\(\int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1108]

Optimal result

Integrand size = 27, antiderivative size = 116 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {3 a b x}{4}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {b^2 \cos ^5(c+d x)}{5 d}+\frac {3 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a b \cos ^3(c+d x) \sin (c+d x)}{2 d} \]

[Out]

3/4*a*b*x-a^2*arctanh(cos(d*x+c))/d+a^2*cos(d*x+c)/d+1/3*a^2*cos(d*x+c)^3/d-1/5*b^2*cos(d*x+c)^5/d+3/4*a*b*cos
(d*x+c)*sin(d*x+c)/d+1/2*a*b*cos(d*x+c)^3*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.64, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2974, 3128, 3112, 3102, 2814, 3855} \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \sin (c+d x) \cos (c+d x)}{60 b d}-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^3}{5 b d}+\frac {3 a b x}{4} \]

[In]

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

[Out]

(3*a*b*x)/4 - (a^2*ArcTanh[Cos[c + d*x]])/d - ((a^4 - 14*a^2*b^2 + 3*b^4)*Cos[c + d*x])/(15*b^2*d) - (a*(2*a^2
- 27*b^2)*Cos[c + d*x]*Sin[c + d*x])/(60*b*d) - ((a^2 - 12*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(30*b^2*
d) + (a*Cos[c + d*x]*(a + b*Sin[c + d*x])^3)/(10*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Sin[c + d*x])^3)/(
5*b*d)

Rule 2814

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 2974

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 3102

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 3112

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 3128

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x))^2 \left (-20 b^2+2 a b \sin (c+d x)-2 \left (a^2-12 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{20 b^2} \\ & = -\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) (a+b \sin (c+d x)) \left (-60 a b^2+2 b \left (a^2-6 b^2\right ) \sin (c+d x)-2 a \left (2 a^2-27 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{60 b^2} \\ & = -\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)-8 \left (a^4-14 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)\right ) \, dx}{120 b^2} \\ & = -\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}-\frac {\int \csc (c+d x) \left (-120 a^2 b^2-90 a b^3 \sin (c+d x)\right ) \, dx}{120 b^2} \\ & = \frac {3 a b x}{4}-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d}+a^2 \int \csc (c+d x) \, dx \\ & = \frac {3 a b x}{4}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {\left (a^4-14 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^2 d}-\frac {a \left (2 a^2-27 b^2\right ) \cos (c+d x) \sin (c+d x)}{60 b d}-\frac {\left (a^2-12 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{30 b^2 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^3}{10 b^2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^3}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.08 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {30 \left (10 a^2-b^2\right ) \cos (c+d x)+5 \left (4 a^2-3 b^2\right ) \cos (3 (c+d x))-3 b^2 \cos (5 (c+d x))+15 a \left (4 \left (3 b (c+d x)-4 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 b \sin (2 (c+d x))+b \sin (4 (c+d x))\right )}{240 d} \]

[In]

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

[Out]

(30*(10*a^2 - b^2)*Cos[c + d*x] + 5*(4*a^2 - 3*b^2)*Cos[3*(c + d*x)] - 3*b^2*Cos[5*(c + d*x)] + 15*a*(4*(3*b*(
c + d*x) - 4*a*Log[Cos[(c + d*x)/2]] + 4*a*Log[Sin[(c + d*x)/2]]) + 8*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])
)/(240*d)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}}{d}\)	\(93\)
default	\(\frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+2 a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}}{d}\)	\(93\)
parallelrisch	\(\frac {180 a b x d +240 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 \cos \left (5 d x +5 c \right ) b^{2}+15 a b \sin \left (4 d x +4 c \right )+120 a b \sin \left (2 d x +2 c \right )+20 \cos \left (3 d x +3 c \right ) a^{2}-15 \cos \left (3 d x +3 c \right ) b^{2}+300 \cos \left (d x +c \right ) a^{2}-30 \cos \left (d x +c \right ) b^{2}+320 a^{2}-48 b^{2}}{240 d}\)	\(128\)
risch	\(\frac {3 a b x}{4}+\frac {5 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {b^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {5 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{16 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {\cos \left (5 d x +5 c \right ) b^{2}}{80 d}+\frac {a b \sin \left (4 d x +4 c \right )}{16 d}+\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{16 d}+\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\)	\(197\)
norman	\(\frac {\frac {\left (4 a^{2}-2 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {40 a^{2}-6 b^{2}}{15 d}+\frac {3 a b x}{4}+\frac {12 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (22 a^{2}-6 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {5 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a b x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a b x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 a b x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(297\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {12 \, b^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a b d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(12*b^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 - 45*a*b*d*x - 60*a^2*cos(d*x + c) + 30*a^2*log(1/2*cos(d
*x + c) + 1/2) - 30*a^2*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a*b*cos(d*x + c)^3 + 3*a*b*cos(d*x + c))*sin(d*x
+ c))/d

Sympy [F]

\[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Integral((a + b*sin(c + d*x))**2*cos(c + d*x)**4*csc(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {48 \, b^{2} \cos \left (d x + c\right )^{5} - 40 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b}{240 \, d} \]

[In]

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

[Out]

-1/240*(48*b^2*cos(d*x + c)^5 - 40*(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 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (106) = 212\).

Time = 0.34 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.84 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {45 \, {\left (d x + c\right )} a b + 60 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 80 \, a^{2} + 12 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(45*(d*x + c)*a*b + 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(75*a*b*tan(1/2*d*x + 1/2*c)^9 - 120*a^2*ta
n(1/2*d*x + 1/2*c)^8 + 60*b^2*tan(1/2*d*x + 1/2*c)^8 + 30*a*b*tan(1/2*d*x + 1/2*c)^7 - 360*a^2*tan(1/2*d*x + 1
/2*c)^6 - 440*a^2*tan(1/2*d*x + 1/2*c)^4 + 120*b^2*tan(1/2*d*x + 1/2*c)^4 - 30*a*b*tan(1/2*d*x + 1/2*c)^3 - 28
0*a^2*tan(1/2*d*x + 1/2*c)^2 - 75*a*b*tan(1/2*d*x + 1/2*c) - 80*a^2 + 12*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/
d

Mupad [B] (verification not implemented)

Time = 12.36 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.75 \[ \int \cos ^3(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (4\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {44\,a^2}{3}-4\,b^2\right )+\frac {28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {8\,a^2}{3}-\frac {2\,b^2}{5}+a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {5\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {5\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,b\,\mathrm {atan}\left (\frac {9\,a^2\,b^2}{4\,\left (3\,a^3\,b-\frac {9\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )}+\frac {3\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3\,a^3\,b-\frac {9\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}\right )}{2\,d} \]

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^8*(4*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*((44*a^2)/3 - 4
*b^2) + (28*a^2*tan(c/2 + (d*x)/2)^2)/3 + 12*a^2*tan(c/2 + (d*x)/2)^6 + (8*a^2)/3 - (2*b^2)/5 + a*b*tan(c/2 +
(d*x)/2)^3 - a*b*tan(c/2 + (d*x)/2)^7 - (5*a*b*tan(c/2 + (d*x)/2)^9)/2 + (5*a*b*tan(c/2 + (d*x)/2))/2)/(d*(5*t
an(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (
d*x)/2)^10 + 1)) + (3*a*b*atan((9*a^2*b^2)/(4*(3*a^3*b - (9*a^2*b^2*tan(c/2 + (d*x)/2))/4)) + (3*a^3*b*tan(c/2
+ (d*x)/2))/(3*a^3*b - (9*a^2*b^2*tan(c/2 + (d*x)/2))/4)))/(2*d)