\(\int \frac {(c+\frac {d}{x})^{3/2} (a+b x)}{x^4} \, dx\) [9]

Optimal result

Integrand size = 20, antiderivative size = 74 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {2 c (a c-b d) \left (c+\frac {d}{x}\right )^{5/2}}{5 d^3}+\frac {2 (2 a c-b d) \left (c+\frac {d}{x}\right )^{7/2}}{7 d^3}-\frac {2 a \left (c+\frac {d}{x}\right )^{9/2}}{9 d^3} \] Output:

-2/5*c*(a*c-b*d)*(c+d/x)^(5/2)/d^3+2/7*(2*a*c-b*d)*(c+d/x)^(7/2)/d^3-2/9*a 
*(c+d/x)^(9/2)/d^3
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {2 \sqrt {c+\frac {d}{x}} (d+c x)^2 \left (9 b d x (-5 d+2 c x)+a \left (-35 d^2+20 c d x-8 c^2 x^2\right )\right )}{315 d^3 x^4} \] Input:

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

Output:

(2*Sqrt[c + d/x]*(d + c*x)^2*(9*b*d*x*(-5*d + 2*c*x) + a*(-35*d^2 + 20*c*d 
*x - 8*c^2*x^2)))/(315*d^3*x^4)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1016, 948, 86, 2009}

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.

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

\(\Big \downarrow \) 1016

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (c+\frac {d}{x}\right )^{7/2} (2 a c-b d)}{7 d^3}-\frac {2 c \left (c+\frac {d}{x}\right )^{5/2} (a c-b d)}{5 d^3}-\frac {2 a \left (c+\frac {d}{x}\right )^{9/2}}{9 d^3}\)

Input:

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

Output:

(-2*c*(a*c - b*d)*(c + d/x)^(5/2))/(5*d^3) + (2*(2*a*c - b*d)*(c + d/x)^(7 
/2))/(7*d^3) - (2*a*(c + d/x)^(9/2))/(9*d^3)
 

Defintions of rubi rules used

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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]]
 

rule 1016

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^( 
p_.), x_Symbol] :> Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ 
[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !I 
ntegerQ[p])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81

method	result	size
orering	\(-\frac {2 \left (8 a \,c^{2} x^{2}-18 b c d \,x^{2}-20 a d x c +45 b x \,d^{2}+35 a \,d^{2}\right ) \left (c x +d \right ) \left (c +\frac {d}{x}\right )^{\frac {3}{2}}}{315 d^{3} x^{3}}\)	\(60\)
gosper	\(-\frac {2 \left (c x +d \right ) \left (8 a \,c^{2} x^{2}-18 b c d \,x^{2}-20 a d x c +45 b x \,d^{2}+35 a \,d^{2}\right ) \left (\frac {c x +d}{x}\right )^{\frac {3}{2}}}{315 d^{3} x^{3}}\)	\(62\)
default	\(-\frac {2 \sqrt {\frac {c x +d}{x}}\, \left (c \,x^{2}+d x \right )^{\frac {3}{2}} \left (8 a \,c^{3} x^{3}-18 b \,c^{2} d \,x^{3}-12 a \,c^{2} d \,x^{2}+27 b c \,d^{2} x^{2}+15 a c \,d^{2} x +45 b \,d^{3} x +35 a \,d^{3}\right )}{315 x^{5} d^{3} \sqrt {\left (c x +d \right ) x}}\)	\(101\)
risch	\(-\frac {2 \sqrt {\frac {c x +d}{x}}\, \left (8 a \,c^{4} x^{4}-18 b \,c^{3} d \,x^{4}-4 a \,c^{3} d \,x^{3}+9 b \,c^{2} d^{2} x^{3}+3 a \,c^{2} d^{2} x^{2}+72 b c \,d^{3} x^{2}+50 a c \,d^{3} x +45 b \,d^{4} x +35 a \,d^{4}\right )}{315 x^{4} d^{3}}\)	\(105\)
trager	\(-\frac {2 \left (8 a \,c^{4} x^{4}-18 b \,c^{3} d \,x^{4}-4 a \,c^{3} d \,x^{3}+9 b \,c^{2} d^{2} x^{3}+3 a \,c^{2} d^{2} x^{2}+72 b c \,d^{3} x^{2}+50 a c \,d^{3} x +45 b \,d^{4} x +35 a \,d^{4}\right ) \sqrt {-\frac {-c x -d}{x}}}{315 x^{4} d^{3}}\)	\(109\)

Input:

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

Output:

-2/315*(8*a*c^2*x^2-18*b*c*d*x^2-20*a*c*d*x+45*b*d^2*x+35*a*d^2)/d^3/x^3*( 
c*x+d)*(c+d/x)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {2 \, {\left (35 \, a d^{4} + 2 \, {\left (4 \, a c^{4} - 9 \, b c^{3} d\right )} x^{4} - {\left (4 \, a c^{3} d - 9 \, b c^{2} d^{2}\right )} x^{3} + 3 \, {\left (a c^{2} d^{2} + 24 \, b c d^{3}\right )} x^{2} + 5 \, {\left (10 \, a c d^{3} + 9 \, b d^{4}\right )} x\right )} \sqrt {\frac {c x + d}{x}}}{315 \, d^{3} x^{4}} \] Input:

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

Output:

-2/315*(35*a*d^4 + 2*(4*a*c^4 - 9*b*c^3*d)*x^4 - (4*a*c^3*d - 9*b*c^2*d^2) 
*x^3 + 3*(a*c^2*d^2 + 24*b*c*d^3)*x^2 + 5*(10*a*c*d^3 + 9*b*d^4)*x)*sqrt(( 
c*x + d)/x)/(d^3*x^4)
 

Sympy [A] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.12 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=- a c \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) - a d \left (\begin {cases} \frac {2 \left (- \frac {c^{3} \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} + \frac {3 c^{2} \left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5} - \frac {3 c \left (c + \frac {d}{x}\right )^{\frac {7}{2}}}{7} + \frac {\left (c + \frac {d}{x}\right )^{\frac {9}{2}}}{9}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{4}} & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{2}} & \text {otherwise} \end {cases}\right ) - b d \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a*c*Piecewise((2*(c**2*(c + d/x)**(3/2)/3 - 2*c*(c + d/x)**(5/2)/5 + (c + 
 d/x)**(7/2)/7)/d**3, Ne(d, 0)), (sqrt(c)/(3*x**3), True)) - a*d*Piecewise 
((2*(-c**3*(c + d/x)**(3/2)/3 + 3*c**2*(c + d/x)**(5/2)/5 - 3*c*(c + d/x)* 
*(7/2)/7 + (c + d/x)**(9/2)/9)/d**4, Ne(d, 0)), (sqrt(c)/(4*x**4), True)) 
- b*c*Piecewise((2*(-c*(c + d/x)**(3/2)/3 + (c + d/x)**(5/2)/5)/d**2, Ne(d 
, 0)), (sqrt(c)/(2*x**2), True)) - b*d*Piecewise((2*(c**2*(c + d/x)**(3/2) 
/3 - 2*c*(c + d/x)**(5/2)/5 + (c + d/x)**(7/2)/7)/d**3, Ne(d, 0)), (sqrt(c 
)/(3*x**3), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=-\frac {2}{315} \, {\left (\frac {35 \, {\left (c + \frac {d}{x}\right )}^{\frac {9}{2}}}{d^{3}} - \frac {90 \, {\left (c + \frac {d}{x}\right )}^{\frac {7}{2}} c}{d^{3}} + \frac {63 \, {\left (c + \frac {d}{x}\right )}^{\frac {5}{2}} c^{2}}{d^{3}}\right )} a - \frac {2}{35} \, {\left (\frac {5 \, {\left (c + \frac {d}{x}\right )}^{\frac {7}{2}}}{d^{2}} - \frac {7 \, {\left (c + \frac {d}{x}\right )}^{\frac {5}{2}} c}{d^{2}}\right )} b \] Input:

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

Output:

-2/315*(35*(c + d/x)^(9/2)/d^3 - 90*(c + d/x)^(7/2)*c/d^3 + 63*(c + d/x)^( 
5/2)*c^2/d^3)*a - 2/35*(5*(c + d/x)^(7/2)/d^2 - 7*(c + d/x)^(5/2)*c/d^2)*b
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (62) = 124\).

Time = 0.63 (sec) , antiderivative size = 397, normalized size of antiderivative = 5.36 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {2 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{7} b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 420 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{6} a c^{3} \mathrm {sgn}\left (x\right ) + 945 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{6} b c^{2} d \mathrm {sgn}\left (x\right ) + 1575 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{5} a c^{\frac {5}{2}} d \mathrm {sgn}\left (x\right ) + 1260 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{5} b c^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 2583 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{4} a c^{2} d^{2} \mathrm {sgn}\left (x\right ) + 882 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{4} b c d^{3} \mathrm {sgn}\left (x\right ) + 2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{3} a c^{\frac {3}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{3} b \sqrt {c} d^{4} \mathrm {sgn}\left (x\right ) + 1170 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{2} a c d^{4} \mathrm {sgn}\left (x\right ) + 45 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{2} b d^{5} \mathrm {sgn}\left (x\right ) + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )} a \sqrt {c} d^{5} \mathrm {sgn}\left (x\right ) + 35 \, a d^{6} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{9}} \] Input:

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

Output:

2/315*(315*(sqrt(c)*x - sqrt(c*x^2 + d*x))^7*b*c^(5/2)*sgn(x) + 420*(sqrt( 
c)*x - sqrt(c*x^2 + d*x))^6*a*c^3*sgn(x) + 945*(sqrt(c)*x - sqrt(c*x^2 + d 
*x))^6*b*c^2*d*sgn(x) + 1575*(sqrt(c)*x - sqrt(c*x^2 + d*x))^5*a*c^(5/2)*d 
*sgn(x) + 1260*(sqrt(c)*x - sqrt(c*x^2 + d*x))^5*b*c^(3/2)*d^2*sgn(x) + 25 
83*(sqrt(c)*x - sqrt(c*x^2 + d*x))^4*a*c^2*d^2*sgn(x) + 882*(sqrt(c)*x - s 
qrt(c*x^2 + d*x))^4*b*c*d^3*sgn(x) + 2310*(sqrt(c)*x - sqrt(c*x^2 + d*x))^ 
3*a*c^(3/2)*d^3*sgn(x) + 315*(sqrt(c)*x - sqrt(c*x^2 + d*x))^3*b*sqrt(c)*d 
^4*sgn(x) + 1170*(sqrt(c)*x - sqrt(c*x^2 + d*x))^2*a*c*d^4*sgn(x) + 45*(sq 
rt(c)*x - sqrt(c*x^2 + d*x))^2*b*d^5*sgn(x) + 315*(sqrt(c)*x - sqrt(c*x^2 
+ d*x))*a*sqrt(c)*d^5*sgn(x) + 35*a*d^6*sgn(x))/(sqrt(c)*x - sqrt(c*x^2 + 
d*x))^9
 

Mupad [B] (verification not implemented)

Time = 7.70 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.22 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {4\,b\,c^3\,\sqrt {c+\frac {d}{x}}}{35\,d^2}-\frac {16\,a\,c^4\,\sqrt {c+\frac {d}{x}}}{315\,d^3}-\frac {20\,a\,c\,\sqrt {c+\frac {d}{x}}}{63\,x^3}-\frac {16\,b\,c\,\sqrt {c+\frac {d}{x}}}{35\,x^2}-\frac {2\,a\,d\,\sqrt {c+\frac {d}{x}}}{9\,x^4}-\frac {2\,b\,d\,\sqrt {c+\frac {d}{x}}}{7\,x^3}-\frac {2\,a\,c^2\,\sqrt {c+\frac {d}{x}}}{105\,d\,x^2}+\frac {8\,a\,c^3\,\sqrt {c+\frac {d}{x}}}{315\,d^2\,x}-\frac {2\,b\,c^2\,\sqrt {c+\frac {d}{x}}}{35\,d\,x} \] Input:

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

Output:

(4*b*c^3*(c + d/x)^(1/2))/(35*d^2) - (16*a*c^4*(c + d/x)^(1/2))/(315*d^3) 
- (20*a*c*(c + d/x)^(1/2))/(63*x^3) - (16*b*c*(c + d/x)^(1/2))/(35*x^2) - 
(2*a*d*(c + d/x)^(1/2))/(9*x^4) - (2*b*d*(c + d/x)^(1/2))/(7*x^3) - (2*a*c 
^2*(c + d/x)^(1/2))/(105*d*x^2) + (8*a*c^3*(c + d/x)^(1/2))/(315*d^2*x) - 
(2*b*c^2*(c + d/x)^(1/2))/(35*d*x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.54 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^4} \, dx=\frac {-\frac {16 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{4} x^{4}}{315}+\frac {8 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{3} d \,x^{3}}{315}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{2} d^{2} x^{2}}{105}-\frac {20 \sqrt {x}\, \sqrt {c x +d}\, a c \,d^{3} x}{63}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,d^{4}}{9}+\frac {4 \sqrt {x}\, \sqrt {c x +d}\, b \,c^{3} d \,x^{4}}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, b \,c^{2} d^{2} x^{3}}{35}-\frac {16 \sqrt {x}\, \sqrt {c x +d}\, b c \,d^{3} x^{2}}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, b \,d^{4} x}{7}+\frac {16 \sqrt {c}\, a \,c^{4} x^{5}}{315}-\frac {4 \sqrt {c}\, b \,c^{3} d \,x^{5}}{35}}{d^{3} x^{5}} \] Input:

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

Output:

(2*( - 8*sqrt(x)*sqrt(c*x + d)*a*c**4*x**4 + 4*sqrt(x)*sqrt(c*x + d)*a*c** 
3*d*x**3 - 3*sqrt(x)*sqrt(c*x + d)*a*c**2*d**2*x**2 - 50*sqrt(x)*sqrt(c*x 
+ d)*a*c*d**3*x - 35*sqrt(x)*sqrt(c*x + d)*a*d**4 + 18*sqrt(x)*sqrt(c*x + 
d)*b*c**3*d*x**4 - 9*sqrt(x)*sqrt(c*x + d)*b*c**2*d**2*x**3 - 72*sqrt(x)*s 
qrt(c*x + d)*b*c*d**3*x**2 - 45*sqrt(x)*sqrt(c*x + d)*b*d**4*x + 8*sqrt(c) 
*a*c**4*x**5 - 18*sqrt(c)*b*c**3*d*x**5))/(315*d**3*x**5)