Integrand size = 20, antiderivative size = 105 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\frac {a^2 c^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^3 (1+n) x}-\frac {2 a c^2 \sqrt {c x^2} (a+b x)^{2+n}}{b^3 (2+n) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{3+n}}{b^3 (3+n) x} \]
a^2*c^2*(b*x+a)^(1+n)*(c*x^2)^(1/2)/b^3/(1+n)/x-2*a*c^2*(b*x+a)^(2+n)*(c*x ^2)^(1/2)/b^3/(2+n)/x+c^2*(b*x+a)^(3+n)*(c*x^2)^(1/2)/b^3/(3+n)/x
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\frac {c^3 x (a+b x)^{1+n} \left (2 a^2-2 a b (1+n) x+b^2 \left (2+3 n+n^2\right ) x^2\right )}{b^3 (1+n) (2+n) (3+n) \sqrt {c x^2}} \]
(c^3*x*(a + b*x)^(1 + n)*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^ 2))/(b^3*(1 + n)*(2 + n)*(3 + n)*Sqrt[c*x^2])
Time = 0.19 (sec) , antiderivative size = 76, 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.
\(\displaystyle \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int x^2 (a+b x)^ndx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int \left (\frac {a^2 (a+b x)^n}{b^2}-\frac {2 a (a+b x)^{n+1}}{b^2}+\frac {(a+b x)^{n+2}}{b^2}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \left (\frac {a^2 (a+b x)^{n+1}}{b^3 (n+1)}-\frac {2 a (a+b x)^{n+2}}{b^3 (n+2)}+\frac {(a+b x)^{n+3}}{b^3 (n+3)}\right )}{x}\) |
(c^2*Sqrt[c*x^2]*((a^2*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*(a + b*x)^( 2 + n))/(b^3*(2 + n)) + (a + b*x)^(3 + n)/(b^3*(3 + n))))/x
3.10.41.3.1 Defintions of rubi rules used
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 = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79
method | result | size |
gosper | \(\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{1+n} \left (b^{2} n^{2} x^{2}+3 b^{2} n \,x^{2}-2 a b n x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} x^{5} \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(83\) |
risch | \(\frac {c^{2} \sqrt {c \,x^{2}}\, \left (b^{3} n^{2} x^{3}+a \,b^{2} n^{2} x^{2}+3 b^{3} n \,x^{3}+a \,b^{2} n \,x^{2}+2 b^{3} x^{3}-2 a^{2} b n x +2 a^{3}\right ) \left (b x +a \right )^{n}}{x \left (2+n \right ) \left (3+n \right ) \left (1+n \right ) b^{3}}\) | \(101\) |
1/b^3/x^5*(c*x^2)^(5/2)*(b*x+a)^(1+n)/(n^3+6*n^2+11*n+6)*(b^2*n^2*x^2+3*b^ 2*n*x^2-2*a*b*n*x+2*b^2*x^2-2*a*b*x+2*a^2)
Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=-\frac {{\left (2 \, a^{2} b c^{2} n x - 2 \, a^{3} c^{2} - {\left (b^{3} c^{2} n^{2} + 3 \, b^{3} c^{2} n + 2 \, b^{3} c^{2}\right )} x^{3} - {\left (a b^{2} c^{2} n^{2} + a b^{2} c^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}\right )} x} \]
-(2*a^2*b*c^2*n*x - 2*a^3*c^2 - (b^3*c^2*n^2 + 3*b^3*c^2*n + 2*b^3*c^2)*x^ 3 - (a*b^2*c^2*n^2 + a*b^2*c^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3)*x)
\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{n}}{x^{3}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} c^{\frac {5}{2}} x^{3} + {\left (n^{2} + n\right )} a b^{2} c^{\frac {5}{2}} x^{2} - 2 \, a^{2} b c^{\frac {5}{2}} n x + 2 \, a^{3} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} \]
((n^2 + 3*n + 2)*b^3*c^(5/2)*x^3 + (n^2 + n)*a*b^2*c^(5/2)*x^2 - 2*a^2*b*c ^(5/2)*n*x + 2*a^3*c^(5/2))*(b*x + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b^3)
\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\int { \frac {\left (c x^{2}\right )^{\frac {5}{2}} {\left (b x + a\right )}^{n}}{x^{3}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.47 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {2\,a^3\,c^2\,\sqrt {c\,x^2}}{b^3\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {c^2\,x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{n^3+6\,n^2+11\,n+6}-\frac {2\,a^2\,c^2\,n\,x\,\sqrt {c\,x^2}}{b^2\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {a\,c^2\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b\,\left (n^3+6\,n^2+11\,n+6\right )}\right )}{x} \]
((a + b*x)^n*((2*a^3*c^2*(c*x^2)^(1/2))/(b^3*(11*n + 6*n^2 + n^3 + 6)) + ( c^2*x^3*(c*x^2)^(1/2)*(3*n + n^2 + 2))/(11*n + 6*n^2 + n^3 + 6) - (2*a^2*c ^2*n*x*(c*x^2)^(1/2))/(b^2*(11*n + 6*n^2 + n^3 + 6)) + (a*c^2*n*x^2*(c*x^2 )^(1/2)*(n + 1))/(b*(11*n + 6*n^2 + n^3 + 6))))/x
\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^n}{x^3} \, dx=\frac {\sqrt {c}\, c^{2} \left (-2 \left (b x +a \right )^{n} {| x |} a^{2} n +\left (b x +a \right )^{n} {| x |} a b \,n^{2} x +\left (b x +a \right )^{n} {| x |} a b n x +\left (b x +a \right )^{n} {| x |} b^{2} n^{2} x^{2}+3 \left (b x +a \right )^{n} {| x |} b^{2} n \,x^{2}+2 \left (b x +a \right )^{n} {| x |} b^{2} x^{2}+2 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{3} x^{2}+a \,n^{3} x +6 b \,n^{2} x^{2}+6 a \,n^{2} x +11 b n \,x^{2}+11 a n x +6 b \,x^{2}+6 a x}d x \right ) a^{3} n^{4}+12 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{3} x^{2}+a \,n^{3} x +6 b \,n^{2} x^{2}+6 a \,n^{2} x +11 b n \,x^{2}+11 a n x +6 b \,x^{2}+6 a x}d x \right ) a^{3} n^{3}+22 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{3} x^{2}+a \,n^{3} x +6 b \,n^{2} x^{2}+6 a \,n^{2} x +11 b n \,x^{2}+11 a n x +6 b \,x^{2}+6 a x}d x \right ) a^{3} n^{2}+12 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{3} x^{2}+a \,n^{3} x +6 b \,n^{2} x^{2}+6 a \,n^{2} x +11 b n \,x^{2}+11 a n x +6 b \,x^{2}+6 a x}d x \right ) a^{3} n \right )}{b^{2} \left (n^{3}+6 n^{2}+11 n +6\right )} \]
(sqrt(c)*c**2*( - 2*(a + b*x)**n*abs(x)*a**2*n + (a + b*x)**n*abs(x)*a*b*n **2*x + (a + b*x)**n*abs(x)*a*b*n*x + (a + b*x)**n*abs(x)*b**2*n**2*x**2 + 3*(a + b*x)**n*abs(x)*b**2*n*x**2 + 2*(a + b*x)**n*abs(x)*b**2*x**2 + 2*i nt(((a + b*x)**n*abs(x))/(a*n**3*x + 6*a*n**2*x + 11*a*n*x + 6*a*x + b*n** 3*x**2 + 6*b*n**2*x**2 + 11*b*n*x**2 + 6*b*x**2),x)*a**3*n**4 + 12*int(((a + b*x)**n*abs(x))/(a*n**3*x + 6*a*n**2*x + 11*a*n*x + 6*a*x + b*n**3*x**2 + 6*b*n**2*x**2 + 11*b*n*x**2 + 6*b*x**2),x)*a**3*n**3 + 22*int(((a + b*x )**n*abs(x))/(a*n**3*x + 6*a*n**2*x + 11*a*n*x + 6*a*x + b*n**3*x**2 + 6*b *n**2*x**2 + 11*b*n*x**2 + 6*b*x**2),x)*a**3*n**2 + 12*int(((a + b*x)**n*a bs(x))/(a*n**3*x + 6*a*n**2*x + 11*a*n*x + 6*a*x + b*n**3*x**2 + 6*b*n**2* x**2 + 11*b*n*x**2 + 6*b*x**2),x)*a**3*n))/(b**2*(n**3 + 6*n**2 + 11*n + 6 ))