3.33 \(\int \csc ^2(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ \frac{a^2 \tan (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{2 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Cot[c + d*x])/d - (2*a^2*Csc[c + d*x])/d + (a^2*Tan[c + d*x])/d

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Rubi [A]  time = 0.246487, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2873, 3767, 8, 2621, 321, 207, 2620, 14} \[ \frac{a^2 \tan (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{2 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Cot[c + d*x])/d - (2*a^2*Csc[c + d*x])/d + (a^2*Tan[c + d*x])/d

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [B]  time = 6.15428, size = 401, normalized size = 7.04 \[ \frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{4 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{4 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}-\frac{\cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{2 d}+\frac{\cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{2 d}+\frac{\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \csc \left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(2*
d) + (Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)
/(2*d) + (Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])
/(2*d) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(4*d*(Cos[c/2] - Sin[c/2])*
(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(
d*x)/2])/(4*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.045, size = 77, normalized size = 1.4 \begin{align*} -3\,{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{\frac{{a}^{2}}{d\sin \left ( dx+c \right ) }}+2\,{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a^2*cot(d*x+c)/d-2/d*a^2/sin(d*x+c)+2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2/sin(d*x+c)/cos(d*x+c)

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Maxima [A]  time = 1.0009, size = 100, normalized size = 1.75 \begin{align*} -\frac{a^{2}{\left (\frac{2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + a^{2}{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac{a^{2}}{\tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*(2/sin(d*x + c) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + a^2*(1/tan(d*x + c) - tan(d*x + c)) +
a^2/tan(d*x + c))/d

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Fricas [A]  time = 1.74643, size = 257, normalized size = 4.51 \begin{align*} \frac{a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}{d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*cos(d*x + c)*log(sin(d*x + c) + 1)*sin(d*x + c) - a^2*cos(d*x + c)*log(-sin(d*x + c) + 1)*sin(d*x + c) -
3*a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)/(d*cos(d*x + c)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \csc ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*csc(c + d*x)**2*sec(c + d*x), x) + Integral(csc(c + d*x)**2*sec(c + d*x)**2, x) + Integral(cs
c(c + d*x)**2, x))

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Giac [A]  time = 1.43561, size = 122, normalized size = 2.14 \begin{align*} \frac{2 \,{\left (a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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