\(\int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [577]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {15 a x}{8}-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d} \]

[Out]

-15/8*a*x-a*arctanh(cos(d*x+c))/d+a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d+1/5*a*cos(d*x+c)^5/d-15/8*a*cot(d*x+c)/d
+5/8*a*cos(d*x+c)^2*cot(d*x+c)/d+1/4*a*cos(d*x+c)^4*cot(d*x+c)/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2917, 2671, 294, 327, 209, 2672, 308, 212} \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x)}{d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}-\frac {15 a x}{8} \]

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

(-15*a*x)/8 - (a*ArcTanh[Cos[c + d*x]])/d + (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d) + (a*Cos[c + d*x]^5)
/(5*d) - (15*a*Cot[c + d*x])/(8*d) + (5*a*Cos[c + d*x]^2*Cot[c + d*x])/(8*d) + (a*Cos[c + d*x]^4*Cot[c + d*x])
/(4*d)

Rule 209

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

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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 2671

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

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 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \cos ^5(c+d x) \cot (c+d x) \, dx+a \int \cos ^4(c+d x) \cot ^2(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \text {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {(5 a) \text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{4 d} \\ & = \frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}-\frac {a \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac {(15 a) \text {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d} \\ & = -\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d}+\frac {(15 a) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d} \\ & = -\frac {15 a x}{8}-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}-\frac {15 a \cot (c+d x)}{8 d}+\frac {5 a \cos ^2(c+d x) \cot (c+d x)}{8 d}+\frac {a \cos ^4(c+d x) \cot (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \left (900 c+900 d x-660 \cos (c+d x)-70 \cos (3 (c+d x))-6 \cos (5 (c+d x))+480 \cot (c+d x)+480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 \sin (2 (c+d x))+15 \sin (4 (c+d x))\right )}{480 d} \]

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

-1/480*(a*(900*c + 900*d*x - 660*Cos[c + d*x] - 70*Cos[3*(c + d*x)] - 6*Cos[5*(c + d*x)] + 480*Cot[c + d*x] +
480*Log[Cos[(c + d*x)/2]] - 480*Log[Sin[(c + d*x)/2]] + 240*Sin[2*(c + d*x)] + 15*Sin[4*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\)	\(114\)
default	\(\frac {a \left (\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\)	\(114\)
parallelrisch	\(-\frac {17 \left (-\frac {16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{17}+\left (\cos \left (d x +c \right )-\frac {9 \cos \left (2 d x +2 c \right )}{17}+\frac {\cos \left (3 d x +3 c \right )}{17}-\frac {\cos \left (4 d x +4 c \right )}{34}+\frac {15}{34}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{17}+\frac {30 d x}{17}-\frac {22 \cos \left (d x +c \right )}{17}-\frac {7 \cos \left (3 d x +3 c \right )}{51}-\frac {\cos \left (5 d x +5 c \right )}{85}-\frac {368}{255}\right ) a}{16 d}\)	\(126\)
risch	\(-\frac {15 a x}{8}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {11 a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {11 a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {2 i a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (4 d x +4 c \right )}{32 d}+\frac {7 a \cos \left (3 d x +3 c \right )}{48 d}\)	\(168\)
norman	\(\frac {\frac {6 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a}{2 d}-\frac {17 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {17 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {15 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}-\frac {75 a x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {75 a x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {75 a x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {75 a x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {15 a x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {12 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {56 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {46 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(307\)

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/5*cos(d*x+c)^5+1/3*cos(d*x+c)^3+cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+a*(-1/sin(d*x+c)*cos(d*x+c)^7-
(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)-15/8*d*x-15/8*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.07 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {30 \, a \cos \left (d x + c\right )^{5} + 75 \, a \cos \left (d x + c\right )^{3} - 60 \, a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 60 \, a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 225 \, a \cos \left (d x + c\right ) + {\left (24 \, a \cos \left (d x + c\right )^{5} + 40 \, a \cos \left (d x + c\right )^{3} - 225 \, a d x + 120 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(30*a*cos(d*x + c)^5 + 75*a*cos(d*x + c)^3 - 60*a*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 60*a*log(-1
/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 225*a*cos(d*x + c) + (24*a*cos(d*x + c)^5 + 40*a*cos(d*x + c)^3 - 225*a*
d*x + 120*a*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {4 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a - 15 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a}{120 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/120*(4*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x +
c) - 1))*a - 15*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^5 + 2*tan(d*x + c)
^3 + tan(d*x + c)))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.64 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {225 \, {\left (d x + c\right )} a - 120 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {60 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (135 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 150 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 150 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 560 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 135 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 184 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(225*(d*x + c)*a - 120*a*log(abs(tan(1/2*d*x + 1/2*c))) - 60*a*tan(1/2*d*x + 1/2*c) + 60*(2*a*tan(1/2*d
*x + 1/2*c) + a)/tan(1/2*d*x + 1/2*c) - 2*(135*a*tan(1/2*d*x + 1/2*c)^9 + 360*a*tan(1/2*d*x + 1/2*c)^8 + 150*a
*tan(1/2*d*x + 1/2*c)^7 + 720*a*tan(1/2*d*x + 1/2*c)^6 + 1120*a*tan(1/2*d*x + 1/2*c)^4 - 150*a*tan(1/2*d*x + 1
/2*c)^3 + 560*a*tan(1/2*d*x + 1/2*c)^2 - 135*a*tan(1/2*d*x + 1/2*c) + 184*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 10.38 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.59 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {92\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-a}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {15\,a\,\mathrm {atan}\left (\frac {225\,a^2}{16\,\left (\frac {15\,a^2}{2}+\frac {225\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {15\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {15\,a^2}{2}+\frac {225\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d} \]

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x)))/sin(c + d*x)^2,x)

[Out]

((92*a*tan(c/2 + (d*x)/2))/15 - a - (19*a*tan(c/2 + (d*x)/2)^2)/2 + (56*a*tan(c/2 + (d*x)/2)^3)/3 - 15*a*tan(c
/2 + (d*x)/2)^4 + (112*a*tan(c/2 + (d*x)/2)^5)/3 - 10*a*tan(c/2 + (d*x)/2)^6 + 24*a*tan(c/2 + (d*x)/2)^7 + 12*
a*tan(c/2 + (d*x)/2)^9 + (7*a*tan(c/2 + (d*x)/2)^10)/2)/(d*(2*tan(c/2 + (d*x)/2) + 10*tan(c/2 + (d*x)/2)^3 + 2
0*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*ta
n(c/2 + (d*x)/2))/(2*d) + (a*log(tan(c/2 + (d*x)/2)))/d + (15*a*atan((225*a^2)/(16*((15*a^2)/2 + (225*a^2*tan(
c/2 + (d*x)/2))/16)) - (15*a^2*tan(c/2 + (d*x)/2))/(2*((15*a^2)/2 + (225*a^2*tan(c/2 + (d*x)/2))/16))))/(4*d)