3.1125 \(\int \cot ^4(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=303 \[ -\frac {a \left (2 a^2+21 b^2\right ) \cot (c+d x)}{35 d}-\frac {3 b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}-\frac {b \left (105 a^4-116 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d} \]

[Out]

-3/16*b*(a^2+2*b^2)*arctanh(cos(d*x+c))/d-1/35*a*(2*a^2+21*b^2)*cot(d*x+c)/d-1/560*b*(105*a^4-116*a^2*b^2+12*b
^4)*cot(d*x+c)*csc(d*x+c)/a^2/d-1/140*(4*a^4-19*a^2*b^2+2*b^4)*cot(d*x+c)*csc(d*x+c)^2/a/d+1/280*b*(53*a^2-6*b
^2)*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^2/a^2/d+1/35*(8*a^2-b^2)*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))
^3/a^2/d+1/14*b*cot(d*x+c)*csc(d*x+c)^5*(a+b*sin(d*x+c))^4/a^2/d-1/7*cot(d*x+c)*csc(d*x+c)^6*(a+b*sin(d*x+c))^
4/a/d

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Rubi [A]  time = 0.86, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2893, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ -\frac {a \left (2 a^2+21 b^2\right ) \cot (c+d x)}{35 d}-\frac {3 b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {\left (-19 a^2 b^2+4 a^4+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}-\frac {b \left (-116 a^2 b^2+105 a^4+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*b*(a^2 + 2*b^2)*ArcTanh[Cos[c + d*x]])/(16*d) - (a*(2*a^2 + 21*b^2)*Cot[c + d*x])/(35*d) - (b*(105*a^4 - 1
16*a^2*b^2 + 12*b^4)*Cot[c + d*x]*Csc[c + d*x])/(560*a^2*d) - ((4*a^4 - 19*a^2*b^2 + 2*b^4)*Cot[c + d*x]*Csc[c
+ d*x]^2)/(140*a*d) + (b*(53*a^2 - 6*b^2)*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(280*a^2*d) + (
(8*a^2 - b^2)*Cot[c + d*x]*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(35*a^2*d) + (b*Cot[c + d*x]*Csc[c + d*x]^5*
(a + b*Sin[c + d*x])^4)/(14*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^6*(a + b*Sin[c + d*x])^4)/(7*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 ^4(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}-\frac {\int \csc ^6(c+d x) (a+b \sin (c+d x))^3 \left (6 \left (8 a^2-b^2\right )+3 a b \sin (c+d x)-3 \left (14 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx}{42 a^2}\\ &=\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}-\frac {\int \csc ^5(c+d x) (a+b \sin (c+d x))^2 \left (3 b \left (53 a^2-6 b^2\right )-6 a \left (3 a^2-b^2\right ) \sin (c+d x)-9 b \left (18 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx}{210 a^2}\\ &=\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}-\frac {\int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (-18 \left (4 a^4-19 a^2 b^2+2 b^4\right )-3 a b \left (81 a^2-2 b^2\right ) \sin (c+d x)-3 b^2 \left (163 a^2-6 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{840 a^2}\\ &=-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}+\frac {\int \csc ^3(c+d x) \left (9 b \left (105 a^4-116 a^2 b^2+12 b^4\right )+72 a^3 \left (2 a^2+21 b^2\right ) \sin (c+d x)+9 b^3 \left (163 a^2-6 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{2520 a^2}\\ &=-\frac {b \left (105 a^4-116 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}+\frac {\int \csc ^2(c+d x) \left (144 a^3 \left (2 a^2+21 b^2\right )+945 a^2 b \left (a^2+2 b^2\right ) \sin (c+d x)\right ) \, dx}{5040 a^2}\\ &=-\frac {b \left (105 a^4-116 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}+\frac {1}{16} \left (3 b \left (a^2+2 b^2\right )\right ) \int \csc (c+d x) \, dx+\frac {1}{35} \left (a \left (2 a^2+21 b^2\right )\right ) \int \csc ^2(c+d x) \, dx\\ &=-\frac {3 b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {b \left (105 a^4-116 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}-\frac {\left (a \left (2 a^2+21 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{35 d}\\ &=-\frac {3 b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a \left (2 a^2+21 b^2\right ) \cot (c+d x)}{35 d}-\frac {b \left (105 a^4-116 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{560 a^2 d}-\frac {\left (4 a^4-19 a^2 b^2+2 b^4\right ) \cot (c+d x) \csc ^2(c+d x)}{140 a d}+\frac {b \left (53 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{280 a^2 d}+\frac {\left (8 a^2-b^2\right ) \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{35 a^2 d}+\frac {b \cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^4}{14 a^2 d}-\frac {\cot (c+d x) \csc ^6(c+d x) (a+b \sin (c+d x))^4}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.97, size = 324, normalized size = 1.07 \[ -\frac {112 a^3 \cos (5 (c+d x)) \csc ^7(c+d x)-16 a^3 \cos (7 (c+d x)) \csc ^7(c+d x)+56 a \left (14 a^2-3 b^2\right ) \cos (3 (c+d x)) \csc ^7(c+d x)+70 \cot (c+d x) \csc ^6(c+d x) \left (b \left (31 a^2-18 b^2\right ) \sin (c+d x)+12 a \left (2 a^2+b^2\right )\right )-3360 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3360 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1540 a^2 b \sin (4 (c+d x)) \csc ^7(c+d x)+105 a^2 b \sin (6 (c+d x)) \csc ^7(c+d x)-504 a b^2 \cos (5 (c+d x)) \csc ^7(c+d x)-168 a b^2 \cos (7 (c+d x)) \csc ^7(c+d x)-6720 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6720 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+840 b^3 \sin (4 (c+d x)) \csc ^7(c+d x)-350 b^3 \sin (6 (c+d x)) \csc ^7(c+d x)}{17920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

-1/17920*(56*a*(14*a^2 - 3*b^2)*Cos[3*(c + d*x)]*Csc[c + d*x]^7 + 112*a^3*Cos[5*(c + d*x)]*Csc[c + d*x]^7 - 50
4*a*b^2*Cos[5*(c + d*x)]*Csc[c + d*x]^7 - 16*a^3*Cos[7*(c + d*x)]*Csc[c + d*x]^7 - 168*a*b^2*Cos[7*(c + d*x)]*
Csc[c + d*x]^7 + 3360*a^2*b*Log[Cos[(c + d*x)/2]] + 6720*b^3*Log[Cos[(c + d*x)/2]] - 3360*a^2*b*Log[Sin[(c + d
*x)/2]] - 6720*b^3*Log[Sin[(c + d*x)/2]] + 70*Cot[c + d*x]*Csc[c + d*x]^6*(12*a*(2*a^2 + b^2) + b*(31*a^2 - 18
*b^2)*Sin[c + d*x]) + 1540*a^2*b*Csc[c + d*x]^7*Sin[4*(c + d*x)] + 840*b^3*Csc[c + d*x]^7*Sin[4*(c + d*x)] + 1
05*a^2*b*Csc[c + d*x]^7*Sin[6*(c + d*x)] - 350*b^3*Csc[c + d*x]^7*Sin[6*(c + d*x)])/d

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fricas [A]  time = 0.63, size = 347, normalized size = 1.15 \[ -\frac {32 \, {\left (2 \, a^{3} + 21 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} - 224 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 105 \, {\left ({\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{2} b - 2 \, b^{3} + 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 105 \, {\left ({\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{2} b - 2 \, b^{3} + 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 70 \, {\left ({\left (3 \, a^{2} b - 10 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 8 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1120 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/1120*(32*(2*a^3 + 21*a*b^2)*cos(d*x + c)^7 - 224*(a^3 + 3*a*b^2)*cos(d*x + c)^5 + 105*((a^2*b + 2*b^3)*cos(
d*x + c)^6 - 3*(a^2*b + 2*b^3)*cos(d*x + c)^4 - 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) - 105*((a^2*b + 2*b^3)*cos(d*x + c)^6 - 3*(a^2*b + 2*b^3)*cos(d*x + c)^4 - 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) - 70*((3*a^2*b - 10*b^3)
*cos(d*x + c)^5 + 8*(a^2*b + 2*b^3)*cos(d*x + c)^3 - 3*(a^2*b + 2*b^3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x
+ c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.39, size = 456, normalized size = 1.50 \[ \frac {5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 35 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 84 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 70 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 420 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 560 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 105 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 840 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 840 \, {\left (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 {2178 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4356 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 840 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 560 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 420 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 105 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 70 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 84 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{4480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/4480*(5*a^3*tan(1/2*d*x + 1/2*c)^7 + 35*a^2*b*tan(1/2*d*x + 1/2*c)^6 - 7*a^3*tan(1/2*d*x + 1/2*c)^5 + 84*a*b
^2*tan(1/2*d*x + 1/2*c)^5 - 105*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 70*b^3*tan(1/2*d*x + 1/2*c)^4 - 35*a^3*tan(1/2*
d*x + 1/2*c)^3 - 420*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 105*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 560*b^3*tan(1/2*d*x + 1
/2*c)^2 + 105*a^3*tan(1/2*d*x + 1/2*c) + 840*a*b^2*tan(1/2*d*x + 1/2*c) + 840*(a^2*b + 2*b^3)*log(abs(tan(1/2*
d*x + 1/2*c))) - (2178*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 4356*b^3*tan(1/2*d*x + 1/2*c)^7 + 105*a^3*tan(1/2*d*x +
1/2*c)^6 + 840*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 105*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 560*b^3*tan(1/2*d*x + 1/2*c)^
5 - 35*a^3*tan(1/2*d*x + 1/2*c)^4 - 420*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 105*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 70*b
^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c)^2 + 84*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 35*a^2*b*tan(1/2*
d*x + 1/2*c) + 5*a^3)/tan(1/2*d*x + 1/2*c)^7)/d

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maple [A]  time = 0.50, size = 309, normalized size = 1.02 \[ -\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{7 d \sin \left (d x +c \right )^{7}}-\frac {2 a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{35 d \sin \left (d x +c \right )^{5}}-\frac {a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{6}}-\frac {a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{4}}+\frac {a^{2} b \left (\cos ^{5}\left (d x +c \right )\right )}{16 d \sin \left (d x +c \right )^{2}}+\frac {a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{16 d}+\frac {3 a^{2} b \cos \left (d x +c \right )}{16 d}+\frac {3 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16 d}-\frac {3 a \,b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {b^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {b^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {3 b^{3} \cos \left (d x +c \right )}{8 d}+\frac {3 b^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^8*(a+b*sin(d*x+c))^3,x)

[Out]

-1/7/d*a^3/sin(d*x+c)^7*cos(d*x+c)^5-2/35/d*a^3/sin(d*x+c)^5*cos(d*x+c)^5-1/2/d*a^2*b/sin(d*x+c)^6*cos(d*x+c)^
5-1/8/d*a^2*b/sin(d*x+c)^4*cos(d*x+c)^5+1/16/d*a^2*b/sin(d*x+c)^2*cos(d*x+c)^5+1/16/d*a^2*b*cos(d*x+c)^3+3/16*
a^2*b*cos(d*x+c)/d+3/16/d*a^2*b*ln(csc(d*x+c)-cot(d*x+c))-3/5/d*a*b^2/sin(d*x+c)^5*cos(d*x+c)^5-1/4/d*b^3/sin(
d*x+c)^4*cos(d*x+c)^5+1/8/d*b^3/sin(d*x+c)^2*cos(d*x+c)^5+1/8*b^3*cos(d*x+c)^3/d+3/8*b^3*cos(d*x+c)/d+3/8/d*b^
3*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.33, size = 208, normalized size = 0.69 \[ \frac {35 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 70 \, b^{3} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {672 \, a b^{2}}{\tan \left (d x + c\right )^{5}} - \frac {32 \, {\left (7 \, \tan \left (d x + c\right )^{2} + 5\right )} a^{3}}{\tan \left (d x + c\right )^{7}}}{1120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/1120*(35*a^2*b*(2*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4
+ 3*cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 70*b^3*(2*(5*cos(d*x + c)^3 - 3
*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) -
672*a*b^2/tan(d*x + c)^5 - 32*(7*tan(d*x + c)^2 + 5)*a^3/tan(d*x + c)^7)/d

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mupad [B]  time = 9.91, size = 359, normalized size = 1.18 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{896\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3}{128}+\frac {3\,a\,b^2}{32}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,a\,b^2}{160}-\frac {a^3}{640}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a^2\,b}{128}+\frac {b^3}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {3\,a^2\,b}{128}-\frac {b^3}{64}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2\,b}{16}+\frac {3\,b^3}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {12\,a\,b^2}{5}-\frac {a^3}{5}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^3+24\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b-2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^2\,b+16\,b^3\right )+\frac {a^3}{7}+a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a^3}{128}+\frac {3\,a\,b^2}{16}\right )}{d}+\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{128\,d} \]

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)^8,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^7)/(896*d) - (tan(c/2 + (d*x)/2)^3*((3*a*b^2)/32 + a^3/128))/d + (tan(c/2 + (d*x)/2)^5
*((3*a*b^2)/160 - a^3/640))/d - (tan(c/2 + (d*x)/2)^2*((3*a^2*b)/128 + b^3/8))/d - (tan(c/2 + (d*x)/2)^4*((3*a
^2*b)/128 - b^3/64))/d + (log(tan(c/2 + (d*x)/2))*((3*a^2*b)/16 + (3*b^3)/8))/d - (tan(c/2 + (d*x)/2)^2*((12*a
*b^2)/5 - a^3/5) - tan(c/2 + (d*x)/2)^4*(12*a*b^2 + a^3) + tan(c/2 + (d*x)/2)^6*(24*a*b^2 + 3*a^3) - tan(c/2 +
(d*x)/2)^3*(3*a^2*b - 2*b^3) - tan(c/2 + (d*x)/2)^5*(3*a^2*b + 16*b^3) + a^3/7 + a^2*b*tan(c/2 + (d*x)/2))/(1
28*d*tan(c/2 + (d*x)/2)^7) + (tan(c/2 + (d*x)/2)*((3*a*b^2)/16 + (3*a^3)/128))/d + (a^2*b*tan(c/2 + (d*x)/2)^6
)/(128*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)**8*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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