3.847 \(\int \frac{\csc ^2(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=200 \[ \frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{17 \tan ^7(c+d x)}{7 a^3 d}+\frac{28 \tan ^5(c+d x)}{5 a^3 d}+\frac{22 \tan ^3(c+d x)}{3 a^3 d}+\frac{8 \tan (c+d x)}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{3 \sec ^7(c+d x)}{7 a^3 d}-\frac{3 \sec ^5(c+d x)}{5 a^3 d}-\frac{\sec ^3(c+d x)}{a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) - (3*Sec[c + d*x])/(a^3*d) - Sec[c + d*x]^3/(a^3*d) -
(3*Sec[c + d*x]^5)/(5*a^3*d) - (3*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + (8*Tan[c + d*x])
/(a^3*d) + (22*Tan[c + d*x]^3)/(3*a^3*d) + (28*Tan[c + d*x]^5)/(5*a^3*d) + (17*Tan[c + d*x]^7)/(7*a^3*d) + (4*
Tan[c + d*x]^9)/(9*a^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.393419, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2620, 270, 2606, 30} \[ \frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{17 \tan ^7(c+d x)}{7 a^3 d}+\frac{28 \tan ^5(c+d x)}{5 a^3 d}+\frac{22 \tan ^3(c+d x)}{3 a^3 d}+\frac{8 \tan (c+d x)}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{3 \sec ^7(c+d x)}{7 a^3 d}-\frac{3 \sec ^5(c+d x)}{5 a^3 d}-\frac{\sec ^3(c+d x)}{a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) - (3*Sec[c + d*x])/(a^3*d) - Sec[c + d*x]^3/(a^3*d) -
(3*Sec[c + d*x]^5)/(5*a^3*d) - (3*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + (8*Tan[c + d*x])
/(a^3*d) + (22*Tan[c + d*x]^3)/(3*a^3*d) + (28*Tan[c + d*x]^5)/(5*a^3*d) + (17*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 2620

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

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]

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]

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc ^2(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)-3 a^3 \csc (c+d x) \sec ^{10}(c+d x)+a^3 \csc ^2(c+d x) \sec ^{10}(c+d x)-a^3 \sec ^9(c+d x) \tan (c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \csc ^2(c+d x) \sec ^{10}(c+d x) \, dx}{a^3}-\frac{\int \sec ^9(c+d x) \tan (c+d x) \, dx}{a^3}+\frac{3 \int \sec ^{10}(c+d x) \, dx}{a^3}-\frac{3 \int \csc (c+d x) \sec ^{10}(c+d x) \, dx}{a^3}\\ &=-\frac{\operatorname{Subst}\left (\int x^8 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^5}{x^2} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \frac{x^{10}}{-1+x^2} \, 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)}{9 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 (5+\frac{1}{x^2}+10 x^2+10 x^4+5 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac{3 \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{\cot (c+d x)}{a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}-\frac{\sec ^3(c+d x)}{a^3 d}-\frac{3 \sec ^5(c+d x)}{5 a^3 d}-\frac{3 \sec ^7(c+d x)}{7 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{8 \tan (c+d x)}{a^3 d}+\frac{22 \tan ^3(c+d x)}{3 a^3 d}+\frac{28 \tan ^5(c+d x)}{5 a^3 d}+\frac{17 \tan ^7(c+d x)}{7 a^3 d}+\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}-\frac{\sec ^3(c+d x)}{a^3 d}-\frac{3 \sec ^5(c+d x)}{5 a^3 d}-\frac{3 \sec ^7(c+d x)}{7 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{8 \tan (c+d x)}{a^3 d}+\frac{22 \tan ^3(c+d x)}{3 a^3 d}+\frac{28 \tan ^5(c+d x)}{5 a^3 d}+\frac{17 \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 = 0.653642, size = 230, normalized size = 1.15 \[ \frac{-1935360 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+1935360 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{\csc (c+d x) (-707328 \sin (c+d x)+1364182 \sin (2 (c+d x))-1161600 \sin (3 (c+d x))+320984 \sin (4 (c+d x))-329344 \sin (5 (c+d x))-240738 \sin (6 (c+d x))+53248 \sin (7 (c+d x))+1083321 \cos (c+d x)-653248 \cos (2 (c+d x))-601845 \cos (3 (c+d x))+340096 \cos (4 (c+d x))-521599 \cos (5 (c+d x))+259008 \cos (6 (c+d x))+40123 \cos (7 (c+d x))-590976)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^9}}{645120 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(1935360*Log[Cos[(c + d*x)/2]] - 1935360*Log[Sin[(c + d*x)/2]] + (Csc[c + d*x]*(-590976 + 1083321*Cos[c + d*x]
- 653248*Cos[2*(c + d*x)] - 601845*Cos[3*(c + d*x)] + 340096*Cos[4*(c + d*x)] - 521599*Cos[5*(c + d*x)] + 259
008*Cos[6*(c + d*x)] + 40123*Cos[7*(c + d*x)] - 707328*Sin[c + d*x] + 1364182*Sin[2*(c + d*x)] - 1161600*Sin[3
*(c + d*x)] + 320984*Sin[4*(c + d*x)] - 329344*Sin[5*(c + d*x)] - 240738*Sin[6*(c + d*x)] + 53248*Sin[7*(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))/(645120*a^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.171, size = 308, normalized size = 1.5 \begin{align*}{\frac{1}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\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{11}{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{76}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}+{\frac{58}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-6}}-{\frac{267}{10\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+{\frac{111}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}-25\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{67}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{501}{32\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-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)^2-11/32/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)^8-76/7/d/a^3/(tan(
1/2*d*x+1/2*c)+1)^7+58/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^6-267/10/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5+111/4/d/a^3/(tan
(1/2*d*x+1/2*c)+1)^4-25/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3+67/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2-501/32/d/a^3/(tan(1
/2*d*x+1/2*c)+1)-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^3*ln(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.15626, size = 765, normalized size = 3.82 \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)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/630*((8786*sin(d*x + c)/(cos(d*x + c) + 1) + 35076*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 43062*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 - 41753*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 152172*sin(d*x + c)^5/(cos(d*x + c) + 1)
^5 - 99072*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 93324*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 157689*sin(d*x +
c)^8/(cos(d*x + c) + 1)^8 + 44730*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 50820*sin(d*x + c)^10/(cos(d*x + c) +
1)^10 - 42210*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 10395*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 315)/(a^3*
sin(d*x + c)/(cos(d*x + c) + 1) + 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 12*a^3*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 2*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 27*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*a^3*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 36*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 27*a^3*sin(d*x + c)^9/(cos(d
*x + c) + 1)^9 - 2*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 12*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 -
6*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - a^3*sin(d*x + c)^13/(cos(d*x + c) + 1)^13) + 1890*log(sin(d*x +
c)/(cos(d*x + c) + 1))/a^3 - 315*sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.87536, size = 794, normalized size = 3.97 \begin{align*} \frac{8094 \, \cos \left (d x + c\right )^{6} - 9484 \, \cos \left (d x + c\right )^{4} + 620 \, \cos \left (d x + c\right )^{2} + 945 \,{\left (\cos \left (d x + c\right )^{7} - 5 \, \cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{3} -{\left (3 \, \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 ) - 945 \,{\left (\cos \left (d x + c\right )^{7} - 5 \, \cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{3} -{\left (3 \, \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 (1664 \, \cos \left (d x + c\right )^{6} - 4653 \, \cos \left (d x + c\right )^{4} + 285 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) + 140}{630 \,{\left (a^{3} d \cos \left (d x + c\right )^{7} - 5 \, a^{3} d \cos \left (d x + c\right )^{5} + 4 \, a^{3} d \cos \left (d x + c\right )^{3} -{\left (3 \, 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)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/630*(8094*cos(d*x + c)^6 - 9484*cos(d*x + c)^4 + 620*cos(d*x + c)^2 + 945*(cos(d*x + c)^7 - 5*cos(d*x + c)^5
+ 4*cos(d*x + c)^3 - (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) - 945*(c
os(d*x + c)^7 - 5*cos(d*x + c)^5 + 4*cos(d*x + c)^3 - (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*(1664*cos(d*x + c)^6 - 4653*cos(d*x + c)^4 + 285*cos(d*x + c)^2 + 35)*sin(d*x + c
) + 140)/(a^3*d*cos(d*x + c)^7 - 5*a^3*d*cos(d*x + c)^5 + 4*a^3*d*cos(d*x + c)^3 - (3*a^3*d*cos(d*x + c)^5 - 4
*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.3084, size = 311, normalized size = 1.56 \begin{align*} -\frac{\frac{30240 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{5040 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{5040 \,{\left (6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{105 \,{\left (33 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 60 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 31\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{157815 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 1093680 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3488940 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 6524280 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 7788186 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6052704 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2995596 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 864504 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 113591}{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)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/10080*(30240*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5040*tan(1/2*d*x + 1/2*c)/a^3 - 5040*(6*tan(1/2*d*x + 1/2
*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)) + 105*(33*tan(1/2*d*x + 1/2*c)^2 - 60*tan(1/2*d*x + 1/2*c) + 31)/(a^3*(tan
(1/2*d*x + 1/2*c) - 1)^3) + (157815*tan(1/2*d*x + 1/2*c)^8 + 1093680*tan(1/2*d*x + 1/2*c)^7 + 3488940*tan(1/2*
d*x + 1/2*c)^6 + 6524280*tan(1/2*d*x + 1/2*c)^5 + 7788186*tan(1/2*d*x + 1/2*c)^4 + 6052704*tan(1/2*d*x + 1/2*c
)^3 + 2995596*tan(1/2*d*x + 1/2*c)^2 + 864504*tan(1/2*d*x + 1/2*c) + 113591)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9
))/d