2)+2/3*(a/d^2+3*c/e^2) *x*(-c*x^4+a)^(1/2)/(e*x^2+d)^(1/2)-1/6*(4*a*e^2+15*c*d^2)*(e*x^2+d)^(1/2) *(-c*x^4+a)^(1/2)/d^2/e^3/x-1/6*c^(1/2)*(c^(1/2)*d+a^(1/2)*e)*(4*a*e^2+15* 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)*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))/d^2/e^3/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)+1/6*a^( 1/2)*c^(1/2)*(4*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))/d^2/e^2/(e*x^2+d)^(1/2)/ (-c*x^4+a)^(1/2)-5/2*c^2*d*(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)*EllipticPi(1/2*(1-a^(1/2)/c^(1/2)/x^2)^(1/2)*2^ (1/2),2,2^(1/2)*(d/(d+a^(1/2)*e/c^(1/2)))^(1/2))/e^3/(e*x^2+d)^(1/2)/(-c*x ^4+a)^(1/2)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx \] Input:
Integrate[(a - c*x^4)^(3/2)/(d + e*x^2)^(5/2),x]
Output:
Integrate[(a - c*x^4)^(3/2)/(d + e*x^2)^(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.
\(\displaystyle \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1571 |
\(\displaystyle \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}}dx\) |
Input:
Int[(a - c*x^4)^(3/2)/(d + e*x^2)^(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]
\[\int \frac {\left (-c \,x^{4}+a \right )^{\frac {3}{2}}}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Input:
int((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x)
Output:
int((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (-c x^{4} + a\right )}^{\frac {3}{2}}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x, algorithm="fricas")
Output:
integral((-c*x^4 + a)^(3/2)*sqrt(e*x^2 + d)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2 *e*x^2 + d^3), x)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {\left (a - c x^{4}\right )^{\frac {3}{2}}}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((-c*x**4+a)**(3/2)/(e*x**2+d)**(5/2),x)
Output:
Integral((a - c*x**4)**(3/2)/(d + e*x**2)**(5/2), x)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (-c x^{4} + a\right )}^{\frac {3}{2}}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x, algorithm="maxima")
Output:
integrate((-c*x^4 + a)^(3/2)/(e*x^2 + d)^(5/2), x)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (-c x^{4} + a\right )}^{\frac {3}{2}}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x, algorithm="giac")
Output:
integrate((-c*x^4 + a)^(3/2)/(e*x^2 + d)^(5/2), x)
Timed out. \[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {{\left (a-c\,x^4\right )}^{3/2}}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \] Input:
int((a - c*x^4)^(3/2)/(d + e*x^2)^(5/2),x)
Output:
int((a - c*x^4)^(3/2)/(d + e*x^2)^(5/2), x)
\[ \int \frac {\left (a-c x^4\right )^{3/2}}{\left (d+e x^2\right )^{5/2}} \, dx=-\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {-c \,x^{4}+a}\, x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) c +\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {-c \,x^{4}+a}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) a \] Input:
int((-c*x^4+a)^(3/2)/(e*x^2+d)^(5/2),x)
Output:
- int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(d**3 + 3*d**2*e*x**2 + 3* d*e**2*x**4 + e**3*x**6),x)*c + int((sqrt(d + e*x**2)*sqrt(a - c*x**4))/(d **3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*a