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

Optimal. Leaf size=100 \[ \frac{\cot (c+d x)}{4 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\cot (c+d x) \csc (c+d x)}{2 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(4*Sqrt[a]*d) + Cot[c + d*x]/(4*d*Sqrt[a + a*Sin[c +
d*x]]) - (Cot[c + d*x]*Csc[c + d*x])/(2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.316527, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2874, 2980, 2772, 2773, 206} \[ \frac{\cot (c+d x)}{4 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\cot (c+d x) \csc (c+d x)}{2 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(4*Sqrt[a]*d) + Cot[c + d*x]/(4*d*Sqrt[a + a*Sin[c +
d*x]]) - (Cot[c + d*x]*Csc[c + d*x])/(2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

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) \csc (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\int \csc ^3(c+d x) (a-a \sin (c+d x)) \sqrt{a+a \sin (c+d x)} \, dx}{a^2}\\ &=-\frac{\cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{4 a}\\ &=\frac{\cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{8 a}\\ &=\frac{\cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc (c+d x)}{2 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 )}{4 d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{4 \sqrt{a} d}+\frac{\cot (c+d x)}{4 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc (c+d x)}{2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 1.85661, size = 272, normalized size = 2.72 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (4 \tan \left (\frac{1}{4} (c+d x)\right )+4 \cot \left (\frac{1}{4} (c+d x)\right )-\csc ^2\left (\frac{1}{4} (c+d x)\right )+\sec ^2\left (\frac{1}{4} (c+d x)\right )-\frac{8 \sin \left (\frac{1}{4} (c+d x)\right )}{\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )}+\frac{8 \sin \left (\frac{1}{4} (c+d x)\right )}{\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )}+\frac{2}{\left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{2}{\left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^2}+4 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-8\right )}{32 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8 + 4*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 + 4*Log[1 + Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]] - 4*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sec[(c + d*x)/4]^2 + 2/(Cos[(c + d*
x)/4] - Sin[(c + d*x)/4])^2 - (8*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - 2/(Cos[(c + d*x)/4]
+ Sin[(c + d*x)/4])^2 + (8*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 4*Tan[(c + d*x)/4]))/(32
*d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.947, size = 124, normalized size = 1.2 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{{\frac{3}{2}}}{a}^{{\frac{3}{2}}}-{\it Artanh} \left ({\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}+\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{{\frac{5}{2}}} \right ){a}^{-{\frac{7}{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)^3/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)*((-a*(sin(d*x+c)-1))^(3/2)*a^(3/2)-arctanh((-a*(sin(d*x+
c)-1))^(1/2)/a^(1/2))*a^3*sin(d*x+c)^2+(-a*(sin(d*x+c)-1))^(1/2)*a^(5/2))/sin(d*x+c)^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 )^{3}}{\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)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2*csc(d*x + c)^3/sqrt(a*sin(d*x + c) + a), x)

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Fricas [B]  time = 1.74132, size = 861, normalized size = 8.61 \begin{align*} \frac{{\left (\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 )} \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 \,{\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}}{16 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d +{\left (a d \cos \left (d x + c\right )^{2} - 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)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/16*((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)*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*(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))/(a*d*cos(d*x + c)^3 +
a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d + (a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))

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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(cos(d*x+c)**2*csc(d*x+c)**3/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.46748, size = 682, normalized size = 6.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/8*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2/(a*sgn(tan
(1/2*d*x + 1/2*c) + 1))) + (4*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 2*sqrt(2)*sqrt(-a
)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 6*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3*sqrt(-a)*log(sqr
t(2)*sqrt(a) + sqrt(a)) + 14*sqrt(2)*sqrt(-a) + 18*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(2*sqrt(2)*sqrt(-a)
*sqrt(a) + 3*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)) + 2*((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1
/2*d*x + 1/2*c)^2 + a))^3 - 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^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))*a + 2*a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/
2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d