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

Optimal. Leaf size=158 \[ -\frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{2 \tan ^3(c+d x)}{a^2 d}-\frac{6 \tan (c+d x)}{a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{9 \sec ^5(c+d x)}{10 a^2 d}+\frac{3 \sec ^3(c+d x)}{2 a^2 d}+\frac{9 \sec (c+d x)}{2 a^2 d}-\frac{9 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^5(c+d x)}{2 a^2 d} \]

[Out]

(-9*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (9*Sec[c + d*x])/(2*a^2*d) + (3*Sec[c + d*x]
^3)/(2*a^2*d) + (9*Sec[c + d*x]^5)/(10*a^2*d) - (Csc[c + d*x]^2*Sec[c + d*x]^5)/(2*a^2*d) - (6*Tan[c + d*x])/(
a^2*d) - (2*Tan[c + d*x]^3)/(a^2*d) - (2*Tan[c + d*x]^5)/(5*a^2*d)

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Rubi [A]  time = 0.346225, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2622, 302, 207, 2620, 270, 288} \[ -\frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{2 \tan ^3(c+d x)}{a^2 d}-\frac{6 \tan (c+d x)}{a^2 d}+\frac{2 \cot (c+d x)}{a^2 d}+\frac{9 \sec ^5(c+d x)}{10 a^2 d}+\frac{3 \sec ^3(c+d x)}{2 a^2 d}+\frac{9 \sec (c+d x)}{2 a^2 d}-\frac{9 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^5(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (9*Sec[c + d*x])/(2*a^2*d) + (3*Sec[c + d*x]
^3)/(2*a^2*d) + (9*Sec[c + d*x]^5)/(10*a^2*d) - (Csc[c + d*x]^2*Sec[c + d*x]^5)/(2*a^2*d) - (6*Tan[c + d*x])/(
a^2*d) - (2*Tan[c + d*x]^3)/(a^2*d) - (2*Tan[c + d*x]^5)/(5*a^2*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 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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

Mathematica [B]  time = 0.699453, size = 328, normalized size = 2.08 \[ -\frac{\csc ^2(c+d x) \sec (c+d x) \left (-432 \sin (c+d x)+744 \sin (2 (c+d x))-176 \sin (3 (c+d x))-372 \sin (4 (c+d x))+128 \sin (5 (c+d x))+176 \cos (2 (c+d x))-651 \cos (3 (c+d x))+332 \cos (4 (c+d x))+93 \cos (5 (c+d x))-720 \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+360 \sin (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-630 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+90 \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+18 \cos (c+d x) \left (-30 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+30 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+31\right )+630 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-90 \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+720 \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-360 \sin (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-348\right )}{320 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^2*Sec[c + d*x]*(-348 + 176*Cos[2*(c + d*x)] - 651*Cos[3*(c + d*x)] + 332*Cos[4*(c + d*x)] + 93*
Cos[5*(c + d*x)] - 630*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 90*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 18
*Cos[c + d*x]*(31 + 30*Log[Cos[(c + d*x)/2]] - 30*Log[Sin[(c + d*x)/2]]) + 630*Cos[3*(c + d*x)]*Log[Sin[(c + d
*x)/2]] - 90*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 432*Sin[c + d*x] + 744*Sin[2*(c + d*x)] + 720*Log[Cos[(c
+ d*x)/2]]*Sin[2*(c + d*x)] - 720*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 176*Sin[3*(c + d*x)] - 372*Sin[4*(
c + d*x)] - 360*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 360*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] + 128*Sin[
5*(c + d*x)]))/(320*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.141, size = 219, normalized size = 1.4 \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{4}{5\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-2\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+5\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-{\frac{11}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{49}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{9}{2\,d{a}^{2}}\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)^3*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x)

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)-1/4/d/a^2/(tan(1/2*d*x+1/2*c)-1)+4/5/d/a^2/(tan(1/2*
d*x+1/2*c)+1)^5-2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^4+5/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3-11/2/d/a^2/(tan(1/2*d*x+1/2*
c)+1)^2+49/4/d/a^2/(tan(1/2*d*x+1/2*c)+1)-1/8/d/a^2/tan(1/2*d*x+1/2*c)^2+1/d/a^2/tan(1/2*d*x+1/2*c)+9/2/d/a^2*
ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.05704, size = 478, normalized size = 3.03 \begin{align*} \frac{\frac{\frac{20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{567 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1448 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{985 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{820 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{1355 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{520 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 5}{\frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{5 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{5 \,{\left (\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{2}} + \frac{180 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/40*((20*sin(d*x + c)/(cos(d*x + c) + 1) + 567*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1448*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 985*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 820*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1355*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 520*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 5)/(a^2*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 4*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 5*a^2*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^2*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8) - 5*(8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 180*log(sin(d*x
+ c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.14918, size = 699, normalized size = 4.42 \begin{align*} -\frac{166 \, \cos \left (d x + c\right )^{4} - 144 \, \cos \left (d x + c\right )^{2} + 45 \,{\left (\cos \left (d x + c\right )^{5} - 3 \, \cos \left (d x + c\right )^{3} - 2 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\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 ) - 45 \,{\left (\cos \left (d x + c\right )^{5} - 3 \, \cos \left (d x + c\right )^{3} - 2 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\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 (32 \, \cos \left (d x + c\right )^{4} - 35 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 12}{20 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} - 3 \, a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right ) - 2 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )\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)^3*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20*(166*cos(d*x + c)^4 - 144*cos(d*x + c)^2 + 45*(cos(d*x + c)^5 - 3*cos(d*x + c)^3 - 2*(cos(d*x + c)^3 - c
os(d*x + c))*sin(d*x + c) + 2*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 45*(cos(d*x + c)^5 - 3*cos(d*x + c)^
3 - 2*(cos(d*x + c)^3 - cos(d*x + c))*sin(d*x + c) + 2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 4*(32*cos(
d*x + c)^4 - 35*cos(d*x + c)^2 - 2)*sin(d*x + c) - 12)/(a^2*d*cos(d*x + c)^5 - 3*a^2*d*cos(d*x + c)^3 + 2*a^2*
d*cos(d*x + c) - 2*(a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c))*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)**3*sec(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33743, size = 252, normalized size = 1.59 \begin{align*} \frac{\frac{180 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{5 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{4}} - \frac{10}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} - \frac{5 \,{\left (54 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{2 \,{\left (245 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 870 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 810 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 211\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/40*(180*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 5*(a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4
- 10/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) - 5*(54*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1
/2*d*x + 1/2*c)^2) + 2*(245*tan(1/2*d*x + 1/2*c)^4 + 870*tan(1/2*d*x + 1/2*c)^3 + 1240*tan(1/2*d*x + 1/2*c)^2
+ 810*tan(1/2*d*x + 1/2*c) + 211)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d