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

Optimal. Leaf size=37 \[ -\frac{a \cot (c+d x)}{d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0959633, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3872, 2838, 2621, 321, 207, 3767, 8} \[ -\frac{a \cot (c+d x)}{d}-\frac{b \csc (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (b*Csc[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 2838

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

Rubi steps

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

Mathematica [C]  time = 0.0269862, size = 41, normalized size = 1.11 \[ -\frac{b \csc (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\sin ^2(c+d x)\right )}{d}-\frac{a \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cot[c + d*x])/d) - (b*Csc[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[c + d*x]^2])/d

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Maple [A]  time = 0.034, size = 47, normalized size = 1.3 \begin{align*} -{\frac{a\cot \left ( dx+c \right ) }{d}}-{\frac{b}{d\sin \left ( dx+c \right ) }}+{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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)),x)

[Out]

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

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Maxima [A]  time = 0.952336, size = 68, normalized size = 1.84 \begin{align*} -\frac{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 )} + \frac{2 \, a}{\tan \left (d x + 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73875, size = 170, normalized size = 4.59 \begin{align*} \frac{b \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - b \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 2 \, b}{2 \, d \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)),x, algorithm="fricas")

[Out]

1/2*(b*log(sin(d*x + c) + 1)*sin(d*x + c) - b*log(-sin(d*x + c) + 1)*sin(d*x + c) - 2*a*cos(d*x + c) - 2*b)/(d
*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 ) \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)),x)

[Out]

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

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Giac [B]  time = 1.28673, size = 104, normalized size = 2.81 \begin{align*} \frac{2 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{a + b}{\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)),x, algorithm="giac")

[Out]

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