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

Optimal. Leaf size=59 \[ -\frac{\left (a^2+b^2\right ) \cot (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.414007, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3872, 2911, 2621, 321, 207, 14} \[ -\frac{\left (a^2+b^2\right ) \cot (c+d x)}{d}-\frac{2 a b \csc (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*ArcTanh[Sin[c + d*x]])/d - ((a^2 + b^2)*Cot[c + d*x])/d - (2*a*b*Csc[c + d*x])/d + (b^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 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

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

Mathematica [B]  time = 0.46498, size = 138, normalized size = 2.34 \[ -\frac{\csc ^3\left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\left (a^2+2 b^2\right ) \cos (2 (c+d x))+4 a b \cos (c+d x)+a \left (a+2 b \sin (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )\right )}{4 d \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]^3*Sec[(c + d*x)/2]*(4*a*b*Cos[c + d*x] + (a^2 + 2*b^2)*Cos[2*(c + d*x)] + a*(a + 2*b*(Log[C
os[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*Sin[2*(c + d*x)])))/(4*d*(-1 +
Cot[(c + d*x)/2]^2))

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Maple [A]  time = 0.036, size = 89, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{\frac{ab}{d\sin \left ( dx+c \right ) }}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-2\,{\frac{{b}^{2}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02287, size = 99, normalized size = 1.68 \begin{align*} -\frac{a b{\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 )} + b^{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+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7732, size = 267, normalized size = 4.53 \begin{align*} \frac{a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{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+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

(a*b*cos(d*x + c)*log(sin(d*x + c) + 1)*sin(d*x + c) - a*b*cos(d*x + c)*log(-sin(d*x + c) + 1)*sin(d*x + c) -
2*a*b*cos(d*x + c) - (a^2 + 2*b^2)*cos(d*x + c)^2 + b^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*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{2} \csc ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.32974, size = 225, normalized size = 3.81 \begin{align*} \frac{4 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{2} - 2 \, a b - b^{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 )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + a^2*tan(1/2*d*x + 1
/2*c) - 2*a*b*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c) - (a^2*tan(1/2*d*x + 1/2*c)^2 + 2*a*b*tan(1/2*d*
x + 1/2*c)^2 + 5*b^2*tan(1/2*d*x + 1/2*c)^2 - a^2 - 2*a*b - b^2)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c
)))/d