\(\int \frac {(c+\frac {d}{x^2})^{3/2} (a+b x)}{x^6} \, dx\) [26]

Optimal result

Integrand size = 20, antiderivative size = 163 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\frac {b c \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^2}-\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^2}-\frac {a c \sqrt {c+\frac {d}{x^2}}}{16 x^5}-\frac {a \left (c+\frac {d}{x^2}\right )^{3/2}}{8 x^5}-\frac {a c^2 \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {3 a c^3 \sqrt {c+\frac {d}{x^2}}}{128 d^2 x}-\frac {3 a c^4 \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{128 d^{5/2}} \] Output:

1/5*b*c*(c+d/x^2)^(5/2)/d^2-1/7*b*(c+d/x^2)^(7/2)/d^2-1/16*a*c*(c+d/x^2)^( 
1/2)/x^5-1/8*a*(c+d/x^2)^(3/2)/x^5-1/64*a*c^2*(c+d/x^2)^(1/2)/d/x^3+3/128* 
a*c^3*(c+d/x^2)^(1/2)/d^2/x-3/128*a*c^4*arctanh(d^(1/2)/(c+d/x^2)^(1/2)/x) 
/d^(5/2)
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.84 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (\sqrt {d} \left (128 b x \left (d+c x^2\right )^2 \left (-5 d+2 c x^2\right )-35 a \left (16 d^3+24 c d^2 x^2+2 c^2 d x^4-3 c^3 x^6\right )\right )+\frac {210 a c^4 x^8 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {d+c x^2}}{\sqrt {d}}\right )}{\sqrt {d+c x^2}}\right )}{4480 d^{5/2} x^7} \] Input:

Integrate[((c + d/x^2)^(3/2)*(a + b*x))/x^6,x]
 

Output:

(Sqrt[c + d/x^2]*(Sqrt[d]*(128*b*x*(d + c*x^2)^2*(-5*d + 2*c*x^2) - 35*a*( 
16*d^3 + 24*c*d^2*x^2 + 2*c^2*d*x^4 - 3*c^3*x^6)) + (210*a*c^4*x^8*ArcTanh 
[(Sqrt[c]*x - Sqrt[d + c*x^2])/Sqrt[d]])/Sqrt[d + c*x^2]))/(4480*d^(5/2)*x 
^7)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {1892, 1803, 533, 533, 25, 27, 533, 27, 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.

\(\displaystyle \int \frac {(a+b x) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^6} \, dx\)

\(\Big \downarrow \) 1892

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {\int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} \left (3 a c-\frac {8 b d}{x}\right )}{x^2}d\frac {1}{x}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {-\frac {\int -\frac {c d \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {21 a}{x}+16 b\right )}{x}d\frac {1}{x}}{7 d}-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {c d \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {21 a}{x}+16 b\right )}{x}d\frac {1}{x}}{7 d}-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} c \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {21 a}{x}+16 b\right )}{x}d\frac {1}{x}-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {\int 3 \left (c+\frac {d}{x^2}\right )^{3/2} \left (7 a c-\frac {32 b d}{x}\right )d\frac {1}{x}}{6 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {\int \left (c+\frac {d}{x^2}\right )^{3/2} \left (7 a c-\frac {32 b d}{x}\right )d\frac {1}{x}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {7 a c \int \left (c+\frac {d}{x^2}\right )^{3/2}d\frac {1}{x}-\frac {32}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {7 a c \left (\frac {3}{4} c \int \sqrt {c+\frac {d}{x^2}}d\frac {1}{x}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {32}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {7 a c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {32}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {7 a c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {32}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{7} c \left (\frac {7 a \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d x}-\frac {7 a c \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 \sqrt {d}}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {32}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{2 d}\right )-\frac {8 b \left (c+\frac {d}{x^2}\right )^{5/2}}{7 x^2}}{8 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

Input:

Int[((c + d/x^2)^(3/2)*(a + b*x))/x^6,x]
 

Output:

-1/8*(a*(c + d/x^2)^(5/2))/(d*x^3) + ((-8*b*(c + d/x^2)^(5/2))/(7*x^2) + ( 
c*((7*a*(c + d/x^2)^(5/2))/(2*d*x) - ((-32*b*(c + d/x^2)^(5/2))/5 + 7*a*c* 
((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))/(2*d)))/7)/(8*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 533

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

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]
 

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 = 150, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\left (256 b \,c^{3} x^{7}+105 a \,c^{3} x^{6}-128 b \,c^{2} d \,x^{5}-70 a \,c^{2} d \,x^{4}-1024 x^{3} b c \,d^{2}-840 a \,d^{2} x^{2} c -640 b \,d^{3} x -560 a \,d^{3}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{4480 x^{7} d^{2}}-\frac {3 c^{4} a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{128 d^{\frac {5}{2}} \sqrt {c \,x^{2}+d}}\)	\(150\)
default	\(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-35 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{4} x^{8}+105 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) a \,c^{4} x^{8}+35 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{3} x^{6}-105 \sqrt {c \,x^{2}+d}\, a \,c^{4} d \,x^{8}+70 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{2} d \,x^{4}-256 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b c \,d^{2} x^{3}-280 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a c \,d^{2} x^{2}+640 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,d^{3} x +560 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,d^{3}\right )}{4480 x^{5} \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{4}}\)	\(208\)

Input:

int((c+d/x^2)^(3/2)*(b*x+a)/x^6,x,method=_RETURNVERBOSE)
 

Output:

1/4480*(256*b*c^3*x^7+105*a*c^3*x^6-128*b*c^2*d*x^5-70*a*c^2*d*x^4-1024*b* 
c*d^2*x^3-840*a*c*d^2*x^2-640*b*d^3*x-560*a*d^3)/x^7/d^2*((c*x^2+d)/x^2)^( 
1/2)-3/128*c^4*a/d^(5/2)*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 = 0.13 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.77 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\left [\frac {105 \, a c^{4} \sqrt {d} x^{7} \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (256 \, b c^{3} d x^{7} + 105 \, a c^{3} d x^{6} - 128 \, b c^{2} d^{2} x^{5} - 70 \, a c^{2} d^{2} x^{4} - 1024 \, b c d^{3} x^{3} - 840 \, a c d^{3} x^{2} - 640 \, b d^{4} x - 560 \, a d^{4}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8960 \, d^{3} x^{7}}, \frac {105 \, a c^{4} \sqrt {-d} x^{7} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{d}\right ) + {\left (256 \, b c^{3} d x^{7} + 105 \, a c^{3} d x^{6} - 128 \, b c^{2} d^{2} x^{5} - 70 \, a c^{2} d^{2} x^{4} - 1024 \, b c d^{3} x^{3} - 840 \, a c d^{3} x^{2} - 640 \, b d^{4} x - 560 \, a d^{4}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4480 \, d^{3} x^{7}}\right ] \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^6,x, algorithm="fricas")
 

Output:

[1/8960*(105*a*c^4*sqrt(d)*x^7*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/ 
x^2) + 2*d)/x^2) + 2*(256*b*c^3*d*x^7 + 105*a*c^3*d*x^6 - 128*b*c^2*d^2*x^ 
5 - 70*a*c^2*d^2*x^4 - 1024*b*c*d^3*x^3 - 840*a*c*d^3*x^2 - 640*b*d^4*x - 
560*a*d^4)*sqrt((c*x^2 + d)/x^2))/(d^3*x^7), 1/4480*(105*a*c^4*sqrt(-d)*x^ 
7*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/d) + (256*b*c^3*d*x^7 + 105*a*c^ 
3*d*x^6 - 128*b*c^2*d^2*x^5 - 70*a*c^2*d^2*x^4 - 1024*b*c*d^3*x^3 - 840*a* 
c*d^3*x^2 - 640*b*d^4*x - 560*a*d^4)*sqrt((c*x^2 + d)/x^2))/(d^3*x^7)]
 

Sympy [A] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.04 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=- a c \left (\begin {cases} \frac {c^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{16 d^{2}} + \sqrt {c + \frac {d}{x^{2}}} \left (- \frac {c^{2}}{16 d^{2} x} + \frac {c}{24 d x^{3}} + \frac {1}{6 x^{5}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{5 x^{5}} & \text {otherwise} \end {cases}\right ) - a d \left (\begin {cases} - \frac {5 c^{4} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{128 d^{3}} + \sqrt {c + \frac {d}{x^{2}}} \cdot \left (\frac {5 c^{3}}{128 d^{3} x} - \frac {5 c^{2}}{192 d^{2} x^{3}} + \frac {c}{48 d x^{5}} + \frac {1}{8 x^{7}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{7 x^{7}} & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \left (- \frac {2 c^{2}}{15 d^{2}} + \frac {c}{15 d x^{2}} + \frac {1}{5 x^{4}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{4}} & \text {otherwise} \end {cases}\right ) - b d \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \cdot \left (\frac {8 c^{3}}{105 d^{3}} - \frac {4 c^{2}}{105 d^{2} x^{2}} + \frac {c}{35 d x^{4}} + \frac {1}{7 x^{6}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{6 x^{6}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((c+d/x**2)**(3/2)*(b*x+a)/x**6,x)
 

Output:

-a*c*Piecewise((c**3*Piecewise((log(2*sqrt(d)*sqrt(c + d/x**2) + 2*d/x)/sq 
rt(d), Ne(c, 0)), (-log(x)/(x*sqrt(d/x**2)), True))/(16*d**2) + sqrt(c + d 
/x**2)*(-c**2/(16*d**2*x) + c/(24*d*x**3) + 1/(6*x**5)), Ne(d, 0)), (sqrt( 
c)/(5*x**5), True)) - a*d*Piecewise((-5*c**4*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) 
)/(128*d**3) + sqrt(c + d/x**2)*(5*c**3/(128*d**3*x) - 5*c**2/(192*d**2*x* 
*3) + c/(48*d*x**5) + 1/(8*x**7)), Ne(d, 0)), (sqrt(c)/(7*x**7), True)) - 
b*c*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*d*Piecewise((sqrt(c + d/x 
**2)*(8*c**3/(105*d**3) - 4*c**2/(105*d**2*x**2) + c/(35*d*x**4) + 1/(7*x* 
*6)), Ne(d, 0)), (sqrt(c)/(6*x**6), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.40 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\frac {1}{256} \, {\left (\frac {3 \, c^{4} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{4} x^{7} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{4} d x^{5} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{4} d^{2} x^{3} + 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{4} d^{3} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{4} d^{2} x^{8} - 4 \, {\left (c + \frac {d}{x^{2}}\right )}^{3} d^{3} x^{6} + 6 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{4} x^{4} - 4 \, {\left (c + \frac {d}{x^{2}}\right )} d^{5} x^{2} + d^{6}}\right )} a - \frac {1}{35} \, {\left (\frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{2}} - \frac {7 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{2}}\right )} b \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^6,x, algorithm="maxima")
 

Output:

1/256*(3*c^4*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d 
)))/d^(5/2) + 2*(3*(c + d/x^2)^(7/2)*c^4*x^7 - 11*(c + d/x^2)^(5/2)*c^4*d* 
x^5 - 11*(c + d/x^2)^(3/2)*c^4*d^2*x^3 + 3*sqrt(c + d/x^2)*c^4*d^3*x)/((c 
+ d/x^2)^4*d^2*x^8 - 4*(c + d/x^2)^3*d^3*x^6 + 6*(c + d/x^2)^2*d^4*x^4 - 4 
*(c + d/x^2)*d^5*x^2 + d^6))*a - 1/35*(5*(c + d/x^2)^(7/2)/d^2 - 7*(c + d/ 
x^2)^(5/2)*c/d^2)*b
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\text {Timed out} \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^6,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\int \frac {{\left (c+\frac {d}{x^2}\right )}^{3/2}\,\left (a+b\,x\right )}{x^6} \,d x \] Input:

int(((c + d/x^2)^(3/2)*(a + b*x))/x^6,x)
 

Output:

int(((c + d/x^2)^(3/2)*(a + b*x))/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.40 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^6} \, dx=\frac {105 \sqrt {c \,x^{2}+d}\, a \,c^{3} d \,x^{6}-70 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{2} x^{4}-840 \sqrt {c \,x^{2}+d}\, a c \,d^{3} x^{2}-560 \sqrt {c \,x^{2}+d}\, a \,d^{4}+256 \sqrt {c \,x^{2}+d}\, b \,c^{3} d \,x^{7}-128 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{2} x^{5}-1024 \sqrt {c \,x^{2}+d}\, b c \,d^{3} x^{3}-640 \sqrt {c \,x^{2}+d}\, b \,d^{4} x -256 \sqrt {c}\, b \,c^{3} d \,x^{8}+105 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) a \,c^{4} x^{8}-105 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) a \,c^{4} x^{8}}{4480 d^{3} x^{8}} \] Input:

int((c+d/x^2)^(3/2)*(b*x+a)/x^6,x)
 

Output:

(105*sqrt(c*x**2 + d)*a*c**3*d*x**6 - 70*sqrt(c*x**2 + d)*a*c**2*d**2*x**4 
 - 840*sqrt(c*x**2 + d)*a*c*d**3*x**2 - 560*sqrt(c*x**2 + d)*a*d**4 + 256* 
sqrt(c*x**2 + d)*b*c**3*d*x**7 - 128*sqrt(c*x**2 + d)*b*c**2*d**2*x**5 - 1 
024*sqrt(c*x**2 + d)*b*c*d**3*x**3 - 640*sqrt(c*x**2 + d)*b*d**4*x - 256*s 
qrt(c)*b*c**3*d*x**8 + 105*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x - sqr 
t(d))/sqrt(d))*a*c**4*x**8 - 105*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x 
 + sqrt(d))/sqrt(d))*a*c**4*x**8)/(4480*d**3*x**8)