Integrand size = 20, antiderivative size = 99 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\frac {a^2 c \sqrt {c x^2} (a+b x)^{1+p}}{b^3 (1+p) x}-\frac {2 a c \sqrt {c x^2} (a+b x)^{2+p}}{b^3 (2+p) x}+\frac {c \sqrt {c x^2} (a+b x)^{3+p}}{b^3 (3+p) x} \] Output:
a^2*c*(c*x^2)^(1/2)*(b*x+a)^(p+1)/b^3/(p+1)/x-2*a*c*(c*x^2)^(1/2)*(b*x+a)^ (2+p)/b^3/(2+p)/x+c*(c*x^2)^(1/2)*(b*x+a)^(3+p)/b^3/(3+p)/x
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\frac {c^2 x (a+b x)^{1+p} \left (2 a^2-2 a b (1+p) x+b^2 \left (2+3 p+p^2\right ) x^2\right )}{b^3 (1+p) (2+p) (3+p) \sqrt {c x^2}} \] Input:
Integrate[((c*x^2)^(3/2)*(a + b*x)^p)/x,x]
Output:
(c^2*x*(a + b*x)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x + b^2*(2 + 3*p + p^2)*x^ 2))/(b^3*(1 + p)*(2 + p)*(3 + p)*Sqrt[c*x^2])
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.75, 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.
\(\displaystyle \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c \sqrt {c x^2} \int x^2 (a+b x)^pdx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c \sqrt {c x^2} \int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{p+1}}{b^2}+\frac {(a+b x)^{p+2}}{b^2}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c x^2} \left (\frac {a^2 (a+b x)^{p+1}}{b^3 (p+1)}-\frac {2 a (a+b x)^{p+2}}{b^3 (p+2)}+\frac {(a+b x)^{p+3}}{b^3 (p+3)}\right )}{x}\) |
Input:
Int[((c*x^2)^(3/2)*(a + b*x)^p)/x,x]
Output:
(c*Sqrt[c*x^2]*((a^2*(a + b*x)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x)^(2 + p))/(b^3*(2 + p)) + (a + b*x)^(3 + p)/(b^3*(3 + p))))/x
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 = 83, normalized size of antiderivative = 0.84
method | result | size |
gosper | \(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{p +1} \left (b^{2} p^{2} x^{2}+3 b^{2} p \,x^{2}-2 a b p x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} x^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(83\) |
orering | \(\frac {\left (b x +a \right ) \left (b^{2} p^{2} x^{2}+3 b^{2} p \,x^{2}-2 a b p x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{p}}{x^{3} b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(86\) |
risch | \(\frac {c \sqrt {c \,x^{2}}\, \left (b^{3} p^{2} x^{3}+a \,b^{2} p^{2} x^{2}+3 b^{3} p \,x^{3}+a \,b^{2} p \,x^{2}+2 b^{3} x^{3}-2 a^{2} b p x +2 a^{3}\right ) \left (b x +a \right )^{p}}{x \left (2+p \right ) \left (3+p \right ) \left (p +1\right ) b^{3}}\) | \(99\) |
Input:
int((c*x^2)^(3/2)*(b*x+a)^p/x,x,method=_RETURNVERBOSE)
Output:
1/b^3/x^3*(c*x^2)^(3/2)*(b*x+a)^(p+1)/(p^3+6*p^2+11*p+6)*(b^2*p^2*x^2+3*b^ 2*p*x^2-2*a*b*p*x+2*b^2*x^2-2*a*b*x+2*a^2)
Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=-\frac {{\left (2 \, a^{2} b c p x - 2 \, a^{3} c - {\left (b^{3} c p^{2} + 3 \, b^{3} c p + 2 \, b^{3} c\right )} x^{3} - {\left (a b^{2} c p^{2} + a b^{2} c p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )} x} \] Input:
integrate((c*x^2)^(3/2)*(b*x+a)^p/x,x, algorithm="fricas")
Output:
-(2*a^2*b*c*p*x - 2*a^3*c - (b^3*c*p^2 + 3*b^3*c*p + 2*b^3*c)*x^3 - (a*b^2 *c*p^2 + a*b^2*c*p)*x^2)*sqrt(c*x^2)*(b*x + a)^p/((b^3*p^3 + 6*b^3*p^2 + 1 1*b^3*p + 6*b^3)*x)
\[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{p}}{x}\, dx \] Input:
integrate((c*x**2)**(3/2)*(b*x+a)**p/x,x)
Output:
Integral((c*x**2)**(3/2)*(a + b*x)**p/x, x)
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} c^{\frac {3}{2}} x^{3} + {\left (p^{2} + p\right )} a b^{2} c^{\frac {3}{2}} x^{2} - 2 \, a^{2} b c^{\frac {3}{2}} p x + 2 \, a^{3} c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \] Input:
integrate((c*x^2)^(3/2)*(b*x+a)^p/x,x, algorithm="maxima")
Output:
((p^2 + 3*p + 2)*b^3*c^(3/2)*x^3 + (p^2 + p)*a*b^2*c^(3/2)*x^2 - 2*a^2*b*c ^(3/2)*p*x + 2*a^3*c^(3/2))*(b*x + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3)
\[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\int { \frac {\left (c x^{2}\right )^{\frac {3}{2}} {\left (b x + a\right )}^{p}}{x} \,d x } \] Input:
integrate((c*x^2)^(3/2)*(b*x+a)^p/x,x, algorithm="giac")
Output:
integrate((c*x^2)^(3/2)*(b*x + a)^p/x, x)
Time = 22.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.47 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {c\,x^3\,\sqrt {c\,x^2}\,\left (p^2+3\,p+2\right )}{p^3+6\,p^2+11\,p+6}+\frac {2\,a^3\,c\,\sqrt {c\,x^2}}{b^3\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {2\,a^2\,c\,p\,x\,\sqrt {c\,x^2}}{b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,c\,p\,x^2\,\sqrt {c\,x^2}\,\left (p+1\right )}{b\,\left (p^3+6\,p^2+11\,p+6\right )}\right )}{x} \] Input:
int(((c*x^2)^(3/2)*(a + b*x)^p)/x,x)
Output:
((a + b*x)^p*((c*x^3*(c*x^2)^(1/2)*(3*p + p^2 + 2))/(11*p + 6*p^2 + p^3 + 6) + (2*a^3*c*(c*x^2)^(1/2))/(b^3*(11*p + 6*p^2 + p^3 + 6)) - (2*a^2*c*p*x *(c*x^2)^(1/2))/(b^2*(11*p + 6*p^2 + p^3 + 6)) + (a*c*p*x^2*(c*x^2)^(1/2)* (p + 1))/(b*(11*p + 6*p^2 + p^3 + 6))))/x
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^p}{x} \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} c \left (b^{3} p^{2} x^{3}+a \,b^{2} p^{2} x^{2}+3 b^{3} p \,x^{3}+a \,b^{2} p \,x^{2}+2 b^{3} x^{3}-2 a^{2} b p x +2 a^{3}\right )}{b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int((c*x^2)^(3/2)*(b*x+a)^p/x,x)
Output:
(sqrt(c)*(a + b*x)**p*c*(2*a**3 - 2*a**2*b*p*x + a*b**2*p**2*x**2 + a*b**2 *p*x**2 + b**3*p**2*x**3 + 3*b**3*p*x**3 + 2*b**3*x**3))/(b**3*(p**3 + 6*p **2 + 11*p + 6))