Integrand size = 31, antiderivative size = 65 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {-c+d x} \sqrt {c+d x}}-\frac {a \arctan \left (\frac {\sqrt {-c+d x} \sqrt {c+d x}}{c}\right )}{c^3} \] Output:
-(a/c^2+b/d^2)/(d*x-c)^(1/2)/(d*x+c)^(1/2)-a*arctan((d*x-c)^(1/2)*(d*x+c)^ (1/2)/c)/c^3
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {-\frac {2 \left (b c^3+a c d^2\right )}{d^2 \sqrt {-c+d x} \sqrt {c+d x}}+4 a \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{2 c^3} \] Input:
Integrate[(a + b*x^2)/(x*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
Output:
((-2*(b*c^3 + a*c*d^2))/(d^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) + 4*a*ArcTan[Sq rt[c + d*x]/Sqrt[-c + d*x]])/(2*c^3)
Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {958, 103, 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.
\(\displaystyle \int \frac {a+b x^2}{x (d x-c)^{3/2} (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 958 |
\(\displaystyle -\frac {a \int \frac {1}{x \sqrt {d x-c} \sqrt {c+d x}}dx}{c^2}-\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}}\) |
\(\Big \downarrow \) 103 |
\(\displaystyle -\frac {a d \int \frac {1}{d c^2+d (d x-c) (c+d x)}d\left (\sqrt {d x-c} \sqrt {c+d x}\right )}{c^2}-\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {a \arctan \left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{c^3}-\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}}\) |
Input:
Int[(a + b*x^2)/(x*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
Output:
-((a/c^2 + b/d^2)/(Sqrt[-c + d*x]*Sqrt[c + d*x])) - (a*ArcTan[(Sqrt[-c + d *x]*Sqrt[c + d*x])/c])/c^3
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ ))), x_] :> Simp[b*f Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d *e - f*(b*c + a*d), 0]
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]
Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) *(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b1*b2* c - a1*a2*d))*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(a1*a2*b1*b2*e*n*(p + 1))), x] - Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*b1*b2*n*(p + 1)) Int[(e*x)^m*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1), x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && LtQ[p, -1] && (( !I ntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[ p + 1/2, 0] && LeQ[-1, m, (-n)*(p + 1)]))
Leaf count of result is larger than twice the leaf count of optimal. \(187\) vs. \(2(57)=114\).
Time = 0.13 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.89
method | result | size |
default | \(\frac {\ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,d^{4} x^{2}-\ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,c^{2} d^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2}-b \,c^{2} \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}}{\sqrt {d x -c}\, \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, c^{2} \sqrt {-c^{2}}\, d^{2}}\) | \(188\) |
Input:
int((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
(ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*a*d^4*x^2-ln(-2*(c^2-(-c^ 2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*a*c^2*d^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2 )*a*d^2-b*c^2*(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/(d*x-c)^(1/2)/(d^2*x^2-c^2 )^(1/2)/(d*x+c)^(1/2)/c^2/(-c^2)^(1/2)/d^2
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.55 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {{\left (b c^{3} + a c d^{2}\right )} \sqrt {d x + c} \sqrt {d x - c} + 2 \, {\left (a d^{4} x^{2} - a c^{2} d^{2}\right )} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right )}{c^{3} d^{4} x^{2} - c^{5} d^{2}} \] Input:
integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
-((b*c^3 + a*c*d^2)*sqrt(d*x + c)*sqrt(d*x - c) + 2*(a*d^4*x^2 - a*c^2*d^2 )*arctan(-(d*x - sqrt(d*x + c)*sqrt(d*x - c))/c))/(c^3*d^4*x^2 - c^5*d^2)
Result contains complex when optimal does not.
Time = 177.65 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.65 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=a \left (- \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & 1, 2, \frac {5}{2} \\\frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2, \frac {5}{2} & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{3}} - \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, 1 & \\\frac {3}{4}, \frac {5}{4} & 0, \frac {1}{2}, \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{3}}\right ) + b \left (- \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4}, 1 & 0, 1, \frac {3}{2} \\\frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c d^{2}} - \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, 1 & \\- \frac {1}{4}, \frac {1}{4} & -1, - \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c d^{2}}\right ) \] Input:
integrate((b*x**2+a)/x/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)
Output:
a*(-meijerg(((5/4, 7/4, 1), (1, 2, 5/2)), ((5/4, 3/2, 7/4, 2, 5/2), (0,)), c**2/(d**2*x**2))/(2*pi**(3/2)*c**3) - I*meijerg(((0, 1/2, 3/4, 1, 5/4, 1 ), ()), ((3/4, 5/4), (0, 1/2, 3/2, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2) )/(2*pi**(3/2)*c**3)) + b*(-meijerg(((1/4, 3/4, 1), (0, 1, 3/2)), ((1/4, 1 /2, 3/4, 1, 3/2), (0,)), c**2/(d**2*x**2))/(2*pi**(3/2)*c*d**2) - I*meijer g(((-1, -1/2, -1/4, 0, 1/4, 1), ()), ((-1/4, 1/4), (-1, -1/2, 1/2, 0)), c* *2*exp_polar(2*I*pi)/(d**2*x**2))/(2*pi**(3/2)*c*d**2))
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {a \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{c^{3}} - \frac {a}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} d^{2}} \] Input:
integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")
Output:
a*arcsin(c/(d*abs(x)))/c^3 - a/(sqrt(d^2*x^2 - c^2)*c^2) - b/(sqrt(d^2*x^2 - c^2)*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (57) = 114\).
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.77 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {2 \, a \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right )}{c^{3}} - \frac {{\left (b c^{2} + a d^{2}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{3} d^{2}} + \frac {2 \, {\left (b c^{2} + a d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c^{2} d^{2}} \] Input:
integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
2*a*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c)/c^3 - 1/2*(b*c^2 + a*d ^2)*sqrt(d*x + c)/(sqrt(d*x - c)*c^3*d^2) + 2*(b*c^2 + a*d^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*c)*c^2*d^2)
Timed out. \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {b\,x^2+a}{x\,{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \] Input:
int((a + b*x^2)/(x*(c + d*x)^(3/2)*(d*x - c)^(3/2)),x)
Output:
int((a + b*x^2)/(x*(c + d*x)^(3/2)*(d*x - c)^(3/2)), x)
Time = 0.23 (sec) , antiderivative size = 206, normalized size of antiderivative = 3.17 \[ \int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {-2 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) a c \,d^{2}-2 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) a \,d^{3} x +2 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) a c \,d^{2}+2 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) a \,d^{3} x -\sqrt {d x +c}\, a c \,d^{2}-\sqrt {d x +c}\, b \,c^{3}}{\sqrt {d x -c}\, c^{3} d^{2} \left (d x +c \right )} \] Input:
int((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)
Output:
( - 2*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt(c + d*x) - sqrt(c))/s qrt(c))*a*c*d**2 - 2*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt(c + d* x) - sqrt(c))/sqrt(c))*a*d**3*x + 2*sqrt( - c + d*x)*atan((sqrt( - c + d*x ) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*a*c*d**2 + 2*sqrt( - c + d*x)*atan(( sqrt( - c + d*x) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*a*d**3*x - sqrt(c + d *x)*a*c*d**2 - sqrt(c + d*x)*b*c**3)/(sqrt( - c + d*x)*c**3*d**2*(c + d*x) )