\(\int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx\) [295]

Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {5 a^4 x}{2}-\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^4 \cos (c+d x)}{d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos ^5(c+d x)}{5 d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d} \]

[Out]

5/2*a^4*x-a^4*arctanh(cos(d*x+c))/d+a^4*cos(d*x+c)/d-7/3*a^4*cos(d*x+c)^3/d+1/5*a^4*cos(d*x+c)^5/d+5/2*a^4*cos
(d*x+c)*sin(d*x+c)/d-a^4*cos(d*x+c)^3*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2952, 2715, 8, 2672, 327, 212, 2645, 30, 2648, 14} \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^4 \cos ^5(c+d x)}{5 d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos (c+d x)}{d}-\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{d}+\frac {5 a^4 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {5 a^4 x}{2} \]

[In]

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

[Out]

(5*a^4*x)/2 - (a^4*ArcTanh[Cos[c + d*x]])/d + (a^4*Cos[c + d*x])/d - (7*a^4*Cos[c + d*x]^3)/(3*d) + (a^4*Cos[c
+ d*x]^5)/(5*d) + (5*a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (a^4*Cos[c + d*x]^3*Sin[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (4 a^4 \cos ^2(c+d x)+a^4 \cos (c+d x) \cot (c+d x)+6 a^4 \cos ^2(c+d x) \sin (c+d x)+4 a^4 \cos ^2(c+d x) \sin ^2(c+d x)+a^4 \cos ^2(c+d x) \sin ^3(c+d x)\right ) \, dx \\ & = a^4 \int \cos (c+d x) \cot (c+d x) \, dx+a^4 \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^2(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^2(c+d x) \sin (c+d x) \, dx \\ & = \frac {2 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}+a^4 \int \cos ^2(c+d x) \, dx+\left (2 a^4\right ) \int 1 \, dx-\frac {a^4 \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^4 \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (6 a^4\right ) \text {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{d} \\ & = 2 a^4 x+\frac {a^4 \cos (c+d x)}{d}-\frac {2 a^4 \cos ^3(c+d x)}{d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d}+\frac {1}{2} a^4 \int 1 \, dx-\frac {a^4 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^4 \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {5 a^4 x}{2}-\frac {a^4 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^4 \cos (c+d x)}{d}-\frac {7 a^4 \cos ^3(c+d x)}{3 d}+\frac {a^4 \cos ^5(c+d x)}{5 d}+\frac {5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^4 \cos ^3(c+d x) \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \left (-150 \cos (c+d x)-125 \cos (3 (c+d x))+3 \cos (5 (c+d x))+30 \left (20 c+20 d x-8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 \sin (2 (c+d x))-\sin (4 (c+d x))\right )\right )}{240 d} \]

[In]

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

[Out]

(a^4*(-150*Cos[c + d*x] - 125*Cos[3*(c + d*x)] + 3*Cos[5*(c + d*x)] + 30*(20*c + 20*d*x - 8*Log[Cos[(c + d*x)/
2]] + 8*Log[Sin[(c + d*x)/2]] + 8*Sin[2*(c + d*x)] - Sin[4*(c + d*x)])))/(240*d)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(-\frac {a^{4} \left (-600 d x +150 \cos \left (d x +c \right )+30 \sin \left (4 d x +4 c \right )-240 \sin \left (2 d x +2 c \right )-3 \cos \left (5 d x +5 c \right )+125 \cos \left (3 d x +3 c \right )-240 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+272\right )}{240 d}\)	\(79\)
risch	\(\frac {5 a^{4} x}{2}-\frac {5 a^{4} {\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {5 a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{4} \cos \left (5 d x +5 c \right )}{80 d}-\frac {a^{4} \sin \left (4 d x +4 c \right )}{8 d}-\frac {25 a^{4} \cos \left (3 d x +3 c \right )}{48 d}+\frac {a^{4} \sin \left (2 d x +2 c \right )}{d}\)	\(148\)
derivativedivides	\(\frac {a^{4} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+4 a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-2 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(149\)
default	\(\frac {a^{4} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+4 a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-2 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(149\)
norman	\(\frac {\frac {5 a^{4} x}{2}-\frac {34 a^{4}}{15 d}+\frac {3 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {14 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {25 a^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+25 a^{4} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+25 a^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {25 a^{4} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{4} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {20 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {10 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(285\)

[In]

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

[Out]

-1/240*a^4*(-600*d*x+150*cos(d*x+c)+30*sin(4*d*x+4*c)-240*sin(2*d*x+2*c)-3*cos(5*d*x+5*c)+125*cos(3*d*x+3*c)-2
40*ln(tan(1/2*d*x+1/2*c))+272)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {6 \, a^{4} \cos \left (d x + c\right )^{5} - 70 \, a^{4} \cos \left (d x + c\right )^{3} + 75 \, a^{4} d x + 30 \, a^{4} \cos \left (d x + c\right ) - 15 \, a^{4} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, a^{4} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{3} - 5 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(6*a^4*cos(d*x + c)^5 - 70*a^4*cos(d*x + c)^3 + 75*a^4*d*x + 30*a^4*cos(d*x + c) - 15*a^4*log(1/2*cos(d*x
+ c) + 1/2) + 15*a^4*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a^4*cos(d*x + c)^3 - 5*a^4*cos(d*x + c))*sin(d*x +
c))/d

Sympy [F]

\[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=a^{4} \left (\int \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 4 \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**4*(Integral(cos(c + d*x)**2*csc(c + d*x), x) + Integral(4*sin(c + d*x)*cos(c + d*x)**2*csc(c + d*x), x) + I
ntegral(6*sin(c + d*x)**2*cos(c + d*x)**2*csc(c + d*x), x) + Integral(4*sin(c + d*x)**3*cos(c + d*x)**2*csc(c
+ d*x), x) + Integral(sin(c + d*x)**4*cos(c + d*x)**2*csc(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {240 \, a^{4} \cos \left (d x + c\right )^{3} - 8 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{4} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 60 \, a^{4} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

[In]

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

[Out]

-1/120*(240*a^4*cos(d*x + c)^3 - 8*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^4 - 15*(4*d*x + 4*c - sin(4*d*x + 4
*c))*a^4 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 - 60*a^4*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos
(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.55 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {75 \, {\left (d x + c\right )} a^{4} + 30 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 210 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 300 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 210 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 34 \, a^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{30 \, d} \]

[In]

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

[Out]

1/30*(75*(d*x + c)*a^4 + 30*a^4*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(45*a^4*tan(1/2*d*x + 1/2*c)^9 + 150*a^4*ta
n(1/2*d*x + 1/2*c)^8 + 210*a^4*tan(1/2*d*x + 1/2*c)^7 + 300*a^4*tan(1/2*d*x + 1/2*c)^6 + 40*a^4*tan(1/2*d*x +
1/2*c)^4 - 210*a^4*tan(1/2*d*x + 1/2*c)^3 + 20*a^4*tan(1/2*d*x + 1/2*c)^2 - 45*a^4*tan(1/2*d*x + 1/2*c) + 34*a
^4)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 11.18 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.52 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a^4\,\mathrm {atan}\left (\frac {25\,a^8}{10\,a^8-25\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {10\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{10\,a^8-25\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+14\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-14\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-3\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {34\,a^4}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(a^4*log(tan(c/2 + (d*x)/2)))/d + (5*a^4*atan((25*a^8)/(10*a^8 - 25*a^8*tan(c/2 + (d*x)/2)) + (10*a^8*tan(c/2
+ (d*x)/2))/(10*a^8 - 25*a^8*tan(c/2 + (d*x)/2))))/d - ((4*a^4*tan(c/2 + (d*x)/2)^2)/3 - 14*a^4*tan(c/2 + (d*x
)/2)^3 + (8*a^4*tan(c/2 + (d*x)/2)^4)/3 + 20*a^4*tan(c/2 + (d*x)/2)^6 + 14*a^4*tan(c/2 + (d*x)/2)^7 + 10*a^4*t
an(c/2 + (d*x)/2)^8 + 3*a^4*tan(c/2 + (d*x)/2)^9 + (34*a^4)/15 - 3*a^4*tan(c/2 + (d*x)/2))/(d*(5*tan(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))