3.1075 \(\int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=152 \[ \frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2+12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}+b^3 (-x) \]

[Out]

-b^3*x+1/8*a*(a^2+12*b^2)*arctanh(cos(d*x+c))/d+1/2*b*(2*a^2-b^2)*cot(d*x+c)/d+1/8*a*(a^2-2*b^2)*cot(d*x+c)*cs
c(d*x+c)/d-1/4*b*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.51, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2889, 3048, 3047, 3031, 3021, 2735, 3770} \[ \frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2+12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}+b^3 (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^3*x) + (a*(a^2 + 12*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) + (b*(2*a^2 - b^2)*Cot[c + d*x])/(2*d) + (a*(a^2 - 2
*b^2)*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (b*Cot[c + d*x]*Csc[c + d*x]^2*(a + b*Sin[c + d*x])^2)/(4*d) - (Cot[c
+ d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^3)/(4*d)

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 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 3021

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

Rule 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -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 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 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 ^3(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc ^5(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 ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{4} \int \csc ^4(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 {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{12} \int \csc ^3(c+d x) (a+b \sin (c+d x)) \left (-3 \left (a^2-2 b^2\right )-9 a b \sin (c+d x)-12 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {1}{24} \int \csc ^2(c+d x) \left (12 b \left (2 a^2-b^2\right )+3 a \left (a^2+12 b^2\right ) \sin (c+d x)+24 b^3 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {1}{24} \int \csc (c+d x) \left (3 a \left (a^2+12 b^2\right )+24 b^3 \sin (c+d x)\right ) \, dx\\ &=-b^3 x+\frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {1}{8} \left (a \left (a^2+12 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=-b^3 x+\frac {a \left (a^2+12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {b \left (2 a^2-b^2\right ) \cot (c+d x)}{2 d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^3}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.23, size = 690, normalized size = 4.54 \[ \frac {\left (a^3-12 a b^2\right ) \sin ^3(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{32 d (a+b \sin (c+d x))^3}+\frac {\left (-a^3-12 a b^2\right ) \sin ^3(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}+\frac {\left (12 a b^2-a^3\right ) \sin ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{32 d (a+b \sin (c+d x))^3}+\frac {\left (a^3+12 a b^2\right ) \sin ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}-\frac {a^3 \sin ^3(c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{64 d (a+b \sin (c+d x))^3}+\frac {a^3 \sin ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{64 d (a+b \sin (c+d x))^3}+\frac {\sin ^3(c+d x) \csc \left (\frac {1}{2} (c+d x)\right ) \left (a^2 b \cos \left (\frac {1}{2} (c+d x)\right )-b^3 \cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{2 d (a+b \sin (c+d x))^3}+\frac {\sin ^3(c+d x) \sec \left (\frac {1}{2} (c+d x)\right ) \left (b^3 \sin \left (\frac {1}{2} (c+d x)\right )-a^2 b \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{2 d (a+b \sin (c+d x))^3}-\frac {a^2 b \sin ^3(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}+\frac {a^2 b \sin ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}-\frac {b^3 (c+d x) \sin ^3(c+d x) (a \csc (c+d x)+b)^3}{d (a+b \sin (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((b^3*(c + d*x)*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d*(a + b*Sin[c + d*x])^3)) + ((a^2*b*Cos[(c + d*x)/2]
- b^3*Cos[(c + d*x)/2])*Csc[(c + d*x)/2]*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3)
+ ((a^3 - 12*a*b^2)*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(32*d*(a + b*Sin[c + d*x])^3) -
(a^2*b*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3)
- (a^3*Csc[(c + d*x)/2]^4*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(64*d*(a + b*Sin[c + d*x])^3) + ((a^3 + 12*a
*b^2)*(b + a*Csc[c + d*x])^3*Log[Cos[(c + d*x)/2]]*Sin[c + d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3) + ((-a^3 - 12*
a*b^2)*(b + a*Csc[c + d*x])^3*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3) + ((-a^3 + 12
*a*b^2)*(b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3)/(32*d*(a + b*Sin[c + d*x])^3) + (a^3*(b + a*
Csc[c + d*x])^3*Sec[(c + d*x)/2]^4*Sin[c + d*x]^3)/(64*d*(a + b*Sin[c + d*x])^3) + ((b + a*Csc[c + d*x])^3*Sec
[(c + d*x)/2]*(-(a^2*b*Sin[(c + d*x)/2]) + b^3*Sin[(c + d*x)/2])*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3)
+ (a^2*b*(b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3*Tan[(c + d*x)/2])/(8*d*(a + b*Sin[c + d*x])^
3)

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 265, normalized size = 1.74 \[ -\frac {16 \, b^{3} d x \cos \left (d x + c\right )^{4} - 32 \, b^{3} d x \cos \left (d x + c\right )^{2} + 16 \, b^{3} d x + 2 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} + 12 \, a b^{2} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} + 12 \, a b^{2} - 2 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16 \, {\left (b^{3} \cos \left (d x + c\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*(16*b^3*d*x*cos(d*x + c)^4 - 32*b^3*d*x*cos(d*x + c)^2 + 16*b^3*d*x + 2*(a^3 - 12*a*b^2)*cos(d*x + c)^3
+ 2*(a^3 + 12*a*b^2)*cos(d*x + c) - ((a^3 + 12*a*b^2)*cos(d*x + c)^4 + a^3 + 12*a*b^2 - 2*(a^3 + 12*a*b^2)*cos
(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + ((a^3 + 12*a*b^2)*cos(d*x + c)^4 + a^3 + 12*a*b^2 - 2*(a^3 + 12*a*b
^2)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) + 16*(b^3*cos(d*x + c) + (a^2*b - b^3)*cos(d*x + c)^3)*sin(d*
x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 234, normalized size = 1.54 \[ \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 192 \, {\left (d x + c\right )} b^{3} - 72 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, {\left (a^{3} + 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {50 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 72 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/192*(3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 192*
(d*x + c)*b^3 - 72*a^2*b*tan(1/2*d*x + 1/2*c) + 96*b^3*tan(1/2*d*x + 1/2*c) - 24*(a^3 + 12*a*b^2)*log(abs(tan(
1/2*d*x + 1/2*c))) + (50*a^3*tan(1/2*d*x + 1/2*c)^4 + 600*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 72*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 - 96*b^3*tan(1/2*d*x + 1/2*c)^3 - 72*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 24*a^2*b*tan(1/2*d*x + 1/2*c) -
3*a^3)/tan(1/2*d*x + 1/2*c)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 207, normalized size = 1.36 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a^{3} \cos \left (d x +c \right )}{8 d}-\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}-\frac {3 a \,b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \cos \left (d x +c \right )}{2 d}-\frac {3 a \,b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-b^{3} x -\frac {\cot \left (d x +c \right ) b^{3}}{d}-\frac {b^{3} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x)

[Out]

-1/4/d*a^3/sin(d*x+c)^4*cos(d*x+c)^3-1/8/d*a^3/sin(d*x+c)^2*cos(d*x+c)^3-1/8*a^3*cos(d*x+c)/d-1/8/d*a^3*ln(csc
(d*x+c)-cot(d*x+c))-1/d*a^2*b/sin(d*x+c)^3*cos(d*x+c)^3-3/2/d*a*b^2/sin(d*x+c)^2*cos(d*x+c)^3-3/2*a*b^2*cos(d*
x+c)/d-3/2/d*a*b^2*ln(csc(d*x+c)-cot(d*x+c))-b^3*x-1/d*cot(d*x+c)*b^3-1/d*b^3*c

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 149, normalized size = 0.98 \[ -\frac {16 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} b^{3} + a^{3} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, a^{2} b}{\tan \left (d x + c\right )^{3}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/16*(16*(d*x + c + 1/tan(d*x + c))*b^3 + a^3*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x
+ c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a*b^2*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) +
log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 16*a^2*b/tan(d*x + c)^3)/d

________________________________________________________________________________________

mupad [B]  time = 9.87, size = 348, normalized size = 2.29 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8\,d}-\frac {b^3\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2+8\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+12\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )}{d}-\frac {3\,a\,b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,d}+\frac {3\,a^2\,b\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {3\,a\,b^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a^2\,b\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(a + b*sin(c + d*x))^3)/sin(c + d*x)^5,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) - (a^3*cot(c/2 + (d*x)/2)^4)/(64*d) - (a^3*log(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2)))/(8*d) - (b^3*cot(c/2 + (d*x)/2))/(2*d) + (b^3*tan(c/2 + (d*x)/2))/(2*d) - (2*b^3*atan((8*b^3*cos(c/
2 + (d*x)/2) + a^3*sin(c/2 + (d*x)/2) + 12*a*b^2*sin(c/2 + (d*x)/2))/(a^3*cos(c/2 + (d*x)/2) - 8*b^3*sin(c/2 +
(d*x)/2) + 12*a*b^2*cos(c/2 + (d*x)/2))))/d - (3*a*b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*d) + (3
*a^2*b*cot(c/2 + (d*x)/2))/(8*d) - (3*a^2*b*tan(c/2 + (d*x)/2))/(8*d) - (3*a*b^2*cot(c/2 + (d*x)/2)^2)/(8*d) -
(a^2*b*cot(c/2 + (d*x)/2)^3)/(8*d) + (3*a*b^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*b*tan(c/2 + (d*x)/2)^3)/(8*d
)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________