Integrand size = 24, antiderivative size = 94 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=-\frac {2 a (b c-a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^3 x^5}+\frac {2 (b c-2 a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^3 x^7}+\frac {2 d \left (a x^2+b x^3\right )^{9/2}}{9 b^3 x^9} \] Output:
-2/5*a*(-a*d+b*c)*(b*x^3+a*x^2)^(5/2)/b^3/x^5+2/7*(-2*a*d+b*c)*(b*x^3+a*x^ 2)^(7/2)/b^3/x^7+2/9*d*(b*x^3+a*x^2)^(9/2)/b^3/x^9
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 x (a+b x)^3 \left (8 a^2 d+5 b^2 x (9 c+7 d x)-2 a b (9 c+10 d x)\right )}{315 b^3 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^2,x]
Output:
(2*x*(a + b*x)^3*(8*a^2*d + 5*b^2*x*(9*c + 7*d*x) - 2*a*b*(9*c + 10*d*x))) /(315*b^3*Sqrt[x^2*(a + b*x)])
Time = 0.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1945, 1922, 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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(9 b c-4 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^2}dx}{9 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(9 b c-4 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}\right )}{9 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}\right ) (9 b c-4 a d)}{9 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^2,x]
Output:
(2*d*(a*x^2 + b*x^3)^(5/2))/(9*b*x^3) + ((9*b*c - 4*a*d)*((-4*a*(a*x^2 + b *x^3)^(5/2))/(35*b^2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)))/(9*b)
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[((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*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || 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 = 0.54 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (35 b^{2} d \,x^{2}-20 a b d x +45 b^{2} c x +8 a^{2} d -18 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{3}}\) | \(61\) |
default | \(\frac {2 \left (b x +a \right ) \left (35 b^{2} d \,x^{2}-20 a b d x +45 b^{2} c x +8 a^{2} d -18 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{3}}\) | \(61\) |
orering | \(\frac {2 \left (b x +a \right ) \left (35 b^{2} d \,x^{2}-20 a b d x +45 b^{2} c x +8 a^{2} d -18 a b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{3}}\) | \(61\) |
pseudoelliptic | \(-\frac {2 \left (a x \left (a d +\frac {3 b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\left (\left (\frac {4 d x}{3}-\frac {c}{2}\right ) a^{\frac {3}{2}}+b x \sqrt {a}\, \left (\frac {d x}{3}+c \right )\right ) \sqrt {b x +a}\right )}{\sqrt {a}\, x}\) | \(67\) |
risch | \(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (35 d \,x^{4} b^{4}+50 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+3 a^{2} b^{2} d \,x^{2}+72 a \,b^{3} c \,x^{2}-4 a^{3} b d x +9 a^{2} b^{2} c x +8 a^{4} d -18 a^{3} b c \right )}{315 x \,b^{3}}\) | \(102\) |
trager | \(\frac {2 \left (35 d \,x^{4} b^{4}+50 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+3 a^{2} b^{2} d \,x^{2}+72 a \,b^{3} c \,x^{2}-4 a^{3} b d x +9 a^{2} b^{2} c x +8 a^{4} d -18 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{3} x}\) | \(104\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^2,x,method=_RETURNVERBOSE)
Output:
2/315*(b*x+a)*(35*b^2*d*x^2-20*a*b*d*x+45*b^2*c*x+8*a^2*d-18*a*b*c)*(b*x^3 +a*x^2)^(3/2)/b^3/x^3
Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (35 \, b^{4} d x^{4} - 18 \, a^{3} b c + 8 \, a^{4} d + 5 \, {\left (9 \, b^{4} c + 10 \, a b^{3} d\right )} x^{3} + 3 \, {\left (24 \, a b^{3} c + a^{2} b^{2} d\right )} x^{2} + {\left (9 \, a^{2} b^{2} c - 4 \, a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{315 \, b^{3} x} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^2,x, algorithm="fricas")
Output:
2/315*(35*b^4*d*x^4 - 18*a^3*b*c + 8*a^4*d + 5*(9*b^4*c + 10*a*b^3*d)*x^3 + 3*(24*a*b^3*c + a^2*b^2*d)*x^2 + (9*a^2*b^2*c - 4*a^3*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^3*x)
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{2}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/x**2,x)
Output:
Integral((x**2*(a + b*x))**(3/2)*(c + d*x)/x**2, x)
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a} c}{35 \, b^{2}} + \frac {2 \, {\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a} d}{315 \, b^{3}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^2,x, algorithm="maxima")
Output:
2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)*c/b^2 + 2/3 15*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b* x + a)*d/b^3
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (82) = 164\).
Time = 0.14 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.29 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (\frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} c \mathrm {sgn}\left (x\right )}{b} + \frac {42 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a c \mathrm {sgn}\left (x\right )}{b} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2} d \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} c \mathrm {sgn}\left (x\right )}{b} + \frac {18 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a d \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} d \mathrm {sgn}\left (x\right )}{b^{2}}\right )}}{315 \, b} + \frac {4 \, {\left (9 \, a^{\frac {7}{2}} b c - 4 \, a^{\frac {9}{2}} d\right )} \mathrm {sgn}\left (x\right )}{315 \, b^{3}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^2,x, algorithm="giac")
Output:
2/315*(105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2*c*sgn(x)/b + 42*(3*(b *x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a*c*sgn(x)/b + 21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^2 *d*sgn(x)/b^2 + 9*(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)*c*sgn(x)/b + 18*(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)*a*d*sg n(x)/b^2 + (35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/ 2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*d*sgn(x)/b^2)/b + 4/315*(9*a^(7/2)*b*c - 4*a^(9/2)*d)*sgn(x)/b^3
Time = 9.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {16\,a^4\,d-36\,a^3\,b\,c}{315\,b^3}+\frac {x^3\,\left (90\,c\,b^4+100\,a\,d\,b^3\right )}{315\,b^3}+\frac {2\,b\,d\,x^4}{9}-\frac {2\,a^2\,x\,\left (4\,a\,d-9\,b\,c\right )}{315\,b^2}+\frac {2\,a\,x^2\,\left (a\,d+24\,b\,c\right )}{105\,b}\right )}{x} \] Input:
int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^2,x)
Output:
((a*x^2 + b*x^3)^(1/2)*((16*a^4*d - 36*a^3*b*c)/(315*b^3) + (x^3*(90*b^4*c + 100*a*b^3*d))/(315*b^3) + (2*b*d*x^4)/9 - (2*a^2*x*(4*a*d - 9*b*c))/(31 5*b^2) + (2*a*x^2*(a*d + 24*b*c))/(105*b)))/x
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \sqrt {b x +a}\, \left (35 b^{4} d \,x^{4}+50 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+3 a^{2} b^{2} d \,x^{2}+72 a \,b^{3} c \,x^{2}-4 a^{3} b d x +9 a^{2} b^{2} c x +8 a^{4} d -18 a^{3} b c \right )}{315 b^{3}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^2,x)
Output:
(2*sqrt(a + b*x)*(8*a**4*d - 18*a**3*b*c - 4*a**3*b*d*x + 9*a**2*b**2*c*x + 3*a**2*b**2*d*x**2 + 72*a*b**3*c*x**2 + 50*a*b**3*d*x**3 + 45*b**4*c*x** 3 + 35*b**4*d*x**4))/(315*b**3)