\(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [468]

Optimal result

Integrand size = 29, antiderivative size = 125 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {9 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

9/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)-2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/4*cot
(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2960, 2830, 2728, 212, 3123, 3063, 3064, 2852} \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {9 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 \sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}+\frac {\cot (c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(9*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*Sqrt[a]*d) - (2*Cos[c + d*x])/(d*Sqrt[a + a*Si
n[c + d*x]]) + 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 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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

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 2960

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/d^4, Int[(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^m, x], x] + Int[(d*Sin[e + f*x])^
n*(a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&
!IGtQ[m, 0]

Rule 3063

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

Rule 3064

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

Rule 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x]
)^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n
+ 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2,
0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx+\int \frac {\csc ^3(c+d x) \left (1-2 \sin ^2(c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {2 \cos (c+d x)}{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 \frac {\csc ^2(c+d x) \left (-\frac {a}{2}-\frac {5}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a}-\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\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 \frac {\csc (c+d x) \left (-\frac {9 a^2}{4}-\frac {1}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^2}+\frac {2 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d} \\ & = \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\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 {9 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a}+\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\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 {2 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {9 \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 {9 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(296\) vs. \(2(125)=250\).

Time = 2.91 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.37 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-8-64 \cos \left (\frac {1}{2} (c+d x)\right )+4 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )+36 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-36 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}+64 \sin \left (\frac {1}{2} (c+d x)\right )+4 \tan \left (\frac {1}{4} (c+d x)\right )\right )}{32 d \sqrt {a (1+\sin (c+d x))}} \]

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8 - 64*Cos[(c + d*x)/2] + 4*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 + 3
6*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 36*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]) +
64*Sin[(c + d*x)/2] + 4*Tan[(c + d*x)/4]))/(32*d*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.20

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (8 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (\sin ^{2}\left (d x +c \right )\right )-9 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+\left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}+\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right )}{4 a^{\frac {5}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(150\)

[In]

int(cos(d*x+c)^4*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^(5/2)*(8*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)^2-9*arct
anh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)+(-a*(sin(d*x+c)-1))^
(1/2)*a^(3/2))/sin(d*x+c)^2/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. 346 vs. \(2 (107) = 214\).

Time = 0.29 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.77 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {9 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (8 \, \cos \left (d x + c\right )^{3} + 9 \, \cos \left (d x + c\right )^{2} - {\left (8 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 11\right )} \sin \left (d x + c\right ) - 10 \, \cos \left (d x + c\right ) - 11\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d + {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/16*(9*(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*(8*cos(d
*x + c)^3 + 9*cos(d*x + c)^2 - (8*cos(d*x + c)^2 - cos(d*x + c) - 11)*sin(d*x + c) - 10*cos(d*x + c) - 11)*sqr
t(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))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**3/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{3}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*csc(d*x + c)^3/sqrt(a*sin(d*x + c) + a), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{[37748736,0]:[1,0,-2]%%},[1]%%%},0]:[1,0,%%%{
-1,[1]%%%}]

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(1/2)), x)