3.138 \(\int \cot ^2(e+f x) (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=89 \[ \frac{2 \sqrt{2} \sqrt{1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{m+2} F_1\left (m+\frac{3}{2};-\frac{1}{2},2;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )}{a^2 f (2 m+3)} \]

[Out]

(2*Sqrt[2]*AppellF1[3/2 + m, -1/2, 2, 5/2 + m, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sec[e + f*x]*Sqrt[1 - S
in[e + f*x]]*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(3 + 2*m))

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Rubi [A]  time = 0.10163, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2719, 137, 136} \[ \frac{2 \sqrt{2} \sqrt{1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{m+2} F_1\left (m+\frac{3}{2};-\frac{1}{2},2;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )}{a^2 f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^m,x]

[Out]

(2*Sqrt[2]*AppellF1[3/2 + m, -1/2, 2, 5/2 + m, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sec[e + f*x]*Sqrt[1 - S
in[e + f*x]]*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(3 + 2*m))

Rule 2719

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[(Sqrt[a + b*Si
n[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]

Rule 137

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 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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])

Rubi steps

\begin{align*} \int \cot ^2(e+f x) (a+a \sin (e+f x))^m \, dx &=\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a-x} (a+x)^{\frac{1}{2}+m}}{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 ) \operatorname{Subst}\left (\int \frac{(a+x)^{\frac{1}{2}+m} \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{2 \sqrt{2} F_1\left (\frac{3}{2}+m;-\frac{1}{2},2;\frac{5}{2}+m;\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))^{2+m}}{a^2 f (3+2 m)}\\ \end{align*}

Mathematica [C]  time = 99.9376, size = 47487, normalized size = 533.56 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^m,x]

[Out]

Result too large to show

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Maple [F]  time = 0.231, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{2} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2*(a+a*sin(f*x+e))^m,x)

[Out]

int(cot(f*x+e)^2*(a+a*sin(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*cot(f*x + e)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*cot(f*x + e)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \cot ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2*(a+a*sin(f*x+e))**m,x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*cot(e + f*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*cot(f*x + e)^2, x)