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

Optimal. Leaf size=135 \[ \frac{\cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 \sqrt{a} d}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}+\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(8*Sqrt[a]*d) + Cot[c + d*x]/(8*d*Sqrt[a + a*Sin[c +
d*x]]) + (Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*d*Sqrt
[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.403147, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2874, 2980, 2772, 2773, 206} \[ \frac{\cot (c+d x)}{8 d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 \sqrt{a} d}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}+\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[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]]]/(8*Sqrt[a]*d) + Cot[c + d*x]/(8*d*Sqrt[a + a*Sin[c +
d*x]]) + (Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*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 ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\int \csc ^4(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 ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{6 a}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{8 a}\\ &=\frac{\cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}+\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}-\frac{\int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{16 a}\\ &=\frac{\cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}+\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 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 )}{8 d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 \sqrt{a} d}+\frac{\cot (c+d x)}{8 d \sqrt{a+a \sin (c+d x)}}+\frac{\cot (c+d x) \csc (c+d x)}{12 d \sqrt{a+a \sin (c+d x)}}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 0.696361, size = 292, normalized size = 2.16 \[ \frac{\csc ^9\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (60 \sin \left (\frac{1}{2} (c+d x)\right )+2 \sin \left (\frac{3}{2} (c+d x)\right )+6 \sin \left (\frac{5}{2} (c+d x)\right )-60 \cos \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{3}{2} (c+d x)\right )-6 \cos \left (\frac{5}{2} (c+d x)\right )+9 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-9 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+3 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d \sqrt{a (\sin (c+d x)+1)} \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(c + d*x)/2]^9*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-60*Cos[(c + d*x)/2] + 2*Cos[(3*(c + d*x))/2] - 6*C
os[(5*(c + d*x))/2] + 60*Sin[(c + d*x)/2] + 9*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Lo
g[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 2*Sin[(3*(c + d*x))/2] + 6*Sin[(5*(c + d*x))/2] - 3*
Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
*Sin[3*(c + d*x)]))/(24*d*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 1.062, size = 144, normalized size = 1.1 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{24\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 3\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{9/2}+8\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}{a}^{7/2}-3\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{5/2}{a}^{5/2}-3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ){a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ){a}^{-{\frac{11}{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)^4/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*(-a*(sin(d*x+c)-1))^(1/2)*a^(9/2)+8*(-a*(sin(d*x+c)-1))^(3/2
)*a^(7/2)-3*(-a*(sin(d*x+c)-1))^(5/2)*a^(5/2)-3*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^5*sin(d*x+c)^3)/a
^(11/2)/sin(d*x+c)^3/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 )^{4}}{\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)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2*csc(d*x + c)^4/sqrt(a*sin(d*x + c) + a), x)

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Fricas [B]  time = 1.73309, size = 975, normalized size = 7.22 \begin{align*} \frac{3 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \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 \,{\left (3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 7\right )} \sin \left (d x + c\right ) + 5 \, \cos \left (d x + c\right ) + 7\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{96 \,{\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d -{\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\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)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/96*(3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + 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*(3*cos(d*x + c)^3 + cos(d*x + c)^2 - (3*cos(d*x + c)^2 + 2*cos(d*x + c) + 7)*sin(d*x + c) + 5*cos(d
*x + c) + 7)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d - (a*d*cos(d*x + c)^3
+ a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - 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)**4/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.44294, size = 737, normalized size = 5.46 \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)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/48*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 3/(a*sgn
(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) + 2/(a*sgn(tan(1/2*d*x + 1/2*c) + 1))) + (30*sqrt(2)*a^(3/2)
*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 15*sqrt(2)*sqrt(-a)*a*log(sqrt(2)*sqrt(a) + sqrt(a)) + 42*a^(3
/2)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 21*sqrt(-a)*a*log(sqrt(2)*sqrt(a) + sqrt(a)) - 88*sqrt(2)*s
qrt(-a)*a - 126*sqrt(-a)*a)*sgn(tan(1/2*d*x + 1/2*c) + 1)/(5*sqrt(2)*sqrt(-a)*a^(3/2) + 7*sqrt(-a)*a^(3/2)) -
6*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)) + 3*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)*s
gn(tan(1/2*d*x + 1/2*c) + 1)) - 2*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5 - 6
*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(a) - 3*(sqrt(a)*tan(1/2*d*x + 1/2*
c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^2 - 2*a^(5/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))^2 - a)^3*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d