Integrand size = 20, antiderivative size = 69 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=-\frac {a c^2 \sqrt {c x^2} (a+b x)^{1+p}}{b^2 (1+p) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{2+p}}{b^2 (2+p) x} \] Output:
-a*c^2*(c*x^2)^(1/2)*(b*x+a)^(p+1)/b^2/(p+1)/x+c^2*(c*x^2)^(1/2)*(b*x+a)^( 2+p)/b^2/(2+p)/x
Time = 0.00 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\frac {c^3 x (a+b x)^{1+p} (-a+b (1+p) x)}{b^2 (1+p) (2+p) \sqrt {c x^2}} \] Input:
Integrate[((c*x^2)^(5/2)*(a + b*x)^p)/x^4,x]
Output:
(c^3*x*(a + b*x)^(1 + p)*(-a + b*(1 + p)*x))/(b^2*(1 + p)*(2 + p)*Sqrt[c*x ^2])
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80, 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.
\(\displaystyle \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int x (a+b x)^pdx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int \left (\frac {(a+b x)^{p+1}}{b}-\frac {a (a+b x)^p}{b}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \left (\frac {(a+b x)^{p+2}}{b^2 (p+2)}-\frac {a (a+b x)^{p+1}}{b^2 (p+1)}\right )}{x}\) |
Input:
Int[((c*x^2)^(5/2)*(a + b*x)^p)/x^4,x]
Output:
(c^2*Sqrt[c*x^2]*(-((a*(a + b*x)^(1 + p))/(b^2*(1 + p))) + (a + b*x)^(2 + p)/(b^2*(2 + 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 = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.67
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{p +1} \left (-b p x -b x +a \right )}{b^{2} x^{5} \left (p^{2}+3 p +2\right )}\) | \(46\) |
orering | \(-\frac {\left (b x +a \right )^{p} \left (c \,x^{2}\right )^{\frac {5}{2}} \left (-b p x -b x +a \right ) \left (b x +a \right )}{x^{5} b^{2} \left (p^{2}+3 p +2\right )}\) | \(49\) |
risch | \(-\frac {c^{2} \sqrt {c \,x^{2}}\, \left (-b^{2} p \,x^{2}-a b p x -b^{2} x^{2}+a^{2}\right ) \left (b x +a \right )^{p}}{x \,b^{2} \left (2+p \right ) \left (p +1\right )}\) | \(63\) |
Input:
int((c*x^2)^(5/2)*(b*x+a)^p/x^4,x,method=_RETURNVERBOSE)
Output:
-1/b^2/x^5*(c*x^2)^(5/2)*(b*x+a)^(p+1)/(p^2+3*p+2)*(-b*p*x-b*x+a)
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\frac {{\left (a b c^{2} p x - a^{2} c^{2} + {\left (b^{2} c^{2} p + b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}\right )} x} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^4,x, algorithm="fricas")
Output:
(a*b*c^2*p*x - a^2*c^2 + (b^2*c^2*p + b^2*c^2)*x^2)*sqrt(c*x^2)*(b*x + a)^ p/((b^2*p^2 + 3*b^2*p + 2*b^2)*x)
\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\begin {cases} \frac {a^{p} \left (c x^{2}\right )^{\frac {5}{2}}}{2 x^{3}} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )^{2}}\, dx & \text {for}\: p = -2 \\\int \frac {\left (c x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )}\, dx & \text {for}\: p = -1 \\- \frac {a^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}}{b^{2} p^{2} x^{5} + 3 b^{2} p x^{5} + 2 b^{2} x^{5}} + \frac {a b p x \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}}{b^{2} p^{2} x^{5} + 3 b^{2} p x^{5} + 2 b^{2} x^{5}} + \frac {b^{2} p x^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}}{b^{2} p^{2} x^{5} + 3 b^{2} p x^{5} + 2 b^{2} x^{5}} + \frac {b^{2} x^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}}{b^{2} p^{2} x^{5} + 3 b^{2} p x^{5} + 2 b^{2} x^{5}} & \text {otherwise} \end {cases} \] Input:
integrate((c*x**2)**(5/2)*(b*x+a)**p/x**4,x)
Output:
Piecewise((a**p*(c*x**2)**(5/2)/(2*x**3), Eq(b, 0)), (Integral((c*x**2)**( 5/2)/(x**4*(a + b*x)**2), x), Eq(p, -2)), (Integral((c*x**2)**(5/2)/(x**4* (a + b*x)), x), Eq(p, -1)), (-a**2*(c*x**2)**(5/2)*(a + b*x)**p/(b**2*p**2 *x**5 + 3*b**2*p*x**5 + 2*b**2*x**5) + a*b*p*x*(c*x**2)**(5/2)*(a + b*x)** p/(b**2*p**2*x**5 + 3*b**2*p*x**5 + 2*b**2*x**5) + b**2*p*x**2*(c*x**2)**( 5/2)*(a + b*x)**p/(b**2*p**2*x**5 + 3*b**2*p*x**5 + 2*b**2*x**5) + b**2*x* *2*(c*x**2)**(5/2)*(a + b*x)**p/(b**2*p**2*x**5 + 3*b**2*p*x**5 + 2*b**2*x **5), True))
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\frac {{\left (b^{2} c^{\frac {5}{2}} {\left (p + 1\right )} x^{2} + a b c^{\frac {5}{2}} p x - a^{2} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^4,x, algorithm="maxima")
Output:
(b^2*c^(5/2)*(p + 1)*x^2 + a*b*c^(5/2)*p*x - a^2*c^(5/2))*(b*x + a)^p/((p^ 2 + 3*p + 2)*b^2)
Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.72 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx={\left (\frac {a^{2} a^{p} \mathrm {sgn}\left (x\right )}{b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}} + \frac {{\left (b x + a\right )}^{p} b^{2} p x^{2} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{p} a b p x \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{p} b^{2} x^{2} \mathrm {sgn}\left (x\right ) - {\left (b x + a\right )}^{p} a^{2} \mathrm {sgn}\left (x\right )}{b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}}\right )} c^{\frac {5}{2}} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^4,x, algorithm="giac")
Output:
(a^2*a^p*sgn(x)/(b^2*p^2 + 3*b^2*p + 2*b^2) + ((b*x + a)^p*b^2*p*x^2*sgn(x ) + (b*x + a)^p*a*b*p*x*sgn(x) + (b*x + a)^p*b^2*x^2*sgn(x) - (b*x + a)^p* a^2*sgn(x))/(b^2*p^2 + 3*b^2*p + 2*b^2))*c^(5/2)
Time = 22.53 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {c^2\,x^2\,\sqrt {c\,x^2}\,\left (p+1\right )}{p^2+3\,p+2}-\frac {a^2\,c^2\,\sqrt {c\,x^2}}{b^2\,\left (p^2+3\,p+2\right )}+\frac {a\,c^2\,p\,x\,\sqrt {c\,x^2}}{b\,\left (p^2+3\,p+2\right )}\right )}{x} \] Input:
int(((c*x^2)^(5/2)*(a + b*x)^p)/x^4,x)
Output:
((a + b*x)^p*((c^2*x^2*(c*x^2)^(1/2)*(p + 1))/(3*p + p^2 + 2) - (a^2*c^2*( c*x^2)^(1/2))/(b^2*(3*p + p^2 + 2)) + (a*c^2*p*x*(c*x^2)^(1/2))/(b*(3*p + p^2 + 2))))/x
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^4} \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} c^{2} \left (b^{2} p \,x^{2}+a b p x +b^{2} x^{2}-a^{2}\right )}{b^{2} \left (p^{2}+3 p +2\right )} \] Input:
int((c*x^2)^(5/2)*(b*x+a)^p/x^4,x)
Output:
(sqrt(c)*(a + b*x)**p*c**2*( - a**2 + a*b*p*x + b**2*p*x**2 + b**2*x**2))/ (b**2*(p**2 + 3*p + 2))