Integrand size = 24, antiderivative size = 60 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 (b c-a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^2 x^5}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{7 b^2 x^7} \] Output:
2/5*(-a*d+b*c)*(b*x^3+a*x^2)^(5/2)/b^2/x^5+2/7*d*(b*x^3+a*x^2)^(7/2)/b^2/x ^7
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 x (a+b x)^3 (7 b c-2 a d+5 b d x)}{35 b^2 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^3,x]
Output:
(2*x*(a + b*x)^3*(7*b*c - 2*a*d + 5*b*d*x))/(35*b^2*Sqrt[x^2*(a + b*x)])
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1945, 1920}
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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{x^3} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(7 b c-2 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2} (7 b c-2 a d)}{35 b^2 x^5}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^3,x]
Output:
(2*(7*b*c - 2*a*d)*(a*x^2 + b*x^3)^(5/2))/(35*b^2*x^5) + (2*d*(a*x^2 + b*x ^3)^(5/2))/(7*b*x^4)
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.46 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-5 b d x +2 a d -7 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}}\) | \(41\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-5 b d x +2 a d -7 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}}\) | \(41\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-5 b d x +2 a d -7 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}}\) | \(41\) |
pseudoelliptic | \(-\frac {3 \left (b \,x^{2} \left (a d +\frac {b c}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {5 \left (\frac {2 \left (2 d x +c \right ) a^{\frac {3}{2}}}{5}+b x \sqrt {a}\, \left (-\frac {8 d x}{5}+c \right )\right ) \sqrt {b x +a}}{12}\right )}{\sqrt {a}\, x^{2}}\) | \(68\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-5 b^{3} d \,x^{3}-8 a \,b^{2} d \,x^{2}-7 b^{3} c \,x^{2}-a^{2} b d x -14 a \,b^{2} c x +2 a^{3} d -7 c \,a^{2} b \right )}{35 x \,b^{2}}\) | \(78\) |
trager | \(-\frac {2 \left (-5 b^{3} d \,x^{3}-8 a \,b^{2} d \,x^{2}-7 b^{3} c \,x^{2}-a^{2} b d x -14 a \,b^{2} c x +2 a^{3} d -7 c \,a^{2} b \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{35 b^{2} x}\) | \(80\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-2/35*(b*x+a)*(-5*b*d*x+2*a*d-7*b*c)*(b*x^3+a*x^2)^(3/2)/b^2/x^3
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 \, {\left (5 \, b^{3} d x^{3} + 7 \, a^{2} b c - 2 \, a^{3} d + {\left (7 \, b^{3} c + 8 \, a b^{2} d\right )} x^{2} + {\left (14 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, b^{2} x} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^3,x, algorithm="fricas")
Output:
2/35*(5*b^3*d*x^3 + 7*a^2*b*c - 2*a^3*d + (7*b^3*c + 8*a*b^2*d)*x^2 + (14* a*b^2*c + a^2*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^2*x)
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{3}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/x**3,x)
Output:
Integral((x**2*(a + b*x))**(3/2)*(c + d*x)/x**3, x)
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a} c}{5 \, b} + \frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a} d}{35 \, b^{2}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^3,x, algorithm="maxima")
Output:
2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)*c/b + 2/35*(5*b^3*x^3 + 8*a*b^ 2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)*d/b^2
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (52) = 104\).
Time = 0.13 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.77 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 \, {\left (105 \, \sqrt {b x + a} a^{2} c \mathrm {sgn}\left (x\right ) + 70 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a c \mathrm {sgn}\left (x\right ) + \frac {35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} d \mathrm {sgn}\left (x\right )}{b} + 7 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} c \mathrm {sgn}\left (x\right ) + \frac {14 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a d \mathrm {sgn}\left (x\right )}{b} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} d \mathrm {sgn}\left (x\right )}{b}\right )}}{105 \, b} - \frac {2 \, {\left (7 \, a^{\frac {5}{2}} b c - 2 \, a^{\frac {7}{2}} d\right )} \mathrm {sgn}\left (x\right )}{35 \, b^{2}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^3,x, algorithm="giac")
Output:
2/105*(105*sqrt(b*x + a)*a^2*c*sgn(x) + 70*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a*c*sgn(x) + 35*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2*d*sgn(x)/ b + 7*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*c* sgn(x) + 14*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a ^2)*a*d*sgn(x)/b + 3*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*d*sgn(x)/b)/b - 2/35*(7*a^(5/2)*b*c - 2*a^(7/2)*d)*sgn(x)/b^2
Time = 9.50 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (7\,b\,c-2\,a\,d+5\,b\,d\,x\right )}{35\,b^2\,x} \] Input:
int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^3,x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(7*b*c - 2*a*d + 5*b*d*x))/(35*b^2*x)
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx=\frac {2 \sqrt {b x +a}\, \left (5 b^{3} d \,x^{3}+8 a \,b^{2} d \,x^{2}+7 b^{3} c \,x^{2}+a^{2} b d x +14 a \,b^{2} c x -2 a^{3} d +7 a^{2} b c \right )}{35 b^{2}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^3,x)
Output:
(2*sqrt(a + b*x)*( - 2*a**3*d + 7*a**2*b*c + a**2*b*d*x + 14*a*b**2*c*x + 8*a*b**2*d*x**2 + 7*b**3*c*x**2 + 5*b**3*d*x**3))/(35*b**2)