\(\int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [327]

Optimal result

Integrand size = 29, antiderivative size = 101 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}-\frac {a \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d} \]

[Out]

5/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d-1/4*a*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2
*cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2953, 3054, 3059, 2852, 212} \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 d}-\frac {a \cot (c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d} \]

[In]

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

[Out]

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

Rule 212

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

Rule 2852

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 2953

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 3054

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3059

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{3/2} \, dx}{a^2} \\ & = -\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d}+\frac {\int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {a^2}{2}-\frac {3}{2} a^2 \sin (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {a \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d}-\frac {5}{8} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {a \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d}+\frac {(5 a) \text {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 {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}-\frac {a \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(101)=202\).

Time = 0.93 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.47 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\csc ^7\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+6 \cos \left (\frac {3}{2} (c+d x)\right )-5 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+5 \cos (2 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+5 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-5 \cos (2 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )+6 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{4 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^2} \]

[In]

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

[Out]

-1/4*(Csc[(c + d*x)/2]^7*Sqrt[a*(1 + Sin[c + d*x])]*(2*Cos[(c + d*x)/2] + 6*Cos[(3*(c + d*x))/2] - 5*Log[1 + C
os[(c + d*x)/2] - Sin[(c + d*x)/2]] + 5*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 5*Log[
1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 5*Cos[2*(c + d*x)]*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2
*Sin[(c + d*x)/2] + 6*Sin[(3*(c + d*x))/2]))/(d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^
2)^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (5 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+3 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}-5 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right )}{4 a^{\frac {3}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(126\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(3/2)*(5*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^
2*a^2+3*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)-5*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))/sin(d*x+c)^2/cos(d*x+c)/(a+a*si
n(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (85) = 170\).

Time = 0.29 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.16 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \]

[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*(5*(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*(3*cos(d
*x + c)^2 + (3*cos(d*x + c) + 1)*sin(d*x + c) + 2*cos(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^
3 + d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c)^2 - d)*sin(d*x + c) - d)

Sympy [F]

\[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \csc \left (d x + c\right )^{3} \,d x } \]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.53 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (5 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {4 \, {\left (6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \]

[In]

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

[Out]

1/16*sqrt(2)*(5*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi +
1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 4*(6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi
+ 1/2*d*x + 1/2*c)^3 - 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(2*sin(-1/4*pi +
1/2*d*x + 1/2*c)^2 - 1)^2)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\sin \left (c+d\,x\right )}^3} \,d x \]

[In]

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

[Out]

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