3.1118 \(\int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=250 \[ -\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}+\frac {1}{16} b x \left (18 a^2+b^2\right )-\frac {a \left (2 a^4-43 a^2 b^2+36 b^4\right ) \cos (c+d x)}{60 b^2 d}-\frac {\left (4 a^4-84 a^2 b^2+15 b^4\right ) \sin (c+d x) \cos (c+d x)}{240 b d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^4}{6 b d} \]

[Out]

1/16*b*(18*a^2+b^2)*x-a^3*arctanh(cos(d*x+c))/d-1/60*a*(2*a^4-43*a^2*b^2+36*b^4)*cos(d*x+c)/b^2/d-1/240*(4*a^4
-84*a^2*b^2+15*b^4)*cos(d*x+c)*sin(d*x+c)/b/d-1/120*a*(2*a^2-39*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^2/b^2/d-1/120
*(2*a^2-35*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^3/b^2/d+1/15*a*cos(d*x+c)*(a+b*sin(d*x+c))^4/b^2/d-1/6*cos(d*x+c)*
sin(d*x+c)*(a+b*sin(d*x+c))^4/b/d

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Rubi [A]  time = 0.66, antiderivative size = 250, 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 = {2895, 3049, 3033, 3023, 2735, 3770} \[ -\frac {a \left (-43 a^2 b^2+2 a^4+36 b^4\right ) \cos (c+d x)}{60 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (-84 a^2 b^2+4 a^4+15 b^4\right ) \sin (c+d x) \cos (c+d x)}{240 b d}+\frac {1}{16} b x \left (18 a^2+b^2\right )-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^4}{6 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(18*a^2 + b^2)*x)/16 - (a^3*ArcTanh[Cos[c + d*x]])/d - (a*(2*a^4 - 43*a^2*b^2 + 36*b^4)*Cos[c + d*x])/(60*b
^2*d) - ((4*a^4 - 84*a^2*b^2 + 15*b^4)*Cos[c + d*x]*Sin[c + d*x])/(240*b*d) - (a*(2*a^2 - 39*b^2)*Cos[c + d*x]
*(a + b*Sin[c + d*x])^2)/(120*b^2*d) - ((2*a^2 - 35*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^3)/(120*b^2*d) + (a
*Cos[c + d*x]*(a + b*Sin[c + d*x])^4)/(15*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^4)/(6*b*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 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[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 ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^3 \left (-30 b^2+3 a b \sin (c+d x)-\left (2 a^2-35 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{30 b^2}\\ &=-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-120 a b^2+3 b \left (2 a^2-5 b^2\right ) \sin (c+d x)-3 a \left (2 a^2-39 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{120 b^2}\\ &=-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (-360 a^2 b^2+3 a b \left (2 a^2-57 b^2\right ) \sin (c+d x)-3 \left (4 a^4-84 a^2 b^2+15 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{360 b^2}\\ &=-\frac {\left (4 a^4-84 a^2 b^2+15 b^4\right ) \cos (c+d x) \sin (c+d x)}{240 b d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}-\frac {\int \csc (c+d x) \left (-720 a^3 b^2-45 b^3 \left (18 a^2+b^2\right ) \sin (c+d x)-12 a \left (2 a^4-43 a^2 b^2+36 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{720 b^2}\\ &=-\frac {a \left (2 a^4-43 a^2 b^2+36 b^4\right ) \cos (c+d x)}{60 b^2 d}-\frac {\left (4 a^4-84 a^2 b^2+15 b^4\right ) \cos (c+d x) \sin (c+d x)}{240 b d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}-\frac {\int \csc (c+d x) \left (-720 a^3 b^2-45 b^3 \left (18 a^2+b^2\right ) \sin (c+d x)\right ) \, dx}{720 b^2}\\ &=\frac {1}{16} b \left (18 a^2+b^2\right ) x-\frac {a \left (2 a^4-43 a^2 b^2+36 b^4\right ) \cos (c+d x)}{60 b^2 d}-\frac {\left (4 a^4-84 a^2 b^2+15 b^4\right ) \cos (c+d x) \sin (c+d x)}{240 b d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}+a^3 \int \csc (c+d x) \, dx\\ &=\frac {1}{16} b \left (18 a^2+b^2\right ) x-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a \left (2 a^4-43 a^2 b^2+36 b^4\right ) \cos (c+d x)}{60 b^2 d}-\frac {\left (4 a^4-84 a^2 b^2+15 b^4\right ) \cos (c+d x) \sin (c+d x)}{240 b d}-\frac {a \left (2 a^2-39 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{120 b^2 d}-\frac {\left (2 a^2-35 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{120 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^4}{15 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^4}{6 b d}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 191, normalized size = 0.76 \[ \frac {20 \left (4 a^3-9 a b^2\right ) \cos (3 (c+d x))+960 a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-960 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 a \left (10 a^2-3 b^2\right ) \cos (c+d x)+720 a^2 b \sin (2 (c+d x))+90 a^2 b \sin (4 (c+d x))+1080 a^2 b c+1080 a^2 b d x-36 a b^2 \cos (5 (c+d x))+15 b^3 \sin (2 (c+d x))-15 b^3 \sin (4 (c+d x))-5 b^3 \sin (6 (c+d x))+60 b^3 c+60 b^3 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1080*a^2*b*c + 60*b^3*c + 1080*a^2*b*d*x + 60*b^3*d*x + 120*a*(10*a^2 - 3*b^2)*Cos[c + d*x] + 20*(4*a^3 - 9*a
*b^2)*Cos[3*(c + d*x)] - 36*a*b^2*Cos[5*(c + d*x)] - 960*a^3*Log[Cos[(c + d*x)/2]] + 960*a^3*Log[Sin[(c + d*x)
/2]] + 720*a^2*b*Sin[2*(c + d*x)] + 15*b^3*Sin[2*(c + d*x)] + 90*a^2*b*Sin[4*(c + d*x)] - 15*b^3*Sin[4*(c + d*
x)] - 5*b^3*Sin[6*(c + d*x)])/(960*d)

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fricas [A]  time = 0.98, size = 150, normalized size = 0.60 \[ -\frac {144 \, a b^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{3} \cos \left (d x + c\right )^{3} - 240 \, a^{3} \cos \left (d x + c\right ) + 120 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 120 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (18 \, a^{2} b + b^{3}\right )} d x + 5 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} - 2 \, {\left (18 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (18 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(144*a*b^2*cos(d*x + c)^5 - 80*a^3*cos(d*x + c)^3 - 240*a^3*cos(d*x + c) + 120*a^3*log(1/2*cos(d*x + c)
+ 1/2) - 120*a^3*log(-1/2*cos(d*x + c) + 1/2) - 15*(18*a^2*b + b^3)*d*x + 5*(8*b^3*cos(d*x + c)^5 - 2*(18*a^2
*b + b^3)*cos(d*x + c)^3 - 3*(18*a^2*b + b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.27, size = 427, normalized size = 1.71 \[ \frac {240 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 15 \, {\left (18 \, a^{2} b + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (450 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 480 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 720 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 630 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 235 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1920 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 720 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 180 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 390 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3200 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1440 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 180 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 390 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2880 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1440 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 630 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 235 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1440 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 144 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 450 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 320 \, a^{3} + 144 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(240*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 15*(18*a^2*b + b^3)*(d*x + c) - 2*(450*a^2*b*tan(1/2*d*x + 1/2
*c)^11 - 15*b^3*tan(1/2*d*x + 1/2*c)^11 - 480*a^3*tan(1/2*d*x + 1/2*c)^10 + 720*a*b^2*tan(1/2*d*x + 1/2*c)^10
+ 630*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 235*b^3*tan(1/2*d*x + 1/2*c)^9 - 1920*a^3*tan(1/2*d*x + 1/2*c)^8 + 720*a*
b^2*tan(1/2*d*x + 1/2*c)^8 + 180*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 390*b^3*tan(1/2*d*x + 1/2*c)^7 - 3200*a^3*tan(
1/2*d*x + 1/2*c)^6 + 1440*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 180*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 390*b^3*tan(1/2*d*
x + 1/2*c)^5 - 2880*a^3*tan(1/2*d*x + 1/2*c)^4 + 1440*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 630*a^2*b*tan(1/2*d*x + 1
/2*c)^3 - 235*b^3*tan(1/2*d*x + 1/2*c)^3 - 1440*a^3*tan(1/2*d*x + 1/2*c)^2 + 144*a*b^2*tan(1/2*d*x + 1/2*c)^2
- 450*a^2*b*tan(1/2*d*x + 1/2*c) + 15*b^3*tan(1/2*d*x + 1/2*c) - 320*a^3 + 144*a*b^2)/(tan(1/2*d*x + 1/2*c)^2
+ 1)^6)/d

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maple [A]  time = 0.58, size = 211, normalized size = 0.84 \[ \frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} \cos \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {9 a^{2} b x}{8}+\frac {9 a^{2} b c}{8 d}-\frac {3 \left (\cos ^{5}\left (d x +c \right )\right ) a \,b^{2}}{5 d}-\frac {b^{3} \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6 d}+\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{24 d}+\frac {b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}+\frac {b^{3} x}{16}+\frac {b^{3} c}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^3*cos(d*x+c)^3/d+a^3*cos(d*x+c)/d+1/d*a^3*ln(csc(d*x+c)-cot(d*x+c))+3/4/d*a^2*b*sin(d*x+c)*cos(d*x+c)^3+
9/8/d*a^2*b*cos(d*x+c)*sin(d*x+c)+9/8*a^2*b*x+9/8/d*a^2*b*c-3/5/d*cos(d*x+c)^5*a*b^2-1/6/d*b^3*sin(d*x+c)*cos(
d*x+c)^5+1/24/d*b^3*cos(d*x+c)^3*sin(d*x+c)+1/16*b^3*cos(d*x+c)*sin(d*x+c)/d+1/16*b^3*x+1/16/d*b^3*c

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maxima [A]  time = 0.43, size = 137, normalized size = 0.55 \[ -\frac {576 \, a b^{2} \cos \left (d x + c\right )^{5} - 160 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(576*a*b^2*cos(d*x + c)^5 - 160*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(co
s(d*x + c) - 1))*a^3 - 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2*b - 5*(4*sin(2*d*x + 2*c
)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*b^3)/d

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mupad [B]  time = 11.57, size = 690, normalized size = 2.76 \[ \frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\frac {6\,a\,b^2}{5}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {15\,a^2\,b}{4}-\frac {b^3}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (6\,a\,b^2-4\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {6\,a\,b^2}{5}-12\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (6\,a\,b^2-16\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a\,b^2-24\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (12\,a\,b^2-\frac {80\,a^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,a^2\,b}{2}-\frac {13\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {3\,a^2\,b}{2}-\frac {13\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {15\,a^2\,b}{4}-\frac {b^3}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {21\,a^2\,b}{4}+\frac {47\,b^3}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {21\,a^2\,b}{4}+\frac {47\,b^3}{24}\right )-\frac {8\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\left (18\,a^2+b^2\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{8}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2+b^2\right )\,3{}\mathrm {i}}{8}\right )}{16}+\frac {b\,\left (18\,a^2+b^2\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{8}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2+b^2\right )\,3{}\mathrm {i}}{8}\right )}{16}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {81\,a^4\,b^2}{16}+\frac {9\,a^2\,b^4}{16}+\frac {b^6}{64}\right )+\frac {9\,a^5\,b}{2}+\frac {a^3\,b^3}{4}-\frac {b\,\left (18\,a^2+b^2\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{8}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2+b^2\right )\,3{}\mathrm {i}}{8}\right )\,1{}\mathrm {i}}{16}+\frac {b\,\left (18\,a^2+b^2\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{8}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2+b^2\right )\,3{}\mathrm {i}}{8}\right )\,1{}\mathrm {i}}{16}}\right )\,\left (18\,a^2+b^2\right )}{8\,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),x)

[Out]

(a^3*log(tan(c/2 + (d*x)/2)))/d - ((6*a*b^2)/5 - tan(c/2 + (d*x)/2)*((15*a^2*b)/4 - b^3/8) + tan(c/2 + (d*x)/2
)^10*(6*a*b^2 - 4*a^3) + tan(c/2 + (d*x)/2)^2*((6*a*b^2)/5 - 12*a^3) + tan(c/2 + (d*x)/2)^8*(6*a*b^2 - 16*a^3)
+ tan(c/2 + (d*x)/2)^4*(12*a*b^2 - 24*a^3) + tan(c/2 + (d*x)/2)^6*(12*a*b^2 - (80*a^3)/3) - tan(c/2 + (d*x)/2
)^5*((3*a^2*b)/2 - (13*b^3)/4) + tan(c/2 + (d*x)/2)^7*((3*a^2*b)/2 - (13*b^3)/4) + tan(c/2 + (d*x)/2)^11*((15*
a^2*b)/4 - b^3/8) - tan(c/2 + (d*x)/2)^3*((21*a^2*b)/4 + (47*b^3)/24) + tan(c/2 + (d*x)/2)^9*((21*a^2*b)/4 + (
47*b^3)/24) - (8*a^3)/3)/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*t
an(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (b*atan(((b*(18*a^2 + b^2)*((9*a
^2*b)/4 + b^3/8 + 2*a^3*tan(c/2 + (d*x)/2) - (b*tan(c/2 + (d*x)/2)*(18*a^2 + b^2)*3i)/8))/16 + (b*(18*a^2 + b^
2)*((9*a^2*b)/4 + b^3/8 + 2*a^3*tan(c/2 + (d*x)/2) + (b*tan(c/2 + (d*x)/2)*(18*a^2 + b^2)*3i)/8))/16)/(2*tan(c
/2 + (d*x)/2)*(b^6/64 + (9*a^2*b^4)/16 + (81*a^4*b^2)/16) + (9*a^5*b)/2 + (a^3*b^3)/4 - (b*(18*a^2 + b^2)*((9*
a^2*b)/4 + b^3/8 + 2*a^3*tan(c/2 + (d*x)/2) - (b*tan(c/2 + (d*x)/2)*(18*a^2 + b^2)*3i)/8)*1i)/16 + (b*(18*a^2
+ b^2)*((9*a^2*b)/4 + b^3/8 + 2*a^3*tan(c/2 + (d*x)/2) + (b*tan(c/2 + (d*x)/2)*(18*a^2 + b^2)*3i)/8)*1i)/16))*
(18*a^2 + b^2))/(8*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)*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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