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

Optimal. Leaf size=130 \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{5 \tan ^3(c+d x)}{3 a^2 d}+\frac{4 \tan (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{2 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Sec[c + d*x])/(a^2*d) - (2*Sec[c + d*x]^3)/(3*a^
2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + (4*Tan[c + d*x])/(a^2*d) + (5*Tan[c + d*x]^3)/(3*a^2*d) + (2*Tan[c + d*x
]^5)/(5*a^2*d)

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Rubi [A]  time = 0.310337, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2620, 270} \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{5 \tan ^3(c+d x)}{3 a^2 d}+\frac{4 \tan (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{2 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Sec[c + d*x])/(a^2*d) - (2*Sec[c + d*x]^3)/(3*a^
2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + (4*Tan[c + d*x])/(a^2*d) + (5*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 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]

Rubi steps

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

Mathematica [B]  time = 0.731926, size = 289, normalized size = 2.22 \[ -\frac{\csc ^3(c+d x) \left (58 \sin (c+d x)-168 \sin (2 (c+d x))+82 \sin (3 (c+d x))+28 \sin (4 (c+d x))+48 \cos (2 (c+d x))+112 \cos (3 (c+d x))-28 \cos (4 (c+d x))+90 \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 \sin (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+60 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 \cos (c+d x) \left (-15 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+28\right )-60 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-90 \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 \sin (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+40\right )}{15 a^2 d (\sin (c+d x)+1)^2 \left (\csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^3*(40 + 48*Cos[2*(c + d*x)] + 112*Cos[3*(c + d*x)] - 28*Cos[4*(c + d*x)] + 60*Cos[3*(c + d*x)]*
Log[Cos[(c + d*x)/2]] - 4*Cos[c + d*x]*(28 + 15*Log[Cos[(c + d*x)/2]] - 15*Log[Sin[(c + d*x)/2]]) - 60*Cos[3*(
c + d*x)]*Log[Sin[(c + d*x)/2]] + 58*Sin[c + d*x] - 168*Sin[2*(c + d*x)] - 90*Log[Cos[(c + d*x)/2]]*Sin[2*(c +
d*x)] + 90*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 82*Sin[3*(c + d*x)] + 28*Sin[4*(c + d*x)] + 15*Log[Cos[(c
+ d*x)/2]]*Sin[4*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]))/(15*a^2*d*(Csc[(c + d*x)/2]^2 - Sec
[(c + d*x)/2]^2)*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.13, size = 182, normalized size = 1.4 \begin{align*}{\frac{1}{2\,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}}}-{\frac{13}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{9}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{31}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/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)^4-13/3/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3+9/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2-31/4/d/a^2/(tan(1/2
*d*x+1/2*c)+1)-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^2*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.07389, size = 417, normalized size = 3.21 \begin{align*} -\frac{\frac{\frac{244 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{571 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{320 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{475 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{660 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{255 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 15}{\frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{4 \, 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 )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{5 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{15 \, \sin \left (d x + c\right )}{a^{2}{\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((244*sin(d*x + c)/(cos(d*x + c) + 1) + 571*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 320*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 - 475*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 660*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 255*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 5*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 4*a^2*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 - a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) + 60*log(sin(d*x + c)/(cos(d*x + c
) + 1))/a^2 - 15*sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.14054, size = 591, normalized size = 4.55 \begin{align*} -\frac{56 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (2 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{3} - 2 \, \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 ) + 15 \,{\left (2 \, \cos \left (d x + c\right )^{3} +{\left (\cos \left (d x + c\right )^{3} - 2 \, \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 ) - 2 \,{\left (41 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 9}{15 \,{\left (2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) +{\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, 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)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/15*(56*cos(d*x + c)^4 - 80*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^3 + (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*(2*cos(d*x + c)^3 + (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) - 2*(41*cos(d*x + c)^2 - 3)*sin(d*x + c) + 9)/(2
*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) + (a^2*d*cos(d*x + c)^3 - 2*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)**2*sec(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27618, size = 217, normalized size = 1.67 \begin{align*} -\frac{\frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{15 \,{\left (4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} a^{2}} + \frac{465 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1590 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1450 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 383}{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)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 30*tan(1/2*d*x + 1/2*c)/a^2 - 15*(4*tan(1/2*d*x + 1/2*c)^2 - 7
*tan(1/2*d*x + 1/2*c) + 2)/((tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a^2) + (465*tan(1/2*d*x + 1/2*c)^4
+ 1590*tan(1/2*d*x + 1/2*c)^3 + 2240*tan(1/2*d*x + 1/2*c)^2 + 1450*tan(1/2*d*x + 1/2*c) + 383)/(a^2*(tan(1/2*
d*x + 1/2*c) + 1)^5))/d