\(\int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [325]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2953, 3055, 3060, 2852, 212} \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cos[c + d*x]*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (2*a*Cos[c + d*x])/(3*d*Sqrt[a + a*S
in[c + d*x]]) + (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*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 3055

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

Rule 3060

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{3/2} \, dx}{a^2} \\ & = \frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {2 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a^2}{2}-\frac {1}{2} a^2 \sin (c+d x)\right ) \, dx}{3 a^2} \\ & = \frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {(2 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 )}{d} \\ & = -\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.54 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )-3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(3*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] - 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/(3*d*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-3 \sqrt {a -a \sin \left (d x +c \right )}\, a \right )}{3 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(103\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (79) = 158\).

Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.69 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/6*(3*sqrt(a)*(cos(d*x + c) + sin(d*x + c) + 1)*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) - 1)*sin(d*x + c) + 2*cos(d*x +
c) + 1)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F]

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

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)*(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), x)

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(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), x)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.42 \[ \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (8 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \]

[In]

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

[Out]

1/6*sqrt(2)*(8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 3*sqrt(2)*log(abs(-2*sqr
t(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)) - 12*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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