Integrand size = 35, antiderivative size = 1564 \[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx =\text {Too large to display} \] Output:
1/3840*(10*A*c*e*(15*c^3*d^3-9*b^3*e^3-c^2*d*e*(-68*a*e+31*b*d)+3*b*c*e^2* (20*a*e+3*b*d))-B*(105*c^4*d^4-45*b^4*e^4-2*c^3*d^2*e*(-166*a*e+95*b*d)+30 *b^2*c*e^3*(10*a*e+b*d)+12*c^2*e^2*(-32*a^2*e^2-18*a*b*d*e+3*b^2*d^2)))*(e *x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/c^3/e^4/x-1/1920*(10*A*c*e*(5*c^2*d^2- 3*b^2*e^2-10*c*e*(6*a*e+b*d))-B*(35*c^3*d^3-15*b^3*e^3-c^2*d*e*(-108*a*e+6 1*b*d)+3*b*c*e^2*(28*a*e+3*b*d)))*x*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/ c^2/e^3+1/480*(10*A*c*e*(9*b*e+c*d)-B*(7*c^2*d^2-3*b^2*e^2-12*c*e*(8*a*e+b *d)))*x^3*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/c/e^2+1/80*(10*A*c*e+11*B* b*e+B*c*d)*x^5*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/e+1/10*B*c*x^7*(e*x^2 +d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)-1/7680*(-4*a*c+b^2)^(1/2)*(10*A*c*e*(15*c^ 3*d^3-9*b^3*e^3-c^2*d*e*(-68*a*e+31*b*d)+3*b*c*e^2*(20*a*e+3*b*d))-B*(105* c^4*d^4-45*b^4*e^4-2*c^3*d^2*e*(-166*a*e+95*b*d)+30*b^2*c*e^3*(10*a*e+b*d) +12*c^2*e^2*(-32*a^2*e^2-18*a*b*d*e+3*b^2*d^2)))*(-a*(c+a/x^4+b/x^2)/(-4*a *c+b^2))^(1/2)*x*(e*x^2+d)^(1/2)*EllipticE(1/2*(1+(b+2*a/x^2)/(-4*a*c+b^2) ^(1/2))^(1/2)*2^(1/2),2^(1/2)*((-4*a*c+b^2)^(1/2)*d/(b*d+(-4*a*c+b^2)^(1/2 )*d-2*a*e))^(1/2))*2^(1/2)/c^3/e^4/(-a*(e+d/x^2)/((b+(-4*a*c+b^2)^(1/2))*d -2*a*e))^(1/2)/(c*x^4+b*x^2+a)^(1/2)+1/3840*(-4*a*c+b^2)^(1/2)*(10*A*c*e*( 5*c^3*d^3-9*b^3*e^3+15*b*c*e^2*(4*a*e+b*d)-c^2*d*e*(196*a*e+11*b*d))-B*(35 *c^4*d^4-45*b^4*e^4-4*c^3*d^2*e*(-29*a*e+17*b*d)+60*b^2*c*e^3*(5*a*e+b*d)+ 6*c^2*e^2*(-64*a^2*e^2-64*a*b*d*e+3*b^2*d^2)))*(-a*(c+a/x^4+b/x^2)/(-4*...
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx \] Input:
Integrate[(A + B*x^2)*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)^(3/2),x]
Output:
Integrate[(A + B*x^2)*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)^(3/2), 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.
\(\displaystyle \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2260 |
\(\displaystyle \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2}dx\) |
Input:
Int[(A + B*x^2)*Sqrt[d + e*x^2]*(a + b*x^2 + c*x^4)^(3/2),x]
Output:
$Aborted
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_.), x_Symbol] :> Unintegrable[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x] /; FreeQ[{a, b, c, d, e, p, q}, x] && PolyQ[Px, x]
\[\int \left (B \,x^{2}+A \right ) \sqrt {e \,x^{2}+d}\, \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x)
Output:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x)
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="fr icas")
Output:
integral((B*c*x^6 + (B*b + A*c)*x^4 + (B*a + A*b)*x^2 + A*a)*sqrt(c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d), x)
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \left (A + B x^{2}\right ) \sqrt {d + e x^{2}} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((B*x**2+A)*(e*x**2+d)**(1/2)*(c*x**4+b*x**2+a)**(3/2),x)
Output:
Integral((A + B*x**2)*sqrt(d + e*x**2)*(a + b*x**2 + c*x**4)**(3/2), x)
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="ma xima")
Output:
integrate((c*x^4 + b*x^2 + a)^(3/2)*(B*x^2 + A)*sqrt(e*x^2 + d), x)
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x, algorithm="gi ac")
Output:
integrate((c*x^4 + b*x^2 + a)^(3/2)*(B*x^2 + A)*sqrt(e*x^2 + d), x)
Timed out. \[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \left (B\,x^2+A\right )\,\sqrt {e\,x^2+d}\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \] Input:
int((A + B*x^2)*(d + e*x^2)^(1/2)*(a + b*x^2 + c*x^4)^(3/2),x)
Output:
int((A + B*x^2)*(d + e*x^2)^(1/2)*(a + b*x^2 + c*x^4)^(3/2), x)
\[ \int \left (A+B x^2\right ) \sqrt {d+e x^2} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \left (B \,x^{2}+A \right ) \sqrt {e \,x^{2}+d}\, \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}d x \] Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x)
Output:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(3/2),x)