3.1076 \(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=183 \[ \frac {a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac {b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d} \]

[Out]

1/8*b*(3*a^2+4*b^2)*arctanh(cos(d*x+c))/d+1/15*a*(2*a^2+15*b^2)*cot(d*x+c)/d+3/40*b*(5*a^2-2*b^2)*cot(d*x+c)*c
sc(d*x+c)/d+1/30*a*(2*a^2-3*b^2)*cot(d*x+c)*csc(d*x+c)^2/d-3/20*b*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^2/d
-1/5*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^3/d

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Rubi [A]  time = 0.57, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2889, 3048, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac {b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(b*(3*a^2 + 4*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) + (a*(2*a^2 + 15*b^2)*Cot[c + d*x])/(15*d) + (3*b*(5*a^2 - 2*b
^2)*Cot[c + d*x]*Csc[c + d*x])/(40*d) + (a*(2*a^2 - 3*b^2)*Cot[c + d*x]*Csc[c + d*x]^2)/(30*d) - (3*b*Cot[c +
d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(20*d) - (Cot[c + d*x]*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3)/(5*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 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 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 ^2(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc ^6(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 ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac {1}{5} \int \csc ^5(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) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac {1}{20} \int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (-2 \left (2 a^2-3 b^2\right )-11 a b \sin (c+d x)-13 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac {1}{60} \int \csc ^3(c+d x) \left (9 b \left (5 a^2-2 b^2\right )+4 a \left (2 a^2+15 b^2\right ) \sin (c+d x)+39 b^3 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac {1}{120} \int \csc ^2(c+d x) \left (8 a \left (2 a^2+15 b^2\right )+15 b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}-\frac {1}{8} \left (b \left (3 a^2+4 b^2\right )\right ) \int \csc (c+d x) \, dx-\frac {1}{15} \left (a \left (2 a^2+15 b^2\right )\right ) \int \csc ^2(c+d x) \, dx\\ &=\frac {b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}+\frac {\left (a \left (2 a^2+15 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{15 d}\\ &=\frac {b \left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a \left (2 a^2+15 b^2\right ) \cot (c+d x)}{15 d}+\frac {3 b \left (5 a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{40 d}+\frac {a \left (2 a^2-3 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{30 d}-\frac {3 b \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{20 d}-\frac {\cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.29, size = 344, normalized size = 1.88 \[ \frac {32 \left (2 a^3+15 a b^2\right ) \cot \left (\frac {1}{2} (c+d x)\right )-64 a^3 \tan \left (\frac {1}{2} (c+d x)\right )-3 a^3 \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )-16 a^3 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+6 a^3 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right )+30 \left (3 a^2 b-4 b^3\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )+a \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (\left (a^2-60 b^2\right ) \sin (c+d x)-45 a b\right )+45 a^2 b \sec ^4\left (\frac {1}{2} (c+d x)\right )-90 a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )-360 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+360 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-480 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )+960 a b^2 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+120 b^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )-480 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+480 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(32*(2*a^3 + 15*a*b^2)*Cot[(c + d*x)/2] + 30*(3*a^2*b - 4*b^3)*Csc[(c + d*x)/2]^2 + 360*a^2*b*Log[Cos[(c + d*x
)/2]] + 480*b^3*Log[Cos[(c + d*x)/2]] - 360*a^2*b*Log[Sin[(c + d*x)/2]] - 480*b^3*Log[Sin[(c + d*x)/2]] - 90*a
^2*b*Sec[(c + d*x)/2]^2 + 120*b^3*Sec[(c + d*x)/2]^2 + 45*a^2*b*Sec[(c + d*x)/2]^4 - 16*a^3*Csc[c + d*x]^3*Sin
[(c + d*x)/2]^4 + 960*a*b^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 3*a^3*Csc[(c + d*x)/2]^6*Sin[c + d*x] + a*Csc[
(c + d*x)/2]^4*(-45*a*b + (a^2 - 60*b^2)*Sin[c + d*x]) - 64*a^3*Tan[(c + d*x)/2] - 480*a*b^2*Tan[(c + d*x)/2]
+ 6*a^3*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(960*d)

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fricas [A]  time = 0.72, size = 275, normalized size = 1.50 \[ \frac {16 \, {\left (2 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 80 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left ({\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{2} b + 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (3 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(16*(2*a^3 + 15*a*b^2)*cos(d*x + c)^5 - 80*(a^3 + 3*a*b^2)*cos(d*x + c)^3 + 15*((3*a^2*b + 4*b^3)*cos(d*
x + c)^4 + 3*a^2*b + 4*b^3 - 2*(3*a^2*b + 4*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15
*((3*a^2*b + 4*b^3)*cos(d*x + c)^4 + 3*a^2*b + 4*b^3 - 2*(3*a^2*b + 4*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x +
c) + 1/2)*sin(d*x + c) - 30*((3*a^2*b - 4*b^3)*cos(d*x + c)^3 + (3*a^2*b + 4*b^3)*cos(d*x + c))*sin(d*x + c))/
((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [A]  time = 0.27, size = 290, normalized size = 1.58 \[ \frac {6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, {\left (3 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {822 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1096 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(6*a^3*tan(1/2*d*x + 1/2*c)^5 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 10*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*a*
b^2*tan(1/2*d*x + 1/2*c)^3 + 120*b^3*tan(1/2*d*x + 1/2*c)^2 - 60*a^3*tan(1/2*d*x + 1/2*c) - 360*a*b^2*tan(1/2*
d*x + 1/2*c) - 120*(3*a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) + (822*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1096
*b^3*tan(1/2*d*x + 1/2*c)^5 + 60*a^3*tan(1/2*d*x + 1/2*c)^4 + 360*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*b^3*tan(1
/2*d*x + 1/2*c)^3 - 10*a^3*tan(1/2*d*x + 1/2*c)^2 - 120*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 45*a^2*b*tan(1/2*d*x +
1/2*c) - 6*a^3)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.49, size = 227, normalized size = 1.24 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {2 a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {3 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}-\frac {3 a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {3 a^{2} b \cos \left (d x +c \right )}{8 d}-\frac {3 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a \,b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}-\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {b^{3} \cos \left (d x +c \right )}{2 d}-\frac {b^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^6*(a+b*sin(d*x+c))^3,x)

[Out]

-1/5/d*a^3/sin(d*x+c)^5*cos(d*x+c)^3-2/15/d*a^3/sin(d*x+c)^3*cos(d*x+c)^3-3/4/d*a^2*b/sin(d*x+c)^4*cos(d*x+c)^
3-3/8/d*a^2*b/sin(d*x+c)^2*cos(d*x+c)^3-3/8*a^2*b*cos(d*x+c)/d-3/8/d*a^2*b*ln(csc(d*x+c)-cot(d*x+c))-1/d*a*b^2
/sin(d*x+c)^3*cos(d*x+c)^3-1/2/d*b^3/sin(d*x+c)^2*cos(d*x+c)^3-1/2*b^3*cos(d*x+c)/d-1/2/d*b^3*ln(csc(d*x+c)-co
t(d*x+c))

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maxima [A]  time = 0.33, size = 157, normalized size = 0.86 \[ -\frac {45 \, a^{2} b {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, b^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a b^{2}}{\tan \left (d x + c\right )^{3}} + \frac {16 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/240*(45*a^2*b*(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)) - 60*b^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos
(d*x + c) - 1)) + 240*a*b^2/tan(d*x + c)^3 + 16*(5*tan(d*x + c)^2 + 3)*a^3/tan(d*x + c)^5)/d

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mupad [B]  time = 9.44, size = 241, normalized size = 1.32 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3}{96}+\frac {a\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2\,b}{8}+\frac {b^3}{2}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (4\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3}{3}+4\,a\,b^2\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^3+12\,a\,b^2\right )+\frac {a^3}{5}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{32\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3}{16}+\frac {3\,a\,b^2}{8}\right )}{d}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d} \]

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)^6,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (b^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (tan(c/2 + (d*x)/2)^3*((a*b^2)/8 + a^3
/96))/d - (log(tan(c/2 + (d*x)/2))*((3*a^2*b)/8 + b^3/2))/d - (cot(c/2 + (d*x)/2)^5*(4*b^3*tan(c/2 + (d*x)/2)^
3 + tan(c/2 + (d*x)/2)^2*(4*a*b^2 + a^3/3) - tan(c/2 + (d*x)/2)^4*(12*a*b^2 + 2*a^3) + a^3/5 + (3*a^2*b*tan(c/
2 + (d*x)/2))/2))/(32*d) - (tan(c/2 + (d*x)/2)*((3*a*b^2)/8 + a^3/16))/d + (3*a^2*b*tan(c/2 + (d*x)/2)^4)/(64*
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)**2*csc(d*x+c)**6*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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