3.846 \(\int \frac{\csc (c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=187 \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{15 \tan ^7(c+d x)}{7 a^3 d}-\frac{21 \tan ^5(c+d x)}{5 a^3 d}-\frac{13 \tan ^3(c+d x)}{3 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{\sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^3*d)) + Sec[c + d*x]/(a^3*d) + Sec[c + d*x]^3/(3*a^3*d) + Sec[c + d*x]^5/(5*a^3*d)
+ Sec[c + d*x]^7/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (13*Tan[c + d*x]^3)/(3*
a^3*d) - (21*Tan[c + d*x]^5)/(5*a^3*d) - (15*Tan[c + d*x]^7)/(7*a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

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Rubi [A]  time = 0.360108, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2606, 30, 2607, 270} \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{15 \tan ^7(c+d x)}{7 a^3 d}-\frac{21 \tan ^5(c+d x)}{5 a^3 d}-\frac{13 \tan ^3(c+d x)}{3 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{\sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^3*d)) + Sec[c + d*x]/(a^3*d) + Sec[c + d*x]^3/(3*a^3*d) + Sec[c + d*x]^5/(5*a^3*d)
+ Sec[c + d*x]^7/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (13*Tan[c + d*x]^3)/(3*
a^3*d) - (21*Tan[c + d*x]^5)/(5*a^3*d) - (15*Tan[c + d*x]^7)/(7*a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

Rule 2875

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 2873

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 3767

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 2622

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 302

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 207

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

Rule 2606

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 30

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

Rule 2607

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

Rule 270

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

Rubi steps

\begin{align*} \int \frac{\csc (c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc (c+d x) \sec ^{10}(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-3 a^3 \sec ^{10}(c+d x)+a^3 \csc (c+d x) \sec ^{10}(c+d x)+3 a^3 \sec ^9(c+d x) \tan (c+d x)-a^3 \sec ^8(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \csc (c+d x) \sec ^{10}(c+d x) \, dx}{a^3}-\frac{\int \sec ^8(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^{10}(c+d x) \, dx}{a^3}+\frac{3 \int \sec ^9(c+d x) \tan (c+d x) \, dx}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^{10}}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int x^8 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (c+d x)\right )}{a^3 d}\\ &=\frac{\sec ^9(c+d x)}{3 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{4 \tan ^3(c+d x)}{a^3 d}-\frac{18 \tan ^5(c+d x)}{5 a^3 d}-\frac{12 \tan ^7(c+d x)}{7 a^3 d}-\frac{\tan ^9(c+d x)}{3 a^3 d}-\frac{\operatorname{Subst}\left (\int \left (x^2+3 x^4+3 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{\operatorname{Subst}\left (\int \left (1+x^2+x^4+x^6+x^8+\frac{1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac{\sec (c+d x)}{a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{13 \tan ^3(c+d x)}{3 a^3 d}-\frac{21 \tan ^5(c+d x)}{5 a^3 d}-\frac{15 \tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\sec (c+d x)}{a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^7(c+d x)}{7 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{13 \tan ^3(c+d x)}{3 a^3 d}-\frac{21 \tan ^5(c+d x)}{5 a^3 d}-\frac{15 \tan ^7(c+d x)}{7 a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.39842, size = 204, normalized size = 1.09 \[ \frac{322560 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-322560 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{196992 \sin (c+d x)-383157 \sin (2 (c+d x))+211648 \sin (3 (c+d x))-170292 \sin (4 (c+d x))+50496 \sin (5 (c+d x))+14191 \sin (6 (c+d x))-510876 \cos (c+d x)+317952 \cos (2 (c+d x))-28382 \cos (3 (c+d x))+20352 \cos (4 (c+d x))+85146 \cos (5 (c+d x))-11776 \cos (6 (c+d x))+357504}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^9}}{322560 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-322560*Log[Cos[(c + d*x)/2]] + 322560*Log[Sin[(c + d*x)/2]] + (357504 - 510876*Cos[c + d*x] + 317952*Cos[2*(
c + d*x)] - 28382*Cos[3*(c + d*x)] + 20352*Cos[4*(c + d*x)] + 85146*Cos[5*(c + d*x)] - 11776*Cos[6*(c + d*x)]
+ 196992*Sin[c + d*x] - 383157*Sin[2*(c + d*x)] + 211648*Sin[3*(c + d*x)] - 170292*Sin[4*(c + d*x)] + 50496*Si
n[5*(c + d*x)] + 14191*Sin[6*(c + d*x)])/((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^9))/(322560*a^3*d)

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Maple [A]  time = 0.158, size = 271, normalized size = 1.5 \begin{align*} -{\frac{1}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{16\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{9}{32\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{8}{9\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-9}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{8}}}+{\frac{72}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}-{\frac{52}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-6}}+{\frac{219}{10\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-{\frac{83}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{193}{12\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{75}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{201}{32\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d/a^3/(tan(1/2*d*x+1/2*c)-1)^3-1/16/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2-9/32/d/a^3/(tan(1/2*d*x+1/2*c)-1)+8/9
/d/a^3/(tan(1/2*d*x+1/2*c)+1)^9-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^8+72/7/d/a^3/(tan(1/2*d*x+1/2*c)+1)^7-52/3/d/a^
3/(tan(1/2*d*x+1/2*c)+1)^6+219/10/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5-83/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^4+193/12/d/
a^3/(tan(1/2*d*x+1/2*c)+1)^3-75/8/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+201/32/d/a^3/(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*l
n(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.10388, size = 686, normalized size = 3.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(2*(3063*sin(d*x + c)/(cos(d*x + c) + 1) + 4866*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1289*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 11736*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10566*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 5292*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 13482*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 6300*sin(d*x + c)^8/
(cos(d*x + c) + 1)^8 - 2625*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 3150*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 -
945*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 668)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*si
n(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*
x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 31
5*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 2.0486, size = 676, normalized size = 3.61 \begin{align*} \frac{736 \, \cos \left (d x + c\right )^{6} - 1422 \, \cos \left (d x + c\right )^{4} - 510 \, \cos \left (d x + c\right )^{2} - 315 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 315 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{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 (789 \, \cos \left (d x + c\right )^{4} + 235 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) - 140}{630 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} +{\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/630*(736*cos(d*x + c)^6 - 1422*cos(d*x + c)^4 - 510*cos(d*x + c)^2 - 315*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^
3 + (cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 315*(3*cos(d*x + c)^5 - 4*
cos(d*x + c)^3 + (cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(789*cos(d
*x + c)^4 + 235*cos(d*x + c)^2 + 35)*sin(d*x + c) - 140)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a
^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.24968, size = 252, normalized size = 1.35 \begin{align*} \frac{\frac{10080 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{105 \,{\left (27 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{63315 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 412020 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1273440 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 2324700 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2731302 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2097228 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1032552 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 297828 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 40127}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10080*(10080*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 105*(27*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x + 1/2*c) +
25)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) + (63315*tan(1/2*d*x + 1/2*c)^8 + 412020*tan(1/2*d*x + 1/2*c)^7 + 1273
440*tan(1/2*d*x + 1/2*c)^6 + 2324700*tan(1/2*d*x + 1/2*c)^5 + 2731302*tan(1/2*d*x + 1/2*c)^4 + 2097228*tan(1/2
*d*x + 1/2*c)^3 + 1032552*tan(1/2*d*x + 1/2*c)^2 + 297828*tan(1/2*d*x + 1/2*c) + 40127)/(a^3*(tan(1/2*d*x + 1/
2*c) + 1)^9))/d