\(\int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [334]

Optimal result

Integrand size = 29, antiderivative size = 131 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}+\frac {13 a^2 \cos (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {3 a \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d} \]

[Out]

1/4*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-1/2*cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(3
/2)/d+13/4*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-3/4*a*cot(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2953, 3054, 3060, 2852, 212} \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {13 a^2 \cos (c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {3 a \cot (c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d} \]

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*d) + (13*a^2*Cos[c + d*x])/(4*d*Sqrt[a +
a*Sin[c + d*x]]) - (3*a*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(4*d) - (Cot[c + d*x]*Csc[c + d*x]*(a + a*Sin[
c + d*x])^(3/2))/(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 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 ^3(c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{5/2} \, dx}{a^2} \\ & = -\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac {\int \csc ^2(c+d x) (a+a \sin (c+d x))^{3/2} \left (\frac {3 a^2}{2}-\frac {5}{2} a^2 \sin (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {3 a \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (-\frac {a^3}{4}-\frac {13}{4} a^3 \sin (c+d x)\right ) \, dx}{2 a^2} \\ & = \frac {13 a^2 \cos (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {3 a \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}-\frac {1}{8} a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {13 a^2 \cos (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {3 a \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d}+\frac {a^2 \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 {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}+\frac {13 a^2 \cos (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {3 a \cot (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2}}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(271\) vs. \(2(131)=262\).

Time = 1.81 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.07 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \csc ^7\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-22 \cos \left (\frac {1}{2} (c+d x)\right )+22 \cos \left (\frac {3}{2} (c+d x)\right )+8 \cos \left (\frac {5}{2} (c+d x)\right )-\log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (2 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\cos (2 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+22 \sin \left (\frac {1}{2} (c+d x)\right )+22 \sin \left (\frac {3}{2} (c+d x)\right )-8 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{4 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^2} \]

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/4*(a*Csc[(c + d*x)/2]^7*Sqrt[a*(1 + Sin[c + d*x])]*(-22*Cos[(c + d*x)/2] + 22*Cos[(3*(c + d*x))/2] + 8*Cos[
(5*(c + d*x))/2] - Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - Cos[2*(c + d*x)]*Log[1 - Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] + 22*Sin[(c + d*x)/2] + 22*Sin[(3*(c + d*x))/2] - 8*Sin[(5*(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 = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (8 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (\sin ^{2}\left (d x +c \right )\right )+\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+7 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}-9 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right )}{4 \sin \left (d x +c \right )^{2} \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(151\)

[In]

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

[Out]

1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(8*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)^2+arctanh((-a*(si
n(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+7*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)-9*(-a*(sin(d*x+c)-1))^(1/2)*a
^(3/2))/sin(d*x+c)^2/a^(1/2)/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. 359 vs. \(2 (111) = 222\).

Time = 0.29 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.74 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (8 \, a \cos \left (d x + c\right )^{3} + 15 \, a \cos \left (d x + c\right )^{2} - 6 \, a \cos \left (d x + c\right ) - {\left (8 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right ) - 13 \, a\right )} \sin \left (d x + c\right ) - 13 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \]

[In]

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

[Out]

1/16*((a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) + (a*cos(d*x + c)^2 - a)*sin(d*x + c) - a)*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
*a*cos(d*x + c)^3 + 15*a*cos(d*x + c)^2 - 6*a*cos(d*x + c) - (8*a*cos(d*x + c)^2 - 7*a*cos(d*x + c) - 13*a)*si
n(d*x + c) - 13*a)*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(-1)]

Timed out. \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.41 \[ \int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (\sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 32 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {4 \, {\left (14 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \]

[In]

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

[Out]

1/16*sqrt(2)*(sqrt(2)*a*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)) - 32*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi
+ 1/2*d*x + 1/2*c) - 4*(14*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 9*a*sgn(c
os(-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)*sqr
t(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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