\(\int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [784]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\sec (c+d x)}{a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}-\frac {4 \tan ^3(c+d x)}{3 a^2 d}-\frac {2 \tan ^5(c+d x)}{5 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2954, 2952, 3852, 2702, 308, 213, 2686, 30} \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {4 \tan ^3(c+d x)}{3 a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}+\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {\sec (c+d x)}{a^2 d} \]

[In]

Int[(Csc[c + d*x]*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 30

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

Rule 213

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

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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2702

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

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + 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] && ILtQ[m, 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]

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

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.70 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec (c+d x) \left (280+136 \cos (2 (c+d x))+79 \cos (3 (c+d x))+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-5 \cos (c+d x) \left (79+60 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-60 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+160 \sin (c+d x)-316 \sin (2 (c+d x))-240 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+240 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+64 \sin (3 (c+d x))\right )}{240 a^2 d (1+\sin (c+d x))^2} \]

[In]

Integrate[(Csc[c + d*x]*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sec[c + d*x]*(280 + 136*Cos[2*(c + d*x)] + 79*Cos[3*(c + d*x)] + 60*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] -
5*Cos[c + d*x]*(79 + 60*Log[Cos[(c + d*x)/2]] - 60*Log[Sin[(c + d*x)/2]]) - 60*Cos[3*(c + d*x)]*Log[Sin[(c + d
*x)/2]] + 160*Sin[c + d*x] - 316*Sin[2*(c + d*x)] - 240*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 240*Log[Sin[(
c + d*x)/2]]*Sin[2*(c + d*x)] + 64*Sin[3*(c + d*x)]))/(240*a^2*d*(1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {11}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{2}}\)	\(109\)
default	\(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {11}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{2}}\)	\(109\)
parallelrisch	\(\frac {15 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-140 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-148 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-52}{15 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(136\)
risch	\(\frac {8 i {\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{5 i \left (d x +c \right )}+\frac {8 i {\mathrm e}^{2 i \left (d x +c \right )}}{3}-\frac {28 \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}-\frac {32 i}{15}-\frac {98 \,{\mathrm e}^{i \left (d x +c \right )}}{15}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\)	\(137\)
norman	\(\frac {\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {52}{15 a d}+\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {28 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {148 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(151\)

[In]

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

[Out]

1/d/a^2*(-1/4/(tan(1/2*d*x+1/2*c)-1)+ln(tan(1/2*d*x+1/2*c))+4/5/(tan(1/2*d*x+1/2*c)+1)^5-2/(tan(1/2*d*x+1/2*c)
+1)^4+11/3/(tan(1/2*d*x+1/2*c)+1)^3-7/2/(tan(1/2*d*x+1/2*c)+1)^2+17/4/(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.46 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {34 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, {\left (8 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right ) + 18}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/30*(34*cos(d*x + c)^2 + 15*(cos(d*x + c)^3 - 2*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))*log(1/2*cos(d*x
+ c) + 1/2) - 15*(cos(d*x + c)^3 - 2*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2)
+ 4*(8*cos(d*x + c)^2 + 3)*sin(d*x + c) + 18)/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c)*sin(d*x + c) - 2*a^
2*d*cos(d*x + c))

Sympy [F]

\[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)*sec(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (107) = 214\).

Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.17 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (\frac {37 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {30 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 13\right )}}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{15 \, d} \]

[In]

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

[Out]

1/15*(4*(37*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 30*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 13)/(a^2 +
4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*a^2*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 - 4*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 15*l
og(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {15}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {255 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 810 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 710 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 193}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 15/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) + (255*tan(1/2*d*x + 1/2*c)^
4 + 810*tan(1/2*d*x + 1/2*c)^3 + 1120*tan(1/2*d*x + 1/2*c)^2 + 710*tan(1/2*d*x + 1/2*c) + 193)/(a^2*(tan(1/2*d
*x + 1/2*c) + 1)^5))/d

Mupad [B] (verification not implemented)

Time = 13.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {148\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {52}{15}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)*(a + a*sin(c + d*x))^2),x)

[Out]

log(tan(c/2 + (d*x)/2))/(a^2*d) - ((148*tan(c/2 + (d*x)/2))/15 + (28*tan(c/2 + (d*x)/2)^2)/3 - (8*tan(c/2 + (d
*x)/2)^3)/3 - 8*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^5 + 52/15)/(a^2*d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/
2 + (d*x)/2) + 1)^5)