Integrand size = 15, antiderivative size = 108 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=-\frac {81 a^3 \left (a x^3+b x^6\right )^{14/3}}{54740 b^4 x^{14}}+\frac {27 a^2 \left (a x^3+b x^6\right )^{14/3}}{3910 b^3 x^{11}}-\frac {9 a \left (a x^3+b x^6\right )^{14/3}}{460 b^2 x^8}+\frac {\left (a x^3+b x^6\right )^{14/3}}{23 b x^5} \] Output:
-81/54740*a^3*(b*x^6+a*x^3)^(14/3)/b^4/x^14+27/3910*a^2*(b*x^6+a*x^3)^(14/ 3)/b^3/x^11-9/460*a*(b*x^6+a*x^3)^(14/3)/b^2/x^8+1/23*(b*x^6+a*x^3)^(14/3) /b/x^5
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.59 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\frac {\left (a+b x^3\right ) \left (x^3 \left (a+b x^3\right )\right )^{11/3} \left (-81 a^3+378 a^2 b x^3-1071 a b^2 x^6+2380 b^3 x^9\right )}{54740 b^4 x^{11}} \] Input:
Integrate[(a*x^3 + b*x^6)^(11/3),x]
Output:
((a + b*x^3)*(x^3*(a + b*x^3))^(11/3)*(-81*a^3 + 378*a^2*b*x^3 - 1071*a*b^ 2*x^6 + 2380*b^3*x^9))/(54740*b^4*x^11)
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1908, 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.
\(\displaystyle \int \left (a x^3+b x^6\right )^{11/3} \, dx\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {\left (a x^3+b x^6\right )^{14/3}}{23 b x^5}-\frac {9 a \int \frac {\left (b x^6+a x^3\right )^{11/3}}{x^3}dx}{23 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {\left (a x^3+b x^6\right )^{14/3}}{23 b x^5}-\frac {9 a \left (\frac {\left (a x^3+b x^6\right )^{14/3}}{20 b x^8}-\frac {3 a \int \frac {\left (b x^6+a x^3\right )^{11/3}}{x^6}dx}{10 b}\right )}{23 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {\left (a x^3+b x^6\right )^{14/3}}{23 b x^5}-\frac {9 a \left (\frac {\left (a x^3+b x^6\right )^{14/3}}{20 b x^8}-\frac {3 a \left (\frac {\left (a x^3+b x^6\right )^{14/3}}{17 b x^{11}}-\frac {3 a \int \frac {\left (b x^6+a x^3\right )^{11/3}}{x^9}dx}{17 b}\right )}{10 b}\right )}{23 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (a x^3+b x^6\right )^{14/3}}{23 b x^5}-\frac {9 a \left (\frac {\left (a x^3+b x^6\right )^{14/3}}{20 b x^8}-\frac {3 a \left (\frac {\left (a x^3+b x^6\right )^{14/3}}{17 b x^{11}}-\frac {3 a \left (a x^3+b x^6\right )^{14/3}}{238 b^2 x^{14}}\right )}{10 b}\right )}{23 b}\) |
Input:
Int[(a*x^3 + b*x^6)^(11/3),x]
Output:
(a*x^3 + b*x^6)^(14/3)/(23*b*x^5) - (9*a*((a*x^3 + b*x^6)^(14/3)/(20*b*x^8 ) - (3*a*((-3*a*(a*x^3 + b*x^6)^(14/3))/(238*b^2*x^14) + (a*x^3 + b*x^6)^( 14/3)/(17*b*x^11)))/(10*b)))/(23*b)
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j - 1)), x] - Simp[b*((n*p + n - j + 1)/(a*( j*p + 1))) Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 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*(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])
Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right ) \left (-2380 x^{9} b^{3}+1071 a \,x^{6} b^{2}-378 a^{2} x^{3} b +81 a^{3}\right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {11}{3}}}{54740 b^{4} x^{11}}\) | \(61\) |
orering | \(-\frac {\left (b \,x^{3}+a \right ) \left (-2380 x^{9} b^{3}+1071 a \,x^{6} b^{2}-378 a^{2} x^{3} b +81 a^{3}\right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {11}{3}}}{54740 b^{4} x^{11}}\) | \(61\) |
trager | \(-\frac {\left (-2380 b^{7} x^{21}-8449 a \,b^{6} x^{18}-10374 a^{2} b^{5} x^{15}-4525 a^{3} b^{4} x^{12}-40 a^{4} x^{9} b^{3}+45 a^{5} x^{6} b^{2}-54 a^{6} b \,x^{3}+81 a^{7}\right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {2}{3}}}{54740 b^{4} x^{2}}\) | \(98\) |
risch | \(-\frac {\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}} \left (-2380 b^{7} x^{21}-8449 a \,b^{6} x^{18}-10374 a^{2} b^{5} x^{15}-4525 a^{3} b^{4} x^{12}-40 a^{4} x^{9} b^{3}+45 a^{5} x^{6} b^{2}-54 a^{6} b \,x^{3}+81 a^{7}\right )}{54740 x^{2} b^{4}}\) | \(98\) |
Input:
int((b*x^6+a*x^3)^(11/3),x,method=_RETURNVERBOSE)
Output:
-1/54740*(b*x^3+a)*(-2380*b^3*x^9+1071*a*b^2*x^6-378*a^2*b*x^3+81*a^3)*(b* x^6+a*x^3)^(11/3)/b^4/x^11
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\frac {{\left (2380 \, b^{7} x^{21} + 8449 \, a b^{6} x^{18} + 10374 \, a^{2} b^{5} x^{15} + 4525 \, a^{3} b^{4} x^{12} + 40 \, a^{4} b^{3} x^{9} - 45 \, a^{5} b^{2} x^{6} + 54 \, a^{6} b x^{3} - 81 \, a^{7}\right )} {\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}}}{54740 \, b^{4} x^{2}} \] Input:
integrate((b*x^6+a*x^3)^(11/3),x, algorithm="fricas")
Output:
1/54740*(2380*b^7*x^21 + 8449*a*b^6*x^18 + 10374*a^2*b^5*x^15 + 4525*a^3*b ^4*x^12 + 40*a^4*b^3*x^9 - 45*a^5*b^2*x^6 + 54*a^6*b*x^3 - 81*a^7)*(b*x^6 + a*x^3)^(2/3)/(b^4*x^2)
\[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\int \left (a x^{3} + b x^{6}\right )^{\frac {11}{3}}\, dx \] Input:
integrate((b*x**6+a*x**3)**(11/3),x)
Output:
Integral((a*x**3 + b*x**6)**(11/3), x)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\frac {{\left (2380 \, b^{7} x^{21} + 8449 \, a b^{6} x^{18} + 10374 \, a^{2} b^{5} x^{15} + 4525 \, a^{3} b^{4} x^{12} + 40 \, a^{4} b^{3} x^{9} - 45 \, a^{5} b^{2} x^{6} + 54 \, a^{6} b x^{3} - 81 \, a^{7}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{54740 \, b^{4}} \] Input:
integrate((b*x^6+a*x^3)^(11/3),x, algorithm="maxima")
Output:
1/54740*(2380*b^7*x^21 + 8449*a*b^6*x^18 + 10374*a^2*b^5*x^15 + 4525*a^3*b ^4*x^12 + 40*a^4*b^3*x^9 - 45*a^5*b^2*x^6 + 54*a^6*b*x^3 - 81*a^7)*(b*x^3 + a)^(2/3)/b^4
\[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\int { {\left (b x^{6} + a x^{3}\right )}^{\frac {11}{3}} \,d x } \] Input:
integrate((b*x^6+a*x^3)^(11/3),x, algorithm="giac")
Output:
integrate((b*x^6 + a*x^3)^(11/3), x)
Time = 20.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.57 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=-\frac {{\left (b\,x^3+a\right )}^4\,{\left (b\,x^6+a\,x^3\right )}^{2/3}\,\left (81\,a^3-378\,a^2\,b\,x^3+1071\,a\,b^2\,x^6-2380\,b^3\,x^9\right )}{54740\,b^4\,x^2} \] Input:
int((a*x^3 + b*x^6)^(11/3),x)
Output:
-((a + b*x^3)^4*(a*x^3 + b*x^6)^(2/3)*(81*a^3 - 2380*b^3*x^9 - 378*a^2*b*x ^3 + 1071*a*b^2*x^6))/(54740*b^4*x^2)
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \left (a x^3+b x^6\right )^{11/3} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (2380 b^{7} x^{21}+8449 a \,b^{6} x^{18}+10374 a^{2} b^{5} x^{15}+4525 a^{3} b^{4} x^{12}+40 a^{4} b^{3} x^{9}-45 a^{5} b^{2} x^{6}+54 a^{6} b \,x^{3}-81 a^{7}\right )}{54740 b^{4}} \] Input:
int((b*x^6+a*x^3)^(11/3),x)
Output:
((a + b*x**3)**(2/3)*( - 81*a**7 + 54*a**6*b*x**3 - 45*a**5*b**2*x**6 + 40 *a**4*b**3*x**9 + 4525*a**3*b**4*x**12 + 10374*a**2*b**5*x**15 + 8449*a*b* *6*x**18 + 2380*b**7*x**21))/(54740*b**4)