3.1782 \(\int \frac {(a+b x)^{5/6}}{(c+d x)^{17/6}} \, dx\)

Optimal. Leaf size=32 \[ \frac {6 (a+b x)^{11/6}}{11 (c+d x)^{11/6} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \frac {6 (a+b x)^{11/6}}{11 (c+d x)^{11/6} (b c-a d)} \]

Antiderivative was successfully verified.

[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*} \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}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.00 \[ \frac {6 (a+b x)^{11/6}}{11 (c+d x)^{11/6} (b c-a d)} \]

Antiderivative was successfully verified.

[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))

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fricas [B]  time = 0.79, size = 65, normalized size = 2.03 \[ \frac {6 \, {\left (b x + a\right )}^{\frac {11}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{11 \, {\left (b c^{3} - a c^{2} d + {\left (b c d^{2} - a d^{3}\right )} x^{2} + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {5}{6}}}{{\left (d x + c\right )}^{\frac {17}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.00, size = 27, normalized size = 0.84 \[ -\frac {6 \left (b x +a \right )^{\frac {11}{6}}}{11 \left (d x +c \right )^{\frac {11}{6}} \left (a d -b c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {5}{6}}}{{\left (d x + c\right )}^{\frac {17}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.59, size = 130, normalized size = 4.06 \[ -\frac {\left (\frac {6\,a\,{\left (a+b\,x\right )}^{5/6}}{11\,a\,d^3-11\,b\,c\,d^2}+\frac {6\,b\,x\,{\left (a+b\,x\right )}^{5/6}}{11\,a\,d^3-11\,b\,c\,d^2}\right )\,{\left (c+d\,x\right )}^{1/6}}{x^2-\frac {11\,b\,c^3-11\,a\,c^2\,d}{11\,a\,d^3-11\,b\,c\,d^2}+\frac {22\,c\,d\,x\,\left (a\,d-b\,c\right )}{11\,a\,d^3-11\,b\,c\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)
)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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