Integrand size = 20, antiderivative size = 46 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {2 (a c-b d) \left (c+\frac {d}{x}\right )^{5/2}}{5 d^2}-\frac {2 a \left (c+\frac {d}{x}\right )^{7/2}}{7 d^2} \] Output:
2/5*(a*c-b*d)*(c+d/x)^(5/2)/d^2-2/7*a*(c+d/x)^(7/2)/d^2
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {2 (d+c x)^2 \sqrt {\frac {d+c x}{x}} (5 a d-2 a c x+7 b d x)}{35 d^2 x^3} \] Input:
Integrate[((c + d/x)^(3/2)*(a + b*x))/x^3,x]
Output:
(-2*(d + c*x)^2*Sqrt[(d + c*x)/x]*(5*a*d - 2*a*c*x + 7*b*d*x))/(35*d^2*x^3 )
Time = 0.34 (sec) , antiderivative size = 46, 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, 946, 53, 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.
\(\displaystyle \int \frac {(a+b x) \left (c+\frac {d}{x}\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 1016 |
\(\displaystyle \int \frac {\left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2}}{x^2}dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle -\int \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2}d\frac {1}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\int \left (\frac {a \left (c+\frac {d}{x}\right )^{5/2}}{d}+\frac {(b d-a c) \left (c+\frac {d}{x}\right )^{3/2}}{d}\right )d\frac {1}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (c+\frac {d}{x}\right )^{5/2} (a c-b d)}{5 d^2}-\frac {2 a \left (c+\frac {d}{x}\right )^{7/2}}{7 d^2}\) |
Input:
Int[((c + d/x)^(3/2)*(a + b*x))/x^3,x]
Output:
(2*(a*c - b*d)*(c + d/x)^(5/2))/(5*d^2) - (2*a*(c + d/x)^(7/2))/(7*d^2)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[(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] && EqQ[m - n + 1, 0]
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])
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83
method | result | size |
orering | \(\frac {2 \left (2 a c x -7 b d x -5 a d \right ) \left (c x +d \right ) \left (c +\frac {d}{x}\right )^{\frac {3}{2}}}{35 d^{2} x^{2}}\) | \(38\) |
gosper | \(\frac {2 \left (c x +d \right ) \left (2 a c x -7 b d x -5 a d \right ) \left (\frac {c x +d}{x}\right )^{\frac {3}{2}}}{35 d^{2} x^{2}}\) | \(40\) |
default | \(\frac {2 \sqrt {\frac {c x +d}{x}}\, \left (c \,x^{2}+d x \right )^{\frac {3}{2}} \left (2 a \,c^{2} x^{2}-7 b c d \,x^{2}-3 a d x c -7 b x \,d^{2}-5 a \,d^{2}\right )}{35 x^{4} d^{2} \sqrt {\left (c x +d \right ) x}}\) | \(77\) |
risch | \(\frac {2 \sqrt {\frac {c x +d}{x}}\, \left (2 a \,c^{3} x^{3}-7 b \,c^{2} d \,x^{3}-a \,c^{2} d \,x^{2}-14 b c \,d^{2} x^{2}-8 a c \,d^{2} x -7 b \,d^{3} x -5 a \,d^{3}\right )}{35 x^{3} d^{2}}\) | \(81\) |
trager | \(\frac {2 \left (2 a \,c^{3} x^{3}-7 b \,c^{2} d \,x^{3}-a \,c^{2} d \,x^{2}-14 b c \,d^{2} x^{2}-8 a c \,d^{2} x -7 b \,d^{3} x -5 a \,d^{3}\right ) \sqrt {-\frac {-c x -d}{x}}}{35 x^{3} d^{2}}\) | \(85\) |
Input:
int((c+d/x)^(3/2)*(b*x+a)/x^3,x,method=_RETURNVERBOSE)
Output:
2/35*(2*a*c*x-7*b*d*x-5*a*d)/d^2/x^2*(c*x+d)*(c+d/x)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (38) = 76\).
Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.72 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {2 \, {\left (5 \, a d^{3} - {\left (2 \, a c^{3} - 7 \, b c^{2} d\right )} x^{3} + {\left (a c^{2} d + 14 \, b c d^{2}\right )} x^{2} + {\left (8 \, a c d^{2} + 7 \, b d^{3}\right )} x\right )} \sqrt {\frac {c x + d}{x}}}{35 \, d^{2} x^{3}} \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^3,x, algorithm="fricas")
Output:
-2/35*(5*a*d^3 - (2*a*c^3 - 7*b*c^2*d)*x^3 + (a*c^2*d + 14*b*c*d^2)*x^2 + (8*a*c*d^2 + 7*b*d^3)*x)*sqrt((c*x + d)/x)/(d^2*x^3)
Time = 2.06 (sec) , antiderivative size = 167, normalized size of antiderivative = 3.63 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=- a c \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{2}} & \text {otherwise} \end {cases}\right ) - a d \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) - b c \left (\begin {cases} \frac {2 \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{x} & \text {otherwise} \end {cases}\right ) - b d \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{2}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((c+d/x)**(3/2)*(b*x+a)/x**3,x)
Output:
-a*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)) - a*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)) - b*c*Piecewise((2*(c + d/x)**(3/2)/(3*d), Ne(d, 0)), (s qrt(c)/x, True)) - b*d*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))
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=-\frac {2 \, b {\left (c + \frac {d}{x}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {2}{35} \, {\left (\frac {5 \, {\left (c + \frac {d}{x}\right )}^{\frac {7}{2}}}{d^{2}} - \frac {7 \, {\left (c + \frac {d}{x}\right )}^{\frac {5}{2}} c}{d^{2}}\right )} a \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^3,x, algorithm="maxima")
Output:
-2/5*b*(c + d/x)^(5/2)/d - 2/35*(5*(c + d/x)^(7/2)/d^2 - 7*(c + d/x)^(5/2) *c/d^2)*a
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (38) = 76\).
Time = 0.53 (sec) , antiderivative size = 333, normalized size of antiderivative = 7.24 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{6} b c^{2} \mathrm {sgn}\left (x\right ) + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{5} a c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{5} b c^{\frac {3}{2}} d \mathrm {sgn}\left (x\right ) + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{4} a c^{2} d \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{4} b c d^{2} \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{3} a c^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{3} b \sqrt {c} d^{3} \mathrm {sgn}\left (x\right ) + 98 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{2} a c d^{3} \mathrm {sgn}\left (x\right ) + 7 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{2} b d^{4} \mathrm {sgn}\left (x\right ) + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )} a \sqrt {c} d^{4} \mathrm {sgn}\left (x\right ) + 5 \, a d^{5} \mathrm {sgn}\left (x\right )\right )}}{35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d x}\right )}^{7}} \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^3,x, algorithm="giac")
Output:
2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + d*x))^6*b*c^2*sgn(x) + 35*(sqrt(c)*x - sqrt(c*x^2 + d*x))^5*a*c^(5/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d*x)) ^5*b*c^(3/2)*d*sgn(x) + 105*(sqrt(c)*x - sqrt(c*x^2 + d*x))^4*a*c^2*d*sgn( x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d*x))^4*b*c*d^2*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + d*x))^3*a*c^(3/2)*d^2*sgn(x) + 35*(sqrt(c)*x - sqrt(c*x^2 + d*x))^3*b*sqrt(c)*d^3*sgn(x) + 98*(sqrt(c)*x - sqrt(c*x^2 + d*x))^2*a*c*d ^3*sgn(x) + 7*(sqrt(c)*x - sqrt(c*x^2 + d*x))^2*b*d^4*sgn(x) + 35*(sqrt(c) *x - sqrt(c*x^2 + d*x))*a*sqrt(c)*d^4*sgn(x) + 5*a*d^5*sgn(x))/(sqrt(c)*x - sqrt(c*x^2 + d*x))^7
Time = 7.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.65 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {4\,a\,c^3\,\sqrt {c+\frac {d}{x}}}{35\,d^2}-\frac {2\,b\,c^2\,\sqrt {c+\frac {d}{x}}}{5\,d}-\frac {16\,a\,c\,\sqrt {c+\frac {d}{x}}}{35\,x^2}-\frac {4\,b\,c\,\sqrt {c+\frac {d}{x}}}{5\,x}-\frac {2\,a\,d\,\sqrt {c+\frac {d}{x}}}{7\,x^3}-\frac {2\,b\,d\,\sqrt {c+\frac {d}{x}}}{5\,x^2}-\frac {2\,a\,c^2\,\sqrt {c+\frac {d}{x}}}{35\,d\,x} \] Input:
int(((c + d/x)^(3/2)*(a + b*x))/x^3,x)
Output:
(4*a*c^3*(c + d/x)^(1/2))/(35*d^2) - (2*b*c^2*(c + d/x)^(1/2))/(5*d) - (16 *a*c*(c + d/x)^(1/2))/(35*x^2) - (4*b*c*(c + d/x)^(1/2))/(5*x) - (2*a*d*(c + d/x)^(1/2))/(7*x^3) - (2*b*d*(c + d/x)^(1/2))/(5*x^2) - (2*a*c^2*(c + d /x)^(1/2))/(35*d*x)
Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 3.22 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^3} \, dx=\frac {\frac {4 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{3} x^{3}}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{2} d \,x^{2}}{35}-\frac {16 \sqrt {x}\, \sqrt {c x +d}\, a c \,d^{2} x}{35}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,d^{3}}{7}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, b \,c^{2} d \,x^{3}}{5}-\frac {4 \sqrt {x}\, \sqrt {c x +d}\, b c \,d^{2} x^{2}}{5}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, b \,d^{3} x}{5}-\frac {4 \sqrt {c}\, a \,c^{3} x^{4}}{35}-\frac {6 \sqrt {c}\, b \,c^{2} d \,x^{4}}{35}}{d^{2} x^{4}} \] Input:
int((c+d/x)^(3/2)*(b*x+a)/x^3,x)
Output:
(2*(2*sqrt(x)*sqrt(c*x + d)*a*c**3*x**3 - sqrt(x)*sqrt(c*x + d)*a*c**2*d*x **2 - 8*sqrt(x)*sqrt(c*x + d)*a*c*d**2*x - 5*sqrt(x)*sqrt(c*x + d)*a*d**3 - 7*sqrt(x)*sqrt(c*x + d)*b*c**2*d*x**3 - 14*sqrt(x)*sqrt(c*x + d)*b*c*d** 2*x**2 - 7*sqrt(x)*sqrt(c*x + d)*b*d**3*x - 2*sqrt(c)*a*c**3*x**4 - 3*sqrt (c)*b*c**2*d*x**4))/(35*d**2*x**4)