\(\int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{a d}+\frac {2 \tan (c+d x)}{a d}-\frac {\tan (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

-arctanh(sin(d*x+c))/a/d+2*tan(d*x+c)/a/d-tan(d*x+c)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2847, 2827, 3852, 8, 3855} \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{a d}+\frac {2 \tan (c+d x)}{a d}-\frac {\tan (c+d x)}{d (a \cos (c+d x)+a)} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2847

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b
^2)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x]))), x] + Dist[d/(a*(b*c -
a*d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne
Q[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

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]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(188\) vs. \(2(53)=106\).

Time = 0.78 (sec) , antiderivative size = 188, normalized size of antiderivative = 3.55 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[
Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2]
- Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))))/(a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(74\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(74\)
parallelrisch	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2 \cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \cos \left (d x +c \right )}\)	\(82\)
norman	\(\frac {\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}}{\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}\)	\(93\)
risch	\(\frac {2 i \left ({\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+2\right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.83 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(sec(c + d*x)**2/(cos(c + d*x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (53) = 106\).

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.25 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.58 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.80 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

[In]

int(1/(cos(c + d*x)^2*(a + a*cos(c + d*x))),x)

[Out]

(2*tan(c/2 + (d*x)/2))/(d*(a - a*tan(c/2 + (d*x)/2)^2)) - (2*atanh(tan(c/2 + (d*x)/2)))/(a*d) + tan(c/2 + (d*x
)/2)/(a*d)