3.1122 \(\int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=187 \[ -\frac {b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac {3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {3}{2} b x \left (2 a^2-b^2\right )+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {13 b^3 \sin (c+d x) \cos (c+d x)}{4 d} \]

[Out]

3/2*b*(2*a^2-b^2)*x-3/8*a*(a^2-12*b^2)*arctanh(cos(d*x+c))/d-1/8*b^2*(73*a^2-2*b^2)*cos(d*x+c)/a/d-13/4*b^3*co
s(d*x+c)*sin(d*x+c)/d+17/8*b*cot(d*x+c)*(a+b*sin(d*x+c))^2/d+5/8*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^3/d-1/
4*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^4/a/d

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Rubi [A]  time = 0.66, antiderivative size = 187, 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, 3047, 3033, 3023, 2735, 3770} \[ -\frac {b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac {3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {3}{2} b x \left (2 a^2-b^2\right )+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {13 b^3 \sin (c+d x) \cos (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(2*a^2 - b^2)*x)/2 - (3*a*(a^2 - 12*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*(73*a^2 - 2*b^2)*Cos[c + d*x
])/(8*a*d) - (13*b^3*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (17*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(8*d) + (5*
Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^3)/(8*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^4)
/(4*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 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 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 (c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac {\int \csc ^3(c+d x) (a+b \sin (c+d x))^3 \left (15 a^2+3 a b \sin (c+d x)-12 a^2 \sin ^2(c+d x)\right ) \, dx}{12 a^2}\\ &=\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac {\int \csc ^2(c+d x) (a+b \sin (c+d x))^2 \left (51 a^2 b-3 a \left (3 a^2-2 b^2\right ) \sin (c+d x)-54 a^2 b \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (-9 a^2 \left (a^2-12 b^2\right )-3 a b \left (21 a^2-2 b^2\right ) \sin (c+d x)-156 a^2 b^2 \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=-\frac {13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac {\int \csc (c+d x) \left (-18 a^3 \left (a^2-12 b^2\right )-72 a^2 b \left (2 a^2-b^2\right ) \sin (c+d x)-6 a b^2 \left (73 a^2-2 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{48 a^2}\\ &=-\frac {b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac {13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac {\int \csc (c+d x) \left (-18 a^3 \left (a^2-12 b^2\right )-72 a^2 b \left (2 a^2-b^2\right ) \sin (c+d x)\right ) \, dx}{48 a^2}\\ &=\frac {3}{2} b \left (2 a^2-b^2\right ) x-\frac {b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac {13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac {1}{8} \left (3 a \left (a^2-12 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=\frac {3}{2} b \left (2 a^2-b^2\right ) x-\frac {3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac {13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac {5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}\\ \end {align*}

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Mathematica [B]  time = 6.29, size = 381, normalized size = 2.04 \[ \frac {\left (5 a^3-12 a b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {\left (12 a b^2-5 a^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {3 \left (a^3-12 a b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {3 \left (a^3-12 a b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {a^3 \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a^3 \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (4 a^2 b \cos \left (\frac {1}{2} (c+d x)\right )-b^3 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b^3 \sin \left (\frac {1}{2} (c+d x)\right )-4 a^2 b \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {3 b \left (b^2-2 a^2\right ) (c+d x)}{2 d}-\frac {a^2 b \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a^2 b \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {3 a b^2 \cos (c+d x)}{d}-\frac {b^3 \sin (2 (c+d x))}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*b*(-2*a^2 + b^2)*(c + d*x))/(2*d) - (3*a*b^2*Cos[c + d*x])/d + ((4*a^2*b*Cos[(c + d*x)/2] - b^3*Cos[(c + d
*x)/2])*Csc[(c + d*x)/2])/(2*d) + ((5*a^3 - 12*a*b^2)*Csc[(c + d*x)/2]^2)/(32*d) - (a^2*b*Cot[(c + d*x)/2]*Csc
[(c + d*x)/2]^2)/(8*d) - (a^3*Csc[(c + d*x)/2]^4)/(64*d) - (3*(a^3 - 12*a*b^2)*Log[Cos[(c + d*x)/2]])/(8*d) +
(3*(a^3 - 12*a*b^2)*Log[Sin[(c + d*x)/2]])/(8*d) + ((-5*a^3 + 12*a*b^2)*Sec[(c + d*x)/2]^2)/(32*d) + (a^3*Sec[
(c + d*x)/2]^4)/(64*d) + (Sec[(c + d*x)/2]*(-4*a^2*b*Sin[(c + d*x)/2] + b^3*Sin[(c + d*x)/2]))/(2*d) - (b^3*Si
n[2*(c + d*x)])/(4*d) + (a^2*b*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(8*d)

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fricas [A]  time = 0.89, size = 331, normalized size = 1.77 \[ -\frac {48 \, a b^{2} \cos \left (d x + c\right )^{5} - 24 \, {\left (2 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{4} + 48 \, {\left (2 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 10 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 24 \, {\left (2 \, a^{2} b - b^{3}\right )} d x - 6 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left ({\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 12 \, a b^{2} - 2 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 12 \, a b^{2} - 2 \, {\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 8 \, {\left (b^{3} \cos \left (d x + c\right )^{5} + 4 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*(48*a*b^2*cos(d*x + c)^5 - 24*(2*a^2*b - b^3)*d*x*cos(d*x + c)^4 + 48*(2*a^2*b - b^3)*d*x*cos(d*x + c)^2
+ 10*(a^3 - 12*a*b^2)*cos(d*x + c)^3 - 24*(2*a^2*b - b^3)*d*x - 6*(a^3 - 12*a*b^2)*cos(d*x + c) + 3*((a^3 - 1
2*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) - 3*
((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) + 8*(b^3*cos(d*x + c)^5 + 4*(2*a^2*b - b^3)*cos(d*x + c)^3 - 3*(2*a^2*b - b^3)*cos(d*x + c))*sin(d*x + c
))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.33, size = 343, normalized size = 1.83 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} + 24 \, {\left (a^{3} - 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {64 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {50 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 600 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/64*(a^3*tan(1/2*d*x + 1/2*c)^4 + 8*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*ta
n(1/2*d*x + 1/2*c)^2 - 120*a^2*b*tan(1/2*d*x + 1/2*c) + 32*b^3*tan(1/2*d*x + 1/2*c) + 96*(2*a^2*b - b^3)*(d*x
+ c) + 24*(a^3 - 12*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 64*(b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d
*x + 1/2*c)^2 - b^3*tan(1/2*d*x + 1/2*c) - 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - (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 - 120*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 32*b^3*tan(1/2*d*x + 1/2*c)^3 -
8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d
*x + 1/2*c)^4)/d

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maple [A]  time = 0.64, size = 316, normalized size = 1.69 \[ -\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {3 a^{3} \cos \left (d x +c \right )}{8 d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a^{2} b \left (\cot ^{3}\left (d x +c \right )\right )}{d}+3 a^{2} b x +\frac {3 a^{2} b \cot \left (d x +c \right )}{d}+\frac {3 a^{2} b c}{d}-\frac {3 a \,b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d}-\frac {9 a \,b^{2} \cos \left (d x +c \right )}{2 d}-\frac {9 a \,b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {b^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {3 b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {3 b^{3} x}{2}-\frac {3 b^{3} c}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.65, size = 212, normalized size = 1.13 \[ \frac {16 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} b - 8 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} b^{3} - a^{3} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(16*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2*b - 8*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)
/(tan(d*x + c)^3 + tan(d*x + c)))*b^3 - a^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)) + 12*a*b^2*(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 = 9.50, size = 699, normalized size = 3.74 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a\,b^2}{8}-\frac {a^3}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a\,b^2-\frac {3\,a^3}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (102\,a\,b^2-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (108\,a\,b^2-\frac {15\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (26\,a^2\,b-8\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (30\,a^2\,b+8\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (58\,a^2\,b-32\,b^3\right )+\frac {a^3}{4}+2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+32\,{\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\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {15\,a^2\,b}{8}-\frac {b^3}{2}\right )}{d}+\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-12\,b^2\right )}{8\,d}+\frac {a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,d}-\frac {3\,b\,\mathrm {atan}\left (\frac {\frac {3\,b\,\left (2\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a\,b^2-\frac {3\,a^3}{4}\right )-6\,a^2\,b+3\,b^3-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\,9{}\mathrm {i}\right )}{2}+\frac {3\,b\,\left (2\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a\,b^2-\frac {3\,a^3}{4}\right )-6\,a^2\,b+3\,b^3+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\,9{}\mathrm {i}\right )}{2}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (36\,a^4\,b^2-36\,a^2\,b^4+9\,b^6\right )+27\,a\,b^5+\frac {9\,a^5\,b}{2}-\frac {225\,a^3\,b^3}{4}-\frac {b\,\left (2\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a\,b^2-\frac {3\,a^3}{4}\right )-6\,a^2\,b+3\,b^3-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\,9{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {b\,\left (2\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a\,b^2-\frac {3\,a^3}{4}\right )-6\,a^2\,b+3\,b^3+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-b^2\right )\,9{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}}\right )\,\left (2\,a^2-b^2\right )}{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)^5,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^4)/(64*d) + (tan(c/2 + (d*x)/2)^2*((3*a*b^2)/8 - a^3/8))/d - (tan(c/2 + (d*x)/2)^2*(6*
a*b^2 - (3*a^3)/2) + tan(c/2 + (d*x)/2)^6*(102*a*b^2 - 2*a^3) + tan(c/2 + (d*x)/2)^4*(108*a*b^2 - (15*a^3)/4)
- tan(c/2 + (d*x)/2)^3*(26*a^2*b - 8*b^3) - tan(c/2 + (d*x)/2)^7*(30*a^2*b + 8*b^3) - tan(c/2 + (d*x)/2)^5*(58
*a^2*b - 32*b^3) + a^3/4 + 2*a^2*b*tan(c/2 + (d*x)/2))/(d*(16*tan(c/2 + (d*x)/2)^4 + 32*tan(c/2 + (d*x)/2)^6 +
16*tan(c/2 + (d*x)/2)^8)) - (tan(c/2 + (d*x)/2)*((15*a^2*b)/8 - b^3/2))/d + (3*a*log(tan(c/2 + (d*x)/2))*(a^2
- 12*b^2))/(8*d) + (a^2*b*tan(c/2 + (d*x)/2)^3)/(8*d) - (3*b*atan(((3*b*(2*a^2 - b^2)*(tan(c/2 + (d*x)/2)*(9*
a*b^2 - (3*a^3)/4) - 6*a^2*b + 3*b^3 - b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*9i))/2 + (3*b*(2*a^2 - b^2)*(tan(c/2
+ (d*x)/2)*(9*a*b^2 - (3*a^3)/4) - 6*a^2*b + 3*b^3 + b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*9i))/2)/(2*tan(c/2 +
(d*x)/2)*(9*b^6 - 36*a^2*b^4 + 36*a^4*b^2) + 27*a*b^5 + (9*a^5*b)/2 - (225*a^3*b^3)/4 - (b*(2*a^2 - b^2)*(tan(
c/2 + (d*x)/2)*(9*a*b^2 - (3*a^3)/4) - 6*a^2*b + 3*b^3 - b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*9i)*3i)/2 + (b*(2*
a^2 - b^2)*(tan(c/2 + (d*x)/2)*(9*a*b^2 - (3*a^3)/4) - 6*a^2*b + 3*b^3 + b*tan(c/2 + (d*x)/2)*(2*a^2 - b^2)*9i
)*3i)/2))*(2*a^2 - b^2))/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)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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