3.1120 \(\int \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=231 \[ -\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3}{8} b x \left (12 a^2-b^2\right )-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d} \]

[Out]

-3/8*b*(12*a^2-b^2)*x+3/2*a*(a^2-2*b^2)*arctanh(cos(d*x+c))/d-1/2*a*(a^2-17*b^2)*cos(d*x+c)/d-1/8*b*(2*a^2-21*
b^2)*cos(d*x+c)*sin(d*x+c)/d-1/4*(a^2-6*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^2/a/d-1/4*(a^2-4*b^2)*cos(d*x+c)*(a+b
*sin(d*x+c))^3/a^2/d-b*cot(d*x+c)*(a+b*sin(d*x+c))^4/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^4/a/d

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Rubi [A]  time = 0.69, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2893, 3049, 3033, 3023, 2735, 3770} \[ -\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {b \left (2 a^2-21 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 a \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3}{8} b x \left (12 a^2-b^2\right )-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(12*a^2 - b^2)*x)/8 + (3*a*(a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) - (a*(a^2 - 17*b^2)*Cos[c + d*x])/
(2*d) - (b*(2*a^2 - 21*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) - ((a^2 - 6*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x]
)^2)/(4*a*d) - ((a^2 - 4*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^3)/(4*a^2*d) - (b*Cot[c + d*x]*(a + b*Sin[c +
d*x])^4)/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^4)/(2*a*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 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

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 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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 \cos (c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^3 \left (3 \left (a^2-2 b^2\right )+3 a b \sin (c+d x)-2 \left (a^2-4 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (12 a \left (a^2-2 b^2\right )+18 a^2 b \sin (c+d x)-6 a \left (a^2-6 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{8 a^2}\\ &=-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (36 a^2 \left (a^2-2 b^2\right )+78 a^3 b \sin (c+d x)-6 a^2 \left (2 a^2-21 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=-\frac {b \left (2 a^2-21 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {\int \csc (c+d x) \left (72 a^3 \left (a^2-2 b^2\right )+18 a^2 b \left (12 a^2-b^2\right ) \sin (c+d x)-24 a^3 \left (a^2-17 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{48 a^2}\\ &=-\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {b \left (2 a^2-21 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {\int \csc (c+d x) \left (72 a^3 \left (a^2-2 b^2\right )+18 a^2 b \left (12 a^2-b^2\right ) \sin (c+d x)\right ) \, dx}{48 a^2}\\ &=-\frac {3}{8} b \left (12 a^2-b^2\right ) x-\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {b \left (2 a^2-21 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}-\frac {1}{2} \left (3 a \left (a^2-2 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=-\frac {3}{8} b \left (12 a^2-b^2\right ) x+\frac {3 a \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a \left (a^2-17 b^2\right ) \cos (c+d x)}{2 d}-\frac {b \left (2 a^2-21 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {\left (a^2-6 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{4 a d}-\frac {\left (a^2-4 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{4 a^2 d}-\frac {b \cot (c+d x) (a+b \sin (c+d x))^4}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^4}{2 a d}\\ \end {align*}

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Mathematica [A]  time = 6.16, size = 252, normalized size = 1.09 \[ -\frac {3 \left (a^3-2 a b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {3 \left (a^3-2 a b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {a^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {3 b \left (b^2-12 a^2\right ) (c+d x)}{8 d}+\frac {b \left (b^2-3 a^2\right ) \sin (2 (c+d x))}{4 d}-\frac {a \left (4 a^2-15 b^2\right ) \cos (c+d x)}{4 d}+\frac {3 a^2 b \tan \left (\frac {1}{2} (c+d x)\right )}{2 d}-\frac {3 a^2 b \cot \left (\frac {1}{2} (c+d x)\right )}{2 d}+\frac {a b^2 \cos (3 (c+d x))}{4 d}+\frac {b^3 \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(-12*a^2 + b^2)*(c + d*x))/(8*d) - (a*(4*a^2 - 15*b^2)*Cos[c + d*x])/(4*d) + (a*b^2*Cos[3*(c + d*x)])/(4*
d) - (3*a^2*b*Cot[(c + d*x)/2])/(2*d) - (a^3*Csc[(c + d*x)/2]^2)/(8*d) + (3*(a^3 - 2*a*b^2)*Log[Cos[(c + d*x)/
2]])/(2*d) - (3*(a^3 - 2*a*b^2)*Log[Sin[(c + d*x)/2]])/(2*d) + (a^3*Sec[(c + d*x)/2]^2)/(8*d) + (b*(-3*a^2 + b
^2)*Sin[2*(c + d*x)])/(4*d) + (b^3*Sin[4*(c + d*x)])/(32*d) + (3*a^2*b*Tan[(c + d*x)/2])/(2*d)

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fricas [A]  time = 0.64, size = 260, normalized size = 1.13 \[ \frac {8 \, a b^{2} \cos \left (d x + c\right )^{5} - 3 \, {\left (12 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 8 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, a^{2} b - b^{3}\right )} d x + 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right ) - 6 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 6 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, b^{3} \cos \left (d x + c\right )^{5} - {\left (12 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*a*b^2*cos(d*x + c)^5 - 3*(12*a^2*b - b^3)*d*x*cos(d*x + c)^2 - 8*(a^3 - 2*a*b^2)*cos(d*x + c)^3 + 3*(12
*a^2*b - b^3)*d*x + 12*(a^3 - 2*a*b^2)*cos(d*x + c) - 6*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2)*log(1
/2*cos(d*x + c) + 1/2) + 6*(a^3 - 2*a*b^2 - (a^3 - 2*a*b^2)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) + (2*
b^3*cos(d*x + c)^5 - (12*a^2*b - b^3)*cos(d*x + c)^3 + 3*(12*a^2*b - b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d
*x + c)^2 - d)

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giac [A]  time = 0.31, size = 400, normalized size = 1.73 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (12 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} - 12 \, {\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {18 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 48 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 96 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 80 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} + 32 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*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) - 3*(12*a^2*b - b^3)*(d*x + c) - 12*(a^3 - 2*a
*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (18*a^3*tan(1/2*d*x + 1/2*c)^2 - 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)^2 + 2*(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 5*b^3*tan(1/2*d
*x + 1/2*c)^7 - 8*a^3*tan(1/2*d*x + 1/2*c)^6 + 48*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 + 3*b^3*tan(1/2*d*x + 1/2*c)^5 - 24*a^3*tan(1/2*d*x + 1/2*c)^4 + 96*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b
*tan(1/2*d*x + 1/2*c)^3 - 3*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*a^3*tan(1/2*d*x + 1/2*c)^2 + 80*a*b^2*tan(1/2*d*x
+ 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 5*b^3*tan(1/2*d*x + 1/2*c) - 8*a^3 + 32*a*b^2)/(tan(1/2*d*x + 1/2
*c)^2 + 1)^4)/d

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maple [A]  time = 0.66, size = 279, normalized size = 1.21 \[ -\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \cos \left (d x +c \right )}{2 d}-\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {3 a^{2} b \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{d}-\frac {9 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {9 a^{2} b x}{2}-\frac {9 a^{2} b c}{2 d}+\frac {a \,b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \cos \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}+\frac {3 b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 b^{3} x}{8}+\frac {3 b^{3} c}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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)^5-1/2*a^3*cos(d*x+c)^3/d-3/2*a^3*cos(d*x+c)/d-3/2/d*a^3*ln(csc(d*x+c)-cot(d
*x+c))-3/d*a^2*b/sin(d*x+c)*cos(d*x+c)^5-3/d*a^2*b*sin(d*x+c)*cos(d*x+c)^3-9/2/d*a^2*b*cos(d*x+c)*sin(d*x+c)-9
/2*a^2*b*x-9/2/d*a^2*b*c+a*b^2*cos(d*x+c)^3/d+3*a*b^2*cos(d*x+c)/d+3/d*a*b^2*ln(csc(d*x+c)-cot(d*x+c))+1/4/d*b
^3*cos(d*x+c)^3*sin(d*x+c)+3/8*b^3*cos(d*x+c)*sin(d*x+c)/d+3/8*b^3*x+3/8/d*b^3*c

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maxima [A]  time = 0.46, size = 186, normalized size = 0.81 \[ -\frac {48 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} b - 16 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a b^{2} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} - 8 \, a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(48*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^2*b - 16*(2*cos(d*x + c)^3
+ 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a*b^2 - (12*d*x + 12*c + sin(4*d*x + 4*c
) + 8*sin(2*d*x + 2*c))*b^3 - 8*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c)
+ 1) - 3*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 9.59, size = 718, normalized size = 3.11 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (32\,a\,b^2-10\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (48\,a\,b^2-\frac {17\,a^3}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (80\,a\,b^2-27\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (96\,a\,b^2-26\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (6\,a^2\,b-5\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (12\,a^2\,b-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (36\,a^2\,b-5\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (48\,a^2\,b+3\,b^3\right )-\frac {a^3}{2}-6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a\,b^2-\frac {3\,a^3}{2}\right )}{d}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {3\,b\,\mathrm {atan}\left (\frac {\frac {3\,b\,\left (12\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-3\,a^3\right )-9\,a^2\,b+\frac {3\,b^3}{4}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2-b^2\right )\,9{}\mathrm {i}}{4}\right )}{8}+\frac {3\,b\,\left (12\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-3\,a^3\right )-9\,a^2\,b+\frac {3\,b^3}{4}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2-b^2\right )\,9{}\mathrm {i}}{4}\right )}{8}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (81\,a^4\,b^2-\frac {27\,a^2\,b^4}{2}+\frac {9\,b^6}{16}\right )+\frac {9\,a\,b^5}{2}+27\,a^5\,b-\frac {225\,a^3\,b^3}{4}-\frac {b\,\left (12\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-3\,a^3\right )-9\,a^2\,b+\frac {3\,b^3}{4}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2-b^2\right )\,9{}\mathrm {i}}{4}\right )\,3{}\mathrm {i}}{8}+\frac {b\,\left (12\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-3\,a^3\right )-9\,a^2\,b+\frac {3\,b^3}{4}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2-b^2\right )\,9{}\mathrm {i}}{4}\right )\,3{}\mathrm {i}}{8}}\right )\,\left (12\,a^2-b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (tan(c/2 + (d*x)/2)^2*(32*a*b^2 - 10*a^3) + tan(c/2 + (d*x)/2)^8*(48*a*b^2
- (17*a^3)/2) + tan(c/2 + (d*x)/2)^4*(80*a*b^2 - 27*a^3) + tan(c/2 + (d*x)/2)^6*(96*a*b^2 - 26*a^3) + tan(c/2
+ (d*x)/2)^9*(6*a^2*b - 5*b^3) - tan(c/2 + (d*x)/2)^7*(12*a^2*b - 3*b^3) - tan(c/2 + (d*x)/2)^3*(36*a^2*b - 5*
b^3) - tan(c/2 + (d*x)/2)^5*(48*a^2*b + 3*b^3) - a^3/2 - 6*a^2*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^
2 + 16*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) +
(log(tan(c/2 + (d*x)/2))*(3*a*b^2 - (3*a^3)/2))/d + (3*a^2*b*tan(c/2 + (d*x)/2))/(2*d) + (3*b*atan(((3*b*(12*a
^2 - b^2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 - (b*tan(c/2 + (d*x)/2)*(12*a^2 - b^2)*9
i)/4))/8 + (3*b*(12*a^2 - b^2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 + (b*tan(c/2 + (d*x
)/2)*(12*a^2 - b^2)*9i)/4))/8)/(2*tan(c/2 + (d*x)/2)*((9*b^6)/16 - (27*a^2*b^4)/2 + 81*a^4*b^2) + (9*a*b^5)/2
+ 27*a^5*b - (225*a^3*b^3)/4 - (b*(12*a^2 - b^2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) - 9*a^2*b + (3*b^3)/4 -
(b*tan(c/2 + (d*x)/2)*(12*a^2 - b^2)*9i)/4)*3i)/8 + (b*(12*a^2 - b^2)*(tan(c/2 + (d*x)/2)*(6*a*b^2 - 3*a^3) -
9*a^2*b + (3*b^3)/4 + (b*tan(c/2 + (d*x)/2)*(12*a^2 - b^2)*9i)/4)*3i)/8))*(12*a^2 - b^2))/(4*d)

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

[Out]

Timed out

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