Integrand size = 26, antiderivative size = 99 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=-\frac {3}{2} b^2 c x \sqrt {a+b x} \sqrt {a c-b c x}-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}-3 a^2 b c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right ) \] Output:
-3/2*b^2*c*x*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)-(b*x+a)^(3/2)*(-b*c*x+a*c)^( 3/2)/x-3*a^2*b*c^(3/2)*arctan(c^(1/2)*(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2))
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=-\frac {(c (a-b x))^{3/2} \left (\sqrt {a-b x} \sqrt {a+b x} \left (2 a^2+b^2 x^2\right )+6 a^2 b x \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )\right )}{2 x (a-b x)^{3/2}} \] Input:
Integrate[((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/x^2,x]
Output:
-1/2*((c*(a - b*x))^(3/2)*(Sqrt[a - b*x]*Sqrt[a + b*x]*(2*a^2 + b^2*x^2) + 6*a^2*b*x*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]]))/(x*(a - b*x)^(3/2))
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {108, 27, 40, 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.
\(\displaystyle \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 108 |
\(\displaystyle \int -3 b^2 c \sqrt {a+b x} \sqrt {a c-b c x}dx-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -3 b^2 c \int \sqrt {a+b x} \sqrt {a c-b c x}dx-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle -3 b^2 c \left (\frac {1}{2} a^2 c \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}}dx+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle -3 b^2 c \left (a^2 c \int \frac {1}{\frac {c (a+b x) b}{a c-b c x}+b}d\frac {\sqrt {a+b x}}{\sqrt {a c-b c x}}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -3 b^2 c \left (\frac {a^2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x}\) |
Input:
Int[((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/x^2,x]
Output:
-(((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/x) - 3*b^2*c*((x*Sqrt[a + b*x]*Sqr t[a*c - b*c*x])/2 + (a^2*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*c - b*c*x]])/b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
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]
Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30
method | result | size |
risch | \(-\frac {\left (-b x +a \right ) \sqrt {b x +a}\, \left (b^{2} x^{2}+2 a^{2}\right ) c^{2}}{2 x \sqrt {-c \left (b x -a \right )}}-\frac {3 a^{2} b^{2} \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}\, c^{2}}{2 \sqrt {b^{2} c}\, \sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}}\) | \(129\) |
default | \(-\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, c \left (3 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{2} b^{2} c x +\sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, b^{2} x^{2}+2 a^{2} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\right )}{2 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, x \sqrt {b^{2} c}}\) | \(142\) |
Input:
int((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/2*(-b*x+a)*(b*x+a)^(1/2)*(b^2*x^2+2*a^2)/x/(-c*(b*x-a))^(1/2)*c^2-3/2*a ^2*b^2/(b^2*c)^(1/2)*arctan((b^2*c)^(1/2)*x/(-b^2*c*x^2+a^2*c)^(1/2))*(-(b *x+a)*c*(b*x-a))^(1/2)/(b*x+a)^(1/2)/(-c*(b*x-a))^(1/2)*c^2
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.94 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=\left [\frac {3 \, a^{2} b \sqrt {-c} c x \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) - 2 \, {\left (b^{2} c x^{2} + 2 \, a^{2} c\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{4 \, x}, \frac {3 \, a^{2} b c^{\frac {3}{2}} x \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) - {\left (b^{2} c x^{2} + 2 \, a^{2} c\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{2 \, x}\right ] \] Input:
integrate((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^2,x, algorithm="fricas")
Output:
[1/4*(3*a^2*b*sqrt(-c)*c*x*log(2*b^2*c*x^2 - 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c) - 2*(b^2*c*x^2 + 2*a^2*c)*sqrt(-b*c*x + a*c)*s qrt(b*x + a))/x, 1/2*(3*a^2*b*c^(3/2)*x*arctan(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c)) - (b^2*c*x^2 + 2*a^2*c)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/x]
\[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=\int \frac {\left (- c \left (- a + b x\right )\right )^{\frac {3}{2}} \left (a + b x\right )^{\frac {3}{2}}}{x^{2}}\, dx \] Input:
integrate((b*x+a)**(3/2)*(-b*c*x+a*c)**(3/2)/x**2,x)
Output:
Integral((-c*(-a + b*x))**(3/2)*(a + b*x)**(3/2)/x**2, x)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=-\frac {3}{2} \, a^{2} b c^{\frac {3}{2}} \arcsin \left (\frac {b x}{a}\right ) - \frac {3}{2} \, \sqrt {-b^{2} c x^{2} + a^{2} c} b^{2} c x - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}}}{x} \] Input:
integrate((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^2,x, algorithm="maxima")
Output:
-3/2*a^2*b*c^(3/2)*arcsin(b*x/a) - 3/2*sqrt(-b^2*c*x^2 + a^2*c)*b^2*c*x - (-b^2*c*x^2 + a^2*c)^(3/2)/x
Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=-\frac {\frac {32 \, a^{4} b^{2} \sqrt {-c} c^{3}}{{\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} + 4 \, a^{2} c^{2}} + 3 \, a^{2} b^{2} \sqrt {-c} c \log \left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4}\right ) + 2 \, {\left ({\left (b x + a\right )} b^{2} c - a b^{2} c\right )} \sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a}}{4 \, b} \] Input:
integrate((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^2,x, algorithm="giac")
Output:
-1/4*(32*a^4*b^2*sqrt(-c)*c^3/((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 + 4*a^2*c^2) + 3*a^2*b^2*sqrt(-c)*c*log((sqrt(b*x + a)*sqrt(- c) - sqrt(-(b*x + a)*c + 2*a*c))^4) + 2*((b*x + a)*b^2*c - a*b^2*c)*sqrt(- (b*x + a)*c + 2*a*c)*sqrt(b*x + a))/b
Timed out. \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=\int \frac {{\left (a\,c-b\,c\,x\right )}^{3/2}\,{\left (a+b\,x\right )}^{3/2}}{x^2} \,d x \] Input:
int(((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2))/x^2,x)
Output:
int(((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2))/x^2, x)
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^2} \, dx=\frac {\sqrt {c}\, c \left (6 \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{2} b x -2 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{2}-\sqrt {b x +a}\, \sqrt {-b x +a}\, b^{2} x^{2}\right )}{2 x} \] Input:
int((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^2,x)
Output:
(sqrt(c)*c*(6*asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))*a**2*b*x - 2*sqrt(a + b*x)*sqrt(a - b*x)*a**2 - sqrt(a + b*x)*sqrt(a - b*x)*b**2*x**2))/(2*x)