3.1073 \(\int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac{a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{1}{2} b x \left (6 a^2-b^2\right )+\frac{15 a b^2 \cos (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{5 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

-(b*(6*a^2 - b^2)*x)/2 + (a*(a^2 - 6*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) + (15*a*b^2*Cos[c + d*x])/(2*d) + (5*b^
3*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (3*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*d) - (Cot[c + d*x]*Csc[c + d
*x]*(a + b*Sin[c + d*x])^3)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.465199, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2889, 3048, 3047, 3033, 3023, 2735, 3770} \[ \frac{a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{1}{2} b x \left (6 a^2-b^2\right )+\frac{15 a b^2 \cos (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{5 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(6*a^2 - b^2)*x)/2 + (a*(a^2 - 6*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) + (15*a*b^2*Cos[c + d*x])/(2*d) + (5*b^
3*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (3*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*d) - (Cot[c + d*x]*Csc[c + d
*x]*(a + b*Sin[c + d*x])^3)/(2*d)

Rule 2889

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

Rule 3048

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

Rule 3047

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

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc ^3(c+d x) (a+b \sin (c+d x))^3 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x))^2 \left (3 b-a \sin (c+d x)-4 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{1}{2} \int \csc (c+d x) (a+b \sin (c+d x)) \left (-a^2+6 b^2-5 a b \sin (c+d x)-10 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{1}{4} \int \csc (c+d x) \left (-2 a \left (a^2-6 b^2\right )-2 b \left (6 a^2-b^2\right ) \sin (c+d x)-30 a b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac{15 a b^2 \cos (c+d x)}{2 d}+\frac{5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac{1}{4} \int \csc (c+d x) \left (-2 a \left (a^2-6 b^2\right )-2 b \left (6 a^2-b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{1}{2} b \left (6 a^2-b^2\right ) x+\frac{15 a b^2 \cos (c+d x)}{2 d}+\frac{5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac{1}{2} \left (a \left (a^2-6 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=-\frac{1}{2} b \left (6 a^2-b^2\right ) x+\frac{a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac{15 a b^2 \cos (c+d x)}{2 d}+\frac{5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac{\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.33709, size = 192, normalized size = 1.39 \[ \frac{12 a^2 b \tan \left (\frac{1}{2} (c+d x)\right )-12 a^2 b \cot \left (\frac{1}{2} (c+d x)\right )-24 a^2 b c-24 a^2 b d x+a^3 \left (-\csc ^2\left (\frac{1}{2} (c+d x)\right )\right )+a^3 \sec ^2\left (\frac{1}{2} (c+d x)\right )-4 a^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+24 a b^2 \cos (c+d x)+24 a b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-24 a b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 b^3 \sin (2 (c+d x))+4 b^3 c+4 b^3 d x}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^2*b*c + 4*b^3*c - 24*a^2*b*d*x + 4*b^3*d*x + 24*a*b^2*Cos[c + d*x] - 12*a^2*b*Cot[(c + d*x)/2] - a^3*Cs
c[(c + d*x)/2]^2 + 4*a^3*Log[Cos[(c + d*x)/2]] - 24*a*b^2*Log[Cos[(c + d*x)/2]] - 4*a^3*Log[Sin[(c + d*x)/2]]
+ 24*a*b^2*Log[Sin[(c + d*x)/2]] + a^3*Sec[(c + d*x)/2]^2 + 2*b^3*Sin[2*(c + d*x)] + 12*a^2*b*Tan[(c + d*x)/2]
)/(8*d)

________________________________________________________________________________________

Maple [A]  time = 0.098, size = 171, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-3\,{a}^{2}bx-3\,{\frac{{a}^{2}b\cot \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{2}bc}{d}}+3\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}x}{2}}+{\frac{{b}^{3}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a^3/sin(d*x+c)^2*cos(d*x+c)^3-1/2*a^3*cos(d*x+c)/d-1/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-3*a^2*b*x-3*a^2*
b*cot(d*x+c)/d-3/d*a^2*b*c+3*a*b^2*cos(d*x+c)/d+3/d*a*b^2*ln(csc(d*x+c)-cot(d*x+c))+1/2*b^3*cos(d*x+c)*sin(d*x
+c)/d+1/2*b^3*x+1/2/d*b^3*c

________________________________________________________________________________________

Maxima [A]  time = 1.63022, size = 173, normalized size = 1.25 \begin{align*} -\frac{12 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{2} b -{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} - a^{3}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a b^{2}{\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*(d*x + c + 1/tan(d*x + c))*a^2*b - (2*d*x + 2*c + sin(2*d*x + 2*c))*b^3 - a^3*(2*cos(d*x + c)/(cos(d*
x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 6*a*b^2*(2*cos(d*x + c) - log(cos(d*x + c) +
1) + log(cos(d*x + c) - 1)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.69171, size = 510, normalized size = 3.7 \begin{align*} \frac{12 \, a b^{2} \cos \left (d x + c\right )^{3} - 2 \,{\left (6 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{2} b - b^{3}\right )} d x + 2 \,{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right ) -{\left (a^{3} - 6 \, a b^{2} -{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{3} - 6 \, a b^{2} -{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (b^{3} \cos \left (d x + c\right )^{3} +{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*a*b^2*cos(d*x + c)^3 - 2*(6*a^2*b - b^3)*d*x*cos(d*x + c)^2 + 2*(6*a^2*b - b^3)*d*x + 2*(a^3 - 6*a*b^2
)*cos(d*x + c) - (a^3 - 6*a*b^2 - (a^3 - 6*a*b^2)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (a^3 - 6*a*b^2
- (a^3 - 6*a*b^2)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) + 2*(b^3*cos(d*x + c)^3 + (6*a^2*b - b^3)*cos(
d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

________________________________________________________________________________________

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)**3*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.27662, size = 367, normalized size = 2.66 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \,{\left (6 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} - 4 \,{\left (a^{3} - 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 8 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 24 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c) - 4*(6*a^2*b - b^3)*(d*x + c) - 4*(a^3 - 6*a*b
^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (2*a^3*tan(1/2*d*x + 1/2*c)^6 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 12*a^2*
b*tan(1/2*d*x + 1/2*c)^5 - 8*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a*b^2*tan(1/2*d*x
+ 1/2*c)^4 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 8*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2
- 12*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2)/d