Integrand size = 20, antiderivative size = 90 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {a^2 x (a+b x)^{1+p}}{b^3 (1+p) \sqrt {c x^2}}-\frac {2 a x (a+b x)^{2+p}}{b^3 (2+p) \sqrt {c x^2}}+\frac {x (a+b x)^{3+p}}{b^3 (3+p) \sqrt {c x^2}} \] Output:
a^2*x*(b*x+a)^(p+1)/b^3/(p+1)/(c*x^2)^(1/2)-2*a*x*(b*x+a)^(2+p)/b^3/(2+p)/ (c*x^2)^(1/2)+x*(b*x+a)^(3+p)/b^3/(3+p)/(c*x^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {x (a+b x)^{1+p} \left (2 a^2-2 a b (1+p) x+b^2 \left (2+3 p+p^2\right ) x^2\right )}{b^3 (1+p) (2+p) (3+p) \sqrt {c x^2}} \] Input:
Integrate[(x^3*(a + b*x)^p)/Sqrt[c*x^2],x]
Output:
(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 = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79, 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.
\(\displaystyle \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {x \int x^2 (a+b x)^pdx}{\sqrt {c x^2}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {x \int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{p+1}}{b^2}+\frac {(a+b x)^{p+2}}{b^2}\right )dx}{\sqrt {c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (\frac {a^2 (a+b x)^{p+1}}{b^3 (p+1)}-\frac {2 a (a+b x)^{p+2}}{b^3 (p+2)}+\frac {(a+b x)^{p+3}}{b^3 (p+3)}\right )}{\sqrt {c x^2}}\) |
Input:
Int[(x^3*(a + b*x)^p)/Sqrt[c*x^2],x]
Output:
(x*((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))))/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 = 81, normalized size of antiderivative = 0.90
method | result | size |
gosper | \(\frac {x \left (b x +a \right )^{p +1} \left (b^{2} p^{2} x^{2}+3 b^{2} p \,x^{2}-2 a b p x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} \sqrt {c \,x^{2}}\, \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(81\) |
orering | \(\frac {\left (b x +a \right ) \left (b^{2} p^{2} x^{2}+3 b^{2} p \,x^{2}-2 a b p x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right ) x \left (b x +a \right )^{p}}{\sqrt {c \,x^{2}}\, b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(84\) |
risch | \(\frac {x \left (b^{3} p^{2} x^{3}+a \,b^{2} p^{2} x^{2}+3 b^{3} p \,x^{3}+a \,b^{2} p \,x^{2}+2 b^{3} x^{3}-2 a^{2} b p x +2 a^{3}\right ) \left (b x +a \right )^{p}}{\sqrt {c \,x^{2}}\, \left (2+p \right ) \left (3+p \right ) \left (p +1\right ) b^{3}}\) | \(96\) |
Input:
int(x^3*(b*x+a)^p/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
x/b^3/(c*x^2)^(1/2)*(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 = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=-\frac {{\left (2 \, a^{2} b p x - {\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} - {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{p}}{{\left (b^{3} c p^{3} + 6 \, b^{3} c p^{2} + 11 \, b^{3} c p + 6 \, b^{3} c\right )} x} \] Input:
integrate(x^3*(b*x+a)^p/(c*x^2)^(1/2),x, algorithm="fricas")
Output:
-(2*a^2*b*p*x - (b^3*p^2 + 3*b^3*p + 2*b^3)*x^3 - 2*a^3 - (a*b^2*p^2 + a*b ^2*p)*x^2)*sqrt(c*x^2)*(b*x + a)^p/((b^3*c*p^3 + 6*b^3*c*p^2 + 11*b^3*c*p + 6*b^3*c)*x)
\[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\begin {cases} \frac {a^{p} x^{4}}{3 \sqrt {c x^{2}}} & \text {for}\: b = 0 \\\int \frac {x^{3}}{\sqrt {c x^{2}} \left (a + b x\right )^{3}}\, dx & \text {for}\: p = -3 \\\int \frac {x^{3}}{\sqrt {c x^{2}} \left (a + b x\right )^{2}}\, dx & \text {for}\: p = -2 \\\int \frac {x^{3}}{\sqrt {c x^{2}} \left (a + b x\right )}\, dx & \text {for}\: p = -1 \\\frac {2 a^{3} x \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} - \frac {2 a^{2} b p x^{2} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} + \frac {a b^{2} p^{2} x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} + \frac {a b^{2} p x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} + \frac {b^{3} p^{2} x^{4} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} + \frac {3 b^{3} p x^{4} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} + \frac {2 b^{3} x^{4} \left (a + b x\right )^{p}}{b^{3} p^{3} \sqrt {c x^{2}} + 6 b^{3} p^{2} \sqrt {c x^{2}} + 11 b^{3} p \sqrt {c x^{2}} + 6 b^{3} \sqrt {c x^{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(b*x+a)**p/(c*x**2)**(1/2),x)
Output:
Piecewise((a**p*x**4/(3*sqrt(c*x**2)), Eq(b, 0)), (Integral(x**3/(sqrt(c*x **2)*(a + b*x)**3), x), Eq(p, -3)), (Integral(x**3/(sqrt(c*x**2)*(a + b*x) **2), x), Eq(p, -2)), (Integral(x**3/(sqrt(c*x**2)*(a + b*x)), x), Eq(p, - 1)), (2*a**3*x*(a + b*x)**p/(b**3*p**3*sqrt(c*x**2) + 6*b**3*p**2*sqrt(c*x **2) + 11*b**3*p*sqrt(c*x**2) + 6*b**3*sqrt(c*x**2)) - 2*a**2*b*p*x**2*(a + b*x)**p/(b**3*p**3*sqrt(c*x**2) + 6*b**3*p**2*sqrt(c*x**2) + 11*b**3*p*s qrt(c*x**2) + 6*b**3*sqrt(c*x**2)) + a*b**2*p**2*x**3*(a + b*x)**p/(b**3*p **3*sqrt(c*x**2) + 6*b**3*p**2*sqrt(c*x**2) + 11*b**3*p*sqrt(c*x**2) + 6*b **3*sqrt(c*x**2)) + a*b**2*p*x**3*(a + b*x)**p/(b**3*p**3*sqrt(c*x**2) + 6 *b**3*p**2*sqrt(c*x**2) + 11*b**3*p*sqrt(c*x**2) + 6*b**3*sqrt(c*x**2)) + b**3*p**2*x**4*(a + b*x)**p/(b**3*p**3*sqrt(c*x**2) + 6*b**3*p**2*sqrt(c*x **2) + 11*b**3*p*sqrt(c*x**2) + 6*b**3*sqrt(c*x**2)) + 3*b**3*p*x**4*(a + b*x)**p/(b**3*p**3*sqrt(c*x**2) + 6*b**3*p**2*sqrt(c*x**2) + 11*b**3*p*sqr t(c*x**2) + 6*b**3*sqrt(c*x**2)) + 2*b**3*x**4*(a + b*x)**p/(b**3*p**3*sqr t(c*x**2) + 6*b**3*p**2*sqrt(c*x**2) + 11*b**3*p*sqrt(c*x**2) + 6*b**3*sqr t(c*x**2)), True))
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} \sqrt {c} x^{3} + {\left (p^{2} + p\right )} a b^{2} \sqrt {c} x^{2} - 2 \, a^{2} b \sqrt {c} p x + 2 \, a^{3} \sqrt {c}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3} c} \] Input:
integrate(x^3*(b*x+a)^p/(c*x^2)^(1/2),x, algorithm="maxima")
Output:
((p^2 + 3*p + 2)*b^3*sqrt(c)*x^3 + (p^2 + p)*a*b^2*sqrt(c)*x^2 - 2*a^2*b*s qrt(c)*p*x + 2*a^3*sqrt(c))*(b*x + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3*c)
Exception generated. \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3*(b*x+a)^p/(c*x^2)^(1/2),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,[0,2,1,0,0]%%%} / %%%{1,[0,0,0,1,1]%%%} Error: Bad Argum ent Value
Time = 22.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {{\left (a+b\,x\right )}^p\,\left (\frac {x^4\,\left (p^2+3\,p+2\right )}{p^3+6\,p^2+11\,p+6}+\frac {2\,a^3\,x}{b^3\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {2\,a^2\,p\,x^2}{b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^3\,\left (p+1\right )}{b\,\left (p^3+6\,p^2+11\,p+6\right )}\right )}{\sqrt {c\,x^2}} \] Input:
int((x^3*(a + b*x)^p)/(c*x^2)^(1/2),x)
Output:
((a + b*x)^p*((x^4*(3*p + p^2 + 2))/(11*p + 6*p^2 + p^3 + 6) + (2*a^3*x)/( b^3*(11*p + 6*p^2 + p^3 + 6)) - (2*a^2*p*x^2)/(b^2*(11*p + 6*p^2 + p^3 + 6 )) + (a*p*x^3*(p + 1))/(b*(11*p + 6*p^2 + p^3 + 6))))/(c*x^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 (a+b x)^p}{\sqrt {c x^2}} \, dx=\frac {\sqrt {c}\, \left (b x +a \right )^{p} \left (b^{3} p^{2} x^{3}+a \,b^{2} p^{2} x^{2}+3 b^{3} p \,x^{3}+a \,b^{2} p \,x^{2}+2 b^{3} x^{3}-2 a^{2} b p x +2 a^{3}\right )}{b^{3} c \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^3*(b*x+a)^p/(c*x^2)^(1/2),x)
Output:
(sqrt(c)*(a + b*x)**p*(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*c*(p**3 + 6*p **2 + 11*p + 6))