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

Optimal. Leaf size=93 \[ \frac{\tan ^3(c+d x)}{3 a d}+\frac{2 \tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d}-\frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.194132, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2839, 2620, 270, 2622, 302, 207} \[ \frac{\tan ^3(c+d x)}{3 a d}+\frac{2 \tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d}-\frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 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 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])

Rubi steps

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

Mathematica [B]  time = 0.586477, size = 245, normalized size = 2.63 \[ -\frac{\csc ^3(c+d x) \left (4 \sin (c+d x)-16 \sin (2 (c+d x))+8 \sin (3 (c+d x))+10 \cos (2 (c+d x))+8 \cos (3 (c+d x))+6 \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-3 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-8\right )-6 \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2\right )}{3 a d (\sin (c+d x)+1) \left (\csc \left (\frac{1}{2} (c+d x)\right )-\sec \left (\frac{1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^3*(2 + 10*Cos[2*(c + d*x)] + 8*Cos[3*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*
Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-8 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]])
+ 4*Sin[c + d*x] - 16*Sin[2*(c + d*x)] - 6*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 6*Log[Sin[(c + d*x)/2]]*S
in[2*(c + d*x)] + 8*Sin[3*(c + d*x)]))/(3*a*d*(Csc[(c + d*x)/2] - Sec[(c + d*x)/2])*(Csc[(c + d*x)/2] + Sec[(c
+ d*x)/2])*(1 + Sin[c + d*x]))

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Maple [A]  time = 0.092, size = 139, normalized size = 1.5 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{2}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{7}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\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)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

1/2/d/a*tan(1/2*d*x+1/2*c)-1/2/d/a/(tan(1/2*d*x+1/2*c)-1)-2/3/d/a/(tan(1/2*d*x+1/2*c)+1)^3+1/d/a/(tan(1/2*d*x+
1/2*c)+1)^2-7/2/a/d/(tan(1/2*d*x+1/2*c)+1)-1/2/d/a/tan(1/2*d*x+1/2*c)-1/d/a*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.0261, size = 290, normalized size = 3.12 \begin{align*} -\frac{\frac{\frac{22 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 3}{\frac{a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{3 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, 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)),x, algorithm="maxima")

[Out]

-1/6*((22*sin(d*x + c)/(cos(d*x + c) + 1) + 8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 30*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 27*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 2*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) +
6*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.15789, size = 440, normalized size = 4.73 \begin{align*} \frac{10 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (8 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4}{6 \,{\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a 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)^2*sec(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 - cos(d*x + c)*sin(d*x + c) - cos(d*x + c))*log(1/2*cos(d*x + c) +
1/2) - 3*(cos(d*x + c)^3 - cos(d*x + c)*sin(d*x + c) - cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(8*cos(d
*x + c)^2 - 1)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^3 - a*d*cos(d*x + c)*sin(d*x + c) - a*d*cos(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)),x)

[Out]

Timed out

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Giac [A]  time = 1.28112, size = 180, normalized size = 1.94 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{3 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\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} + \frac{21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 19}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, 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)),x, algorithm="giac")

[Out]

-1/6*(6*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*tan(1/2*d*x + 1/2*c)/a - 3*(tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*
x + 1/2*c) + 1)/((tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a) + (21*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*
d*x + 1/2*c) + 19)/(a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d