3.1119 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=229 \[ \frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {a \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}-\frac {3}{8} a x \left (4 a^2-3 b^2\right )-\frac {3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}+\frac {\left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{10 b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d} \]

[Out]

-3/8*a*(4*a^2-3*b^2)*x-3*a^2*b*arctanh(cos(d*x+c))/d+1/10*(a^4+56*a^2*b^2-2*b^4)*cos(d*x+c)/b/d+1/40*a*(2*a^2+
83*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/20*(a^2+28*b^2)*cos(d*x+c)*(a+b*sin(d*x+c))^2/b/d+1/20*(a^2+20*b^2)*cos(d*x+
c)*(a+b*sin(d*x+c))^3/a/b/d-1/5*cos(d*x+c)*(a+b*sin(d*x+c))^4/b/d-cot(d*x+c)*(a+b*sin(d*x+c))^4/a/d

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Rubi [A]  time = 0.68, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2894, 3049, 3033, 3023, 2735, 3770} \[ \frac {\left (56 a^2 b^2+a^4-2 b^4\right ) \cos (c+d x)}{10 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {a \left (2 a^2+83 b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}-\frac {3}{8} a x \left (4 a^2-3 b^2\right )-\frac {3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*(4*a^2 - 3*b^2)*x)/8 - (3*a^2*b*ArcTanh[Cos[c + d*x]])/d + ((a^4 + 56*a^2*b^2 - 2*b^4)*Cos[c + d*x])/(10
*b*d) + (a*(2*a^2 + 83*b^2)*Cos[c + d*x]*Sin[c + d*x])/(40*d) + ((a^2 + 28*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*
x])^2)/(20*b*d) + ((a^2 + 20*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^3)/(20*a*b*d) - (Cos[c + d*x]*(a + b*Sin[c
+ d*x])^4)/(5*b*d) - (Cot[c + d*x]*(a + b*Sin[c + d*x])^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 2894

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*b*d*(n + 1)*(m + n + 4)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 1)*Simp[a^2*(n + 1)*(n
+ 2) - b^2*(m + n + 2)*(m + n + 4) + a*b*(m + 3)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n
+ 4))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/(
b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m
, 2*n]) &&  !m < -1 && LtQ[n, -1] && 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 ^2(c+d x) \cot ^2(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^3 \left (-15 b^2+6 a b \sin (c+d x)+\left (a^2+20 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{5 a b}\\ &=\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-60 a b^2+27 a^2 b \sin (c+d x)+3 a \left (a^2+28 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{20 a b}\\ &=\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (-180 a^2 b^2+3 a b \left (29 a^2-4 b^2\right ) \sin (c+d x)+3 a^2 \left (2 a^2+83 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{60 a b}\\ &=\frac {a \left (2 a^2+83 b^2\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}-\frac {\int \csc (c+d x) \left (-360 a^3 b^2+45 a^2 b \left (4 a^2-3 b^2\right ) \sin (c+d x)+12 a \left (a^4+56 a^2 b^2-2 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{120 a b}\\ &=\frac {\left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{10 b d}+\frac {a \left (2 a^2+83 b^2\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}-\frac {\int \csc (c+d x) \left (-360 a^3 b^2+45 a^2 b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right ) \, dx}{120 a b}\\ &=-\frac {3}{8} a \left (4 a^2-3 b^2\right ) x+\frac {\left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{10 b d}+\frac {a \left (2 a^2+83 b^2\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}+\left (3 a^2 b\right ) \int \csc (c+d x) \, dx\\ &=-\frac {3}{8} a \left (4 a^2-3 b^2\right ) x-\frac {3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}+\frac {\left (a^4+56 a^2 b^2-2 b^4\right ) \cos (c+d x)}{10 b d}+\frac {a \left (2 a^2+83 b^2\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {\left (a^2+28 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{20 b d}+\frac {\left (a^2+20 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{20 a b d}-\frac {\cos (c+d x) (a+b \sin (c+d x))^4}{5 b d}-\frac {\cot (c+d x) (a+b \sin (c+d x))^4}{a d}\\ \end {align*}

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Mathematica [A]  time = 3.04, size = 194, normalized size = 0.85 \[ \frac {-40 a^3 \sin (2 (c+d x))+80 a^3 \tan \left (\frac {1}{2} (c+d x)\right )-80 a^3 \cot \left (\frac {1}{2} (c+d x)\right )-240 a^3 c-240 a^3 d x+10 \left (4 a^2 b-b^3\right ) \cos (3 (c+d x))-20 b \left (b^2-30 a^2\right ) \cos (c+d x)+480 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-480 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 a b^2 \sin (2 (c+d x))+15 a b^2 \sin (4 (c+d x))+180 a b^2 c+180 a b^2 d x-2 b^3 \cos (5 (c+d x))}{160 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-240*a^3*c + 180*a*b^2*c - 240*a^3*d*x + 180*a*b^2*d*x - 20*b*(-30*a^2 + b^2)*Cos[c + d*x] + 10*(4*a^2*b - b^
3)*Cos[3*(c + d*x)] - 2*b^3*Cos[5*(c + d*x)] - 80*a^3*Cot[(c + d*x)/2] - 480*a^2*b*Log[Cos[(c + d*x)/2]] + 480
*a^2*b*Log[Sin[(c + d*x)/2]] - 40*a^3*Sin[2*(c + d*x)] + 120*a*b^2*Sin[2*(c + d*x)] + 15*a*b^2*Sin[4*(c + d*x)
] + 80*a^3*Tan[(c + d*x)/2])/(160*d)

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fricas [A]  time = 0.78, size = 179, normalized size = 0.78 \[ -\frac {30 \, a b^{2} \cos \left (d x + c\right )^{5} + 60 \, a^{2} b \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, a^{2} b \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 5 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right ) + {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b \cos \left (d x + c\right )^{3} - 120 \, a^{2} b \cos \left (d x + c\right ) + 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} d x\right )} \sin \left (d x + c\right )}{40 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/40*(30*a*b^2*cos(d*x + c)^5 + 60*a^2*b*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 60*a^2*b*log(-1/2*cos(d*x
+ c) + 1/2)*sin(d*x + c) - 5*(4*a^3 - 3*a*b^2)*cos(d*x + c)^3 + 15*(4*a^3 - 3*a*b^2)*cos(d*x + c) + (8*b^3*co
s(d*x + c)^5 - 40*a^2*b*cos(d*x + c)^3 - 120*a^2*b*cos(d*x + c) + 15*(4*a^3 - 3*a*b^2)*d*x)*sin(d*x + c))/(d*s
in(d*x + c))

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giac [A]  time = 0.30, size = 345, normalized size = 1.51 \[ \frac {120 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {20 \, {\left (6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 40 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 880 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 160 \, a^{2} b - 8 \, b^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/40*(120*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + 20*a^3*tan(1/2*d*x + 1/2*c) - 15*(4*a^3 - 3*a*b^2)*(d*x + c)
- 20*(6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) + 2*(20*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*a*b^2*t
an(1/2*d*x + 1/2*c)^9 + 240*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 40*b^3*tan(1/2*d*x + 1/2*c)^8 + 40*a^3*tan(1/2*d*x
+ 1/2*c)^7 - 30*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*a^2*b*tan(1/2*d*x + 1/2*c)^6 + 880*a^2*b*tan(1/2*d*x + 1/2*
c)^4 - 80*b^3*tan(1/2*d*x + 1/2*c)^4 - 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 560*a
^2*b*tan(1/2*d*x + 1/2*c)^2 - 20*a^3*tan(1/2*d*x + 1/2*c) + 75*a*b^2*tan(1/2*d*x + 1/2*c) + 160*a^2*b - 8*b^3)
/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 0.54, size = 216, normalized size = 0.94 \[ -\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {3 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {3 a^{3} x}{2}-\frac {3 a^{3} c}{2 d}+\frac {a^{2} b \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \cos \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {9 a \,b^{2} x}{8}+\frac {9 a \,b^{2} c}{8 d}-\frac {\left (\cos ^{5}\left (d x +c \right )\right ) b^{3}}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.51, size = 143, normalized size = 0.62 \[ -\frac {32 \, b^{3} \cos \left (d x + c\right )^{5} + 80 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} - 80 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} b - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2}}{160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/160*(32*b^3*cos(d*x + c)^5 + 80*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^3
- 80*(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^2*b - 15*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b^2)/d

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mupad [B]  time = 9.59, size = 674, normalized size = 2.94 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (16\,a^2\,b-\frac {4\,b^3}{5}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (a^3+3\,a\,b^2\right )-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a\,b^2-14\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {15\,a\,b^2}{2}-7\,a^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {15\,a\,b^2}{2}-a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (24\,a^2\,b-4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (88\,a^2\,b-8\,b^3\right )-a^3+56\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+72\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {3\,a\,\mathrm {atan}\left (\frac {\frac {3\,a\,\left (4\,a^2-3\,b^2\right )\,\left (\frac {9\,a\,b^2}{4}-3\,a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2-3\,b^2\right )\,9{}\mathrm {i}}{4}\right )}{8}+\frac {3\,a\,\left (4\,a^2-3\,b^2\right )\,\left (\frac {9\,a\,b^2}{4}-3\,a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2-3\,b^2\right )\,9{}\mathrm {i}}{4}\right )}{8}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^6-\frac {27\,a^4\,b^2}{2}+\frac {81\,a^2\,b^4}{16}\right )-18\,a^5\,b+\frac {27\,a^3\,b^3}{2}-\frac {a\,\left (4\,a^2-3\,b^2\right )\,\left (\frac {9\,a\,b^2}{4}-3\,a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2-3\,b^2\right )\,9{}\mathrm {i}}{4}\right )\,3{}\mathrm {i}}{8}+\frac {a\,\left (4\,a^2-3\,b^2\right )\,\left (\frac {9\,a\,b^2}{4}-3\,a^3+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2-3\,b^2\right )\,9{}\mathrm {i}}{4}\right )\,3{}\mathrm {i}}{8}}\right )\,\left (4\,a^2-3\,b^2\right )}{4\,d}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\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)^2,x)

[Out]

(tan(c/2 + (d*x)/2)*(16*a^2*b - (4*b^3)/5) - tan(c/2 + (d*x)/2)^8*(3*a*b^2 + a^3) - 10*a^3*tan(c/2 + (d*x)/2)^
6 + tan(c/2 + (d*x)/2)^4*(3*a*b^2 - 14*a^3) + tan(c/2 + (d*x)/2)^2*((15*a*b^2)/2 - 7*a^3) - tan(c/2 + (d*x)/2)
^10*((15*a*b^2)/2 - a^3) + tan(c/2 + (d*x)/2)^9*(24*a^2*b - 4*b^3) + tan(c/2 + (d*x)/2)^5*(88*a^2*b - 8*b^3) -
a^3 + 56*a^2*b*tan(c/2 + (d*x)/2)^3 + 72*a^2*b*tan(c/2 + (d*x)/2)^7)/(d*(2*tan(c/2 + (d*x)/2) + 10*tan(c/2 +
(d*x)/2)^3 + 20*tan(c/2 + (d*x)/2)^5 + 20*tan(c/2 + (d*x)/2)^7 + 10*tan(c/2 + (d*x)/2)^9 + 2*tan(c/2 + (d*x)/2
)^11)) + (a^3*tan(c/2 + (d*x)/2))/(2*d) + (3*a*atan(((3*a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 - 3*a^3 - (a*tan(c/2 +
(d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2)))/8 + (3*a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 - 3*a^3 +
(a*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2)))/8)/(2*tan(c/2 + (d*x)/2)*(9*a^6 + (
81*a^2*b^4)/16 - (27*a^4*b^2)/2) - 18*a^5*b + (27*a^3*b^3)/2 - (a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 - 3*a^3 - (a*ta
n(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2))*3i)/8 + (a*(4*a^2 - 3*b^2)*((9*a*b^2)/4 -
3*a^3 + (a*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*9i)/4 + 6*a^2*b*tan(c/2 + (d*x)/2))*3i)/8))*(4*a^2 - 3*b^2))/(4
*d) + (3*a^2*b*log(tan(c/2 + (d*x)/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)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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