\(\int \frac {(c+\frac {d}{x^2})^{3/2} (a+b x)}{x^4} \, dx\) [24]

Optimal result

Integrand size = 20, antiderivative size = 118 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d}-\frac {a c \sqrt {c+\frac {d}{x^2}}}{8 x^3}-\frac {a \left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}-\frac {a c^2 \sqrt {c+\frac {d}{x^2}}}{16 d x}+\frac {a c^3 \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{16 d^{3/2}} \] Output:

-1/5*b*(c+d/x^2)^(5/2)/d-1/8*a*c*(c+d/x^2)^(1/2)/x^3-1/6*a*(c+d/x^2)^(3/2) 
/x^3-1/16*a*c^2*(c+d/x^2)^(1/2)/d/x+1/16*a*c^3*arctanh(d^(1/2)/(c+d/x^2)^( 
1/2)/x)/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (-\sqrt {d} \left (48 b x \left (d+c x^2\right )^2+5 a \left (8 d^2+14 c d x^2+3 c^2 x^4\right )\right )-\frac {30 a c^3 x^6 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {d+c x^2}}{\sqrt {d}}\right )}{\sqrt {d+c x^2}}\right )}{240 d^{3/2} x^5} \] Input:

Integrate[((c + d/x^2)^(3/2)*(a + b*x))/x^4,x]
 

Output:

(Sqrt[c + d/x^2]*(-(Sqrt[d]*(48*b*x*(d + c*x^2)^2 + 5*a*(8*d^2 + 14*c*d*x^ 
2 + 3*c^2*x^4))) - (30*a*c^3*x^6*ArcTanh[(Sqrt[c]*x - Sqrt[d + c*x^2])/Sqr 
t[d]])/Sqrt[d + c*x^2]))/(240*d^(3/2)*x^5)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1892, 1803, 533, 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.

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

\(\Big \downarrow \) 1892

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

\(\displaystyle \frac {a c \left (\frac {3}{4} c \int \sqrt {c+\frac {d}{x^2}}d\frac {1}{x}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {6}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {a c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {6}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {a c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {6}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a c \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 \sqrt {d}}+\frac {\sqrt {c+\frac {d}{x^2}}}{2 x}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{4 x}\right )-\frac {6}{5} b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d}-\frac {a \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}\)

Input:

Int[((c + d/x^2)^(3/2)*(a + b*x))/x^4,x]
 

Output:

-1/6*(a*(c + d/x^2)^(5/2))/(d*x) + ((-6*b*(c + d/x^2)^(5/2))/5 + a*c*((c + 
 d/x^2)^(3/2)/(4*x) + (3*c*(Sqrt[c + d/x^2]/(2*x) + (c*ArcTanh[Sqrt[d]/(Sq 
rt[c + d/x^2]*x)])/(2*Sqrt[d])))/4))/(6*d)
 

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 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 = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {\left (48 b \,c^{2} x^{5}+15 a \,c^{2} x^{4}+96 b c d \,x^{3}+70 a d \,x^{2} c +48 b x \,d^{2}+40 a \,d^{2}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{240 x^{5} d}+\frac {a \,c^{3} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{16 d^{\frac {3}{2}} \sqrt {c \,x^{2}+d}}\)	\(126\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (-5 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{3} x^{6}+15 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) a \,c^{3} x^{6}+5 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{2} x^{4}-15 \sqrt {c \,x^{2}+d}\, a \,c^{3} d \,x^{6}+10 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a c d \,x^{2}-48 b \left (c \,x^{2}+d \right )^{\frac {5}{2}} d^{2} x -40 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,d^{2}\right )}{240 x^{3} \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{3}}\)	\(168\)

Input:

int((c+d/x^2)^(3/2)*(b*x+a)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/240*(48*b*c^2*x^5+15*a*c^2*x^4+96*b*c*d*x^3+70*a*c*d*x^2+48*b*d^2*x+40* 
a*d^2)/x^5/d*((c*x^2+d)/x^2)^(1/2)+1/16*a*c^3/d^(3/2)*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 = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.03 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=\left [\frac {15 \, a c^{3} \sqrt {d} x^{5} \log \left (-\frac {c x^{2} + 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (48 \, b c^{2} d x^{5} + 15 \, a c^{2} d x^{4} + 96 \, b c d^{2} x^{3} + 70 \, a c d^{2} x^{2} + 48 \, b d^{3} x + 40 \, a d^{3}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{480 \, d^{2} x^{5}}, -\frac {15 \, a c^{3} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{d}\right ) + {\left (48 \, b c^{2} d x^{5} + 15 \, a c^{2} d x^{4} + 96 \, b c d^{2} x^{3} + 70 \, a c d^{2} x^{2} + 48 \, b d^{3} x + 40 \, a d^{3}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{240 \, d^{2} x^{5}}\right ] \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^4,x, algorithm="fricas")
 

Output:

[1/480*(15*a*c^3*sqrt(d)*x^5*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^ 
2) + 2*d)/x^2) - 2*(48*b*c^2*d*x^5 + 15*a*c^2*d*x^4 + 96*b*c*d^2*x^3 + 70* 
a*c*d^2*x^2 + 48*b*d^3*x + 40*a*d^3)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5), -1/ 
240*(15*a*c^3*sqrt(-d)*x^5*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/d) + (4 
8*b*c^2*d*x^5 + 15*a*c^2*d*x^4 + 96*b*c*d^2*x^3 + 70*a*c*d^2*x^2 + 48*b*d^ 
3*x + 40*a*d^3)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5)]
 

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.33 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=- a c \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{8 d} + \sqrt {c + \frac {d}{x^{2}}} \left (\frac {c}{8 d x} + \frac {1}{4 x^{3}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) - a d \left (\begin {cases} \frac {c^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + \frac {d}{x^{2}}} + \frac {2 d}{x} \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\- \frac {\log {\left (x \right )}}{x \sqrt {\frac {d}{x^{2}}}} & \text {otherwise} \end {cases}\right )}{16 d^{2}} + \sqrt {c + \frac {d}{x^{2}}} \left (- \frac {c^{2}}{16 d^{2} x} + \frac {c}{24 d x^{3}} + \frac {1}{6 x^{5}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{5 x^{5}} & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \left (\frac {c}{3 d} + \frac {1}{3 x^{2}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{2}} & \text {otherwise} \end {cases}\right ) - b d \left (\begin {cases} \sqrt {c + \frac {d}{x^{2}}} \left (- \frac {2 c^{2}}{15 d^{2}} + \frac {c}{15 d x^{2}} + \frac {1}{5 x^{4}}\right ) & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{4}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((c+d/x**2)**(3/2)*(b*x+a)/x**4,x)
 

Output:

-a*c*Piecewise((-c**2*Piecewise((log(2*sqrt(d)*sqrt(c + d/x**2) + 2*d/x)/s 
qrt(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)) - a*d*P 
iecewise((c**3*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))/(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)) - b*c*Piecewise((sqrt(c + d/x**2)*(c/(3*d) + 1/(3*x**2)), Ne 
(d, 0)), (sqrt(c)/(2*x**2), True)) - b*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))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.47 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {1}{96} \, {\left (\frac {3 \, c^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} x^{5} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3} d x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{2} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} d x^{6} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} x^{4} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} d^{3} x^{2} - d^{4}}\right )} a \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^4,x, algorithm="maxima")
 

Output:

-1/5*b*(c + d/x^2)^(5/2)/d - 1/96*(3*c^3*log((sqrt(c + d/x^2)*x - sqrt(d)) 
/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(3/2) + 2*(3*(c + d/x^2)^(5/2)*c^3*x^5 + 
 8*(c + d/x^2)^(3/2)*c^3*d*x^3 - 3*sqrt(c + d/x^2)*c^3*d^2*x)/((c + d/x^2) 
^3*d*x^6 - 3*(c + d/x^2)^2*d^2*x^4 + 3*(c + d/x^2)*d^3*x^2 - d^4))*a
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (94) = 188\).

Time = 171.60 (sec) , antiderivative size = 405, normalized size of antiderivative = 3.43 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {a c^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + d}}{\sqrt {-d}}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {-d} d} + \frac {15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{11} a c^{3} \mathrm {sgn}\left (x\right ) + 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} b c^{\frac {5}{2}} d \mathrm {sgn}\left (x\right ) + 235 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{9} a c^{3} d \mathrm {sgn}\left (x\right ) - 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 390 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{7} a c^{3} d^{2} \mathrm {sgn}\left (x\right ) + 480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {5}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 390 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{5} a c^{3} d^{3} \mathrm {sgn}\left (x\right ) - 480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {5}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 235 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{3} a c^{3} d^{4} \mathrm {sgn}\left (x\right ) + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {5}{2}} d^{5} \mathrm {sgn}\left (x\right ) + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )} a c^{3} d^{5} \mathrm {sgn}\left (x\right ) - 48 \, b c^{\frac {5}{2}} d^{6} \mathrm {sgn}\left (x\right )}{120 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{6} d} \] Input:

integrate((c+d/x^2)^(3/2)*(b*x+a)/x^4,x, algorithm="giac")
 

Output:

-1/8*a*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + d))/sqrt(-d))*sgn(x)/(sqrt(-d 
)*d) + 1/120*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^11*a*c^3*sgn(x) + 240*(sqrt 
(c)*x - sqrt(c*x^2 + d))^10*b*c^(5/2)*d*sgn(x) + 235*(sqrt(c)*x - sqrt(c*x 
^2 + d))^9*a*c^3*d*sgn(x) - 240*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(5/2)* 
d^2*sgn(x) + 390*(sqrt(c)*x - sqrt(c*x^2 + d))^7*a*c^3*d^2*sgn(x) + 480*(s 
qrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(5/2)*d^3*sgn(x) + 390*(sqrt(c)*x - sqrt 
(c*x^2 + d))^5*a*c^3*d^3*sgn(x) - 480*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^ 
(5/2)*d^4*sgn(x) + 235*(sqrt(c)*x - sqrt(c*x^2 + d))^3*a*c^3*d^4*sgn(x) + 
48*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(5/2)*d^5*sgn(x) + 15*(sqrt(c)*x - 
sqrt(c*x^2 + d))*a*c^3*d^5*sgn(x) - 48*b*c^(5/2)*d^6*sgn(x))/(((sqrt(c)*x 
- sqrt(c*x^2 + d))^2 - d)^6*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=\int \frac {{\left (c+\frac {d}{x^2}\right )}^{3/2}\,\left (a+b\,x\right )}{x^4} \,d x \] Input:

int(((c + d/x^2)^(3/2)*(a + b*x))/x^4,x)
                                                                                    
                                                                                    
 

Output:

int(((c + d/x^2)^(3/2)*(a + b*x))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.59 \[ \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {-15 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{4}-70 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{2}-40 \sqrt {c \,x^{2}+d}\, a \,d^{3}-48 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{5}-96 \sqrt {c \,x^{2}+d}\, b c \,d^{2} x^{3}-48 \sqrt {c \,x^{2}+d}\, b \,d^{3} x -32 \sqrt {c}\, b \,c^{2} d \,x^{6}-15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) a \,c^{3} x^{6}+15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) a \,c^{3} x^{6}}{240 d^{2} x^{6}} \] Input:

int((c+d/x^2)^(3/2)*(b*x+a)/x^4,x)
 

Output:

( - 15*sqrt(c*x**2 + d)*a*c**2*d*x**4 - 70*sqrt(c*x**2 + d)*a*c*d**2*x**2 
- 40*sqrt(c*x**2 + d)*a*d**3 - 48*sqrt(c*x**2 + d)*b*c**2*d*x**5 - 96*sqrt 
(c*x**2 + d)*b*c*d**2*x**3 - 48*sqrt(c*x**2 + d)*b*d**3*x - 32*sqrt(c)*b*c 
**2*d*x**6 - 15*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x - sqrt(d))/sqrt( 
d))*a*c**3*x**6 + 15*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x + sqrt(d))/ 
sqrt(d))*a*c**3*x**6)/(240*d**2*x**6)