3.2.17 \(\int \cot ^2(e+f x) \sqrt [3]{a+a \sin (e+f x)} \, dx\) [117]
Optimal. Leaf size=80 \[ \frac {6 \sqrt {2} F_1\left (\frac {11}{6};-\frac {1}{2},2;\frac {17}{6};\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{7/3}}{11 a^2 f} \]
[Out]
6/11*AppellF1(11/6,2,-1/2,17/6,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*sec(f*x+e)*(a+a*sin(f*x+e))^(7/3)*2^(1/2)*(1-s
in(f*x+e))^(1/2)/a^2/f
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Rubi [A]
time = 0.08, 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 {6 \sqrt {2} \sqrt {1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{7/3} F_1\left (\frac {11}{6};-\frac {1}{2},2;\frac {17}{6};\frac {1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )}{11 a^2 f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(1/3),x]
[Out]
(6*Sqrt[2]*AppellF1[11/6, -1/2, 2, 17/6, (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])^(7/3))/(11*a^2*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 \cot ^2(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 {\sqrt {a-x} (a+x)^{5/6}}{x^2} \, dx,x,a \sin (e+f x)\right )}{a f}\\ &=\frac {\left (\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)^{5/6} \sqrt {\frac {1}{2}-\frac {x}{2 a}}}{x^2} \, dx,x,a \sin (e+f x)\right )}{a f \sqrt {\frac {a-a \sin (e+f x)}{a}}}\\ &=\frac {6 \sqrt {2} F_1\left (\frac {11}{6};-\frac {1}{2},2;\frac {17}{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))^{7/3}}{11 a^2 f}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 19.00, size = 2692, normalized size = 33.65 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(1/3),x]
[Out]
((15/2 + (15*I)/2)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e +
f*x)/2])]*(a*(1 + Sin[e + f*x]))^(1/3))/(f*((5 + 5*I)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f
*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])] + (AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]
), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])] + I*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1
/2 - I/2)*(1 + Cot[(e + f*x)/2])])*(1 + Cot[(e + f*x)/2]))) + ((-4 - Cot[e + f*x])*(a*(1 + Sin[e + f*x]))^(1/3
))/f + ((5/2 + (5*I)/2)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Tan[(e + f*x)/2]), (1/2 - I/2)*(1 + Tan[
(e + f*x)/2])]*(a*(1 + Sin[e + f*x]))^(1/3))/(f*((5 + 5*I)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Tan[(
e + f*x)/2]), (1/2 - I/2)*(1 + Tan[(e + f*x)/2])] + (AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Tan[(e + f*
x)/2]), (1/2 - I/2)*(1 + Tan[(e + f*x)/2])] + I*AppellF1[5/3, 4/3, 1/3, 8/3, (1/2 + I/2)*(1 + Tan[(e + f*x)/2]
), (1/2 - I/2)*(1 + Tan[(e + f*x)/2])])*(1 + Tan[(e + f*x)/2]))) + (Cos[(3*(e + f*x))/2]*Csc[(e + f*x)/2]*Sec[
(e + f*x)/2]*(a*(1 + Sin[e + f*x]))^(1/3)*((1 + Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x)/2]^2])^(2/3)*(8 + (1 + I)
*2^(2/3)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, ((1
+ I) + (1 - I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])]*(I + Tan[(e + f*x)/2]) - AppellF1[2/3, 1/3, 1/3, 5/
3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(e + f*x
)/2])^(1/3)*((-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*(1 + Tan[(e + f*x)/2])))/(f*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2])*(1 + Tan[(e + f*x)/2])*((-3*Sec[(e + f*x)/2]^2*((1 + Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x)/2]^2])^(2/3)
*(8 + (1 + I)*2^(2/3)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2
/3, 5/3, ((1 + I) + (1 - I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])]*(I + Tan[(e + f*x)/2]) - AppellF1[2/3,
1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*((2 + 2*I) - (2 - 2*I)
*Cot[(e + f*x)/2])^(1/3)*((-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*(1 + Tan[(e + f*x)/2])))/(4*(1 + Tan[(e + f*x
)/2])^2) + ((8 + (1 + I)*2^(2/3)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))^(1/3)*Hypergeometri
c2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])]*(I + Tan[(e + f*x)/2]) - Ap
pellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*((2 + 2*I)
- (2 - 2*I)*Cot[(e + f*x)/2])^(1/3)*((-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*(1 + Tan[(e + f*x)/2]))*(Sqrt[Sec[
(e + f*x)/2]^2]/2 - (Tan[(e + f*x)/2]*(1 + Tan[(e + f*x)/2]))/(2*Sqrt[Sec[(e + f*x)/2]^2])))/((1 + Tan[(e + f*
x)/2])*((1 + Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x)/2]^2])^(1/3)) + (3*((1 + Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x
)/2]^2])^(2/3)*(-1/2*(AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e
+ f*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(e + f*x)/2])^(1/3)*((-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*Sec[(e + f
*x)/2]^2) + ((1 + I)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))^(1/3)*Hypergeometric2F1[1/3, 2/
3, 5/3, ((1 + I) + (1 - I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])]*Sec[(e + f*x)/2]^2)/2^(1/3) + ((1/3 + I
/3)*2^(2/3)*(((1/2 - I/2)*(I + Cot[(e + f*x)/2])*Csc[(e + f*x)/2]^2)/(1 + Cot[(e + f*x)/2])^2 - ((1/2 - I/2)*C
sc[(e + f*x)/2]^2)/(1 + Cot[(e + f*x)/2]))*Hypergeometric2F1[1/3, 2/3, 5/3, ((1 + I) + (1 - I)*Tan[(e + f*x)/2
])/(2 + 2*Tan[(e + f*x)/2])]*(I + Tan[(e + f*x)/2]))/(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))
^(2/3) - ((1/6 + I/6)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e
+ f*x)/2])]*((2 + 2*I) - (2 - 2*I)*Cot[(e + f*x)/2])^(1/3)*Csc[(e + f*x)/2]^2*(1 + Tan[(e + f*x)/2]))/((-1 -
I)*(I + Cot[(e + f*x)/2]))^(2/3) - ((1/3 - I/3)*AppellF1[2/3, 1/3, 1/3, 5/3, (1/2 + I/2)*(1 + Cot[(e + f*x)/2]
), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*((-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*Csc[(e + f*x)/2]^2*(1 + Tan[(e
+ f*x)/2]))/((2 + 2*I) - (2 - 2*I)*Cot[(e + f*x)/2])^(2/3) - ((2 + 2*I) - (2 - 2*I)*Cot[(e + f*x)/2])^(1/3)*((
-1 - I)*(I + Cot[(e + f*x)/2]))^(1/3)*((-1/30 + I/30)*AppellF1[5/3, 1/3, 4/3, 8/3, (1/2 + I/2)*(1 + Cot[(e + f
*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*Csc[(e + f*x)/2]^2 - (1/30 + I/30)*AppellF1[5/3, 4/3, 1/3, 8/3, (
1/2 + I/2)*(1 + Cot[(e + f*x)/2]), (1/2 - I/2)*(1 + Cot[(e + f*x)/2])]*Csc[(e + f*x)/2]^2)*(1 + Tan[(e + f*x)/
2]) + ((2/3 + (2*I)/3)*2^(2/3)*(((1 - I)*(I + Cot[(e + f*x)/2]))/(1 + Cot[(e + f*x)/2]))^(1/3)*(I + Tan[(e + f
*x)/2])*(2 + 2*Tan[(e + f*x)/2])*(-((Sec[(e + f*x)/2]^2*((1 + I) + (1 - I)*Tan[(e + f*x)/2]))/(2 + 2*Tan[(e +
f*x)/2])^2) + ((1/2 - I/2)*Sec[(e + f*x)/2]^2)/(2 + 2*Tan[(e + f*x)/2]))*(-Hypergeometric2F1[1/3, 2/3, 5/3, ((
1 + I) + (1 - I)*Tan[(e + f*x)/2])/(2 + 2*Tan[(e + f*x)/2])] + (1 - ((1 + I) + (1 - I)*Tan[(e + f*x)/2])/(2 +
2*Tan[(e + f*x)/2]))^(-1/3)))/((1 + I) + (1 - I...
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (\cot ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{\frac {1}{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^2*(a+a*sin(f*x+e))^(1/3),x)
[Out]
int(cot(f*x+e)^2*(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)^2*(a+a*sin(f*x+e))^(1/3),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^(1/3)*cot(f*x + e)^2, 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)^2*(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 \sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{2}{\left (e + f x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**2*(a+a*sin(f*x+e))**(1/3),x)
[Out]
Integral((a*(sin(e + f*x) + 1))**(1/3)*cot(e + f*x)**2, 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)^2*(a+a*sin(f*x+e))^(1/3),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^(1/3)*cot(f*x + e)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cot}\left (e+f\,x\right )}^2\,{\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)^2*(a + a*sin(e + f*x))^(1/3),x)
[Out]
int(cot(e + f*x)^2*(a + a*sin(e + f*x))^(1/3), x)
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