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

Optimal. Leaf size=151 \[ -\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{11 \tan ^5(c+d x)}{5 a^3 d}-\frac{10 \tan ^3(c+d x)}{3 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[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)
+ (4*Sec[c + d*x]^7)/(7*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (10*Tan[c + d*x]^3)/(3*a^3*d) - (11*Tan[c + d*x]^5
)/(5*a^3*d) - (4*Tan[c + d*x]^7)/(7*a^3*d)

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Rubi [A]  time = 0.293391, antiderivative size = 151, 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} \[ -\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{11 \tan ^5(c+d x)}{5 a^3 d}-\frac{10 \tan ^3(c+d x)}{3 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d}+\frac{\sec (c+d x)}{a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]*Sec[c + d*x]^2)/(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)
+ (4*Sec[c + d*x]^7)/(7*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (10*Tan[c + d*x]^3)/(3*a^3*d) - (11*Tan[c + d*x]^5
)/(5*a^3*d) - (4*Tan[c + d*x]^7)/(7*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 ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc (c+d x) \sec ^8(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-3 a^3 \sec ^8(c+d x)+a^3 \csc (c+d x) \sec ^8(c+d x)+3 a^3 \sec ^7(c+d x) \tan (c+d x)-a^3 \sec ^6(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \csc (c+d x) \sec ^8(c+d x) \, dx}{a^3}-\frac{\int \sec ^6(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^8(c+d x) \, dx}{a^3}+\frac{3 \int \sec ^7(c+d x) \tan (c+d x) \, dx}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{-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 )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int x^6 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (c+d x)\right )}{a^3 d}\\ &=\frac{3 \sec ^7(c+d x)}{7 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{3 \tan ^3(c+d x)}{a^3 d}-\frac{9 \tan ^5(c+d x)}{5 a^3 d}-\frac{3 \tan ^7(c+d x)}{7 a^3 d}-\frac{\operatorname{Subst}\left (\int \left (x^2+2 x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{\operatorname{Subst}\left (\int \left (1+x^2+x^4+x^6+\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{4 \sec ^7(c+d x)}{7 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{10 \tan ^3(c+d x)}{3 a^3 d}-\frac{11 \tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 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{4 \sec ^7(c+d x)}{7 a^3 d}-\frac{3 \tan (c+d x)}{a^3 d}-\frac{10 \tan ^3(c+d x)}{3 a^3 d}-\frac{11 \tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 a^3 d}\\ \end{align*}

Mathematica [B]  time = 0.407884, size = 341, normalized size = 2.26 \[ \frac{\frac{105 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-2281 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5+353 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4-706 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3+162 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-324 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{120 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-840 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6+840 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6+60}{840 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60 - (120*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 324*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]) + 162*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 706*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^3 + 353*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 2281*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^5 - 840*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6 + 840*Log[Sin[(c + d*x)/2
]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6 + (105*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(C
os[(c + d*x)/2] - Sin[(c + d*x)/2]))/(840*d*(a + a*Sin[c + d*x])^3)

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Maple [A]  time = 0.133, size = 187, normalized size = 1.2 \begin{align*} -{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{8}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}+{\frac{42}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-11\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{67}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{31}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{49}{8\,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)^2/(a+a*sin(d*x+c))^3,x)

[Out]

-1/8/d/a^3/(tan(1/2*d*x+1/2*c)-1)+8/7/d/a^3/(tan(1/2*d*x+1/2*c)+1)^7-4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^6+42/5/d/a
^3/(tan(1/2*d*x+1/2*c)+1)^5-11/d/a^3/(tan(1/2*d*x+1/2*c)+1)^4+67/6/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3-31/4/d/a^3/(
tan(1/2*d*x+1/2*c)+1)^2+49/8/d/a^3/(tan(1/2*d*x+1/2*c)+1)+1/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.06973, size = 454, normalized size = 3.01 \begin{align*} \frac{\frac{2 \,{\left (\frac{1011 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{1939 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{1379 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{525 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{1715 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{1155 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{315 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 221\right )}}{a^{3} + \frac{6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{105 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(2*(1011*sin(d*x + c)/(cos(d*x + c) + 1) + 1939*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1379*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 525*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1715*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
1155*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 315*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 221)/(a^3 + 6*a^3*sin(d*x
+ c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*a^3*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*a^3*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 105*log(sin(d*x + c)/(cos(d*x + c) +
1))/a^3)/d

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Fricas [A]  time = 1.15971, size = 602, normalized size = 3.99 \begin{align*} \frac{272 \, \cos \left (d x + c\right )^{4} - 594 \, \cos \left (d x + c\right )^{2} - 105 \,{\left (3 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 105 \,{\left (3 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 6 \,{\left (101 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) - 120}{210 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) +{\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} 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)*sec(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/210*(272*cos(d*x + c)^4 - 594*cos(d*x + c)^2 - 105*(3*cos(d*x + c)^3 + (cos(d*x + c)^3 - 4*cos(d*x + c))*sin
(d*x + c) - 4*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 105*(3*cos(d*x + c)^3 + (cos(d*x + c)^3 - 4*cos(d*x
+ c))*sin(d*x + c) - 4*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 6*(101*cos(d*x + c)^2 + 15)*sin(d*x + c) -
120)/(3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*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)*sec(d*x+c)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.24654, size = 182, normalized size = 1.21 \begin{align*} \frac{\frac{840 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{105}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} + \frac{5145 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 24360 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 54005 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 66080 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 47691 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 18872 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3431}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 105/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) + (5145*tan(1/2*d*x + 1/2
*c)^6 + 24360*tan(1/2*d*x + 1/2*c)^5 + 54005*tan(1/2*d*x + 1/2*c)^4 + 66080*tan(1/2*d*x + 1/2*c)^3 + 47691*tan
(1/2*d*x + 1/2*c)^2 + 18872*tan(1/2*d*x + 1/2*c) + 3431)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d