Integrand size = 20, antiderivative size = 88 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=-\frac {3 \sqrt [3]{a-b x}}{14 a b (a+b x)^{7/3}}-\frac {9 \sqrt [3]{a-b x}}{56 a^2 b (a+b x)^{4/3}}-\frac {27 \sqrt [3]{a-b x}}{112 a^3 b \sqrt [3]{a+b x}} \] Output:
-3/14*(-b*x+a)^(1/3)/a/b/(b*x+a)^(7/3)-9/56*(-b*x+a)^(1/3)/a^2/b/(b*x+a)^( 4/3)-27/112*(-b*x+a)^(1/3)/a^3/b/(b*x+a)^(1/3)
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=-\frac {3 \sqrt [3]{a-b x} \left (23 a^2+24 a b x+9 b^2 x^2\right )}{112 a^3 b (a+b x)^{7/3}} \] Input:
Integrate[1/((a - b*x)^(2/3)*(a + b*x)^(10/3)),x]
Output:
(-3*(a - b*x)^(1/3)*(23*a^2 + 24*a*b*x + 9*b^2*x^2))/(112*a^3*b*(a + b*x)^ (7/3))
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {55, 55, 48}
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.
\(\displaystyle \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {3 \int \frac {1}{(a-b x)^{2/3} (a+b x)^{7/3}}dx}{7 a}-\frac {3 \sqrt [3]{a-b x}}{14 a b (a+b x)^{7/3}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {3 \left (\frac {3 \int \frac {1}{(a-b x)^{2/3} (a+b x)^{4/3}}dx}{8 a}-\frac {3 \sqrt [3]{a-b x}}{8 a b (a+b x)^{4/3}}\right )}{7 a}-\frac {3 \sqrt [3]{a-b x}}{14 a b (a+b x)^{7/3}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {3 \left (-\frac {9 \sqrt [3]{a-b x}}{16 a^2 b \sqrt [3]{a+b x}}-\frac {3 \sqrt [3]{a-b x}}{8 a b (a+b x)^{4/3}}\right )}{7 a}-\frac {3 \sqrt [3]{a-b x}}{14 a b (a+b x)^{7/3}}\) |
Input:
Int[1/((a - b*x)^(2/3)*(a + b*x)^(10/3)),x]
Output:
(-3*(a - b*x)^(1/3))/(14*a*b*(a + b*x)^(7/3)) + (3*((-3*(a - b*x)^(1/3))/( 8*a*b*(a + b*x)^(4/3)) - (9*(a - b*x)^(1/3))/(16*a^2*b*(a + b*x)^(1/3))))/ (7*a)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.49
method | result | size |
gosper | \(-\frac {3 \left (-b x +a \right )^{\frac {1}{3}} \left (9 b^{2} x^{2}+24 a b x +23 a^{2}\right )}{112 \left (b x +a \right )^{\frac {7}{3}} a^{3} b}\) | \(43\) |
orering | \(-\frac {3 \left (-b x +a \right )^{\frac {1}{3}} \left (9 b^{2} x^{2}+24 a b x +23 a^{2}\right )}{112 \left (b x +a \right )^{\frac {7}{3}} a^{3} b}\) | \(43\) |
Input:
int(1/(-b*x+a)^(2/3)/(b*x+a)^(10/3),x,method=_RETURNVERBOSE)
Output:
-3/112*(-b*x+a)^(1/3)*(9*b^2*x^2+24*a*b*x+23*a^2)/(b*x+a)^(7/3)/a^3/b
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=-\frac {3 \, {\left (9 \, b^{2} x^{2} + 24 \, a b x + 23 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}}{112 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} \] Input:
integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(10/3),x, algorithm="fricas")
Output:
-3/112*(9*b^2*x^2 + 24*a*b*x + 23*a^2)*(b*x + a)^(2/3)*(-b*x + a)^(1/3)/(a ^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b)
\[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=\int \frac {1}{\left (a - b x\right )^{\frac {2}{3}} \left (a + b x\right )^{\frac {10}{3}}}\, dx \] Input:
integrate(1/(-b*x+a)**(2/3)/(b*x+a)**(10/3),x)
Output:
Integral(1/((a - b*x)**(2/3)*(a + b*x)**(10/3)), x)
\[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (-b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(10/3),x, algorithm="maxima")
Output:
integrate(1/((b*x + a)^(10/3)*(-b*x + a)^(2/3)), x)
\[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (-b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(10/3),x, algorithm="giac")
Output:
integrate(1/((b*x + a)^(10/3)*(-b*x + a)^(2/3)), x)
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=-\frac {{\left (a-b\,x\right )}^{1/3}\,\left (\frac {69}{112\,a\,b^3}+\frac {9\,x}{14\,a^2\,b^2}+\frac {27\,x^2}{112\,a^3\,b}\right )}{x^2\,{\left (a+b\,x\right )}^{1/3}+\frac {a^2\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/3}}{b}} \] Input:
int(1/((a + b*x)^(10/3)*(a - b*x)^(2/3)),x)
Output:
-((a - b*x)^(1/3)*(69/(112*a*b^3) + (9*x)/(14*a^2*b^2) + (27*x^2)/(112*a^3 *b)))/(x^2*(a + b*x)^(1/3) + (a^2*(a + b*x)^(1/3))/b^2 + (2*a*x*(a + b*x)^ (1/3))/b)
\[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{10/3}} \, dx=\int \frac {1}{\left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {2}{3}} a^{3}+3 \left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {2}{3}} a^{2} b x +3 \left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {2}{3}} a \,b^{2} x^{2}+\left (b x +a \right )^{\frac {1}{3}} \left (-b x +a \right )^{\frac {2}{3}} b^{3} x^{3}}d x \] Input:
int(1/(-b*x+a)^(2/3)/(b*x+a)^(10/3),x)
Output:
int(1/((a + b*x)**(1/3)*(a - b*x)**(2/3)*a**3 + 3*(a + b*x)**(1/3)*(a - b* x)**(2/3)*a**2*b*x + 3*(a + b*x)**(1/3)*(a - b*x)**(2/3)*a*b**2*x**2 + (a + b*x)**(1/3)*(a - b*x)**(2/3)*b**3*x**3),x)