Integrand size = 22, antiderivative size = 97 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {a^2 c (d x)^{4+m} \sqrt {c x^2}}{d^4 (4+m) x}+\frac {2 a b c (d x)^{5+m} \sqrt {c x^2}}{d^5 (5+m) x}+\frac {b^2 c (d x)^{6+m} \sqrt {c x^2}}{d^6 (6+m) x} \] Output:
a^2*c*(d*x)^(4+m)*(c*x^2)^(1/2)/d^4/(4+m)/x+2*a*b*c*(d*x)^(5+m)*(c*x^2)^(1 /2)/d^5/(5+m)/x+b^2*c*(d*x)^(6+m)*(c*x^2)^(1/2)/d^6/(6+m)/x
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.49 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=x (d x)^m \left (c x^2\right )^{3/2} \left (\frac {a^2}{4+m}+\frac {2 a b x}{5+m}+\frac {b^2 x^2}{6+m}\right ) \] Input:
Integrate[(d*x)^m*(c*x^2)^(3/2)*(a + b*x)^2,x]
Output:
x*(d*x)^m*(c*x^2)^(3/2)*(a^2/(4 + m) + (2*a*b*x)/(5 + m) + (b^2*x^2)/(6 + m))
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.
\(\displaystyle \int \left (c x^2\right )^{3/2} (a+b x)^2 (d x)^m \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c \sqrt {c x^2} \int (d x)^{m+3} (a+b x)^2dx}{d^3 x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c \sqrt {c x^2} \int \left (a^2 (d x)^{m+3}+\frac {2 a b (d x)^{m+4}}{d}+\frac {b^2 (d x)^{m+5}}{d^2}\right )dx}{d^3 x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c x^2} \left (\frac {a^2 (d x)^{m+4}}{d (m+4)}+\frac {2 a b (d x)^{m+5}}{d^2 (m+5)}+\frac {b^2 (d x)^{m+6}}{d^3 (m+6)}\right )}{d^3 x}\) |
Input:
Int[(d*x)^m*(c*x^2)^(3/2)*(a + b*x)^2,x]
Output:
(c*Sqrt[c*x^2]*((a^2*(d*x)^(4 + m))/(d*(4 + m)) + (2*a*b*(d*x)^(5 + m))/(d ^2*(5 + m)) + (b^2*(d*x)^(6 + m))/(d^3*(6 + m))))/(d^3*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 = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98
method | result | size |
gosper | \(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +9 m \,x^{2} b^{2}+a^{2} m^{2}+20 a b m x +20 b^{2} x^{2}+11 a^{2} m +48 a b x +30 a^{2}\right ) \left (d x \right )^{m} \left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (6+m \right ) \left (m +5\right ) \left (4+m \right )}\) | \(95\) |
orering | \(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +9 m \,x^{2} b^{2}+a^{2} m^{2}+20 a b m x +20 b^{2} x^{2}+11 a^{2} m +48 a b x +30 a^{2}\right ) \left (d x \right )^{m} \left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (6+m \right ) \left (m +5\right ) \left (4+m \right )}\) | \(95\) |
risch | \(\frac {c \,x^{3} \sqrt {c \,x^{2}}\, \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +9 m \,x^{2} b^{2}+a^{2} m^{2}+20 a b m x +20 b^{2} x^{2}+11 a^{2} m +48 a b x +30 a^{2}\right ) \left (d x \right )^{m}}{\left (6+m \right ) \left (m +5\right ) \left (4+m \right )}\) | \(98\) |
Input:
int((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
x*(b^2*m^2*x^2+2*a*b*m^2*x+9*b^2*m*x^2+a^2*m^2+20*a*b*m*x+20*b^2*x^2+11*a^ 2*m+48*a*b*x+30*a^2)*(d*x)^m*(c*x^2)^(3/2)/(6+m)/(m+5)/(4+m)
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {{\left ({\left (b^{2} c m^{2} + 9 \, b^{2} c m + 20 \, b^{2} c\right )} x^{5} + 2 \, {\left (a b c m^{2} + 10 \, a b c m + 24 \, a b c\right )} x^{4} + {\left (a^{2} c m^{2} + 11 \, a^{2} c m + 30 \, a^{2} c\right )} x^{3}\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{3} + 15 \, m^{2} + 74 \, m + 120} \] Input:
integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="fricas")
Output:
((b^2*c*m^2 + 9*b^2*c*m + 20*b^2*c)*x^5 + 2*(a*b*c*m^2 + 10*a*b*c*m + 24*a *b*c)*x^4 + (a^2*c*m^2 + 11*a^2*c*m + 30*a^2*c)*x^3)*sqrt(c*x^2)*(d*x)^m/( m^3 + 15*m^2 + 74*m + 120)
Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (87) = 174\).
Time = 3.40 (sec) , antiderivative size = 493, normalized size of antiderivative = 5.08 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\begin {cases} \frac {- \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{2 x^{5}} - \frac {2 a b \left (c x^{2}\right )^{\frac {3}{2}}}{x^{4}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {3}{2}} \log {\left (x \right )}}{x^{3}}}{d^{6}} & \text {for}\: m = -6 \\\frac {- \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{x^{4}} + \frac {2 a b \left (c x^{2}\right )^{\frac {3}{2}} \log {\left (x \right )}}{x^{3}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{x^{2}}}{d^{5}} & \text {for}\: m = -5 \\\frac {\frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}} \log {\left (x \right )}}{x^{3}} + \frac {2 a b \left (c x^{2}\right )^{\frac {3}{2}}}{x^{2}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{2 x}}{d^{4}} & \text {for}\: m = -4 \\\frac {a^{2} m^{2} x \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {11 a^{2} m x \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {30 a^{2} x \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {2 a b m^{2} x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {20 a b m x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {48 a b x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {b^{2} m^{2} x^{3} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {9 b^{2} m x^{3} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} + \frac {20 b^{2} x^{3} \left (c x^{2}\right )^{\frac {3}{2}} \left (d x\right )^{m}}{m^{3} + 15 m^{2} + 74 m + 120} & \text {otherwise} \end {cases} \] Input:
integrate((d*x)**m*(c*x**2)**(3/2)*(b*x+a)**2,x)
Output:
Piecewise(((-a**2*(c*x**2)**(3/2)/(2*x**5) - 2*a*b*(c*x**2)**(3/2)/x**4 + b**2*(c*x**2)**(3/2)*log(x)/x**3)/d**6, Eq(m, -6)), ((-a**2*(c*x**2)**(3/2 )/x**4 + 2*a*b*(c*x**2)**(3/2)*log(x)/x**3 + b**2*(c*x**2)**(3/2)/x**2)/d* *5, Eq(m, -5)), ((a**2*(c*x**2)**(3/2)*log(x)/x**3 + 2*a*b*(c*x**2)**(3/2) /x**2 + b**2*(c*x**2)**(3/2)/(2*x))/d**4, Eq(m, -4)), (a**2*m**2*x*(c*x**2 )**(3/2)*(d*x)**m/(m**3 + 15*m**2 + 74*m + 120) + 11*a**2*m*x*(c*x**2)**(3 /2)*(d*x)**m/(m**3 + 15*m**2 + 74*m + 120) + 30*a**2*x*(c*x**2)**(3/2)*(d* x)**m/(m**3 + 15*m**2 + 74*m + 120) + 2*a*b*m**2*x**2*(c*x**2)**(3/2)*(d*x )**m/(m**3 + 15*m**2 + 74*m + 120) + 20*a*b*m*x**2*(c*x**2)**(3/2)*(d*x)** m/(m**3 + 15*m**2 + 74*m + 120) + 48*a*b*x**2*(c*x**2)**(3/2)*(d*x)**m/(m* *3 + 15*m**2 + 74*m + 120) + b**2*m**2*x**3*(c*x**2)**(3/2)*(d*x)**m/(m**3 + 15*m**2 + 74*m + 120) + 9*b**2*m*x**3*(c*x**2)**(3/2)*(d*x)**m/(m**3 + 15*m**2 + 74*m + 120) + 20*b**2*x**3*(c*x**2)**(3/2)*(d*x)**m/(m**3 + 15*m **2 + 74*m + 120), True))
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {b^{2} c^{\frac {3}{2}} d^{m} x^{6} x^{m}}{m + 6} + \frac {2 \, a b c^{\frac {3}{2}} d^{m} x^{5} x^{m}}{m + 5} + \frac {a^{2} c^{\frac {3}{2}} d^{m} x^{4} x^{m}}{m + 4} \] Input:
integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="maxima")
Output:
b^2*c^(3/2)*d^m*x^6*x^m/(m + 6) + 2*a*b*c^(3/2)*d^m*x^5*x^m/(m + 5) + a^2* c^(3/2)*d^m*x^4*x^m/(m + 4)
Exception generated. \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\text {Exception raised: TypeError} \] Input:
integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 23.56 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx={\left (d\,x\right )}^m\,\left (\frac {a^2\,c\,x^3\,\sqrt {c\,x^2}\,\left (m^2+11\,m+30\right )}{m^3+15\,m^2+74\,m+120}+\frac {b^2\,c\,x^5\,\sqrt {c\,x^2}\,\left (m^2+9\,m+20\right )}{m^3+15\,m^2+74\,m+120}+\frac {2\,a\,b\,c\,x^4\,\sqrt {c\,x^2}\,\left (m^2+10\,m+24\right )}{m^3+15\,m^2+74\,m+120}\right ) \] Input:
int((d*x)^m*(c*x^2)^(3/2)*(a + b*x)^2,x)
Output:
(d*x)^m*((a^2*c*x^3*(c*x^2)^(1/2)*(11*m + m^2 + 30))/(74*m + 15*m^2 + m^3 + 120) + (b^2*c*x^5*(c*x^2)^(1/2)*(9*m + m^2 + 20))/(74*m + 15*m^2 + m^3 + 120) + (2*a*b*c*x^4*(c*x^2)^(1/2)*(10*m + m^2 + 24))/(74*m + 15*m^2 + m^3 + 120))
Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {x^{m} d^{m} \sqrt {c}\, c \,x^{4} \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +9 b^{2} m \,x^{2}+a^{2} m^{2}+20 a b m x +20 b^{2} x^{2}+11 a^{2} m +48 a b x +30 a^{2}\right )}{m^{3}+15 m^{2}+74 m +120} \] Input:
int((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^2,x)
Output:
(x**m*d**m*sqrt(c)*c*x**4*(a**2*m**2 + 11*a**2*m + 30*a**2 + 2*a*b*m**2*x + 20*a*b*m*x + 48*a*b*x + b**2*m**2*x**2 + 9*b**2*m*x**2 + 20*b**2*x**2))/ (m**3 + 15*m**2 + 74*m + 120)