Integrand size = 26, antiderivative size = 70 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {x}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^3 c^{3/2}} \] Output:
x/b^2/c/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2)-2*arctan(c^(1/2)*(b*x+a)^(1/2)/(- b*c*x+a*c)^(1/2))/b^3/c^(3/2)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {b x-2 \sqrt {a-b x} \sqrt {a+b x} \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{b^3 c \sqrt {c (a-b x)} \sqrt {a+b x}} \] Input:
Integrate[x^2/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]
Output:
(b*x - 2*Sqrt[a - b*x]*Sqrt[a + b*x]*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]])/ (b^3*c*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 27, 87, 45, 218}
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.
\(\displaystyle \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\int \frac {a b^2 c x}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{a b^3 c}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {\sqrt {a+b x}}{b^2 c \sqrt {a c-b c x}}-\frac {\int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}}dx}{b c}}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \frac {\frac {\sqrt {a+b x}}{b^2 c \sqrt {a c-b c x}}-\frac {2 \int \frac {1}{\frac {c (a+b x) b}{a c-b c x}+b}d\frac {\sqrt {a+b x}}{\sqrt {a c-b c x}}}{b c}}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\sqrt {a+b x}}{b^2 c \sqrt {a c-b c x}}-\frac {2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^2 c^{3/2}}}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}\) |
Input:
Int[x^2/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]
Output:
-(a/(b^3*c*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])) + (Sqrt[a + b*x]/(b^2*c*Sqrt[ a*c - b*c*x]) - (2*ArcTan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*c - b*c*x]])/(b^2 *c^(3/2)))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(145\) vs. \(2(58)=116\).
Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.09
method | result | size |
default | \(\frac {\left (\arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {\left (b x +a \right ) c \left (-b x +a \right )}}\right ) b^{2} c \,x^{2}-\arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {\left (b x +a \right ) c \left (-b x +a \right )}}\right ) a^{2} c +x \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}\right ) \sqrt {c \left (-b x +a \right )}}{c^{2} \sqrt {b^{2} c}\, \left (-b x +a \right ) \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, b^{2} \sqrt {b x +a}}\) | \(146\) |
Input:
int(x^2/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x,method=_RETURNVERBOSE)
Output:
(arctan((b^2*c)^(1/2)*x/((b*x+a)*c*(-b*x+a))^(1/2))*b^2*c*x^2-arctan((b^2* c)^(1/2)*x/((b*x+a)*c*(-b*x+a))^(1/2))*a^2*c+x*(c*(-b^2*x^2+a^2))^(1/2)*(b ^2*c)^(1/2))*(c*(-b*x+a))^(1/2)/c^2/(b^2*c)^(1/2)/(-b*x+a)/(c*(-b^2*x^2+a^ 2))^(1/2)/b^2/(b*x+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.16 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\left [-\frac {2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b x + {\left (b^{2} x^{2} - a^{2}\right )} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right )}{2 \, {\left (b^{5} c^{2} x^{2} - a^{2} b^{3} c^{2}\right )}}, -\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b x - {\left (b^{2} x^{2} - a^{2}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right )}{b^{5} c^{2} x^{2} - a^{2} b^{3} c^{2}}\right ] \] Input:
integrate(x^2/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="fricas")
Output:
[-1/2*(2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*x + (b^2*x^2 - a^2)*sqrt(-c)*l og(2*b^2*c*x^2 + 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c)) /(b^5*c^2*x^2 - a^2*b^3*c^2), -(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*x - (b^ 2*x^2 - a^2)*sqrt(c)*arctan(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/( b^2*c*x^2 - a^2*c)))/(b^5*c^2*x^2 - a^2*b^3*c^2)]
Result contains complex when optimal does not.
Time = 82.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.53 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {i {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, \frac {1}{2}, 1, 1 \\- \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1, 0 & \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} b^{3} c^{\frac {3}{2}}} + \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 1 & \\- \frac {3}{4}, - \frac {1}{4} & - \frac {3}{2}, -1, 0, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} b^{3} c^{\frac {3}{2}}} \] Input:
integrate(x**2/(b*x+a)**(3/2)/(-b*c*x+a*c)**(3/2),x)
Output:
I*meijerg(((-1/4, 1/4), (-1/2, 1/2, 1, 1)), ((-1/4, 0, 1/4, 1/2, 1, 0), () ), a**2/(b**2*x**2))/(2*pi**(3/2)*b**3*c**(3/2)) + meijerg(((-3/2, -1, -3/ 4, -1/2, -1/4, 1), ()), ((-3/4, -1/4), (-3/2, -1, 0, 0)), a**2*exp_polar(- 2*I*pi)/(b**2*x**2))/(2*pi**(3/2)*b**3*c**(3/2))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.59 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {x}{\sqrt {-b^{2} c x^{2} + a^{2} c} b^{2} c} - \frac {\arcsin \left (\frac {b x}{a}\right )}{b^{3} c^{\frac {3}{2}}} \] Input:
integrate(x^2/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="maxima")
Output:
x/(sqrt(-b^2*c*x^2 + a^2*c)*b^2*c) - arcsin(b*x/a)/(b^3*c^(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (58) = 116\).
Time = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.10 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=-\frac {\frac {4 \, a}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} - 2 \, a c\right )} b^{2} \sqrt {-c}} - \frac {2 \, \log \left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}\right )}{b^{2} \sqrt {-c} c} + \frac {\sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} c - 2 \, a c\right )} b^{2} c}}{2 \, b} \] Input:
integrate(x^2/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="giac")
Output:
-1/2*(4*a/(((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^2 - 2*a* c)*b^2*sqrt(-c)) - 2*log((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a *c))^2)/(b^2*sqrt(-c)*c) + sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)/(((b*x + a)*c - 2*a*c)*b^2*c))/b
Timed out. \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\int \frac {x^2}{{\left (a\,c-b\,c\,x\right )}^{3/2}\,{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:
int(x^2/((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2)),x)
Output:
int(x^2/((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2)), x)
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {\sqrt {c}\, \left (2 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )+b x \right )}{\sqrt {b x +a}\, \sqrt {-b x +a}\, b^{3} c^{2}} \] Input:
int(x^2/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x)
Output:
(sqrt(c)*(2*sqrt(a + b*x)*sqrt(a - b*x)*asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2 ))) + b*x))/(sqrt(a + b*x)*sqrt(a - b*x)*b**3*c**2)