\(\int (a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2} x \, dx\) [155]

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=-\left ((b c+a d) \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{3} b \left (c+\frac {d}{x^2}\right )^{3/2}+\frac {1}{2} a c \sqrt {c+\frac {d}{x^2}} x^2+\frac {1}{2} \sqrt {c} (2 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (-6 a d x^2+3 a c x^4-2 b \left (d+4 c x^2\right )+\frac {6 \sqrt {c} (2 b c+3 a d) x^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {d}+\sqrt {d+c x^2}}\right )}{\sqrt {d+c x^2}}\right )}{6 x^2} \] Input:

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

Output:

(Sqrt[c + d/x^2]*(-6*a*d*x^2 + 3*a*c*x^4 - 2*b*(d + 4*c*x^2) + (6*Sqrt[c]* 
(2*b*c + 3*a*d)*x^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[d] + Sqrt[d + c*x^2])])/Sqr 
t[d + c*x^2]))/(6*x^2)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {948, 87, 60, 60, 73, 221}

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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{5/2}}{c}-\frac {(3 a d+2 b c) \int \left (c+\frac {d}{x^2}\right )^{3/2} x^2d\frac {1}{x^2}}{2 c}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{5/2}}{c}-\frac {(3 a d+2 b c) \left (c \int \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x^2}+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )}{2 c}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{5/2}}{c}-\frac {(3 a d+2 b c) \left (c \left (c \int \frac {x^2}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}+2 \sqrt {c+\frac {d}{x^2}}\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )}{2 c}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{5/2}}{c}-\frac {(3 a d+2 b c) \left (c \left (\frac {2 c \int \frac {1}{\frac {1}{d x^4}-\frac {c}{d}}d\sqrt {c+\frac {d}{x^2}}}{d}+2 \sqrt {c+\frac {d}{x^2}}\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )}{2 c}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{5/2}}{c}-\frac {(3 a d+2 b c) \left (c \left (2 \sqrt {c+\frac {d}{x^2}}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+\frac {d}{x^2}\right )^{3/2}\right )}{2 c}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

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]))))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {\left (3 a c \,x^{4}-6 a d \,x^{2}-8 b c \,x^{2}-2 b d \right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{6 x^{2}}+\frac {\left (3 a d +2 c b \right ) \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}{2 \sqrt {c \,x^{2}+d}}\)	\(99\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (4 c^{\frac {5}{2}} \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,x^{4}+6 c^{\frac {5}{2}} \sqrt {c \,x^{2}+d}\, b d \,x^{4}+6 c^{\frac {3}{2}} \left (c \,x^{2}+d \right )^{\frac {3}{2}} a d \,x^{4}-4 c^{\frac {3}{2}} \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,x^{2}+9 c^{\frac {3}{2}} \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{4}-6 \sqrt {c}\, \left (c \,x^{2}+d \right )^{\frac {5}{2}} a d \,x^{2}+9 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) a c \,d^{3} x^{3}+6 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) b \,c^{2} d^{2} x^{3}-2 \sqrt {c}\, \left (c \,x^{2}+d \right )^{\frac {5}{2}} b d \right )}{6 \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{2} \sqrt {c}}\)	\(216\)

Input:

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

Output:

1/6*(3*a*c*x^4-6*a*d*x^2-8*b*c*x^2-2*b*d)/x^2*((c*x^2+d)/x^2)^(1/2)+1/2*(3 
*a*d+2*b*c)*c^(1/2)*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*((c*x^2+d)/x^2)^(1/2)*x/ 
(c*x^2+d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.10 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\left [\frac {3 \, {\left (2 \, b c + 3 \, a d\right )} \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) + 2 \, {\left (3 \, a c x^{4} - 2 \, {\left (4 \, b c + 3 \, a d\right )} x^{2} - 2 \, b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{12 \, x^{2}}, -\frac {3 \, {\left (2 \, b c + 3 \, a d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (3 \, a c x^{4} - 2 \, {\left (4 \, b c + 3 \, a d\right )} x^{2} - 2 \, b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{6 \, x^{2}}\right ] \] Input:

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

Output:

[1/12*(3*(2*b*c + 3*a*d)*sqrt(c)*x^2*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c* 
x^2 + d)/x^2) - d) + 2*(3*a*c*x^4 - 2*(4*b*c + 3*a*d)*x^2 - 2*b*d)*sqrt((c 
*x^2 + d)/x^2))/x^2, -1/6*(3*(2*b*c + 3*a*d)*sqrt(-c)*x^2*arctan(sqrt(-c)* 
x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (3*a*c*x^4 - 2*(4*b*c + 3*a*d)*x^ 
2 - 2*b*d)*sqrt((c*x^2 + d)/x^2))/x^2]
 

Sympy [A] (verification not implemented)

Time = 23.71 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.01 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\frac {3 a \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2} + \frac {a c \sqrt {d} x \sqrt {\frac {c x^{2}}{d} + 1}}{2} - \frac {a c \sqrt {d} x}{\sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a d^{\frac {3}{2}}}{x \sqrt {\frac {c x^{2}}{d} + 1}} + b c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )} - \frac {b c^{2} x}{\sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {b c \sqrt {d}}{x \sqrt {\frac {c x^{2}}{d} + 1}} + b d \left (\begin {cases} - \frac {\sqrt {c}}{2 x^{2}} & \text {for}\: d = 0 \\- \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

3*a*sqrt(c)*d*asinh(sqrt(c)*x/sqrt(d))/2 + a*c*sqrt(d)*x*sqrt(c*x**2/d + 1 
)/2 - a*c*sqrt(d)*x/sqrt(c*x**2/d + 1) - a*d**(3/2)/(x*sqrt(c*x**2/d + 1)) 
 + b*c**(3/2)*asinh(sqrt(c)*x/sqrt(d)) - b*c**2*x/(sqrt(d)*sqrt(c*x**2/d + 
 1)) - b*c*sqrt(d)/(x*sqrt(c*x**2/d + 1)) + b*d*Piecewise((-sqrt(c)/(2*x** 
2), Eq(d, 0)), (-(c + d/x**2)**(3/2)/(3*d), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\frac {1}{4} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} c x^{2} - 3 \, \sqrt {c} d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) - 4 \, \sqrt {c + \frac {d}{x^{2}}} d\right )} a - \frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) + 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} + 6 \, \sqrt {c + \frac {d}{x^{2}}} c\right )} b \] Input:

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

Output:

1/4*(2*sqrt(c + d/x^2)*c*x^2 - 3*sqrt(c)*d*log((sqrt(c + d/x^2) - sqrt(c)) 
/(sqrt(c + d/x^2) + sqrt(c))) - 4*sqrt(c + d/x^2)*d)*a - 1/6*(3*c^(3/2)*lo 
g((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c))) + 2*(c + d/x^2) 
^(3/2) + 6*sqrt(c + d/x^2)*c)*b
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (75) = 150\).

Time = 0.26 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.42 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\frac {1}{2} \, \sqrt {c x^{2} + d} a c x \mathrm {sgn}\left (x\right ) - \frac {1}{4} \, {\left (2 \, b c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 3 \, a \sqrt {c} d \mathrm {sgn}\left (x\right )\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right ) + \frac {2 \, {\left (6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {3}{2}} d \mathrm {sgn}\left (x\right ) + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a \sqrt {c} d^{2} \mathrm {sgn}\left (x\right ) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a \sqrt {c} d^{3} \mathrm {sgn}\left (x\right ) + 4 \, b c^{\frac {3}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 3 \, a \sqrt {c} d^{4} \mathrm {sgn}\left (x\right )\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{3}} \] Input:

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

Output:

1/2*sqrt(c*x^2 + d)*a*c*x*sgn(x) - 1/4*(2*b*c^(3/2)*sgn(x) + 3*a*sqrt(c)*d 
*sgn(x))*log((sqrt(c)*x - sqrt(c*x^2 + d))^2) + 2/3*(6*(sqrt(c)*x - sqrt(c 
*x^2 + d))^4*b*c^(3/2)*d*sgn(x) + 3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*sqrt 
(c)*d^2*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(3/2)*d^2*sgn(x) - 
6*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*sqrt(c)*d^3*sgn(x) + 4*b*c^(3/2)*d^3*s 
gn(x) + 3*a*sqrt(c)*d^4*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^3
 

Mupad [B] (verification not implemented)

Time = 4.77 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=b\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {b\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{3}-a\,d\,\sqrt {c+\frac {d}{x^2}}-b\,c\,\sqrt {c+\frac {d}{x^2}}+\frac {3\,a\,\sqrt {c}\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2}+\frac {a\,c\,x^2\,\sqrt {c+\frac {d}{x^2}}}{2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.38 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x \, dx=\frac {6 \sqrt {c \,x^{2}+d}\, a c \,x^{4}-12 \sqrt {c \,x^{2}+d}\, a d \,x^{2}-16 \sqrt {c \,x^{2}+d}\, b c \,x^{2}-4 \sqrt {c \,x^{2}+d}\, b d +18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a d \,x^{3}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) b c \,x^{3}+5 \sqrt {c}\, a d \,x^{3}}{12 x^{3}} \] Input:

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

Output:

(6*sqrt(c*x**2 + d)*a*c*x**4 - 12*sqrt(c*x**2 + d)*a*d*x**2 - 16*sqrt(c*x* 
*2 + d)*b*c*x**2 - 4*sqrt(c*x**2 + d)*b*d + 18*sqrt(c)*log((sqrt(c*x**2 + 
d) + sqrt(c)*x)/sqrt(d))*a*d*x**3 + 12*sqrt(c)*log((sqrt(c*x**2 + d) + sqr 
t(c)*x)/sqrt(d))*b*c*x**3 + 5*sqrt(c)*a*d*x**3)/(12*x**3)