\(\int \frac {(c+\frac {d}{x^2})^{3/2} (a+b x)}{x^3} \, dx\) [23]

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}-\frac {3 b c \sqrt {c+\frac {d}{x^2}}}{8 x}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}-\frac {3 b c^2 \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{8 \sqrt {d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.32 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {\sqrt {c+\frac {d}{x^2}} \left (\sqrt {d+c x^2} \left (8 a \left (d+c x^2\right )^2+5 b d x \left (2 d+5 c x^2\right )\right )+15 b c^2 \sqrt {d} x^5 \log (x)-15 b c^2 \sqrt {d} x^5 \log \left (-\sqrt {d}+\sqrt {d+c x^2}\right )\right )}{40 d x^4 \sqrt {d+c x^2}} \] Input:

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

Output:

-1/40*(Sqrt[c + d/x^2]*(Sqrt[d + c*x^2]*(8*a*(d + c*x^2)^2 + 5*b*d*x*(2*d 
+ 5*c*x^2)) + 15*b*c^2*Sqrt[d]*x^5*Log[x] - 15*b*c^2*Sqrt[d]*x^5*Log[-Sqrt 
[d] + Sqrt[d + c*x^2]]))/(d*x^4*Sqrt[d + c*x^2])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1892, 1799, 455, 211, 211, 224, 219}

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.

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

\(\Big \downarrow \) 1892

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

\(\Big \downarrow \) 1799

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/5*(a*(c + d/x^2)^(5/2))/d - b*((c + d/x^2)^(3/2)/(4*x) + (3*c*(Sqrt[c + 
 d/x^2]/(2*x) + (c*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(2*Sqrt[d])))/4)
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 1799

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^ 
n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Simplif 
y[m - n + 1], 0]
 

rule 1892

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^ 
(p_.), x_Symbol] :> Int[x^(m + mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; F 
reeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n 
2] ||  !IntegerQ[p])
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {\left (8 a \,c^{2} x^{4}+25 b c d \,x^{3}+16 a d \,x^{2} c +10 b x \,d^{2}+8 a \,d^{2}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{40 x^{4} d}-\frac {3 b \,c^{2} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{8 \sqrt {d}\, \sqrt {c \,x^{2}+d}}\)	\(117\)
default	\(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-5 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{2} x^{5}+15 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) b \,c^{2} x^{5}+5 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b c \,x^{3}-15 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{5}+10 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b d x +8 a \left (c \,x^{2}+d \right )^{\frac {5}{2}} d \right )}{40 x^{2} \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{2}}\)	\(145\)

Input:

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

Output:

-1/40*(8*a*c^2*x^4+25*b*c*d*x^3+16*a*c*d*x^2+10*b*d^2*x+8*a*d^2)/x^4/d*((c 
*x^2+d)/x^2)^(1/2)-3/8*b*c^2/d^(1/2)*ln((2*d+2*d^(1/2)*(c*x^2+d)^(1/2))/x) 
/(c*x^2+d)^(1/2)*x*((c*x^2+d)/x^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.27 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=\left [\frac {15 \, b c^{2} \sqrt {d} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (8 \, a c^{2} x^{4} + 25 \, b c d x^{3} + 16 \, a c d x^{2} + 10 \, b d^{2} x + 8 \, a d^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{80 \, d x^{4}}, \frac {15 \, b c^{2} \sqrt {-d} x^{4} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{d}\right ) - {\left (8 \, a c^{2} x^{4} + 25 \, b c d x^{3} + 16 \, a c d x^{2} + 10 \, b d^{2} x + 8 \, a d^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{40 \, d x^{4}}\right ] \] Input:

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

Output:

[1/80*(15*b*c^2*sqrt(d)*x^4*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2 
) + 2*d)/x^2) - 2*(8*a*c^2*x^4 + 25*b*c*d*x^3 + 16*a*c*d*x^2 + 10*b*d^2*x 
+ 8*a*d^2)*sqrt((c*x^2 + d)/x^2))/(d*x^4), 1/40*(15*b*c^2*sqrt(-d)*x^4*arc 
tan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/d) - (8*a*c^2*x^4 + 25*b*c*d*x^3 + 16 
*a*c*d*x^2 + 10*b*d^2*x + 8*a*d^2)*sqrt((c*x^2 + d)/x^2))/(d*x^4)]
 

Sympy [A] (verification not implemented)

Time = 1.84 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.63 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=- a c \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \left (\frac {c}{3 d} + \frac {1}{3 x^{2}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{2}} & \text {otherwise} \end {cases}\right ) - a d \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \left (- \frac {2 c^{2}}{15 d^{2}} + \frac {c}{15 d x^{2}} + \frac {1}{5 x^{4}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{4}} & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {\sqrt {c + \frac {d}{x^{2}}}}{2 x} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{x} & \text {otherwise} \end {cases}\right ) - b d \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{8 d} + \sqrt {c + \frac {d}{x^{2}}} \left (\frac {c}{8 d x} + \frac {1}{4 x^{3}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a*c*Piecewise((sqrt(c + d/x**2)*(c/(3*d) + 1/(3*x**2)), Ne(d, 0)), (sqrt( 
c)/(2*x**2), True)) - a*d*Piecewise((sqrt(c + d/x**2)*(-2*c**2/(15*d**2) + 
 c/(15*d*x**2) + 1/(5*x**4)), Ne(d, 0)), (sqrt(c)/(4*x**4), True)) - b*c*P 
iecewise((c*Piecewise((log(2*sqrt(d)*sqrt(c + d/x**2) + 2*d/x)/sqrt(d), Ne 
(c, 0)), (-log(x)/(x*sqrt(d/x**2)), True))/2 + sqrt(c + d/x**2)/(2*x), Ne( 
d, 0)), (sqrt(c)/x, True)) - b*d*Piecewise((-c**2*Piecewise((log(2*sqrt(d) 
*sqrt(c + d/x**2) + 2*d/x)/sqrt(d), Ne(c, 0)), (-log(x)/(x*sqrt(d/x**2)), 
True))/(8*d) + sqrt(c + d/x**2)*(c/(8*d*x) + 1/(4*x**3)), Ne(d, 0)), (sqrt 
(c)/(3*x**3), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.42 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, d} + \frac {1}{16} \, {\left (\frac {3 \, c^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{\sqrt {d}} - \frac {2 \, {\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} d x^{2} + d^{2}}\right )} b \] Input:

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

Output:

-1/5*a*(c + d/x^2)^(5/2)/d + 1/16*(3*c^2*log((sqrt(c + d/x^2)*x - sqrt(d)) 
/(sqrt(c + d/x^2)*x + sqrt(d)))/sqrt(d) - 2*(5*(c + d/x^2)^(3/2)*c^2*x^3 - 
 3*sqrt(c + d/x^2)*c^2*d*x)/((c + d/x^2)^2*x^4 - 2*(c + d/x^2)*d*x^2 + d^2 
))*b
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (73) = 146\).

Time = 81.23 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.67 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {3 \, b c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + d}}{\sqrt {-d}}\right ) \mathrm {sgn}\left (x\right )}{4 \, \sqrt {-d}} + \frac {25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{9} b c^{2} \mathrm {sgn}\left (x\right ) + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{7} b c^{2} d \mathrm {sgn}\left (x\right ) + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{3} b c^{2} d^{3} \mathrm {sgn}\left (x\right ) - 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )} b c^{2} d^{4} \mathrm {sgn}\left (x\right ) + 8 \, a c^{\frac {5}{2}} d^{4} \mathrm {sgn}\left (x\right )}{20 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{5}} \] Input:

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

Output:

3/4*b*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + d))/sqrt(-d))*sgn(x)/sqrt(-d) 
+ 1/20*(25*(sqrt(c)*x - sqrt(c*x^2 + d))^9*b*c^2*sgn(x) + 40*(sqrt(c)*x - 
sqrt(c*x^2 + d))^8*a*c^(5/2)*sgn(x) - 10*(sqrt(c)*x - sqrt(c*x^2 + d))^7*b 
*c^2*d*sgn(x) + 80*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(5/2)*d^2*sgn(x) + 
10*(sqrt(c)*x - sqrt(c*x^2 + d))^3*b*c^2*d^3*sgn(x) - 25*(sqrt(c)*x - sqrt 
(c*x^2 + d))*b*c^2*d^4*sgn(x) + 8*a*c^(5/2)*d^4*sgn(x))/((sqrt(c)*x - sqrt 
(c*x^2 + d))^2 - d)^5
 

Mupad [B] (verification not implemented)

Time = 8.84 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {b\,{\left (c\,x^2+d\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {d}{c\,x^2}\right )}{x\,{\left (\frac {d}{c}+x^2\right )}^{3/2}}-\frac {a\,\sqrt {c+\frac {d}{x^2}}\,{\left (c\,x^2+d\right )}^2}{5\,d\,x^4} \] Input:

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

Output:

- (b*(d + c*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -d/(c*x^2)))/(x*(d/c + 
x^2)^(3/2)) - (a*(c + d/x^2)^(1/2)*(d + c*x^2)^2)/(5*d*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.76 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {-8 \sqrt {c \,x^{2}+d}\, a \,c^{2} x^{4}-16 \sqrt {c \,x^{2}+d}\, a c d \,x^{2}-8 \sqrt {c \,x^{2}+d}\, a \,d^{2}-25 \sqrt {c \,x^{2}+d}\, b c d \,x^{3}-10 \sqrt {c \,x^{2}+d}\, b \,d^{2} x -8 \sqrt {c}\, a \,c^{2} x^{5}+15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) b \,c^{2} x^{5}-15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) b \,c^{2} x^{5}}{40 d \,x^{5}} \] Input:

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

Output:

( - 8*sqrt(c*x**2 + d)*a*c**2*x**4 - 16*sqrt(c*x**2 + d)*a*c*d*x**2 - 8*sq 
rt(c*x**2 + d)*a*d**2 - 25*sqrt(c*x**2 + d)*b*c*d*x**3 - 10*sqrt(c*x**2 + 
d)*b*d**2*x - 8*sqrt(c)*a*c**2*x**5 + 15*sqrt(d)*log((sqrt(c*x**2 + d) + s 
qrt(c)*x - sqrt(d))/sqrt(d))*b*c**2*x**5 - 15*sqrt(d)*log((sqrt(c*x**2 + d 
) + sqrt(c)*x + sqrt(d))/sqrt(d))*b*c**2*x**5)/(40*d*x**5)