\(\int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\) [39]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 b c^2+3 a d^2}{2 c^4 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {a}{2 c^2 x^2 \sqrt {-c+d x} \sqrt {c+d x}}-\frac {\left (2 b c^2+3 a d^2\right ) \arctan \left (\frac {\sqrt {-c+d x} \sqrt {c+d x}}{c}\right )}{2 c^5} \] Output:

-1/2*(3*a*d^2+2*b*c^2)/c^4/(d*x-c)^(1/2)/(d*x+c)^(1/2)+1/2*a/c^2/x^2/(d*x- 
c)^(1/2)/(d*x+c)^(1/2)-1/2*(3*a*d^2+2*b*c^2)*arctan((d*x-c)^(1/2)*(d*x+c)^ 
(1/2)/c)/c^5
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {\frac {-2 b c^3 x^2+a \left (c^3-3 c d^2 x^2\right )}{x^2 \sqrt {-c+d x} \sqrt {c+d x}}+\left (4 b c^2+6 a d^2\right ) \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{2 c^5} \] Input:

Integrate[(a + b*x^2)/(x^3*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

((-2*b*c^3*x^2 + a*(c^3 - 3*c*d^2*x^2))/(x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) 
 + (4*b*c^2 + 6*a*d^2)*ArcTan[Sqrt[c + d*x]/Sqrt[-c + d*x]])/(2*c^5)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {956, 115, 27, 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.

\(\displaystyle \int \frac {a+b x^2}{x^3 (d x-c)^{3/2} (c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 956

\(\displaystyle \frac {1}{2} \left (\frac {3 a d^2}{c^2}+2 b\right ) \int \frac {1}{x (d x-c)^{3/2} (c+d x)^{3/2}}dx+\frac {a}{2 c^2 x^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 115

\(\displaystyle \frac {1}{2} \left (\frac {3 a d^2}{c^2}+2 b\right ) \left (-\frac {\int \frac {d}{x \sqrt {d x-c} \sqrt {c+d x}}dx}{c^2 d}-\frac {1}{c^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {a}{2 c^2 x^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {3 a d^2}{c^2}+2 b\right ) \left (-\frac {\int \frac {1}{x \sqrt {d x-c} \sqrt {c+d x}}dx}{c^2}-\frac {1}{c^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {a}{2 c^2 x^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 103

\(\displaystyle \frac {1}{2} \left (\frac {3 a d^2}{c^2}+2 b\right ) \left (-\frac {d \int \frac {1}{d c^2+d (d x-c) (c+d x)}d\left (\sqrt {d x-c} \sqrt {c+d x}\right )}{c^2}-\frac {1}{c^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {a}{2 c^2 x^2 \sqrt {d x-c} \sqrt {c+d x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{2} \left (\frac {3 a d^2}{c^2}+2 b\right ) \left (-\frac {\arctan \left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{c^3}-\frac {1}{c^2 \sqrt {d x-c} \sqrt {c+d x}}\right )+\frac {a}{2 c^2 x^2 \sqrt {d x-c} \sqrt {c+d x}}\)

Input:

Int[(a + b*x^2)/(x^3*(-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

a/(2*c^2*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) + ((2*b + (3*a*d^2)/c^2)*(-(1/( 
c^2*Sqrt[-c + d*x]*Sqrt[c + d*x])) - ArcTan[(Sqrt[-c + d*x]*Sqrt[c + d*x]) 
/c]/c^3))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 103

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]
 

rule 115

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 
*n, 2*p]
 

rule 218

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]
 

rule 956

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) 
*(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^( 
m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(a1*a2*e*(m + 1 
))), x] + Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*e^n*( 
m + 1))   Int[(e*x)^(m + n)*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] 
 /; FreeQ[{a1, b1, a2, b2, c, d, e, p}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + 
 a1*b2, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || ( 
LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(238\) vs. \(2(99)=198\).

Time = 0.13 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.04

method	result	size
risch	\(\frac {a \left (-d x +c \right ) \sqrt {d x +c}}{2 c^{4} x^{2} \sqrt {d x -c}}-\frac {\left (-\frac {\left (3 a \,d^{2}+2 b \,c^{2}\right ) \ln \left (\frac {-2 c^{2}+2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}}{x}\right )}{\sqrt {-c^{2}}}+\frac {\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}}{d c \left (x -\frac {c}{d}\right )}-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}}{d c \left (x +\frac {c}{d}\right )}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{2 c^{4} \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(239\)
default	\(-\frac {-3 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,d^{4} x^{4}-2 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) b \,c^{2} d^{2} x^{4}+3 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,c^{2} d^{2} x^{2}+2 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) b \,c^{4} x^{2}+3 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2} x^{2}+2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2} x^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,c^{2}}{2 \sqrt {d x -c}\, \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, c^{4} \sqrt {-c^{2}}\, x^{2}}\)	\(316\)

Input:

int((b*x^2+a)/x^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*a*(-d*x+c)*(d*x+c)^(1/2)/c^4/x^2/(d*x-c)^(1/2)-1/2/c^4*(-(3*a*d^2+2*b* 
c^2)/(-c^2)^(1/2)*ln((-2*c^2+2*(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)+(a*d^2 
+b*c^2)/d/c/(x-c/d)*(d^2*(x-c/d)^2+2*c*d*(x-c/d))^(1/2)-(a*d^2+b*c^2)/d/c/ 
(x+c/d)*(d^2*(x+c/d)^2-2*c*d*(x+c/d))^(1/2))*((d*x-c)*(d*x+c))^(1/2)/(d*x- 
c)^(1/2)/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.18 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {{\left (a c^{3} - {\left (2 \, b c^{3} + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c} - 2 \, {\left ({\left (2 \, b c^{2} d^{2} + 3 \, a d^{4}\right )} x^{4} - {\left (2 \, b c^{4} + 3 \, a c^{2} d^{2}\right )} x^{2}\right )} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right )}{2 \, {\left (c^{5} d^{2} x^{4} - c^{7} x^{2}\right )}} \] Input:

integrate((b*x^2+a)/x^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

1/2*((a*c^3 - (2*b*c^3 + 3*a*c*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c) - 2*( 
(2*b*c^2*d^2 + 3*a*d^4)*x^4 - (2*b*c^4 + 3*a*c^2*d^2)*x^2)*arctan(-(d*x - 
sqrt(d*x + c)*sqrt(d*x - c))/c))/(c^5*d^2*x^4 - c^7*x^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)/x**3/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {b \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{c^{3}} + \frac {3 \, a d^{2} \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{2 \, c^{5}} - \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {3 \, a d^{2}}{2 \, \sqrt {d^{2} x^{2} - c^{2}} c^{4}} + \frac {a}{2 \, \sqrt {d^{2} x^{2} - c^{2}} c^{2} x^{2}} \] Input:

integrate((b*x^2+a)/x^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

b*arcsin(c/(d*abs(x)))/c^3 + 3/2*a*d^2*arcsin(c/(d*abs(x)))/c^5 - b/(sqrt( 
d^2*x^2 - c^2)*c^2) - 3/2*a*d^2/(sqrt(d^2*x^2 - c^2)*c^4) + 1/2*a/(sqrt(d^ 
2*x^2 - c^2)*c^2*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (99) = 198\).

Time = 0.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.80 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {{\left (2 \, b c^{2} + 3 \, a d^{2}\right )} \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right )}{c^{5}} - \frac {{\left (b c^{2} + a d^{2}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{5}} + \frac {2 \, {\left (b c^{2} + a d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c^{4}} + \frac {2 \, {\left (a d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{6} - 4 \, a c^{2} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{2} c^{4}} \] Input:

integrate((b*x^2+a)/x^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

(2*b*c^2 + 3*a*d^2)*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c)/c^5 - 
1/2*(b*c^2 + a*d^2)*sqrt(d*x + c)/(sqrt(d*x - c)*c^5) + 2*(b*c^2 + a*d^2)/ 
(((sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*c)*c^4) + 2*(a*d^2*(sqrt(d*x + c) 
- sqrt(d*x - c))^6 - 4*a*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^2)/(((sqr 
t(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^2*c^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {b\,x^2+a}{x^3\,{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \] Input:

int((a + b*x^2)/(x^3*(c + d*x)^(3/2)*(d*x - c)^(3/2)),x)
 

Output:

int((a + b*x^2)/(x^3*(c + d*x)^(3/2)*(d*x - c)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.42 \[ \int \frac {a+b x^2}{x^3 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {-6 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) a c \,d^{2} x^{2}-6 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) a \,d^{3} x^{3}-4 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) b \,c^{3} x^{2}-4 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) b \,c^{2} d \,x^{3}+6 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) a c \,d^{2} x^{2}+6 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) a \,d^{3} x^{3}+4 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) b \,c^{3} x^{2}+4 \sqrt {d x -c}\, \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) b \,c^{2} d \,x^{3}+\sqrt {d x +c}\, a \,c^{3}-3 \sqrt {d x +c}\, a c \,d^{2} x^{2}-2 \sqrt {d x +c}\, b \,c^{3} x^{2}}{2 \sqrt {d x -c}\, c^{5} x^{2} \left (d x +c \right )} \] Input:

int((b*x^2+a)/x^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)
 

Output:

( - 6*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt(c + d*x) - sqrt(c))/s 
qrt(c))*a*c*d**2*x**2 - 6*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt(c 
 + d*x) - sqrt(c))/sqrt(c))*a*d**3*x**3 - 4*sqrt( - c + d*x)*atan((sqrt( - 
 c + d*x) + sqrt(c + d*x) - sqrt(c))/sqrt(c))*b*c**3*x**2 - 4*sqrt( - c + 
d*x)*atan((sqrt( - c + d*x) + sqrt(c + d*x) - sqrt(c))/sqrt(c))*b*c**2*d*x 
**3 + 6*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt(c + d*x) + sqrt(c)) 
/sqrt(c))*a*c*d**2*x**2 + 6*sqrt( - c + d*x)*atan((sqrt( - c + d*x) + sqrt 
(c + d*x) + sqrt(c))/sqrt(c))*a*d**3*x**3 + 4*sqrt( - c + d*x)*atan((sqrt( 
 - c + d*x) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*b*c**3*x**2 + 4*sqrt( - c 
+ d*x)*atan((sqrt( - c + d*x) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*b*c**2*d 
*x**3 + sqrt(c + d*x)*a*c**3 - 3*sqrt(c + d*x)*a*c*d**2*x**2 - 2*sqrt(c + 
d*x)*b*c**3*x**2)/(2*sqrt( - c + d*x)*c**5*x**2*(c + d*x))