\(\int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\) [817]

Optimal result

Integrand size = 29, antiderivative size = 110 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {11 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))} \]

[Out]

-11/2*a^3*arctanh(cos(d*x+c))/d-3*a^3*cot(d*x+c)/d-1/2*a^3*cot(d*x+c)*csc(d*x+c)/d+2/3*a^3*cos(d*x+c)/d/(1-sin
(d*x+c))^2+17/3*a^3*cos(d*x+c)/d/(1-sin(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2951, 3855, 3852, 8, 3853, 2729, 2727} \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {11 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x)}{d}+\frac {17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[In]

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

[Out]

(-11*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (2*a^
3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) + (17*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x]))

Rule 8

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

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

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

Rule 2951

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 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]

Rubi steps \begin{align*} \text {integral}& = a^4 \int \left (\frac {5 \csc (c+d x)}{a}+\frac {3 \csc ^2(c+d x)}{a}+\frac {\csc ^3(c+d x)}{a}+\frac {2}{a (-1+\sin (c+d x))^2}-\frac {5}{a (-1+\sin (c+d x))}\right ) \, dx \\ & = a^3 \int \csc ^3(c+d x) \, dx+\left (2 a^3\right ) \int \frac {1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (5 a^3\right ) \int \csc (c+d x) \, dx-\left (5 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx \\ & = -\frac {5 a^3 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {5 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {1}{2} a^3 \int \csc (c+d x) \, dx-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\frac {\left (3 a^3\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d} \\ & = -\frac {11 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.73 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-36 \cot \left (\frac {1}{2} (c+d x)\right )-3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-132 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+132 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {16}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {32 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {272 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+36 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d} \]

[In]

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

[Out]

(a^3*(-36*Cot[(c + d*x)/2] - 3*Csc[(c + d*x)/2]^2 - 132*Log[Cos[(c + d*x)/2]] + 132*Log[Sin[(c + d*x)/2]] + 3*
Sec[(c + d*x)/2]^2 + 16/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (32*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2])^3 + (272*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 36*Tan[(c + d*x)/2]))/(24*d
)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {\left (44 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-188 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+314 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {502}{3}\right ) a^{3}}{8 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(119\)
risch	\(\frac {a^{3} \left (-99 i {\mathrm e}^{5 i \left (d x +c \right )}+33 \,{\mathrm e}^{6 i \left (d x +c \right )}+210 i {\mathrm e}^{3 i \left (d x +c \right )}-154 \,{\mathrm e}^{4 i \left (d x +c \right )}-123 i {\mathrm e}^{i \left (d x +c \right )}+161 \,{\mathrm e}^{2 i \left (d x +c \right )}-52\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}-\frac {11 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}+\frac {11 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\)	\(148\)
derivativedivides	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+a^{3} \left (\frac {1}{3 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}-\frac {5}{6 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {5}{2 \cos \left (d x +c \right )}+\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(188\)
default	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 a^{3} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+a^{3} \left (\frac {1}{3 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}-\frac {5}{6 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {5}{2 \cos \left (d x +c \right )}+\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(188\)

[In]

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

[Out]

1/8*(44*(tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))+tan(1/2*d*x+1/2*c)^5+9*tan(1/2*d*x+1/2*c)^4+cot(1/2*d*
x+1/2*c)^2-188*tan(1/2*d*x+1/2*c)^2+9*cot(1/2*d*x+1/2*c)+314*tan(1/2*d*x+1/2*c)-502/3)*a^3/d/(tan(1/2*d*x+1/2*
c)-1)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (98) = 196\).

Time = 0.28 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.89 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {104 \, a^{3} \cos \left (d x + c\right )^{4} + 142 \, a^{3} \cos \left (d x + c\right )^{3} - 90 \, a^{3} \cos \left (d x + c\right )^{2} - 136 \, a^{3} \cos \left (d x + c\right ) - 8 \, a^{3} + 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (52 \, a^{3} \cos \left (d x + c\right )^{3} - 19 \, a^{3} \cos \left (d x + c\right )^{2} - 64 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{3} - 3 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

[In]

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

[Out]

-1/12*(104*a^3*cos(d*x + c)^4 + 142*a^3*cos(d*x + c)^3 - 90*a^3*cos(d*x + c)^2 - 136*a^3*cos(d*x + c) - 8*a^3
+ 33*(a^3*cos(d*x + c)^4 - a^3*cos(d*x + c)^3 - 3*a^3*cos(d*x + c)^2 + a^3*cos(d*x + c) + 2*a^3 + (a^3*cos(d*x
+ c)^3 + 2*a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 33*(a^3
*cos(d*x + c)^4 - a^3*cos(d*x + c)^3 - 3*a^3*cos(d*x + c)^2 + a^3*cos(d*x + c) + 2*a^3 + (a^3*cos(d*x + c)^3 +
2*a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(52*a^3*cos(d
*x + c)^3 - 19*a^3*cos(d*x + c)^2 - 64*a^3*cos(d*x + c) + 4*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4 - d*cos(d*x +
c)^3 - 3*d*cos(d*x + c)^2 + d*cos(d*x + c) + (d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c) - 2*d)*s
in(d*x + c) + 2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.65 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {12 \, {\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{3} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} + a^{3} {\left (\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

[In]

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

[Out]

1/12*(12*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 + a^
3*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2)/(cos(d*x + c)^5 - cos(d*x + c)^3) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1)) + 6*a^3*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x + c) + 1) + 3*log
(cos(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.36 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 132 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (66 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {16 \, {\left (21 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*a^3*tan(1/2*d*x + 1/2*c)^2 + 132*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 36*a^3*tan(1/2*d*x + 1/2*c) - 3*
(66*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^2 - 16*(21*a^3*tan(1/
2*d*x + 1/2*c)^2 - 36*a^3*tan(1/2*d*x + 1/2*c) + 19*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 11.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.66 \[ \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {11\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {-62\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {227\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {403\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{6}+\frac {9\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a^3}{2}}{d\,\left (-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (11*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) - ((227*a^3*tan(c/2 + (d*x)/2)^3)/2
- (403*a^3*tan(c/2 + (d*x)/2)^2)/6 - 62*a^3*tan(c/2 + (d*x)/2)^4 + a^3/2 + (9*a^3*tan(c/2 + (d*x)/2))/2)/(d*(4
*tan(c/2 + (d*x)/2)^2 - 12*tan(c/2 + (d*x)/2)^3 + 12*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^5)) + (3*a^3*
tan(c/2 + (d*x)/2))/(2*d)