\(\int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [438]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {11 \text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {5 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d (1+\csc (c+d x))} \]

[Out]

11/2*arctanh(cos(d*x+c))/a^3/d-13*cot(d*x+c)/a^3/d-13/3*cot(d*x+c)^3/a^3/d+11/2*cot(d*x+c)*csc(d*x+c)/a^3/d+4*
cot(d*x+c)*csc(d*x+c)^2/a^3/d/(1+sin(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2788, 3855, 3852, 8, 3853, 3862} \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {11 \text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}-\frac {5 \cot (c+d x)}{a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d (\csc (c+d x)+1)} \]

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

(11*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (5*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a^3*d) + (3*Cot[c + d*x]*C
sc[c + d*x])/(2*a^3*d) - (4*Cot[c + d*x])/(a^3*d*(1 + Csc[c + d*x]))

Rule 8

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

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

Rule 3855

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

Rule 3862

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*
(2*n + 1))), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (4 a-4 a \csc (c+d x)+4 a \csc ^2(c+d x)-3 a \csc ^3(c+d x)+a \csc ^4(c+d x)-\frac {4 a}{1+\csc (c+d x)}\right ) \, dx}{a^4} \\ & = \frac {4 x}{a^3}+\frac {\int \csc ^4(c+d x) \, dx}{a^3}-\frac {3 \int \csc ^3(c+d x) \, dx}{a^3}-\frac {4 \int \csc (c+d x) \, dx}{a^3}+\frac {4 \int \csc ^2(c+d x) \, dx}{a^3}-\frac {4 \int \frac {1}{1+\csc (c+d x)} \, dx}{a^3} \\ & = \frac {4 x}{a^3}+\frac {4 \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac {3 \int \csc (c+d x) \, dx}{2 a^3}+\frac {4 \int -1 \, dx}{a^3}-\frac {\text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac {4 \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d} \\ & = \frac {11 \text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {5 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d (1+\csc (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(251\) vs. \(2(96)=192\).

Time = 3.92 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.61 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right )^5 \csc ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (-4 \sin ^8\left (\frac {1}{2} (c+d x)\right )-8 \sin ^6\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x) (-2+7 \sin (c+d x))+\frac {1}{4} \sin ^4(c+d x) \left (-8+\cot \left (\frac {1}{2} (c+d x)\right )+28 \sin (c+d x)\right )-\frac {1}{2} \sin ^2\left (\frac {1}{2} (c+d x)\right ) \sin ^3(c+d x) \left (9+\left (-28+66 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-66 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)\right )+\sin ^4\left (\frac {1}{2} (c+d x)\right ) \sin ^2(c+d x) \left (9-2 \left (62+33 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-33 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (c+d x)\right )\right )}{12 a^3 d (1+\sin (c+d x))^3} \]

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/12*((1 + Cot[(c + d*x)/2])^5*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2*(-4*Sin[(c + d*x)/2]^8 - 8*Sin[(c + d*x)/2]^
6*Sin[c + d*x]*(-2 + 7*Sin[c + d*x]) + (Sin[c + d*x]^4*(-8 + Cot[(c + d*x)/2] + 28*Sin[c + d*x]))/4 - (Sin[(c
+ d*x)/2]^2*Sin[c + d*x]^3*(9 + (-28 + 66*Log[Cos[(c + d*x)/2]] - 66*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))/2 +
Sin[(c + d*x)/2]^4*Sin[c + d*x]^2*(9 - 2*(62 + 33*Log[Cos[(c + d*x)/2]] - 33*Log[Sin[(c + d*x)/2]])*Sin[c + d
*x])))/(a^3*d*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {19}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-44 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {64}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{3}}\)	\(113\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {19}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-44 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {64}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{3}}\)	\(113\)
parallelrisch	\(\frac {\left (-132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-132\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+306 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(132\)
risch	\(-\frac {33 i {\mathrm e}^{5 i \left (d x +c \right )}-96 \,{\mathrm e}^{4 i \left (d x +c \right )}+33 \,{\mathrm e}^{6 i \left (d x +c \right )}-60 i {\mathrm e}^{3 i \left (d x +c \right )}+123 \,{\mathrm e}^{2 i \left (d x +c \right )}+19 i {\mathrm e}^{i \left (d x +c \right )}-52}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d \,a^{3}}-\frac {11 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{3}}+\frac {11 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{3}}\)	\(148\)
norman	\(\frac {-\frac {107 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {11 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}+\frac {11 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}-\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}-\frac {301 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}-\frac {263 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {863 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}-\frac {2221 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {11 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}\)	\(245\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^3*(1/3*tan(1/2*d*x+1/2*c)^3-3*tan(1/2*d*x+1/2*c)^2+19*tan(1/2*d*x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3+3/ta
n(1/2*d*x+1/2*c)^2-19/tan(1/2*d*x+1/2*c)-44*ln(tan(1/2*d*x+1/2*c))-64/(tan(1/2*d*x+1/2*c)+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (98) = 196\).

Time = 0.28 (sec) , antiderivative size = 302, normalized size of antiderivative = 3.15 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {104 \, \cos \left (d x + c\right )^{4} + 38 \, \cos \left (d x + c\right )^{3} - 156 \, \cos \left (d x + c\right )^{2} + 33 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 33 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (52 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )^{2} - 45 \, \cos \left (d x + c\right ) - 24\right )} \sin \left (d x + c\right ) - 42 \, \cos \left (d x + c\right ) + 48}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/12*(104*cos(d*x + c)^4 + 38*cos(d*x + c)^3 - 156*cos(d*x + c)^2 + 33*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (c
os(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - 33*(cos(d*x
+ c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(-1/2*c
os(d*x + c) + 1/2) + 2*(52*cos(d*x + c)^3 + 33*cos(d*x + c)^2 - 45*cos(d*x + c) - 24)*sin(d*x + c) - 42*cos(d*
x + c) + 48)/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d - (a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x +
c)^2 - a^3*d*cos(d*x + c) - a^3*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (98) = 196\).

Time = 0.21 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.07 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {249 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {57 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac {132 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*((8*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 249*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 1)/(a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (57*s
in(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/
a^3 - 132*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.52 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {132 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {192}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {242 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 57 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 57 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \]

[In]

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

[Out]

-1/24*(132*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 192/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) - (242*tan(1/2*d*x + 1/2*
c)^3 - 57*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)^3) - (a^6*tan(1/2*d*x
+ 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)^2 + 57*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.59 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {11\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}-\frac {83\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d} \]

[In]

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

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - (11*log(tan(c/2 + (d*x)/2)))/(2*a^3*d)
- (16*tan(c/2 + (d*x)/2)^2 - (8*tan(c/2 + (d*x)/2))/3 + 83*tan(c/2 + (d*x)/2)^3 + 1/3)/(d*(8*a^3*tan(c/2 + (d*
x)/2)^3 + 8*a^3*tan(c/2 + (d*x)/2)^4)) + (19*tan(c/2 + (d*x)/2))/(8*a^3*d)