Integrand size = 19, antiderivative size = 136 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2 a^4 \left (a x^2+b x^3\right )^{7/2}}{7 b^5 x^7}-\frac {8 a^3 \left (a x^2+b x^3\right )^{9/2}}{9 b^5 x^9}+\frac {12 a^2 \left (a x^2+b x^3\right )^{11/2}}{11 b^5 x^{11}}-\frac {8 a \left (a x^2+b x^3\right )^{13/2}}{13 b^5 x^{13}}+\frac {2 \left (a x^2+b x^3\right )^{15/2}}{15 b^5 x^{15}} \] Output:
2/7*a^4*(b*x^3+a*x^2)^(7/2)/b^5/x^7-8/9*a^3*(b*x^3+a*x^2)^(9/2)/b^5/x^9+12 /11*a^2*(b*x^3+a*x^2)^(11/2)/b^5/x^11-8/13*a*(b*x^3+a*x^2)^(13/2)/b^5/x^13 +2/15*(b*x^3+a*x^2)^(15/2)/b^5/x^15
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2 x (a+b x)^4 \left (128 a^4-448 a^3 b x+1008 a^2 b^2 x^2-1848 a b^3 x^3+3003 b^4 x^4\right )}{45045 b^5 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(a*x^2 + b*x^3)^(5/2)/x,x]
Output:
(2*x*(a + b*x)^4*(128*a^4 - 448*a^3*b*x + 1008*a^2*b^2*x^2 - 1848*a*b^3*x^ 3 + 3003*b^4*x^4))/(45045*b^5*Sqrt[x^2*(a + b*x)])
Time = 0.57 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1922, 1922, 1922, 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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}-\frac {8 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^2}dx}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^3}dx}{13 b}\right )}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^4}dx}{11 b}\right )}{13 b}\right )}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^5}dx}{9 b}\right )}{11 b}\right )}{13 b}\right )}{15 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{7/2}}{15 b x^3}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{13 b x^4}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{11 b x^5}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{7/2}}{9 b x^6}-\frac {4 a \left (a x^2+b x^3\right )^{7/2}}{63 b^2 x^7}\right )}{11 b}\right )}{13 b}\right )}{15 b}\) |
Input:
Int[(a*x^2 + b*x^3)^(5/2)/x,x]
Output:
(2*(a*x^2 + b*x^3)^(7/2))/(15*b*x^3) - (8*a*((2*(a*x^2 + b*x^3)^(7/2))/(13 *b*x^4) - (6*a*((2*(a*x^2 + b*x^3)^(7/2))/(11*b*x^5) - (4*a*((-4*a*(a*x^2 + b*x^3)^(7/2))/(63*b^2*x^7) + (2*(a*x^2 + b*x^3)^(7/2))/(9*b*x^6)))/(11*b )))/(13*b)))/(15*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])
Result contains higher order function than in optimal. Order 3 vs. order 2.
Time = 0.39 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.35
method | result | size |
pseudoelliptic | \(-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (3 b^{2} x^{2}+11 a b x +23 a^{2}\right )}{15}\) | \(47\) |
gosper | \(\frac {2 \left (b x +a \right ) \left (3003 b^{4} x^{4}-1848 a \,b^{3} x^{3}+1008 a^{2} b^{2} x^{2}-448 a^{3} b x +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(68\) |
default | \(\frac {2 \left (b x +a \right ) \left (3003 b^{4} x^{4}-1848 a \,b^{3} x^{3}+1008 a^{2} b^{2} x^{2}-448 a^{3} b x +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(68\) |
orering | \(\frac {2 \left (b x +a \right ) \left (3003 b^{4} x^{4}-1848 a \,b^{3} x^{3}+1008 a^{2} b^{2} x^{2}-448 a^{3} b x +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}{45045 b^{5} x^{5}}\) | \(68\) |
risch | \(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (3003 x^{7} b^{7}+7161 b^{6} a \,x^{6}+4473 a^{2} x^{5} b^{5}+35 b^{4} x^{4} a^{3}-40 b^{3} x^{3} a^{4}+48 x^{2} b^{2} a^{5}-64 x b \,a^{6}+128 a^{7}\right )}{45045 x \,b^{5}}\) | \(94\) |
trager | \(\frac {2 \left (3003 x^{7} b^{7}+7161 b^{6} a \,x^{6}+4473 a^{2} x^{5} b^{5}+35 b^{4} x^{4} a^{3}-40 b^{3} x^{3} a^{4}+48 x^{2} b^{2} a^{5}-64 x b \,a^{6}+128 a^{7}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{45045 b^{5} x}\) | \(96\) |
Input:
int((b*x^3+a*x^2)^(5/2)/x,x,method=_RETURNVERBOSE)
Output:
-2*a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2/15*(b*x+a)^(1/2)*(3*b^2*x^2+11 *a*b*x+23*a^2)
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2 \, {\left (3003 \, b^{7} x^{7} + 7161 \, a b^{6} x^{6} + 4473 \, a^{2} b^{5} x^{5} + 35 \, a^{3} b^{4} x^{4} - 40 \, a^{4} b^{3} x^{3} + 48 \, a^{5} b^{2} x^{2} - 64 \, a^{6} b x + 128 \, a^{7}\right )} \sqrt {b x^{3} + a x^{2}}}{45045 \, b^{5} x} \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x,x, algorithm="fricas")
Output:
2/45045*(3003*b^7*x^7 + 7161*a*b^6*x^6 + 4473*a^2*b^5*x^5 + 35*a^3*b^4*x^4 - 40*a^4*b^3*x^3 + 48*a^5*b^2*x^2 - 64*a^6*b*x + 128*a^7)*sqrt(b*x^3 + a* x^2)/(b^5*x)
\[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}{x}\, dx \] Input:
integrate((b*x**3+a*x**2)**(5/2)/x,x)
Output:
Integral((x**2*(a + b*x))**(5/2)/x, x)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.63 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2 \, {\left (3003 \, b^{7} x^{7} + 7161 \, a b^{6} x^{6} + 4473 \, a^{2} b^{5} x^{5} + 35 \, a^{3} b^{4} x^{4} - 40 \, a^{4} b^{3} x^{3} + 48 \, a^{5} b^{2} x^{2} - 64 \, a^{6} b x + 128 \, a^{7}\right )} \sqrt {b x + a}}{45045 \, b^{5}} \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x,x, algorithm="maxima")
Output:
2/45045*(3003*b^7*x^7 + 7161*a*b^6*x^6 + 4473*a^2*b^5*x^5 + 35*a^3*b^4*x^4 - 40*a^4*b^3*x^3 + 48*a^5*b^2*x^2 - 64*a^6*b*x + 128*a^7)*sqrt(b*x + a)/b ^5
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (116) = 232\).
Time = 0.13 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.56 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=-\frac {256 \, a^{\frac {15}{2}} \mathrm {sgn}\left (x\right )}{45045 \, b^{5}} + \frac {2 \, {\left (\frac {143 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{3} \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {195 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {45 \, {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} a \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {7 \, {\left (429 \, {\left (b x + a\right )}^{\frac {15}{2}} - 3465 \, {\left (b x + a\right )}^{\frac {13}{2}} a + 12285 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{2} - 25025 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{3} + 32175 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 27027 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + 15015 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} - 6435 \, \sqrt {b x + a} a^{7}\right )} \mathrm {sgn}\left (x\right )}{b^{4}}\right )}}{45045 \, b} \] Input:
integrate((b*x^3+a*x^2)^(5/2)/x,x, algorithm="giac")
Output:
-256/45045*a^(15/2)*sgn(x)/b^5 + 2/45045*(143*(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)*a^3*sgn(x)/b^4 + 195*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b* x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a^2*sgn(x)/b^4 + 45*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 30 03*sqrt(b*x + a)*a^6)*a*sgn(x)/b^4 + 7*(429*(b*x + a)^(15/2) - 3465*(b*x + a)^(13/2)*a + 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/2)*a^3 + 32 175*(b*x + a)^(7/2)*a^4 - 27027*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2 )*a^6 - 6435*sqrt(b*x + a)*a^7)*sgn(x)/b^4)/b
Time = 9.00 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^3\,\left (128\,a^4-448\,a^3\,b\,x+1008\,a^2\,b^2\,x^2-1848\,a\,b^3\,x^3+3003\,b^4\,x^4\right )}{45045\,b^5\,x} \] Input:
int((a*x^2 + b*x^3)^(5/2)/x,x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^3*(128*a^4 + 3003*b^4*x^4 - 1848*a*b^3* x^3 + 1008*a^2*b^2*x^2 - 448*a^3*b*x))/(45045*b^5*x)
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x} \, dx=\frac {2 \sqrt {b x +a}\, \left (3003 b^{7} x^{7}+7161 a \,b^{6} x^{6}+4473 a^{2} b^{5} x^{5}+35 a^{3} b^{4} x^{4}-40 a^{4} b^{3} x^{3}+48 a^{5} b^{2} x^{2}-64 a^{6} b x +128 a^{7}\right )}{45045 b^{5}} \] Input:
int((b*x^3+a*x^2)^(5/2)/x,x)
Output:
(2*sqrt(a + b*x)*(128*a**7 - 64*a**6*b*x + 48*a**5*b**2*x**2 - 40*a**4*b** 3*x**3 + 35*a**3*b**4*x**4 + 4473*a**2*b**5*x**5 + 7161*a*b**6*x**6 + 3003 *b**7*x**7))/(45045*b**5)