3.2.22 \(\int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\) [122]

Optimal. Leaf size=80 \[ \frac {12 \sqrt {2} F_1\left (\frac {13}{6};-\frac {3}{2},4;\frac {19}{6};\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{8/3}}{13 a^3 f} \]

[Out]

12/13*AppellF1(13/6,4,-3/2,19/6,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*sec(f*x+e)*(a+a*sin(f*x+e))^(8/3)*2^(1/2)*(1-
sin(f*x+e))^(1/2)/a^3/f

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Rubi [A]
time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2798, 142, 141} \begin {gather*} \frac {12 \sqrt {2} \sqrt {1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{8/3} F_1\left (\frac {13}{6};-\frac {3}{2},4;\frac {19}{6};\frac {1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )}{13 a^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3),x]

[Out]

(12*Sqrt[2]*AppellF1[13/6, -3/2, 4, 19/6, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sec[e + f*x]*Sqrt[1 - Sin[e
+ f*x]]*(a + a*Sin[e + f*x])^(8/3))/(13*a^3*f)

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 142

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*(b*(c/(b*c -
a*d)) + b*d*(x/(b*c - a*d)))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 2798

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[Sqrt[a + b*Sin
[e + f*x]]*(Sqrt[a - b*Sin[e + f*x]]/(b*f*Cos[e + f*x])), Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p +
1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && Inte
gerQ[p/2]

Rubi steps

\begin {align*} \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx &=\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {(a-x)^{3/2} (a+x)^{7/6}}{x^4} \, dx,x,a \sin (e+f x)\right )}{a f}\\ &=\frac {\left (2 \sqrt {2} \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {(a+x)^{7/6} \left (\frac {1}{2}-\frac {x}{2 a}\right )^{3/2}}{x^4} \, dx,x,a \sin (e+f x)\right )}{f \sqrt {\frac {a-a \sin (e+f x)}{a}}}\\ &=\frac {12 \sqrt {2} F_1\left (\frac {13}{6};-\frac {3}{2},4;\frac {19}{6};\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{8/3}}{13 a^3 f}\\ \end {align*}

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Mathematica [F]
time = 5.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^4(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3),x]

[Out]

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(1/3), x]

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\cot ^{4}\left (f x +e \right )}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x)

[Out]

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate(cot(f*x + e)^4/(a*sin(f*x + e) + a)^(1/3), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{4}{\left (e + f x \right )}}{\sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a+a*sin(f*x+e))**(1/3),x)

[Out]

Integral(cot(e + f*x)**4/(a*(sin(e + f*x) + 1))**(1/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/3),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^4/(a*sin(f*x + e) + a)^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(1/3),x)

[Out]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(1/3), x)

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