Integrand size = 19, antiderivative size = 58 \[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\frac {3 (a+b x)^{11/3} \operatorname {Hypergeometric2F1}\left (1,\frac {10}{3},\frac {14}{3},-\frac {d (a+b x)}{b c-a d}\right )}{11 (b c-a d) \sqrt [3]{c+d x}} \] Output:
3/11*(b*x+a)^(11/3)*hypergeom([1, 10/3],[14/3],-d*(b*x+a)/(-a*d+b*c))/(-a* d+b*c)/(d*x+c)^(1/3)
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\frac {3 (a+b x)^{11/3} \left (\frac {b (c+d x)}{b c-a d}\right )^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {11}{3},\frac {14}{3},\frac {d (a+b x)}{-b c+a d}\right )}{11 b (c+d x)^{4/3}} \] Input:
Integrate[(a + b*x)^(8/3)/(c + d*x)^(4/3),x]
Output:
(3*(a + b*x)^(11/3)*((b*(c + d*x))/(b*c - a*d))^(4/3)*Hypergeometric2F1[4/ 3, 11/3, 14/3, (d*(a + b*x))/(-(b*c) + a*d)])/(11*b*(c + d*x)^(4/3))
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {80, 79}
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.
\(\displaystyle \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {b \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \int \frac {(a+b x)^{8/3}}{\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{4/3}}dx}{\sqrt [3]{c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {3 (a+b x)^{11/3} \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {11}{3},\frac {14}{3},-\frac {d (a+b x)}{b c-a d}\right )}{11 \sqrt [3]{c+d x} (b c-a d)}\) |
Input:
Int[(a + b*x)^(8/3)/(c + d*x)^(4/3),x]
Output:
(3*(a + b*x)^(11/3)*((b*(c + d*x))/(b*c - a*d))^(1/3)*Hypergeometric2F1[4/ 3, 11/3, 14/3, -((d*(a + b*x))/(b*c - a*d))])/(11*(b*c - a*d)*(c + d*x)^(1 /3))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
\[\int \frac {\left (b x +a \right )^{\frac {8}{3}}}{\left (x d +c \right )^{\frac {4}{3}}}d x\]
Input:
int((b*x+a)^(8/3)/(d*x+c)^(4/3),x)
Output:
int((b*x+a)^(8/3)/(d*x+c)^(4/3),x)
\[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="fricas")
Output:
integral((b^2*x^2 + 2*a*b*x + a^2)*(b*x + a)^(2/3)*(d*x + c)^(2/3)/(d^2*x^ 2 + 2*c*d*x + c^2), x)
\[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {8}{3}}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \] Input:
integrate((b*x+a)**(8/3)/(d*x+c)**(4/3),x)
Output:
Integral((a + b*x)**(8/3)/(c + d*x)**(4/3), x)
\[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x + a)^(8/3)/(d*x + c)^(4/3), x)
\[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {8}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="giac")
Output:
integrate((b*x + a)^(8/3)/(d*x + c)^(4/3), x)
Timed out. \[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^{8/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \] Input:
int((a + b*x)^(8/3)/(c + d*x)^(4/3),x)
Output:
int((a + b*x)^(8/3)/(c + d*x)^(4/3), x)
\[ \int \frac {(a+b x)^{8/3}}{(c+d x)^{4/3}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {2}{3}}}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) a^{2}+\left (\int \frac {\left (b x +a \right )^{\frac {2}{3}} x^{2}}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) b^{2}+2 \left (\int \frac {\left (b x +a \right )^{\frac {2}{3}} x}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) a b \] Input:
int((b*x+a)^(8/3)/(d*x+c)^(4/3),x)
Output:
int((a + b*x)**(2/3)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x),x)*a**2 + int(((a + b*x)**(2/3)*x**2)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x),x )*b**2 + 2*int(((a + b*x)**(2/3)*x)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3) *d*x),x)*a*b