Integrand size = 21, antiderivative size = 128 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=-\frac {2 a^3 c^4 \left (a x+b x^2\right )^{3/2}}{3 b^4 (c x)^{3/2}}+\frac {6 a^2 c^5 \left (a x+b x^2\right )^{5/2}}{5 b^4 (c x)^{5/2}}-\frac {6 a c^6 \left (a x+b x^2\right )^{7/2}}{7 b^4 (c x)^{7/2}}+\frac {2 c^7 \left (a x+b x^2\right )^{9/2}}{9 b^4 (c x)^{9/2}} \] Output:
-2/3*a^3*c^4*(b*x^2+a*x)^(3/2)/b^4/(c*x)^(3/2)+6/5*a^2*c^5*(b*x^2+a*x)^(5/ 2)/b^4/(c*x)^(5/2)-6/7*a*c^6*(b*x^2+a*x)^(7/2)/b^4/(c*x)^(7/2)+2/9*c^7*(b* x^2+a*x)^(9/2)/b^4/(c*x)^(9/2)
Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.45 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2 c^4 (x (a+b x))^{3/2} \left (-16 a^3+24 a^2 b x-30 a b^2 x^2+35 b^3 x^3\right )}{315 b^4 (c x)^{3/2}} \] Input:
Integrate[(c*x)^(5/2)*Sqrt[a*x + b*x^2],x]
Output:
(2*c^4*(x*(a + b*x))^(3/2)*(-16*a^3 + 24*a^2*b*x - 30*a*b^2*x^2 + 35*b^3*x ^3))/(315*b^4*(c*x)^(3/2))
Time = 0.43 (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.
\(\displaystyle \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}{9 b}-\frac {2 a c \int (c x)^{3/2} \sqrt {b x^2+a x}dx}{3 b}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}{9 b}-\frac {2 a c \left (\frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{3/2}}{7 b}-\frac {4 a c \int \sqrt {c x} \sqrt {b x^2+a x}dx}{7 b}\right )}{3 b}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 c (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}{9 b}-\frac {2 a c \left (\frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{3/2}}{7 b}-\frac {4 a c \left (\frac {2 c \left (a x+b x^2\right )^{3/2}}{5 b \sqrt {c x}}-\frac {2 a c \int \frac {\sqrt {b x^2+a x}}{\sqrt {c x}}dx}{5 b}\right )}{7 b}\right )}{3 b}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 c (c x)^{3/2} \left (a x+b x^2\right )^{3/2}}{9 b}-\frac {2 a c \left (\frac {2 c \sqrt {c x} \left (a x+b x^2\right )^{3/2}}{7 b}-\frac {4 a c \left (\frac {2 c \left (a x+b x^2\right )^{3/2}}{5 b \sqrt {c x}}-\frac {4 a c^2 \left (a x+b x^2\right )^{3/2}}{15 b^2 (c x)^{3/2}}\right )}{7 b}\right )}{3 b}\) |
Input:
Int[(c*x)^(5/2)*Sqrt[a*x + b*x^2],x]
Output:
(2*c*(c*x)^(3/2)*(a*x + b*x^2)^(3/2))/(9*b) - (2*a*c*((2*c*Sqrt[c*x]*(a*x + b*x^2)^(3/2))/(7*b) - (4*a*c*((-4*a*c^2*(a*x + b*x^2)^(3/2))/(15*b^2*(c* x)^(3/2)) + (2*c*(a*x + b*x^2)^(3/2))/(5*b*Sqrt[c*x])))/(7*b)))/(3*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 = 0.36 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.47
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \left (c x \right )^{\frac {5}{2}} \sqrt {b \,x^{2}+a x}}{315 b^{4} x^{3}}\) | \(60\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \left (c x \right )^{\frac {5}{2}} \sqrt {b \,x^{2}+a x}}{315 b^{4} x^{3}}\) | \(60\) |
default | \(-\frac {2 c^{2} \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right )}{315 x \,b^{4}}\) | \(61\) |
risch | \(-\frac {2 c^{3} x \left (-35 b^{4} x^{4}-5 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}-8 a^{3} b x +16 a^{4}\right ) \left (b x +a \right )}{315 \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, b^{4}}\) | \(70\) |
Input:
int((c*x)^(5/2)*(b*x^2+a*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/315*(b*x+a)*(-35*b^3*x^3+30*a*b^2*x^2-24*a^2*b*x+16*a^3)*(c*x)^(5/2)*(b *x^2+a*x)^(1/2)/b^4/x^3
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2 \, {\left (35 \, b^{4} c^{2} x^{4} + 5 \, a b^{3} c^{2} x^{3} - 6 \, a^{2} b^{2} c^{2} x^{2} + 8 \, a^{3} b c^{2} x - 16 \, a^{4} c^{2}\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{315 \, b^{4} x} \] Input:
integrate((c*x)^(5/2)*(b*x^2+a*x)^(1/2),x, algorithm="fricas")
Output:
2/315*(35*b^4*c^2*x^4 + 5*a*b^3*c^2*x^3 - 6*a^2*b^2*c^2*x^2 + 8*a^3*b*c^2* x - 16*a^4*c^2)*sqrt(b*x^2 + a*x)*sqrt(c*x)/(b^4*x)
\[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\int \left (c x\right )^{\frac {5}{2}} \sqrt {x \left (a + b x\right )}\, dx \] Input:
integrate((c*x)**(5/2)*(b*x**2+a*x)**(1/2),x)
Output:
Integral((c*x)**(5/2)*sqrt(x*(a + b*x)), x)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.53 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2 \, {\left (35 \, b^{4} c^{\frac {5}{2}} x^{4} + 5 \, a b^{3} c^{\frac {5}{2}} x^{3} - 6 \, a^{2} b^{2} c^{\frac {5}{2}} x^{2} + 8 \, a^{3} b c^{\frac {5}{2}} x - 16 \, a^{4} c^{\frac {5}{2}}\right )} \sqrt {b x + a}}{315 \, b^{4}} \] Input:
integrate((c*x)^(5/2)*(b*x^2+a*x)^(1/2),x, algorithm="maxima")
Output:
2/315*(35*b^4*c^(5/2)*x^4 + 5*a*b^3*c^(5/2)*x^3 - 6*a^2*b^2*c^(5/2)*x^2 + 8*a^3*b*c^(5/2)*x - 16*a^4*c^(5/2))*sqrt(b*x + a)/b^4
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.70 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2}{315} \, {\left (\frac {16 \, \sqrt {a c} a^{4} c}{b^{4}} - \frac {105 \, {\left (b c x + a c\right )}^{\frac {3}{2}} a^{3} c^{3} - 189 \, {\left (b c x + a c\right )}^{\frac {5}{2}} a^{2} c^{2} + 135 \, {\left (b c x + a c\right )}^{\frac {7}{2}} a c - 35 \, {\left (b c x + a c\right )}^{\frac {9}{2}}}{b^{4} c^{3}}\right )} {\left | c \right |} \] Input:
integrate((c*x)^(5/2)*(b*x^2+a*x)^(1/2),x, algorithm="giac")
Output:
2/315*(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))*abs(c)
Time = 9.39 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.53 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2\,c^2\,\sqrt {b\,x^2+a\,x}\,\sqrt {c\,x}\,\left (-16\,a^4+8\,a^3\,b\,x-6\,a^2\,b^2\,x^2+5\,a\,b^3\,x^3+35\,b^4\,x^4\right )}{315\,b^4\,x} \] Input:
int((a*x + b*x^2)^(1/2)*(c*x)^(5/2),x)
Output:
(2*c^2*(a*x + b*x^2)^(1/2)*(c*x)^(1/2)*(35*b^4*x^4 - 16*a^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x))/(315*b^4*x)
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.45 \[ \int (c x)^{5/2} \sqrt {a x+b x^2} \, dx=\frac {2 \sqrt {c}\, \sqrt {b x +a}\, c^{2} \left (35 b^{4} x^{4}+5 a \,b^{3} x^{3}-6 a^{2} b^{2} x^{2}+8 a^{3} b x -16 a^{4}\right )}{315 b^{4}} \] Input:
int((c*x)^(5/2)*(b*x^2+a*x)^(1/2),x)
Output:
(2*sqrt(c)*sqrt(a + b*x)*c**2*( - 16*a**4 + 8*a**3*b*x - 6*a**2*b**2*x**2 + 5*a*b**3*x**3 + 35*b**4*x**4))/(315*b**4)