Integrand size = 24, antiderivative size = 60 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 (b c-a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^2 x^7}+\frac {2 d \left (a x^2+b x^3\right )^{9/2}}{9 b^2 x^9} \] Output:
2/7*(-a*d+b*c)*(b*x^3+a*x^2)^(7/2)/b^2/x^7+2/9*d*(b*x^3+a*x^2)^(9/2)/b^2/x ^9
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 x (a+b x)^4 (9 b c-2 a d+7 b d x)}{63 b^2 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^5,x]
Output:
(2*x*(a + b*x)^4*(9*b*c - 2*a*d + 7*b*d*x))/(63*b^2*Sqrt[x^2*(a + b*x)])
Time = 0.35 (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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2} (c+d x)}{x^5} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(9 b c-2 a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^5}dx}{9 b}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2} (9 b c-2 a d)}{63 b^2 x^7}+\frac {2 d \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^5,x]
Output:
(2*(9*b*c - 2*a*d)*(a*x^2 + b*x^3)^(7/2))/(63*b^2*x^7) + (2*d*(a*x^2 + b*x ^3)^(7/2))/(9*b*x^6)
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 = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-7 b d x +2 a d -9 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{63 b^{2} x^{5}}\) | \(41\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-7 b d x +2 a d -9 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{63 b^{2} x^{5}}\) | \(41\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-7 b d x +2 a d -9 b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{63 b^{2} x^{5}}\) | \(41\) |
pseudoelliptic | \(-\frac {5 \left (b^{3} x^{4} \left (a d -\frac {b c}{8}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {17 \sqrt {b x +a}\, \left (\frac {59 x^{2} \left (\frac {132 d x}{59}+c \right ) b^{2} a^{\frac {3}{2}}}{68}+b x \left (\frac {26 d x}{17}+c \right ) a^{\frac {5}{2}}+\frac {2 \left (4 d x +3 c \right ) a^{\frac {7}{2}}}{17}+\frac {15 \sqrt {a}\, b^{3} c \,x^{3}}{136}\right )}{15}\right )}{8 a^{\frac {3}{2}} x^{4}}\) | \(101\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-7 d \,x^{4} b^{4}-19 a \,b^{3} d \,x^{3}-9 b^{4} c \,x^{3}-15 a^{2} b^{2} d \,x^{2}-27 a \,b^{3} c \,x^{2}-a^{3} b d x -27 a^{2} b^{2} c x +2 a^{4} d -9 a^{3} b c \right )}{63 x \,b^{2}}\) | \(102\) |
trager | \(-\frac {2 \left (-7 d \,x^{4} b^{4}-19 a \,b^{3} d \,x^{3}-9 b^{4} c \,x^{3}-15 a^{2} b^{2} d \,x^{2}-27 a \,b^{3} c \,x^{2}-a^{3} b d x -27 a^{2} b^{2} c x +2 a^{4} d -9 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{63 b^{2} x}\) | \(104\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-2/63*(b*x+a)*(-7*b*d*x+2*a*d-9*b*c)*(b*x^3+a*x^2)^(5/2)/b^2/x^5
Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.70 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 \, {\left (7 \, b^{4} d x^{4} + 9 \, a^{3} b c - 2 \, a^{4} d + {\left (9 \, b^{4} c + 19 \, a b^{3} d\right )} x^{3} + 3 \, {\left (9 \, a b^{3} c + 5 \, a^{2} b^{2} d\right )} x^{2} + {\left (27 \, a^{2} b^{2} c + a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{63 \, b^{2} x} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^5,x, algorithm="fricas")
Output:
2/63*(7*b^4*d*x^4 + 9*a^3*b*c - 2*a^4*d + (9*b^4*c + 19*a*b^3*d)*x^3 + 3*( 9*a*b^3*c + 5*a^2*b^2*d)*x^2 + (27*a^2*b^2*c + a^3*b*d)*x)*sqrt(b*x^3 + a* x^2)/(b^2*x)
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{5}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**5,x)
Output:
Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**5, x)
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.57 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b x + a} c}{7 \, b} + \frac {2 \, {\left (7 \, b^{4} x^{4} + 19 \, a b^{3} x^{3} + 15 \, a^{2} b^{2} x^{2} + a^{3} b x - 2 \, a^{4}\right )} \sqrt {b x + a} d}{63 \, b^{2}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^5,x, algorithm="maxima")
Output:
2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)*c/b + 2/63*(7* b^4*x^4 + 19*a*b^3*x^3 + 15*a^2*b^2*x^2 + a^3*b*x - 2*a^4)*sqrt(b*x + a)*d /b^2
Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (52) = 104\).
Time = 0.12 (sec) , antiderivative size = 344, normalized size of antiderivative = 5.73 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 \, {\left (315 \, \sqrt {b x + a} a^{3} c \mathrm {sgn}\left (x\right ) + 315 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} c \mathrm {sgn}\left (x\right ) + \frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{3} d \mathrm {sgn}\left (x\right )}{b} + 63 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a c \mathrm {sgn}\left (x\right ) + \frac {63 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2} d \mathrm {sgn}\left (x\right )}{b} + 9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} c \mathrm {sgn}\left (x\right ) + \frac {27 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a d \mathrm {sgn}\left (x\right )}{b} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} d \mathrm {sgn}\left (x\right )}{b}\right )}}{315 \, b} - \frac {2 \, {\left (9 \, a^{\frac {7}{2}} b c - 2 \, a^{\frac {9}{2}} d\right )} \mathrm {sgn}\left (x\right )}{63 \, b^{2}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^5,x, algorithm="giac")
Output:
2/315*(315*sqrt(b*x + a)*a^3*c*sgn(x) + 315*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2*c*sgn(x) + 105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^3*d*sgn (x)/b + 63*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^ 2)*a*c*sgn(x) + 63*(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 + 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) + 27*(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*sgn(x)/b + (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)/b - 2/63*(9*a^(7/2)*b*c - 2*a^(9/2)*d)*sgn(x)/b^2
Time = 9.56 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.05 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {6\,a^2\,c\,\sqrt {b\,x^3+a\,x^2}}{7}+\frac {2\,b^2\,c\,x^2\,\sqrt {b\,x^3+a\,x^2}}{7}+\frac {2\,a^3\,c\,\sqrt {b\,x^3+a\,x^2}}{7\,b\,x}+\frac {6\,a\,b\,c\,x\,\sqrt {b\,x^3+a\,x^2}}{7}-\frac {2\,d\,\left (2\,a-7\,b\,x\right )\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^3}{63\,b^2\,x} \] Input:
int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^5,x)
Output:
(6*a^2*c*(a*x^2 + b*x^3)^(1/2))/7 + (2*b^2*c*x^2*(a*x^2 + b*x^3)^(1/2))/7 + (2*a^3*c*(a*x^2 + b*x^3)^(1/2))/(7*b*x) + (6*a*b*c*x*(a*x^2 + b*x^3)^(1/ 2))/7 - (2*d*(2*a - 7*b*x)*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^3)/(63*b^2*x)
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.53 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^5} \, dx=\frac {2 \sqrt {b x +a}\, \left (7 b^{4} d \,x^{4}+19 a \,b^{3} d \,x^{3}+9 b^{4} c \,x^{3}+15 a^{2} b^{2} d \,x^{2}+27 a \,b^{3} c \,x^{2}+a^{3} b d x +27 a^{2} b^{2} c x -2 a^{4} d +9 a^{3} b c \right )}{63 b^{2}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^5,x)
Output:
(2*sqrt(a + b*x)*( - 2*a**4*d + 9*a**3*b*c + a**3*b*d*x + 27*a**2*b**2*c*x + 15*a**2*b**2*d*x**2 + 27*a*b**3*c*x**2 + 19*a*b**3*d*x**3 + 9*b**4*c*x* *3 + 7*b**4*d*x**4))/(63*b**2)