\(\int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [755]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 36 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \]

[Out]

-a*arctanh(cos(d*x+c))/d+a*sec(d*x+c)/d+a*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2917, 2702, 327, 213, 3852, 8} \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

[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 8

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

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 2702

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 2917

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 3852

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]

Rubi steps \begin{align*} \text {integral}& = a \int \sec ^2(c+d x) \, dx+a \int \csc (c+d x) \sec ^2(c+d x) \, dx \\ & = -\frac {a \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}+\frac {a \text {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 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.56 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \]

[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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14

method result size
derivativedivides \(\frac {a \tan \left (d x +c \right )+a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) \(41\)
default \(\frac {a \tan \left (d x +c \right )+a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) \(41\)
parallelrisch \(\frac {\left (-2+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )\right ) a}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) \(43\)
risch \(\frac {2 a}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) \(57\)
norman \(\frac {-\frac {2 a}{d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) \(104\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (36) = 72\).

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.00 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, a \sin \left (d x + c\right ) + 2 \, a}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

[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)

Sympy [F]

\[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, a \tan \left (d x + c\right )}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 10.45 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

int((a + a*sin(c + d*x))/(cos(c + d*x)^2*sin(c + d*x)),x)

[Out]

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