\(\int \frac {(A+B x^2) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx\) [166]

Optimal result

Integrand size = 33, antiderivative size = 547 \[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\frac {\left (A c e+B (c d-b e)+\frac {b^2 B e-b c (B d+A e)+2 c (A c d-a B e)}{\sqrt {b^2-4 a c}}\right ) x \left (1+\frac {e x^2}{d}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right ) \left (d+e x^2\right )^{2/3}}+\frac {\left (A c e+B (c d-b e)-\frac {b^2 B e-b c (B d+A e)+2 c (A c d-a B e)}{\sqrt {b^2-4 a c}}\right ) x \left (1+\frac {e x^2}{d}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{c \left (b+\sqrt {b^2-4 a c}\right ) \left (d+e x^2\right )^{2/3}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} B \left (\sqrt [3]{d}-\sqrt [3]{d+e x^2}\right ) \sqrt {\frac {d^{2/3}+\sqrt [3]{d} \sqrt [3]{d+e x^2}+\left (d+e x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{d}-\sqrt [3]{d+e x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{d}-\sqrt [3]{d+e x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{d}-\sqrt [3]{d+e x^2}}\right ),-7+4 \sqrt {3}\right )}{c x \sqrt {-\frac {\sqrt [3]{d} \left (\sqrt [3]{d}-\sqrt [3]{d+e x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{d}-\sqrt [3]{d+e x^2}\right )^2}}} \] Output:

(A*c*e+B*(-b*e+c*d)+(B*b^2*e-b*c*(A*e+B*d)+2*c*(A*c*d-B*a*e))/(-4*a*c+b^2) 
^(1/2))*x*(1+e*x^2/d)^(2/3)*AppellF1(1/2,1,2/3,3/2,-2*c*x^2/(b-(-4*a*c+b^2 
)^(1/2)),-e*x^2/d)/c/(b-(-4*a*c+b^2)^(1/2))/(e*x^2+d)^(2/3)+(A*c*e+B*(-b*e 
+c*d)-(B*b^2*e-b*c*(A*e+B*d)+2*c*(A*c*d-B*a*e))/(-4*a*c+b^2)^(1/2))*x*(1+e 
*x^2/d)^(2/3)*AppellF1(1/2,1,2/3,3/2,-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)),-e*x^ 
2/d)/c/(b+(-4*a*c+b^2)^(1/2))/(e*x^2+d)^(2/3)-3^(3/4)*(1/2*6^(1/2)-1/2*2^( 
1/2))*B*(d^(1/3)-(e*x^2+d)^(1/3))*((d^(2/3)+d^(1/3)*(e*x^2+d)^(1/3)+(e*x^2 
+d)^(2/3))/((1-3^(1/2))*d^(1/3)-(e*x^2+d)^(1/3))^2)^(1/2)*EllipticF(((1+3^ 
(1/2))*d^(1/3)-(e*x^2+d)^(1/3))/((1-3^(1/2))*d^(1/3)-(e*x^2+d)^(1/3)),2*I- 
I*3^(1/2))/c/x/(-d^(1/3)*(d^(1/3)-(e*x^2+d)^(1/3))/((1-3^(1/2))*d^(1/3)-(e 
*x^2+d)^(1/3))^2)^(1/2)
 

Mathematica [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx \] Input:

Integrate[((A + B*x^2)*(d + e*x^2)^(1/3))/(a + b*x^2 + c*x^4),x]
 

Output:

Integrate[((A + B*x^2)*(d + e*x^2)^(1/3))/(a + b*x^2 + c*x^4), x]
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.41, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2256, 2009}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 2256

\(\displaystyle \int \left (\frac {\sqrt [3]{d+e x^2} \left (B-\frac {2 A c-b B}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b+2 c x^2}+\frac {\sqrt [3]{d+e x^2} \left (\frac {2 A c-b B}{\sqrt {b^2-4 a c}}+B\right )}{-\sqrt {b^2-4 a c}+b+2 c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \sqrt [3]{d+e x^2} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \operatorname {AppellF1}\left (\frac {1}{2},1,-\frac {1}{3},\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) \sqrt [3]{\frac {e x^2}{d}+1}}+\frac {x \sqrt [3]{d+e x^2} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) \operatorname {AppellF1}\left (\frac {1}{2},1,-\frac {1}{3},\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{\left (\sqrt {b^2-4 a c}+b\right ) \sqrt [3]{\frac {e x^2}{d}+1}}\)

Input:

Int[((A + B*x^2)*(d + e*x^2)^(1/3))/(a + b*x^2 + c*x^4),x]
 

Output:

((B - (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^(1/3)*AppellF1[1/2, 1 
, -1/3, 3/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/((b - Sqrt 
[b^2 - 4*a*c])*(1 + (e*x^2)/d)^(1/3)) + ((B + (b*B - 2*A*c)/Sqrt[b^2 - 4*a 
*c])*x*(d + e*x^2)^(1/3)*AppellF1[1/2, 1, -1/3, 3/2, (-2*c*x^2)/(b + Sqrt[ 
b^2 - 4*a*c]), -((e*x^2)/d)])/((b + Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^(1/ 
3))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2256

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ 
(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4 
)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

Maple [F]

\[\int \frac {\left (B \,x^{2}+A \right ) \left (e \,x^{2}+d \right )^{\frac {1}{3}}}{c \,x^{4}+b \,x^{2}+a}d x\]

Input:

int((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x)
 

Output:

int((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\text {Timed out} \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt [3]{d + e x^{2}}}{a + b x^{2} + c x^{4}}\, dx \] Input:

integrate((B*x**2+A)*(e*x**2+d)**(1/3)/(c*x**4+b*x**2+a),x)
 

Output:

Integral((A + B*x**2)*(d + e*x**2)**(1/3)/(a + b*x**2 + c*x**4), x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {1}{3}}}{c x^{4} + b x^{2} + a} \,d x } \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*(e*x^2 + d)^(1/3)/(c*x^4 + b*x^2 + a), x)
 

Giac [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {1}{3}}}{c x^{4} + b x^{2} + a} \,d x } \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*(e*x^2 + d)^(1/3)/(c*x^4 + b*x^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x^2+d\right )}^{1/3}}{c\,x^4+b\,x^2+a} \,d x \] Input:

int(((A + B*x^2)*(d + e*x^2)^(1/3))/(a + b*x^2 + c*x^4),x)
 

Output:

int(((A + B*x^2)*(d + e*x^2)^(1/3))/(a + b*x^2 + c*x^4), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt [3]{d+e x^2}}{a+b x^2+c x^4} \, dx=\left (\int \frac {\left (e \,x^{2}+d \right )^{\frac {1}{3}}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a +\left (\int \frac {\left (e \,x^{2}+d \right )^{\frac {1}{3}} x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b \] Input:

int((B*x^2+A)*(e*x^2+d)^(1/3)/(c*x^4+b*x^2+a),x)
 

Output:

int((d + e*x**2)**(1/3)/(a + b*x**2 + c*x**4),x)*a + int(((d + e*x**2)**(1 
/3)*x**2)/(a + b*x**2 + c*x**4),x)*b