\(\int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx\) [31]

Optimal result

Integrand size = 31, antiderivative size = 123 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {a \sqrt {-c+d x} \sqrt {c+d x}}{4 c^2 x^4}+\frac {\left (4 b c^2+3 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x}}{8 c^4 x^2}+\frac {d^2 \left (4 b c^2+3 a d^2\right ) \arctan \left (\frac {\sqrt {-c+d x} \sqrt {c+d x}}{c}\right )}{8 c^5} \] Output:

1/4*a*(d*x-c)^(1/2)*(d*x+c)^(1/2)/c^2/x^4+1/8*(3*a*d^2+4*b*c^2)*(d*x-c)^(1 
/2)*(d*x+c)^(1/2)/c^4/x^2+1/8*d^2*(3*a*d^2+4*b*c^2)*arctan((d*x-c)^(1/2)*( 
d*x+c)^(1/2)/c)/c^5
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {c \sqrt {-c+d x} \sqrt {c+d x} \left (2 a c^2+4 b c^2 x^2+3 a d^2 x^2\right )+2 d^2 \left (4 b c^2+3 a d^2\right ) x^4 \arctan \left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{8 c^5 x^4} \] Input:

Integrate[(a + b*x^2)/(x^5*Sqrt[-c + d*x]*Sqrt[c + d*x]),x]
 

Output:

(c*Sqrt[-c + d*x]*Sqrt[c + d*x]*(2*a*c^2 + 4*b*c^2*x^2 + 3*a*d^2*x^2) + 2* 
d^2*(4*b*c^2 + 3*a*d^2)*x^4*ArcTan[Sqrt[-c + d*x]/Sqrt[c + d*x]])/(8*c^5*x 
^4)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {956, 114, 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.

\(\displaystyle \int \frac {a+b x^2}{x^5 \sqrt {d x-c} \sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 956

\(\displaystyle \frac {1}{4} \left (\frac {3 a d^2}{c^2}+4 b\right ) \int \frac {1}{x^3 \sqrt {d x-c} \sqrt {c+d x}}dx+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{4 c^2 x^4}\)

\(\Big \downarrow \) 114

\(\displaystyle \frac {1}{4} \left (\frac {3 a d^2}{c^2}+4 b\right ) \left (\frac {\int \frac {d^2}{x \sqrt {d x-c} \sqrt {c+d x}}dx}{2 c^2}+\frac {\sqrt {d x-c} \sqrt {c+d x}}{2 c^2 x^2}\right )+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{4 c^2 x^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {3 a d^2}{c^2}+4 b\right ) \left (\frac {d^2 \int \frac {1}{x \sqrt {d x-c} \sqrt {c+d x}}dx}{2 c^2}+\frac {\sqrt {d x-c} \sqrt {c+d x}}{2 c^2 x^2}\right )+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{4 c^2 x^4}\)

\(\Big \downarrow \) 103

\(\displaystyle \frac {1}{4} \left (\frac {3 a d^2}{c^2}+4 b\right ) \left (\frac {d^3 \int \frac {1}{d c^2+d (d x-c) (c+d x)}d\left (\sqrt {d x-c} \sqrt {c+d x}\right )}{2 c^2}+\frac {\sqrt {d x-c} \sqrt {c+d x}}{2 c^2 x^2}\right )+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{4 c^2 x^4}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{4} \left (\frac {3 a d^2}{c^2}+4 b\right ) \left (\frac {d^2 \arctan \left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{2 c^3}+\frac {\sqrt {d x-c} \sqrt {c+d x}}{2 c^2 x^2}\right )+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{4 c^2 x^4}\)

Input:

Int[(a + b*x^2)/(x^5*Sqrt[-c + d*x]*Sqrt[c + d*x]),x]
 

Output:

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

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 114

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] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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 [A] (verified)

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.23

method	result	size
risch	\(-\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (3 a \,d^{2} x^{2}+4 b \,c^{2} x^{2}+2 a \,c^{2}\right )}{8 c^{4} x^{4} \sqrt {d x -c}}-\frac {d^{2} \left (3 a \,d^{2}+4 b \,c^{2}\right ) \ln \left (\frac {-2 c^{2}+2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}}{x}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{8 c^{4} \sqrt {-c^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(151\)
default	\(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (3 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,d^{4} x^{4}+4 \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) b \,c^{2} d^{2} x^{4}-3 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2} x^{2}-4 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2} x^{2}-2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,c^{2}\right )}{8 c^{4} \sqrt {d^{2} x^{2}-c^{2}}\, x^{4} \sqrt {-c^{2}}}\)	\(227\)

Input:

int((b*x^2+a)/x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*(d*x+c)^(1/2)*(-d*x+c)*(3*a*d^2*x^2+4*b*c^2*x^2+2*a*c^2)/c^4/x^4/(d*x 
-c)^(1/2)-1/8*d^2*(3*a*d^2+4*b*c^2)/c^4/(-c^2)^(1/2)*ln((-2*c^2+2*(-c^2)^( 
1/2)*(d^2*x^2-c^2)^(1/2))/x)*((d*x-c)*(d*x+c))^(1/2)/(d*x-c)^(1/2)/(d*x+c) 
^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {2 \, {\left (4 \, b c^{2} d^{2} + 3 \, a d^{4}\right )} x^{4} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right ) + {\left (2 \, a c^{3} + {\left (4 \, b c^{3} + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{8 \, c^{5} x^{4}} \] Input:

integrate((b*x^2+a)/x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

1/8*(2*(4*b*c^2*d^2 + 3*a*d^4)*x^4*arctan(-(d*x - sqrt(d*x + c)*sqrt(d*x - 
 c))/c) + (2*a*c^3 + (4*b*c^3 + 3*a*c*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c 
))/(c^5*x^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)/x**5/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=-\frac {b d^{2} \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{2 \, c^{3}} - \frac {3 \, a d^{4} \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{8 \, c^{5}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} b}{2 \, c^{2} x^{2}} + \frac {3 \, \sqrt {d^{2} x^{2} - c^{2}} a d^{2}}{8 \, c^{4} x^{2}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a}{4 \, c^{2} x^{4}} \] Input:

integrate((b*x^2+a)/x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

-1/2*b*d^2*arcsin(c/(d*abs(x)))/c^3 - 3/8*a*d^4*arcsin(c/(d*abs(x)))/c^5 + 
 1/2*sqrt(d^2*x^2 - c^2)*b/(c^2*x^2) + 3/8*sqrt(d^2*x^2 - c^2)*a*d^2/(c^4* 
x^2) + 1/4*sqrt(d^2*x^2 - c^2)*a/(c^2*x^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (105) = 210\).

Time = 0.15 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.64 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=-\frac {\frac {{\left (4 \, b c^{2} d^{3} + 3 \, a d^{5}\right )} \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right )}{c^{5}} + \frac {2 \, {\left (4 \, b c^{2} d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{14} + 3 \, a d^{5} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{14} + 16 \, b c^{4} d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{10} + 44 \, a c^{2} d^{5} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{10} - 64 \, b c^{6} d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{6} - 176 \, a c^{4} d^{5} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{6} - 256 \, b c^{8} d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} - 192 \, a c^{6} d^{5} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{4} c^{4}}}{4 \, d} \] Input:

integrate((b*x^2+a)/x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

-1/4*((4*b*c^2*d^3 + 3*a*d^5)*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2 
/c)/c^5 + 2*(4*b*c^2*d^3*(sqrt(d*x + c) - sqrt(d*x - c))^14 + 3*a*d^5*(sqr 
t(d*x + c) - sqrt(d*x - c))^14 + 16*b*c^4*d^3*(sqrt(d*x + c) - sqrt(d*x - 
c))^10 + 44*a*c^2*d^5*(sqrt(d*x + c) - sqrt(d*x - c))^10 - 64*b*c^6*d^3*(s 
qrt(d*x + c) - sqrt(d*x - c))^6 - 176*a*c^4*d^5*(sqrt(d*x + c) - sqrt(d*x 
- c))^6 - 256*b*c^8*d^3*(sqrt(d*x + c) - sqrt(d*x - c))^2 - 192*a*c^6*d^5* 
(sqrt(d*x + c) - sqrt(d*x - c))^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4 
*c^2)^4*c^4))/d
 

Mupad [B] (verification not implemented)

Time = 23.03 (sec) , antiderivative size = 1005, normalized size of antiderivative = 8.17 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx =\text {Too large to display} \] Input:

int((a + b*x^2)/(x^5*(c + d*x)^(1/2)*(d*x - c)^(1/2)),x)
 

Output:

(3*a*(-c)^(1/2)*d^4*log(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c 
)^(1/2))))/(8*c^(11/2)) - ((b*(-c)^(3/2)*d^2)/(32*c^(9/2)) + (b*(-c)^(3/2) 
*d^2*((c + d*x)^(1/2) - c^(1/2))^2)/(16*c^(9/2)*((-c)^(1/2) - (d*x - c)^(1 
/2))^2) - (15*b*(-c)^(3/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^4)/(32*c^(9/2)* 
((-c)^(1/2) - (d*x - c)^(1/2))^4))/(((c + d*x)^(1/2) - c^(1/2))^2/((-c)^(1 
/2) - (d*x - c)^(1/2))^2 + (2*((c + d*x)^(1/2) - c^(1/2))^4)/((-c)^(1/2) - 
 (d*x - c)^(1/2))^4 + ((c + d*x)^(1/2) - c^(1/2))^6/((-c)^(1/2) - (d*x - c 
)^(1/2))^6) - ((a*(-c)^(1/2)*d^4)/(1024*c^(11/2)) - (3*a*(-c)^(1/2)*d^4*(( 
c + d*x)^(1/2) - c^(1/2))^2)/(128*c^(11/2)*((-c)^(1/2) - (d*x - c)^(1/2))^ 
2) - (53*a*(-c)^(1/2)*d^4*((c + d*x)^(1/2) - c^(1/2))^4)/(512*c^(11/2)*((- 
c)^(1/2) - (d*x - c)^(1/2))^4) + (87*a*(-c)^(1/2)*d^4*((c + d*x)^(1/2) - c 
^(1/2))^6)/(256*c^(11/2)*((-c)^(1/2) - (d*x - c)^(1/2))^6) + (657*a*(-c)^( 
1/2)*d^4*((c + d*x)^(1/2) - c^(1/2))^8)/(1024*c^(11/2)*((-c)^(1/2) - (d*x 
- c)^(1/2))^8) + (121*a*(-c)^(1/2)*d^4*((c + d*x)^(1/2) - c^(1/2))^10)/(25 
6*c^(11/2)*((-c)^(1/2) - (d*x - c)^(1/2))^10))/(((c + d*x)^(1/2) - c^(1/2) 
)^4/((-c)^(1/2) - (d*x - c)^(1/2))^4 + (4*((c + d*x)^(1/2) - c^(1/2))^6)/( 
(-c)^(1/2) - (d*x - c)^(1/2))^6 + (6*((c + d*x)^(1/2) - c^(1/2))^8)/((-c)^ 
(1/2) - (d*x - c)^(1/2))^8 + (4*((c + d*x)^(1/2) - c^(1/2))^10)/((-c)^(1/2 
) - (d*x - c)^(1/2))^10 + ((c + d*x)^(1/2) - c^(1/2))^12/((-c)^(1/2) - (d* 
x - c)^(1/2))^12) - (b*(-c)^(3/2)*d^2*log(((c + d*x)^(1/2) - c^(1/2))/(...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.74 \[ \int \frac {a+b x^2}{x^5 \sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {6 \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) a \,d^{4} x^{4}+8 \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}-\sqrt {c}}{\sqrt {c}}\right ) b \,c^{2} d^{2} x^{4}-6 \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) a \,d^{4} x^{4}-8 \mathit {atan} \left (\frac {\sqrt {d x -c}+\sqrt {d x +c}+\sqrt {c}}{\sqrt {c}}\right ) b \,c^{2} d^{2} x^{4}+2 \sqrt {d x +c}\, \sqrt {d x -c}\, a \,c^{3}+3 \sqrt {d x +c}\, \sqrt {d x -c}\, a c \,d^{2} x^{2}+4 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,c^{3} x^{2}}{8 c^{5} x^{4}} \] Input:

int((b*x^2+a)/x^5/(d*x-c)^(1/2)/(d*x+c)^(1/2),x)
 

Output:

(6*atan((sqrt( - c + d*x) + sqrt(c + d*x) - sqrt(c))/sqrt(c))*a*d**4*x**4 
+ 8*atan((sqrt( - c + d*x) + sqrt(c + d*x) - sqrt(c))/sqrt(c))*b*c**2*d**2 
*x**4 - 6*atan((sqrt( - c + d*x) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*a*d** 
4*x**4 - 8*atan((sqrt( - c + d*x) + sqrt(c + d*x) + sqrt(c))/sqrt(c))*b*c* 
*2*d**2*x**4 + 2*sqrt(c + d*x)*sqrt( - c + d*x)*a*c**3 + 3*sqrt(c + d*x)*s 
qrt( - c + d*x)*a*c*d**2*x**2 + 4*sqrt(c + d*x)*sqrt( - c + d*x)*b*c**3*x* 
*2)/(8*c**5*x**4)