\(\int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx\) [1823]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\frac {6 b \sqrt [6]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {17}{6},\frac {7}{6},-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^2 (c+d x)^{5/6}} \]

[Out]

6*b*(b*x+a)^(1/6)*(b*(d*x+c)/(-a*d+b*c))^(5/6)*hypergeom([1/6, 17/6],[7/6],-d*(b*x+a)/(-a*d+b*c))/(-a*d+b*c)^2
/(d*x+c)^(5/6)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {72, 71} \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\frac {6 b \sqrt [6]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {17}{6},\frac {7}{6},-\frac {d (a+b x)}{b c-a d}\right )}{(c+d x)^{5/6} (b c-a d)^2} \]

[In]

Int[1/((a + b*x)^(5/6)*(c + d*x)^(17/6)),x]

[Out]

(6*b*(a + b*x)^(1/6)*((b*(c + d*x))/(b*c - a*d))^(5/6)*Hypergeometric2F1[1/6, 17/6, 7/6, -((d*(a + b*x))/(b*c
- a*d))])/((b*c - a*d)^2*(c + d*x)^(5/6))

Rule 71

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6}\right ) \int \frac {1}{(a+b x)^{5/6} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{17/6}} \, dx}{(b c-a d)^2 (c+d x)^{5/6}} \\ & = \frac {6 b \sqrt [6]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \, _2F_1\left (\frac {1}{6},\frac {17}{6};\frac {7}{6};-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^2 (c+d x)^{5/6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\frac {6 b \sqrt [6]{a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {17}{6},\frac {7}{6},\frac {d (a+b x)}{-b c+a d}\right )}{(b c-a d)^2 (c+d x)^{5/6}} \]

[In]

Integrate[1/((a + b*x)^(5/6)*(c + d*x)^(17/6)),x]

[Out]

(6*b*(a + b*x)^(1/6)*((b*(c + d*x))/(b*c - a*d))^(5/6)*Hypergeometric2F1[1/6, 17/6, 7/6, (d*(a + b*x))/(-(b*c)
+ a*d)])/((b*c - a*d)^2*(c + d*x)^(5/6))

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{6}} \left (d x +c \right )^{\frac {17}{6}}}d x\]

[In]

int(1/(b*x+a)^(5/6)/(d*x+c)^(17/6),x)

[Out]

int(1/(b*x+a)^(5/6)/(d*x+c)^(17/6),x)

Fricas [F]

\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {17}{6}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/6)/(d*x+c)^(17/6),x, algorithm="fricas")

[Out]

integral((b*x + a)^(1/6)*(d*x + c)^(1/6)/(b*d^3*x^4 + a*c^3 + (3*b*c*d^2 + a*d^3)*x^3 + 3*(b*c^2*d + a*c*d^2)*
x^2 + (b*c^3 + 3*a*c^2*d)*x), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x+a)**(5/6)/(d*x+c)**(17/6),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {17}{6}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/6)/(d*x+c)^(17/6),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/6)*(d*x + c)^(17/6)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {17}{6}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/6)/(d*x+c)^(17/6),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/6)*(d*x + c)^(17/6)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/6} (c+d x)^{17/6}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/6}\,{\left (c+d\,x\right )}^{17/6}} \,d x \]

[In]

int(1/((a + b*x)^(5/6)*(c + d*x)^(17/6)),x)

[Out]

int(1/((a + b*x)^(5/6)*(c + d*x)^(17/6)), x)