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

Optimal. Leaf size=36 \[ \frac{a \tan (c+d x)}{d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0742867, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2838, 2622, 321, 207, 3767, 8} \[ \frac{a \tan (c+d x)}{d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0322275, size = 56, normalized size = 1.56 \[ \frac{a \tan (c+d x)}{d}+\frac{a \sec (c+d x)}{d}+\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.066, size = 47, normalized size = 1.3 \begin{align*}{\frac{a\tan \left ( dx+c \right ) }{d}}+{\frac{a}{d\cos \left ( dx+c \right ) }}+{\frac{a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*tan(d*x+c)/d+1/d*a/cos(d*x+c)+1/d*a*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.06426, size = 65, normalized size = 1.81 \begin{align*} \frac{a{\left (\frac{2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 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)*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.08253, size = 302, normalized size = 8.39 \begin{align*} \frac{2 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, a}{2 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*sec(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.26279, size = 46, normalized size = 1.28 \begin{align*} \frac{a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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