Integrand size = 25, antiderivative size = 277 \[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=-\frac {d \left (1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (-\frac {1}{3},-p,-p,\frac {2}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{x}+\frac {1}{2} e x^2 \left (1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (\frac {2}{3},-p,-p,\frac {5}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right ) \] Output:
-d*(c*x^6+b*x^3+a)^p*AppellF1(-1/3,-p,-p,2/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2 )),-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))/x/((1+2*c*x^3/(b-(-4*a*c+b^2)^(1/2)))^ p)/((1+2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))^p)+1/2*e*x^2*(c*x^6+b*x^3+a)^p*Appe llF1(2/3,-p,-p,5/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^3/(b+(-4*a*c+b^2 )^(1/2)))/((1+2*c*x^3/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^3/(b+(-4*a*c+b^ 2)^(1/2)))^p)
Time = 0.66 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.84 \[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\frac {\left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^3+c x^6\right )^p \left (-2 d \operatorname {AppellF1}\left (-\frac {1}{3},-p,-p,\frac {2}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+e x^3 \operatorname {AppellF1}\left (\frac {2}{3},-p,-p,\frac {5}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )\right )}{2 x} \] Input:
Integrate[((d + e*x^3)*(a + b*x^3 + c*x^6)^p)/x^2,x]
Output:
((a + b*x^3 + c*x^6)^p*(-2*d*AppellF1[-1/3, -p, -p, 2/3, (-2*c*x^3)/(b + S qrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + e*x^3*AppellF1[2/ 3, -p, -p, 5/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b ^2 - 4*a*c])]))/(2*x*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4* a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 0.44 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1864, 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.
\(\displaystyle \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx\) |
\(\Big \downarrow \) 1864 |
\(\displaystyle \int \left (\frac {d \left (a+b x^3+c x^6\right )^p}{x^2}+e x \left (a+b x^3+c x^6\right )^p\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} e x^2 \left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (\frac {2}{3},-p,-p,\frac {5}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )-\frac {d \left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (-\frac {1}{3},-p,-p,\frac {2}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{x}\) |
Input:
Int[((d + e*x^3)*(a + b*x^3 + c*x^6)^p)/x^2,x]
Output:
-((d*(a + b*x^3 + c*x^6)^p*AppellF1[-1/3, -p, -p, 2/3, (-2*c*x^3)/(b - Sqr t[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(x*(1 + (2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]))^p)) + (e* x^2*(a + b*x^3 + c*x^6)^p*AppellF1[2/3, -p, -p, 5/3, (-2*c*x^3)/(b - Sqrt[ b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(2*(1 + (2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]))^p)
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*( (d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n + c*x^(2*n))^p, (f*x)^m*(d + e*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m , q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IGtQ[q, 0]
\[\int \frac {\left (e \,x^{3}+d \right ) \left (c \,x^{6}+b \,x^{3}+a \right )^{p}}{x^{2}}d x\]
Input:
int((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x)
Output:
int((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x)
\[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\int { \frac {{\left (e x^{3} + d\right )} {\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x, algorithm="fricas")
Output:
integral((e*x^3 + d)*(c*x^6 + b*x^3 + a)^p/x^2, x)
Timed out. \[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\text {Timed out} \] Input:
integrate((e*x**3+d)*(c*x**6+b*x**3+a)**p/x**2,x)
Output:
Timed out
\[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\int { \frac {{\left (e x^{3} + d\right )} {\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x, algorithm="maxima")
Output:
integrate((e*x^3 + d)*(c*x^6 + b*x^3 + a)^p/x^2, x)
\[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\int { \frac {{\left (e x^{3} + d\right )} {\left (c x^{6} + b x^{3} + a\right )}^{p}}{x^{2}} \,d x } \] Input:
integrate((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x, algorithm="giac")
Output:
integrate((e*x^3 + d)*(c*x^6 + b*x^3 + a)^p/x^2, x)
Timed out. \[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx=\int \frac {\left (e\,x^3+d\right )\,{\left (c\,x^6+b\,x^3+a\right )}^p}{x^2} \,d x \] Input:
int(((d + e*x^3)*(a + b*x^3 + c*x^6)^p)/x^2,x)
Output:
int(((d + e*x^3)*(a + b*x^3 + c*x^6)^p)/x^2, x)
\[ \int \frac {\left (d+e x^3\right ) \left (a+b x^3+c x^6\right )^p}{x^2} \, dx =\text {Too large to display} \] Input:
int((e*x^3+d)*(c*x^6+b*x^3+a)^p/x^2,x)
Output:
(6*(a + b*x**3 + c*x**6)**p*a*e*p + 6*(a + b*x**3 + c*x**6)**p*b*d*p + 2*( a + b*x**3 + c*x**6)**p*b*d + 3*(a + b*x**3 + c*x**6)**p*b*e*p*x**3 - (a + b*x**3 + c*x**6)**p*b*e*x**3 + 54*int((a + b*x**3 + c*x**6)**p/(9*a*p**2* x**2 - a*x**2 + 9*b*p**2*x**5 - b*x**5 + 9*c*p**2*x**8 - c*x**8),x)*a**2*e *p**3*x - 6*int((a + b*x**3 + c*x**6)**p/(9*a*p**2*x**2 - a*x**2 + 9*b*p** 2*x**5 - b*x**5 + 9*c*p**2*x**8 - c*x**8),x)*a**2*e*p*x + 162*int((a + b*x **3 + c*x**6)**p/(9*a*p**2*x**2 - a*x**2 + 9*b*p**2*x**5 - b*x**5 + 9*c*p* *2*x**8 - c*x**8),x)*a*b*d*p**4*x + 54*int((a + b*x**3 + c*x**6)**p/(9*a*p **2*x**2 - a*x**2 + 9*b*p**2*x**5 - b*x**5 + 9*c*p**2*x**8 - c*x**8),x)*a* b*d*p**3*x - 18*int((a + b*x**3 + c*x**6)**p/(9*a*p**2*x**2 - a*x**2 + 9*b *p**2*x**5 - b*x**5 + 9*c*p**2*x**8 - c*x**8),x)*a*b*d*p**2*x - 6*int((a + b*x**3 + c*x**6)**p/(9*a*p**2*x**2 - a*x**2 + 9*b*p**2*x**5 - b*x**5 + 9* c*p**2*x**8 - c*x**8),x)*a*b*d*p*x - 324*int(((a + b*x**3 + c*x**6)**p*x** 4)/(9*a*p**2 - a + 9*b*p**2*x**3 - b*x**3 + 9*c*p**2*x**6 - c*x**6),x)*a*c *e*p**4*x + 54*int(((a + b*x**3 + c*x**6)**p*x**4)/(9*a*p**2 - a + 9*b*p** 2*x**3 - b*x**3 + 9*c*p**2*x**6 - c*x**6),x)*a*c*e*p**3*x + 36*int(((a + b *x**3 + c*x**6)**p*x**4)/(9*a*p**2 - a + 9*b*p**2*x**3 - b*x**3 + 9*c*p**2 *x**6 - c*x**6),x)*a*c*e*p**2*x - 6*int(((a + b*x**3 + c*x**6)**p*x**4)/(9 *a*p**2 - a + 9*b*p**2*x**3 - b*x**3 + 9*c*p**2*x**6 - c*x**6),x)*a*c*e*p* x + 81*int(((a + b*x**3 + c*x**6)**p*x**4)/(9*a*p**2 - a + 9*b*p**2*x**...