\(\int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx\) [1782]

Optimal result

Integrand size = 19, antiderivative size = 32 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx=\frac {6 (a+b x)^{11/6}}{11 (b c-a d) (c+d x)^{11/6}} \]

[Out]

6/11*(b*x+a)^(11/6)/(-a*d+b*c)/(d*x+c)^(11/6)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx=\frac {6 (a+b x)^{11/6}}{11 (c+d x)^{11/6} (b c-a d)} \]

[In]

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

[Out]

(6*(a + b*x)^(11/6))/(11*(b*c - a*d)*(c + d*x)^(11/6))

Rule 37

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] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {6 (a+b x)^{11/6}}{11 (b c-a d) (c+d x)^{11/6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx=\frac {6 (a+b x)^{11/6}}{11 (b c-a d) (c+d x)^{11/6}} \]

[In]

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

[Out]

(6*(a + b*x)^(11/6))/(11*(b*c - a*d)*(c + d*x)^(11/6))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
gosper	\(-\frac {6 \left (b x +a \right )^{\frac {11}{6}}}{11 \left (d x +c \right )^{\frac {11}{6}} \left (a d -b c \right )}\)	\(27\)

[In]

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

[Out]

-6/11*(b*x+a)^(11/6)/(d*x+c)^(11/6)/(a*d-b*c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).

Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx=\frac {6 \, {\left (b x + a\right )}^{\frac {11}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{11 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\right )}} \]

[In]

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

[Out]

6/11*(b*x + a)^(11/6)*(d*x + c)^(1/6)/(b*c^3 - a*c^2*d + (b*c*d^2 - a*d^3)*x^2 + 2*(b*c^2*d - a*c*d^2)*x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.06 \[ \int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx=-\frac {\left (\frac {6\,a\,{\left (a+b\,x\right )}^{5/6}}{11\,a\,d^3-11\,b\,c\,d^2}+\frac {6\,b\,x\,{\left (a+b\,x\right )}^{5/6}}{11\,a\,d^3-11\,b\,c\,d^2}\right )\,{\left (c+d\,x\right )}^{1/6}}{x^2-\frac {11\,b\,c^3-11\,a\,c^2\,d}{11\,a\,d^3-11\,b\,c\,d^2}+\frac {22\,c\,d\,x\,\left (a\,d-b\,c\right )}{11\,a\,d^3-11\,b\,c\,d^2}} \]

[In]

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

[Out]

-(((6*a*(a + b*x)^(5/6))/(11*a*d^3 - 11*b*c*d^2) + (6*b*x*(a + b*x)^(5/6))/(11*a*d^3 - 11*b*c*d^2))*(c + d*x)^
(1/6))/(x^2 - (11*b*c^3 - 11*a*c^2*d)/(11*a*d^3 - 11*b*c*d^2) + (22*c*d*x*(a*d - b*c))/(11*a*d^3 - 11*b*c*d^2)
)