Integrand size = 21, antiderivative size = 128 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=-\frac {2 a^3 c^4 \left (a x+b x^2\right )^{5/2}}{5 b^4 (c x)^{5/2}}+\frac {6 a^2 c^5 \left (a x+b x^2\right )^{7/2}}{7 b^4 (c x)^{7/2}}-\frac {2 a c^6 \left (a x+b x^2\right )^{9/2}}{3 b^4 (c x)^{9/2}}+\frac {2 c^7 \left (a x+b x^2\right )^{11/2}}{11 b^4 (c x)^{11/2}} \] Output:
-2/5*a^3*c^4*(b*x^2+a*x)^(5/2)/b^4/(c*x)^(5/2)+6/7*a^2*c^5*(b*x^2+a*x)^(7/ 2)/b^4/(c*x)^(7/2)-2/3*a*c^6*(b*x^2+a*x)^(9/2)/b^4/(c*x)^(9/2)+2/11*c^7*(b *x^2+a*x)^(11/2)/b^4/(c*x)^(11/2)
Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.51 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\frac {2 c^2 (a+b x)^2 \sqrt {x (a+b x)} \left (-16 a^3+40 a^2 b x-70 a b^2 x^2+105 b^3 x^3\right )}{1155 b^4 \sqrt {c x}} \] Input:
Integrate[(c*x)^(3/2)*(a*x + b*x^2)^(3/2),x]
Output:
(2*c^2*(a + b*x)^2*Sqrt[x*(a + b*x)]*(-16*a^3 + 40*a^2*b*x - 70*a*b^2*x^2 + 105*b^3*x^3))/(1155*b^4*Sqrt[c*x])
Time = 0.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1128, 1128, 1128, 1122}
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.
\(\displaystyle \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{5/2}}{11 b}-\frac {6 a c \int \sqrt {c x} \left (b x^2+a x\right )^{3/2}dx}{11 b}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{5/2}}{11 b}-\frac {6 a c \left (\frac {2 c \left (a x+b x^2\right )^{5/2}}{9 b \sqrt {c x}}-\frac {4 a c \int \frac {\left (b x^2+a x\right )^{3/2}}{\sqrt {c x}}dx}{9 b}\right )}{11 b}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{5/2}}{11 b}-\frac {6 a c \left (\frac {2 c \left (a x+b x^2\right )^{5/2}}{9 b \sqrt {c x}}-\frac {4 a c \left (\frac {2 c \left (a x+b x^2\right )^{5/2}}{7 b (c x)^{3/2}}-\frac {2 a c \int \frac {\left (b x^2+a x\right )^{3/2}}{(c x)^{3/2}}dx}{7 b}\right )}{9 b}\right )}{11 b}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{5/2}}{11 b}-\frac {6 a c \left (\frac {2 c \left (a x+b x^2\right )^{5/2}}{9 b \sqrt {c x}}-\frac {4 a c \left (\frac {2 c \left (a x+b x^2\right )^{5/2}}{7 b (c x)^{3/2}}-\frac {4 a c^2 \left (a x+b x^2\right )^{5/2}}{35 b^2 (c x)^{5/2}}\right )}{9 b}\right )}{11 b}\) |
Input:
Int[(c*x)^(3/2)*(a*x + b*x^2)^(3/2),x]
Output:
(2*c*Sqrt[c*x]*(a*x + b*x^2)^(5/2))/(11*b) - (6*a*c*((2*c*(a*x + b*x^2)^(5 /2))/(9*b*Sqrt[c*x]) - (4*a*c*((-4*a*c^2*(a*x + b*x^2)^(5/2))/(35*b^2*(c*x )^(5/2)) + (2*c*(a*x + b*x^2)^(5/2))/(7*b*(c*x)^(3/2))))/(9*b)))/(11*b)
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[m + p], 0]
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.47
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (c x \right )^{\frac {3}{2}} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\) | \(60\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (c x \right )^{\frac {3}{2}} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\) | \(60\) |
default | \(-\frac {2 c \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, \left (b x +a \right )^{2} \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right )}{1155 x \,b^{4}}\) | \(61\) |
risch | \(-\frac {2 c^{2} x \left (-105 b^{5} x^{5}-140 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}+6 a^{3} b^{2} x^{2}-8 a^{4} b x +16 a^{5}\right ) \left (b x +a \right )}{1155 \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, b^{4}}\) | \(81\) |
Input:
int((c*x)^(3/2)*(b*x^2+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/1155*(b*x+a)*(-105*b^3*x^3+70*a*b^2*x^2-40*a^2*b*x+16*a^3)*(c*x)^(3/2)* (b*x^2+a*x)^(3/2)/b^4/x^3
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.64 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} c x^{5} + 140 \, a b^{4} c x^{4} + 5 \, a^{2} b^{3} c x^{3} - 6 \, a^{3} b^{2} c x^{2} + 8 \, a^{4} b c x - 16 \, a^{5} c\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{1155 \, b^{4} x} \] Input:
integrate((c*x)^(3/2)*(b*x^2+a*x)^(3/2),x, algorithm="fricas")
Output:
2/1155*(105*b^5*c*x^5 + 140*a*b^4*c*x^4 + 5*a^2*b^3*c*x^3 - 6*a^3*b^2*c*x^ 2 + 8*a^4*b*c*x - 16*a^5*c)*sqrt(b*x^2 + a*x)*sqrt(c*x)/(b^4*x)
\[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\int \left (c x\right )^{\frac {3}{2}} \left (x \left (a + b x\right )\right )^{\frac {3}{2}}\, dx \] Input:
integrate((c*x)**(3/2)*(b*x**2+a*x)**(3/2),x)
Output:
Integral((c*x)**(3/2)*(x*(a + b*x))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.23 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\frac {2 \, {\left ({\left (315 \, b^{5} c^{\frac {3}{2}} x^{5} + 35 \, a b^{4} c^{\frac {3}{2}} x^{4} - 40 \, a^{2} b^{3} c^{\frac {3}{2}} x^{3} + 48 \, a^{3} b^{2} c^{\frac {3}{2}} x^{2} - 64 \, a^{4} b c^{\frac {3}{2}} x + 128 \, a^{5} c^{\frac {3}{2}}\right )} x^{4} + 11 \, {\left (35 \, a b^{4} c^{\frac {3}{2}} x^{5} + 5 \, a^{2} b^{3} c^{\frac {3}{2}} x^{4} - 6 \, a^{3} b^{2} c^{\frac {3}{2}} x^{3} + 8 \, a^{4} b c^{\frac {3}{2}} x^{2} - 16 \, a^{5} c^{\frac {3}{2}} x\right )} x^{3}\right )} \sqrt {b x + a}}{3465 \, b^{4} x^{4}} \] Input:
integrate((c*x)^(3/2)*(b*x^2+a*x)^(3/2),x, algorithm="maxima")
Output:
2/3465*((315*b^5*c^(3/2)*x^5 + 35*a*b^4*c^(3/2)*x^4 - 40*a^2*b^3*c^(3/2)*x ^3 + 48*a^3*b^2*c^(3/2)*x^2 - 64*a^4*b*c^(3/2)*x + 128*a^5*c^(3/2))*x^4 + 11*(35*a*b^4*c^(3/2)*x^5 + 5*a^2*b^3*c^(3/2)*x^4 - 6*a^3*b^2*c^(3/2)*x^3 + 8*a^4*b*c^(3/2)*x^2 - 16*a^5*c^(3/2)*x)*x^3)*sqrt(b*x + a)/(b^4*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (104) = 208\).
Time = 0.33 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.63 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\frac {2}{3465} \, c {\left (\frac {11 \, {\left (\frac {16 \, \sqrt {a c} a^{4} c}{b^{4}} - \frac {105 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a^{3} c^{3} - 189 \, {\left (b c x + a c\right )}^{\frac {5}{2}} a^{2} c^{2} + 135 \, {\left (b c x + a c\right )}^{\frac {7}{2}} a c - 35 \, {\left (b c x + a c\right )}^{\frac {9}{2}}}{b^{4} c^{3}}\right )} a {\left | c \right |}}{c^{2}} - \frac {{\left (\frac {128 \, \sqrt {a c} a^{5} c^{5}}{b^{5}} - \frac {1155 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a^{4} c^{4} - 2772 \, {\left (b c x + a c\right )}^{\frac {5}{2}} a^{3} c^{3} + 2970 \, {\left (b c x + a c\right )}^{\frac {7}{2}} a^{2} c^{2} - 1540 \, {\left (b c x + a c\right )}^{\frac {9}{2}} a c + 315 \, {\left (b c x + a c\right )}^{\frac {11}{2}}}{b^{5}}\right )} b {\left | c \right |}}{c^{6}}\right )} \] Input:
integrate((c*x)^(3/2)*(b*x^2+a*x)^(3/2),x, algorithm="giac")
Output:
2/3465*c*(11*(16*sqrt(a*c)*a^4*c/b^4 - (105*(b*c*x + a*c)^(3/2)*a^3*c^3 - 189*(b*c*x + a*c)^(5/2)*a^2*c^2 + 135*(b*c*x + a*c)^(7/2)*a*c - 35*(b*c*x + a*c)^(9/2))/(b^4*c^3))*a*abs(c)/c^2 - (128*sqrt(a*c)*a^5*c^5/b^5 - (1155 *(b*c*x + a*c)^(3/2)*a^4*c^4 - 2772*(b*c*x + a*c)^(5/2)*a^3*c^3 + 2970*(b* c*x + a*c)^(7/2)*a^2*c^2 - 1540*(b*c*x + a*c)^(9/2)*a*c + 315*(b*c*x + a*c )^(11/2))/b^5)*b*abs(c)/c^6)
Time = 9.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.48 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=-\frac {2\,c\,\sqrt {b\,x^2+a\,x}\,\sqrt {c\,x}\,{\left (a+b\,x\right )}^2\,\left (16\,a^3-40\,a^2\,b\,x+70\,a\,b^2\,x^2-105\,b^3\,x^3\right )}{1155\,b^4\,x} \] Input:
int((a*x + b*x^2)^(3/2)*(c*x)^(3/2),x)
Output:
-(2*c*(a*x + b*x^2)^(1/2)*(c*x)^(1/2)*(a + b*x)^2*(16*a^3 - 105*b^3*x^3 + 70*a*b^2*x^2 - 40*a^2*b*x))/(1155*b^4*x)
Time = 0.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int (c x)^{3/2} \left (a x+b x^2\right )^{3/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {b x +a}\, c \left (105 b^{5} x^{5}+140 a \,b^{4} x^{4}+5 a^{2} b^{3} x^{3}-6 a^{3} b^{2} x^{2}+8 a^{4} b x -16 a^{5}\right )}{1155 b^{4}} \] Input:
int((c*x)^(3/2)*(b*x^2+a*x)^(3/2),x)
Output:
(2*sqrt(c)*sqrt(a + b*x)*c*( - 16*a**5 + 8*a**4*b*x - 6*a**3*b**2*x**2 + 5 *a**2*b**3*x**3 + 140*a*b**4*x**4 + 105*b**5*x**5))/(1155*b**4)