3.340 \(\int \frac{\cot ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=62 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\cot (c+d x)}{d \sqrt{a \sin (c+d x)+a}} \]

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(Sqrt[a]*d) - Cot[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x]
])

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Rubi [A]  time = 0.100631, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2716, 21, 2773, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\cot (c+d x)}{d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(Sqrt[a]*d) - Cot[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x]
])

Rule 2716

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

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])

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=-\frac{\cot (c+d x)}{d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc (c+d x) \left (-\frac{a}{2}-\frac{1}{2} a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{a}\\ &=-\frac{\cot (c+d x)}{d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{2 a}\\ &=-\frac{\cot (c+d x)}{d \sqrt{a+a \sin (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{\sqrt{a} d}-\frac{\cot (c+d x)}{d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 0.299677, size = 138, normalized size = 2.23 \[ \frac{\left (\tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \csc \left (\frac{1}{4} (c+d x)\right ) \sec \left (\frac{1}{4} (c+d x)\right ) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )-2 \cos \left (\frac{1}{2} (c+d x)\right )+\sin (c+d x) \left (\log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )\right )}{8 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Csc[(c + d*x)/4]*Sec[(c + d*x)/4]*(-2*Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2] + (Log[1 + Cos[(c + d*x)/2] - Sin
[(c + d*x)/2]] - Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Sin[c + d*x])*(1 + Tan[(c + d*x)/2]))/(8*d*Sqrt
[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.898, size = 103, normalized size = 1.7 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( -{\it Artanh} \left ({\sqrt{a-a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) \sin \left ( dx+c \right ) a+\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{a} \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*sin(d*x+c)*a+(a-a*sin(d*x+
c))^(1/2)*a^(1/2))/sin(d*x+c)/a^(3/2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2} \csc \left (d x + c\right )^{2}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2*csc(d*x + c)^2/sqrt(a*sin(d*x + c) + a), x)

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Fricas [B]  time = 1.69811, size = 706, normalized size = 11.39 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \,{\left (\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} - 9 \, a \cos \left (d x + c\right ) +{\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{4 \,{\left (a d \cos \left (d x + c\right )^{2} - a d -{\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*((cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2
+ 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a)
- 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c
)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(
d*x + c) + 1))/(a*d*cos(d*x + c)^2 - a*d - (a*d*cos(d*x + c) + a*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**2/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**2/sqrt(a*(sin(c + d*x) + 1)), x)

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Giac [B]  time = 2.3487, size = 486, normalized size = 7.84 \begin{align*} \frac{\frac{{\left (2 \, \sqrt{2} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{2} \sqrt{-a} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) + 2 \, \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{-a} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) - \sqrt{2} \sqrt{-a} - 3 \, \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{2} \sqrt{-a} \sqrt{a} + \sqrt{-a} \sqrt{a}} - \frac{2 \, \arctan \left (-\frac{\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} + \frac{\log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} + \frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} + \frac{2 \, \sqrt{a}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2*((2*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) +
sqrt(a)) + 2*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) -
sqrt(2)*sqrt(-a) - 3*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(2)*sqrt(-a)*sqrt(a) + sqrt(-a)*sqrt(a)) - 2
*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*d
*x + 1/2*c) + 1)) + log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(a)*sgn(
tan(1/2*d*x + 1/2*c) + 1)) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*sqrt(a)/
(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d