3.113 \(\int \frac{\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=141 \[ \frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{a^{5/2} f}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{2} a^{5/2} f}-\frac{2 \cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}-\frac{\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/(a^(5/2)*f) - (7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(
Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) - (2*Cos[e + f*x])/(a*f*(a + a*Sin[e + f*x])^(3/2)) -
Cot[e + f*x]/(a*f*(a + a*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.346435, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2715, 2978, 2985, 2649, 206, 2773} \[ \frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{a^{5/2} f}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{2} a^{5/2} f}-\frac{2 \cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}}-\frac{\cot (e+f x)}{a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/(a^(5/2)*f) - (7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(
Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) - (2*Cos[e + f*x])/(a*f*(a + a*Sin[e + f*x])^(3/2)) -
Cot[e + f*x]/(a*f*(a + a*Sin[e + f*x])^(3/2))

Rule 2715

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(a*f*Tan[e + f*x]), x] + Dist[1/b^2, Int[((a + b*Sin[e + f*x])^(m + 1)*(b*m - a*(m + 1)*Sin[e + f
*x]))/Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && LtQ[m, -1]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \frac{\cot ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\csc (e+f x) \left (-\frac{5 a}{2}+\frac{3}{2} a \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{a^2}\\ &=-\frac{2 \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac{\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\csc (e+f x) \left (-5 a^2+2 a^2 \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^4}\\ &=-\frac{2 \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac{\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac{5 \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \, dx}{2 a^3}+\frac{7 \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{2 \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac{\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{a^2 f}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{a^2 f}\\ &=\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{a^{5/2} f}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{2} a^{5/2} f}-\frac{2 \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}-\frac{\cot (e+f x)}{a f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.71848, size = 451, normalized size = 3.2 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (8 \sin \left (\frac{1}{2} (e+f x)\right )+\frac{2 \sin \left (\frac{1}{4} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}{\cos \left (\frac{1}{4} (e+f x)\right )-\sin \left (\frac{1}{4} (e+f x)\right )}-\frac{2 \sin \left (\frac{1}{4} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}{\sin \left (\frac{1}{4} (e+f x)\right )+\cos \left (\frac{1}{4} (e+f x)\right )}+2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+10 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-10 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-\tan \left (\frac{1}{4} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-\cot \left (\frac{1}{4} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(28+28 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{4 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(8*Sin[(e + f*x)/2] - 4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (28 + 28*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*
x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - Cot[(e + f*x)/4]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 1
0*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 10*Log[1 - Cos[(e + f
*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (2*Sin[(e + f*x)/4]*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^2)/(Cos[(e + f*x)/4] - Sin[(e + f*x)/4]) - (2*Sin[(e + f*x)/4]*(Cos[(e + f*x)/2] + Sin[(e + f
*x)/2])^2)/(Cos[(e + f*x)/4] + Sin[(e + f*x)/4]) - (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Tan[(e + f*x)/4]))/
(4*f*(a*(1 + Sin[e + f*x]))^(5/2))

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Maple [A]  time = 0.57, size = 219, normalized size = 1.6 \begin{align*} -{\frac{1}{2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) f} \left ( 7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}a-10\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}a+7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a\sin \left ( fx+e \right ) +4\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a}\sin \left ( fx+e \right ) -10\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{\sqrt{a}}} \right ) \sin \left ( fx+e \right ) a+2\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a} \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \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))^(5/2),x)

[Out]

-1/2/a^(7/2)*(7*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a-10*arctanh((-a*
(-1+sin(f*x+e)))^(1/2)/a^(1/2))*sin(f*x+e)^2*a+7*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2
))*a*sin(f*x+e)+4*(-a*(-1+sin(f*x+e)))^(1/2)*a^(1/2)*sin(f*x+e)-10*arctanh((-a*(-1+sin(f*x+e)))^(1/2)/a^(1/2))
*sin(f*x+e)*a+2*(-a*(-1+sin(f*x+e)))^(1/2)*a^(1/2))*(-a*(-1+sin(f*x+e)))^(1/2)/sin(f*x+e)/cos(f*x+e)/(a+a*sin(
f*x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.8593, size = 1436, normalized size = 10.18 \begin{align*} \text{result too large to display} \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))^(5/2),x, algorithm="fricas")

[Out]

1/4*(5*(cos(f*x + e)^3 + 2*cos(f*x + e)^2 + (cos(f*x + e)^2 - cos(f*x + e) - 2)*sin(f*x + e) - cos(f*x + e) -
2)*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e) -
2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e)
- a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) -
1)) + 7*sqrt(2)*(a*cos(f*x + e)^3 + 2*a*cos(f*x + e)^2 - a*cos(f*x + e) + (a*cos(f*x + e)^2 - a*cos(f*x + e) -
2*a)*sin(f*x + e) - 2*a)*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x +
e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*s
in(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 4*(2*cos(f*x + e)^2 + (2*cos(f*x + e) + 1)*sin(f*x + e) + cos(f*x +
e) - 1)*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 2*a^3*f*cos(f*x + e)^2 - a^3*f*cos(f*x + e) - 2*a^3
*f + (a^3*f*cos(f*x + e)^2 - a^3*f*cos(f*x + e) - 2*a^3*f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(5/2),x, algorithm="giac")

[Out]

sage2