3.1074 \(\int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}+\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-3 a b^2 x-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac {11 b^3 \cos (c+d x)}{6 d} \]

[Out]

-3*a*b^2*x+1/2*b*(3*a^2-2*b^2)*arctanh(cos(d*x+c))/d+11/6*b^3*cos(d*x+c)/d+1/3*a*(a^2-3*b^2)*cot(d*x+c)/d-1/2*
b*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^2/d-1/3*cot(d*x+c)*csc(d*x+c)^2*(a+b*sin(d*x+c))^3/d

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Rubi [A]  time = 0.48, antiderivative size = 138, 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, 3023, 2735, 3770} \[ \frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}+\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-3 a b^2 x-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac {11 b^3 \cos (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3*a*b^2*x + (b*(3*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) + (11*b^3*Cos[c + d*x])/(6*d) + (a*(a^2 - 3*b^2)*
Cot[c + d*x])/(3*d) - (b*Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*d) - (Cot[c + d*x]*Csc[c + d*x]^
2*(a + b*Sin[c + d*x])^3)/(3*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 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 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 ^2(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc ^4(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 ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}+\frac {1}{3} \int \csc ^3(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 (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}+\frac {1}{6} \int \csc ^2(c+d x) (a+b \sin (c+d x)) \left (-2 \left (a^2-3 b^2\right )-7 a b \sin (c+d x)-11 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}-\frac {1}{6} \int \csc (c+d x) \left (3 b \left (3 a^2-2 b^2\right )+18 a b^2 \sin (c+d x)+11 b^3 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {11 b^3 \cos (c+d x)}{6 d}+\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}-\frac {1}{6} \int \csc (c+d x) \left (3 b \left (3 a^2-2 b^2\right )+18 a b^2 \sin (c+d x)\right ) \, dx\\ &=-3 a b^2 x+\frac {11 b^3 \cos (c+d x)}{6 d}+\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}-\frac {1}{2} \left (b \left (3 a^2-2 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=-3 a b^2 x+\frac {b \left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {11 b^3 \cos (c+d x)}{6 d}+\frac {a \left (a^2-3 b^2\right ) \cot (c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^3}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 615, normalized size = 4.46 \[ \frac {\sin ^3(c+d x) \csc \left (\frac {1}{2} (c+d x)\right ) \left (a^3 \cos \left (\frac {1}{2} (c+d x)\right )-9 a b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{6 d (a+b \sin (c+d x))^3}+\frac {\sin ^3(c+d x) \sec \left (\frac {1}{2} (c+d x)\right ) \left (9 a b^2 \sin \left (\frac {1}{2} (c+d x)\right )-a^3 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{6 d (a+b \sin (c+d x))^3}-\frac {a^3 \sin ^3(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{24 d (a+b \sin (c+d x))^3}+\frac {a^3 \sin ^3(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{24 d (a+b \sin (c+d x))^3}+\frac {\left (2 b^3-3 a^2 b\right ) \sin ^3(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{2 d (a+b \sin (c+d x))^3}+\frac {\left (3 a^2 b-2 b^3\right ) \sin ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^3}{2 d (a+b \sin (c+d x))^3}-\frac {3 a^2 b \sin ^3(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}+\frac {3 a^2 b \sin ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^3}{8 d (a+b \sin (c+d x))^3}+\frac {b^3 \sin ^3(c+d x) \cos (c+d x) (a \csc (c+d x)+b)^3}{d (a+b \sin (c+d x))^3}-\frac {3 a b^2 (c+d x) \sin ^3(c+d x) (a \csc (c+d x)+b)^3}{d (a+b \sin (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*a*b^2*(c + d*x)*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d*(a + b*Sin[c + d*x])^3) + (b^3*Cos[c + d*x]*(b +
a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(d*(a + b*Sin[c + d*x])^3) + ((a^3*Cos[(c + d*x)/2] - 9*a*b^2*Cos[(c + d*x)
/2])*Csc[(c + d*x)/2]*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(6*d*(a + b*Sin[c + d*x])^3) - (3*a^2*b*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*Cot[(c + d*x)/2]*Csc[(c +
d*x)/2]^2*(b + a*Csc[c + d*x])^3*Sin[c + d*x]^3)/(24*d*(a + b*Sin[c + d*x])^3) + ((3*a^2*b - 2*b^3)*(b + a*Cs
c[c + d*x])^3*Log[Cos[(c + d*x)/2]]*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3) + ((-3*a^2*b + 2*b^3)*(b + a*
Csc[c + d*x])^3*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^3)/(2*d*(a + b*Sin[c + d*x])^3) + (3*a^2*b*(b + a*Csc[c + d
*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3)/(8*d*(a + b*Sin[c + d*x])^3) + ((b + a*Csc[c + d*x])^3*Sec[(c + d*x)
/2]*(-(a^3*Sin[(c + d*x)/2]) + 9*a*b^2*Sin[(c + d*x)/2])*Sin[c + d*x]^3)/(6*d*(a + b*Sin[c + d*x])^3) + (a^3*(
b + a*Csc[c + d*x])^3*Sec[(c + d*x)/2]^2*Sin[c + d*x]^3*Tan[(c + d*x)/2])/(24*d*(a + b*Sin[c + d*x])^3)

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fricas [A]  time = 0.74, size = 231, normalized size = 1.67 \[ \frac {36 \, a b^{2} \cos \left (d x + c\right ) + 4 \, {\left (a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 6 \, {\left (6 \, a b^{2} d x \cos \left (d x + c\right )^{2} - 2 \, b^{3} \cos \left (d x + c\right )^{3} - 6 \, a b^{2} d x - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(36*a*b^2*cos(d*x + c) + 4*(a^3 - 9*a*b^2)*cos(d*x + c)^3 - 3*(3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*
x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*lo
g(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 6*(6*a*b^2*d*x*cos(d*x + c)^2 - 2*b^3*cos(d*x + c)^3 - 6*a*b^2*d*x -
(3*a^2*b - 2*b^3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.25, size = 222, normalized size = 1.61 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, {\left (d x + c\right )} a b^{2} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {48 \, b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - 12 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {66 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a*b^2 - 3*a^3*tan(1/2*d*x + 1
/2*c) + 36*a*b^2*tan(1/2*d*x + 1/2*c) + 48*b^3/(tan(1/2*d*x + 1/2*c)^2 + 1) - 12*(3*a^2*b - 2*b^3)*log(abs(tan
(1/2*d*x + 1/2*c))) + (66*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 44*b^3*tan(1/2*d*x + 1/2*c)^3 + 3*a^3*tan(1/2*d*x + 1
/2*c)^2 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.46, size = 159, normalized size = 1.15 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}-\frac {3 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 a^{2} b \cos \left (d x +c \right )}{2 d}-\frac {3 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-3 a \,b^{2} x -\frac {3 a \,b^{2} \cot \left (d x +c \right )}{d}-\frac {3 a \,b^{2} c}{d}+\frac {b^{3} \cos \left (d x +c \right )}{d}+\frac {b^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 119, normalized size = 0.86 \[ -\frac {36 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a b^{2} - 9 \, a^{2} b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, b^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {4 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(36*(d*x + c + 1/tan(d*x + c))*a*b^2 - 9*a^2*b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) +
1) - log(cos(d*x + c) - 1)) - 6*b^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 4*a^3/
tan(d*x + c)^3)/d

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mupad [B]  time = 10.71, size = 477, normalized size = 3.46 \[ -\frac {\frac {a^3\,\cos \left (c+d\,x\right )}{4}-\frac {3\,b^3\,\sin \left (c+d\,x\right )}{4}+\frac {a^3\,\cos \left (3\,c+3\,d\,x\right )}{12}-\frac {b^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {b^3\,\sin \left (4\,c+4\,d\,x\right )}{8}-\frac {3\,a\,b^2\,\cos \left (3\,c+3\,d\,x\right )}{4}-\frac {3\,b^3\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}+\frac {3\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {b^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 )\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )}{4}-\frac {3\,a^2\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (3\,c+3\,d\,x\right )}{8}-\frac {9\,a\,b^2\,\mathrm {atan}\left (\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\sin \left (c+d\,x\right )}{2}+\frac {3\,a\,b^2\,\mathrm {atan}\left (\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {9\,a^2\,b\,\sin \left (c+d\,x\right )\,\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\,{\sin \left (c+d\,x\right )}^3} \]

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)^4,x)

[Out]

-((a^3*cos(c + d*x))/4 - (3*b^3*sin(c + d*x))/4 + (a^3*cos(3*c + 3*d*x))/12 - (b^3*sin(2*c + 2*d*x))/4 + (b^3*
sin(3*c + 3*d*x))/4 + (b^3*sin(4*c + 4*d*x))/8 - (3*a*b^2*cos(3*c + 3*d*x))/4 - (3*b^3*sin(c + d*x)*log(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (3*a^2*b*sin(2*c + 2*d*x))/4 + (b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2))*sin(3*c + 3*d*x))/4 + (3*a*b^2*cos(c + d*x))/4 - (3*a^2*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*si
n(3*c + 3*d*x))/8 - (9*a*b^2*atan((3*a^2*sin(c/2 + (d*x)/2) - 2*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)
/2))/(2*b^2*cos(c/2 + (d*x)/2) - 3*a^2*cos(c/2 + (d*x)/2) + 6*a*b*sin(c/2 + (d*x)/2)))*sin(c + d*x))/2 + (3*a*
b^2*atan((3*a^2*sin(c/2 + (d*x)/2) - 2*b^2*sin(c/2 + (d*x)/2) + 6*a*b*cos(c/2 + (d*x)/2))/(2*b^2*cos(c/2 + (d*
x)/2) - 3*a^2*cos(c/2 + (d*x)/2) + 6*a*b*sin(c/2 + (d*x)/2)))*sin(3*c + 3*d*x))/2 + (9*a^2*b*sin(c + d*x)*log(
sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8)/(d*sin(c + d*x)^3)

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

[Out]

Timed out

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