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

Optimal. Leaf size=115 \[ -\frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{4 \tan ^3(c+d x)}{3 a^2 d}-\frac{2 \tan (c+d x)}{a^2 d}+\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}+\frac{\sec (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.266254, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2606, 30} \[ -\frac{2 \tan ^5(c+d x)}{5 a^2 d}-\frac{4 \tan ^3(c+d x)}{3 a^2 d}-\frac{2 \tan (c+d x)}{a^2 d}+\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}+\frac{\sec (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[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 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]

Rubi steps

\begin{align*} \int \frac{\csc (c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\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{\operatorname{Subst}\left (\int x^4 \, 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 \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{\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{\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{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac{\tanh ^{-1}(\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]  time = 0.478946, size = 196, normalized size = 1.7 \[ \frac{\sec (c+d x) \left (160 \sin (c+d x)-316 \sin (2 (c+d x))+64 \sin (3 (c+d x))+136 \cos (2 (c+d x))+79 \cos (3 (c+d x))+240 \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-5 \cos (c+d x) \left (-60 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+79\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-240 \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+280\right )}{240 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[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]  time = 0.105, size = 145, normalized size = 1.3 \begin{align*} -{\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}}}+{\frac{11}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{7}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{17}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{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)*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.02998, size = 338, normalized size = 2.94 \begin{align*} \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} \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))^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

________________________________________________________________________________________

Fricas [A]  time = 1.16387, size = 471, normalized size = 4.1 \begin{align*} -\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 )}} \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))^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(-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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.28113, size = 147, normalized size = 1.28 \begin{align*} \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} \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))^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