Integrand size = 20, antiderivative size = 143 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=-\frac {a^3 c^2 \sqrt {c x^2} (a+b x)^{1+p}}{b^4 (1+p) x}+\frac {3 a^2 c^2 \sqrt {c x^2} (a+b x)^{2+p}}{b^4 (2+p) x}-\frac {3 a c^2 \sqrt {c x^2} (a+b x)^{3+p}}{b^4 (3+p) x}+\frac {c^2 \sqrt {c x^2} (a+b x)^{4+p}}{b^4 (4+p) x} \] Output:
-a^3*c^2*(c*x^2)^(1/2)*(b*x+a)^(p+1)/b^4/(p+1)/x+3*a^2*c^2*(c*x^2)^(1/2)*( b*x+a)^(2+p)/b^4/(2+p)/x-3*a*c^2*(c*x^2)^(1/2)*(b*x+a)^(3+p)/b^4/(3+p)/x+c ^2*(c*x^2)^(1/2)*(b*x+a)^(4+p)/b^4/(4+p)/x
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\frac {c \left (c x^2\right )^{3/2} (a+b x)^{1+p} \left (-6 a^3+6 a^2 b (1+p) x-3 a b^2 \left (2+3 p+p^2\right ) x^2+b^3 \left (6+11 p+6 p^2+p^3\right ) x^3\right )}{b^4 (1+p) (2+p) (3+p) (4+p) x^3} \] Input:
Integrate[((c*x^2)^(5/2)*(a + b*x)^p)/x^2,x]
Output:
(c*(c*x^2)^(3/2)*(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*(1 + p)*(2 + p)*(3 + p)*(4 + p)*x^3)
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.69, 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.
\(\displaystyle \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int x^3 (a+b x)^pdx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \int \left (-\frac {a^3 (a+b x)^p}{b^3}+\frac {3 a^2 (a+b x)^{p+1}}{b^3}-\frac {3 a (a+b x)^{p+2}}{b^3}+\frac {(a+b x)^{p+3}}{b^3}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \sqrt {c x^2} \left (-\frac {a^3 (a+b x)^{p+1}}{b^4 (p+1)}+\frac {3 a^2 (a+b x)^{p+2}}{b^4 (p+2)}-\frac {3 a (a+b x)^{p+3}}{b^4 (p+3)}+\frac {(a+b x)^{p+4}}{b^4 (p+4)}\right )}{x}\) |
Input:
Int[((c*x^2)^(5/2)*(a + b*x)^p)/x^2,x]
Output:
(c^2*Sqrt[c*x^2]*(-((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))))/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 = 136, normalized size of antiderivative = 0.95
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (b x +a \right )^{p +1} \left (-b^{3} p^{3} x^{3}-6 b^{3} p^{2} x^{3}+3 a \,b^{2} p^{2} x^{2}-11 b^{3} p \,x^{3}+9 a \,b^{2} p \,x^{2}-6 b^{3} x^{3}-6 a^{2} b p x +6 a \,b^{2} x^{2}-6 b \,a^{2} x +6 a^{3}\right )}{b^{4} x^{5} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(136\) |
orering | \(-\frac {\left (b x +a \right )^{p} \left (c \,x^{2}\right )^{\frac {5}{2}} \left (-b^{3} p^{3} x^{3}-6 b^{3} p^{2} x^{3}+3 a \,b^{2} p^{2} x^{2}-11 b^{3} p \,x^{3}+9 a \,b^{2} p \,x^{2}-6 b^{3} x^{3}-6 a^{2} b p x +6 a \,b^{2} x^{2}-6 b \,a^{2} x +6 a^{3}\right ) \left (b x +a \right )}{x^{5} b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\) | \(139\) |
risch | \(-\frac {c^{2} \sqrt {c \,x^{2}}\, \left (-b^{4} p^{3} x^{4}-a \,b^{3} p^{3} x^{3}-6 b^{4} p^{2} x^{4}-3 a \,b^{3} p^{2} x^{3}-11 b^{4} p \,x^{4}+3 a^{2} b^{2} p^{2} x^{2}-2 x^{3} a p \,b^{3}-6 b^{4} x^{4}+3 a^{2} p \,x^{2} b^{2}-6 a^{3} b p x +6 a^{4}\right ) \left (b x +a \right )^{p}}{x \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (p +1\right ) b^{4}}\) | \(159\) |
Input:
int((c*x^2)^(5/2)*(b*x+a)^p/x^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 = 0.12 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.30 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\frac {{\left (6 \, a^{3} b c^{2} p x - 6 \, a^{4} c^{2} + {\left (b^{4} c^{2} p^{3} + 6 \, b^{4} c^{2} p^{2} + 11 \, b^{4} c^{2} p + 6 \, b^{4} c^{2}\right )} x^{4} + {\left (a b^{3} c^{2} p^{3} + 3 \, a b^{3} c^{2} p^{2} + 2 \, a b^{3} c^{2} p\right )} x^{3} - 3 \, {\left (a^{2} b^{2} c^{2} p^{2} + a^{2} b^{2} c^{2} p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )} x} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^2,x, algorithm="fricas")
Output:
(6*a^3*b*c^2*p*x - 6*a^4*c^2 + (b^4*c^2*p^3 + 6*b^4*c^2*p^2 + 11*b^4*c^2*p + 6*b^4*c^2)*x^4 + (a*b^3*c^2*p^3 + 3*a*b^3*c^2*p^2 + 2*a*b^3*c^2*p)*x^3 - 3*(a^2*b^2*c^2*p^2 + a^2*b^2*c^2*p)*x^2)*sqrt(c*x^2)*(b*x + a)^p/((b^4*p ^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4)*x)
\[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {5}{2}} \left (a + b x\right )^{p}}{x^{2}}\, dx \] Input:
integrate((c*x**2)**(5/2)*(b*x+a)**p/x**2,x)
Output:
Integral((c*x**2)**(5/2)*(a + b*x)**p/x**2, x)
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} c^{\frac {5}{2}} x^{4} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} c^{\frac {5}{2}} x^{3} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} c^{\frac {5}{2}} x^{2} + 6 \, a^{3} b c^{\frac {5}{2}} p x - 6 \, a^{4} c^{\frac {5}{2}}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^2,x, algorithm="maxima")
Output:
((p^3 + 6*p^2 + 11*p + 6)*b^4*c^(5/2)*x^4 + (p^3 + 3*p^2 + 2*p)*a*b^3*c^(5 /2)*x^3 - 3*(p^2 + p)*a^2*b^2*c^(5/2)*x^2 + 6*a^3*b*c^(5/2)*p*x - 6*a^4*c^ (5/2))*(b*x + a)^p/((p^4 + 10*p^3 + 35*p^2 + 50*p + 24)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (135) = 270\).
Time = 0.12 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.10 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx={\left (\frac {6 \, a^{4} a^{p} \mathrm {sgn}\left (x\right )}{b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}} + \frac {{\left (b x + a\right )}^{p} b^{4} p^{3} x^{4} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{p} a b^{3} p^{3} x^{3} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{p} b^{4} p^{2} x^{4} \mathrm {sgn}\left (x\right ) + 3 \, {\left (b x + a\right )}^{p} a b^{3} p^{2} x^{3} \mathrm {sgn}\left (x\right ) + 11 \, {\left (b x + a\right )}^{p} b^{4} p x^{4} \mathrm {sgn}\left (x\right ) - 3 \, {\left (b x + a\right )}^{p} a^{2} b^{2} p^{2} x^{2} \mathrm {sgn}\left (x\right ) + 2 \, {\left (b x + a\right )}^{p} a b^{3} p x^{3} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{p} b^{4} x^{4} \mathrm {sgn}\left (x\right ) - 3 \, {\left (b x + a\right )}^{p} a^{2} b^{2} p x^{2} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{p} a^{3} b p x \mathrm {sgn}\left (x\right ) - 6 \, {\left (b x + a\right )}^{p} a^{4} \mathrm {sgn}\left (x\right )}{b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}}\right )} c^{\frac {5}{2}} \] Input:
integrate((c*x^2)^(5/2)*(b*x+a)^p/x^2,x, algorithm="giac")
Output:
(6*a^4*a^p*sgn(x)/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4) + ((b*x + a)^p*b^4*p^3*x^4*sgn(x) + (b*x + a)^p*a*b^3*p^3*x^3*sgn(x) + 6*( b*x + a)^p*b^4*p^2*x^4*sgn(x) + 3*(b*x + a)^p*a*b^3*p^2*x^3*sgn(x) + 11*(b *x + a)^p*b^4*p*x^4*sgn(x) - 3*(b*x + a)^p*a^2*b^2*p^2*x^2*sgn(x) + 2*(b*x + a)^p*a*b^3*p*x^3*sgn(x) + 6*(b*x + a)^p*b^4*x^4*sgn(x) - 3*(b*x + a)^p* a^2*b^2*p*x^2*sgn(x) + 6*(b*x + a)^p*a^3*b*p*x*sgn(x) - 6*(b*x + a)^p*a^4* sgn(x))/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4))*c^(5/2)
Time = 23.40 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.60 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {c^2\,x^4\,\sqrt {c\,x^2}\,\left (p^3+6\,p^2+11\,p+6\right )}{p^4+10\,p^3+35\,p^2+50\,p+24}-\frac {6\,a^4\,c^2\,\sqrt {c\,x^2}}{b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {6\,a^3\,c^2\,p\,x\,\sqrt {c\,x^2}}{b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,c^2\,p\,x^3\,\sqrt {c\,x^2}\,\left (p^2+3\,p+2\right )}{b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^2\,c^2\,p\,x^2\,\sqrt {c\,x^2}\,\left (p+1\right )}{b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right )}{x} \] Input:
int(((c*x^2)^(5/2)*(a + b*x)^p)/x^2,x)
Output:
((a + b*x)^p*((c^2*x^4*(c*x^2)^(1/2)*(11*p + 6*p^2 + p^3 + 6))/(50*p + 35* p^2 + 10*p^3 + p^4 + 24) - (6*a^4*c^2*(c*x^2)^(1/2))/(b^4*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (6*a^3*c^2*p*x*(c*x^2)^(1/2))/(b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*c^2*p*x^3*(c*x^2)^(1/2)*(3*p + p^2 + 2))/(b*(50* p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^2*c^2*p*x^2*(c*x^2)^(1/2)*(p + 1)) /(b^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24))))/x
Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.03 \[ \int \frac {\left (c x^2\right )^{5/2} (a+b x)^p}{x^2} \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} c^{2} \left (b^{4} p^{3} x^{4}+a \,b^{3} p^{3} x^{3}+6 b^{4} p^{2} x^{4}+3 a \,b^{3} p^{2} x^{3}+11 b^{4} p \,x^{4}-3 a^{2} b^{2} p^{2} x^{2}+2 a \,b^{3} p \,x^{3}+6 b^{4} x^{4}-3 a^{2} b^{2} p \,x^{2}+6 a^{3} b p x -6 a^{4}\right )}{b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:
int((c*x^2)^(5/2)*(b*x+a)^p/x^2,x)
Output:
(sqrt(c)*(a + b*x)**p*c**2*( - 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*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))