Integrand size = 20, antiderivative size = 135 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a^3 x (a+b x)^{1+p}}{b^4 c^2 (1+p) \sqrt {c x^2}}+\frac {3 a^2 x (a+b x)^{2+p}}{b^4 c^2 (2+p) \sqrt {c x^2}}-\frac {3 a x (a+b x)^{3+p}}{b^4 c^2 (3+p) \sqrt {c x^2}}+\frac {x (a+b x)^{4+p}}{b^4 c^2 (4+p) \sqrt {c x^2}} \] Output:
-a^3*x*(b*x+a)^(p+1)/b^4/c^2/(p+1)/(c*x^2)^(1/2)+3*a^2*x*(b*x+a)^(2+p)/b^4 /c^2/(2+p)/(c*x^2)^(1/2)-3*a*x*(b*x+a)^(3+p)/b^4/c^2/(3+p)/(c*x^2)^(1/2)+x *(b*x+a)^(4+p)/b^4/c^2/(4+p)/(c*x^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\frac {x (a+b x)^{1+p} \left (-6 a^3+6 a^2 b (1+p) x-3 a b^2 \left (2+3 p+p^2\right ) x^2+b^3 \left (6+11 p+6 p^2+p^3\right ) x^3\right )}{b^4 c^2 (1+p) (2+p) (3+p) (4+p) \sqrt {c x^2}} \] Input:
Integrate[(x^8*(a + b*x)^p)/(c*x^2)^(5/2),x]
Output:
(x*(a + b*x)^(1 + p)*(-6*a^3 + 6*a^2*b*(1 + p)*x - 3*a*b^2*(2 + 3*p + p^2) *x^2 + b^3*(6 + 11*p + 6*p^2 + p^3)*x^3))/(b^4*c^2*(1 + p)*(2 + p)*(3 + p) *(4 + p)*Sqrt[c*x^2])
Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {30, 53, 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.
\(\displaystyle \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {x \int x^3 (a+b x)^pdx}{c^2 \sqrt {c x^2}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {x \int \left (-\frac {a^3 (a+b x)^p}{b^3}+\frac {3 a^2 (a+b x)^{p+1}}{b^3}-\frac {3 a (a+b x)^{p+2}}{b^3}+\frac {(a+b x)^{p+3}}{b^3}\right )dx}{c^2 \sqrt {c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (-\frac {a^3 (a+b x)^{p+1}}{b^4 (p+1)}+\frac {3 a^2 (a+b x)^{p+2}}{b^4 (p+2)}-\frac {3 a (a+b x)^{p+3}}{b^4 (p+3)}+\frac {(a+b x)^{p+4}}{b^4 (p+4)}\right )}{c^2 \sqrt {c x^2}}\) |
Input:
Int[(x^8*(a + b*x)^p)/(c*x^2)^(5/2),x]
Output:
(x*(-((a^3*(a + b*x)^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*x)^(2 + p))/( b^4*(2 + p)) - (3*a*(a + b*x)^(3 + p))/(b^4*(3 + p)) + (a + b*x)^(4 + p)/( b^4*(4 + p))))/(c^2*Sqrt[c*x^2])
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01
method | result | size |
gosper | \(-\frac {x^{5} \left (b x +a \right )^{p +1} \left (-b^{3} p^{3} x^{3}-6 b^{3} p^{2} x^{3}+3 a \,b^{2} p^{2} x^{2}-11 b^{3} p \,x^{3}+9 a \,b^{2} p \,x^{2}-6 b^{3} x^{3}-6 a^{2} b p x +6 a \,b^{2} x^{2}-6 b \,a^{2} x +6 a^{3}\right )}{b^{4} \left (c \,x^{2}\right )^{\frac {5}{2}} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(136\) |
orering | \(-\frac {\left (b x +a \right )^{p} x^{5} \left (-b^{3} p^{3} x^{3}-6 b^{3} p^{2} x^{3}+3 a \,b^{2} p^{2} x^{2}-11 b^{3} p \,x^{3}+9 a \,b^{2} p \,x^{2}-6 b^{3} x^{3}-6 a^{2} b p x +6 a \,b^{2} x^{2}-6 b \,a^{2} x +6 a^{3}\right ) \left (b x +a \right )}{\left (c \,x^{2}\right )^{\frac {5}{2}} b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(139\) |
risch | \(-\frac {x \left (-b^{4} p^{3} x^{4}-a \,b^{3} p^{3} x^{3}-6 b^{4} p^{2} x^{4}-3 a \,b^{3} p^{2} x^{3}-11 b^{4} p \,x^{4}+3 a^{2} b^{2} p^{2} x^{2}-2 x^{3} a p \,b^{3}-6 b^{4} x^{4}+3 a^{2} p \,x^{2} b^{2}-6 a^{3} b p x +6 a^{4}\right ) \left (b x +a \right )^{p}}{c^{2} \sqrt {c \,x^{2}}\, \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (p +1\right ) b^{4}}\) | \(157\) |
Input:
int(x^8*(b*x+a)^p/(c*x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/b^4*x^5/(c*x^2)^(5/2)*(b*x+a)^(p+1)/(p^4+10*p^3+35*p^2+50*p+24)*(-b^3*p ^3*x^3-6*b^3*p^2*x^3+3*a*b^2*p^2*x^2-11*b^3*p*x^3+9*a*b^2*p*x^2-6*b^3*x^3- 6*a^2*b*p*x+6*a*b^2*x^2-6*a^2*b*x+6*a^3)
Time = 0.13 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\frac {{\left (6 \, a^{3} b p x + {\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x^{3} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{4} c^{3} p^{4} + 10 \, b^{4} c^{3} p^{3} + 35 \, b^{4} c^{3} p^{2} + 50 \, b^{4} c^{3} p + 24 \, b^{4} c^{3}\right )} x} \] Input:
integrate(x^8*(b*x+a)^p/(c*x^2)^(5/2),x, algorithm="fricas")
Output:
(6*a^3*b*p*x + (b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^4 - 6*a^4 + (a*b ^3*p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*x^3 - 3*(a^2*b^2*p^2 + a^2*b^2*p)*x^2)*s qrt(c*x^2)*(b*x + a)^p/((b^4*c^3*p^4 + 10*b^4*c^3*p^3 + 35*b^4*c^3*p^2 + 5 0*b^4*c^3*p + 24*b^4*c^3)*x)
\[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(x**8*(b*x+a)**p/(c*x**2)**(5/2),x)
Output:
Piecewise((a**p*x**9/(4*(c*x**2)**(5/2)), Eq(b, 0)), (Integral(x**8/((c*x* *2)**(5/2)*(a + b*x)**4), x), Eq(p, -4)), (Integral(x**8/((c*x**2)**(5/2)* (a + b*x)**3), x), Eq(p, -3)), (Integral(x**8/((c*x**2)**(5/2)*(a + b*x)** 2), x), Eq(p, -2)), (Integral(x**8/((c*x**2)**(5/2)*(a + b*x)), x), Eq(p, -1)), (-6*a**4*x**5*(a + b*x)**p/(b**4*p**4*(c*x**2)**(5/2) + 10*b**4*p**3 *(c*x**2)**(5/2) + 35*b**4*p**2*(c*x**2)**(5/2) + 50*b**4*p*(c*x**2)**(5/2 ) + 24*b**4*(c*x**2)**(5/2)) + 6*a**3*b*p*x**6*(a + b*x)**p/(b**4*p**4*(c* x**2)**(5/2) + 10*b**4*p**3*(c*x**2)**(5/2) + 35*b**4*p**2*(c*x**2)**(5/2) + 50*b**4*p*(c*x**2)**(5/2) + 24*b**4*(c*x**2)**(5/2)) - 3*a**2*b**2*p**2 *x**7*(a + b*x)**p/(b**4*p**4*(c*x**2)**(5/2) + 10*b**4*p**3*(c*x**2)**(5/ 2) + 35*b**4*p**2*(c*x**2)**(5/2) + 50*b**4*p*(c*x**2)**(5/2) + 24*b**4*(c *x**2)**(5/2)) - 3*a**2*b**2*p*x**7*(a + b*x)**p/(b**4*p**4*(c*x**2)**(5/2 ) + 10*b**4*p**3*(c*x**2)**(5/2) + 35*b**4*p**2*(c*x**2)**(5/2) + 50*b**4* p*(c*x**2)**(5/2) + 24*b**4*(c*x**2)**(5/2)) + a*b**3*p**3*x**8*(a + b*x)* *p/(b**4*p**4*(c*x**2)**(5/2) + 10*b**4*p**3*(c*x**2)**(5/2) + 35*b**4*p** 2*(c*x**2)**(5/2) + 50*b**4*p*(c*x**2)**(5/2) + 24*b**4*(c*x**2)**(5/2)) + 3*a*b**3*p**2*x**8*(a + b*x)**p/(b**4*p**4*(c*x**2)**(5/2) + 10*b**4*p**3 *(c*x**2)**(5/2) + 35*b**4*p**2*(c*x**2)**(5/2) + 50*b**4*p*(c*x**2)**(5/2 ) + 24*b**4*(c*x**2)**(5/2)) + 2*a*b**3*p*x**8*(a + b*x)**p/(b**4*p**4*(c* x**2)**(5/2) + 10*b**4*p**3*(c*x**2)**(5/2) + 35*b**4*p**2*(c*x**2)**(5...
Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.77 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{4} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{3} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b p x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4} c^{\frac {5}{2}}} \] Input:
integrate(x^8*(b*x+a)^p/(c*x^2)^(5/2),x, algorithm="maxima")
Output:
((p^3 + 6*p^2 + 11*p + 6)*b^4*x^4 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^3 - 3*(p^2 + p)*a^2*b^2*x^2 + 6*a^3*b*p*x - 6*a^4)*(b*x + a)^p/((p^4 + 10*p^3 + 35*p ^2 + 50*p + 24)*b^4*c^(5/2))
Exception generated. \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^8*(b*x+a)^p/(c*x^2)^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,3,1,0,0]%%%} / %%%{1,[0,0,0,1,1]%%%} Error: Bad Argum ent Value
Time = 22.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.49 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {x^5\,\left (p^3+6\,p^2+11\,p+6\right )}{c^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {6\,a^4\,x}{b^4\,c^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {6\,a^3\,p\,x^2}{b^3\,c^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^4\,\left (p^2+3\,p+2\right )}{b\,c^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^2\,p\,x^3\,\left (p+1\right )}{b^2\,c^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right )}{\sqrt {c\,x^2}} \] Input:
int((x^8*(a + b*x)^p)/(c*x^2)^(5/2),x)
Output:
((a + b*x)^p*((x^5*(11*p + 6*p^2 + p^3 + 6))/(c^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (6*a^4*x)/(b^4*c^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + ( 6*a^3*p*x^2)/(b^3*c^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^4*(3*p + p^2 + 2))/(b*c^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^2*p*x^3*(p + 1))/(b^2*c^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24))))/(c*x^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09 \[ \int \frac {x^8 (a+b x)^p}{\left (c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} \left (b^{4} p^{3} x^{4}+a \,b^{3} p^{3} x^{3}+6 b^{4} p^{2} x^{4}+3 a \,b^{3} p^{2} x^{3}+11 b^{4} p \,x^{4}-3 a^{2} b^{2} p^{2} x^{2}+2 a \,b^{3} p \,x^{3}+6 b^{4} x^{4}-3 a^{2} b^{2} p \,x^{2}+6 a^{3} b p x -6 a^{4}\right )}{b^{4} c^{3} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:
int(x^8*(b*x+a)^p/(c*x^2)^(5/2),x)
Output:
(sqrt(c)*(a + b*x)**p*( - 6*a**4 + 6*a**3*b*p*x - 3*a**2*b**2*p**2*x**2 - 3*a**2*b**2*p*x**2 + a*b**3*p**3*x**3 + 3*a*b**3*p**2*x**3 + 2*a*b**3*p*x* *3 + b**4*p**3*x**4 + 6*b**4*p**2*x**4 + 11*b**4*p*x**4 + 6*b**4*x**4))/(b **4*c**3*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))