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

Optimal. Leaf size=53 \[ \frac{2 \tan (c+d x)}{a d}-\frac{\tanh ^{-1}(\sin (c+d x))}{a d}-\frac{\tan (c+d x)}{d (a \cos (c+d x)+a)} \]

[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]))

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Rubi [A]  time = 0.0758197, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2768, 2748, 3767, 8, 3770} \[ \frac{2 \tan (c+d x)}{a d}-\frac{\tanh ^{-1}(\sin (c+d x))}{a d}-\frac{\tan (c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[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 2768

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] && NeQ[
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 2748

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 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]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx &=-\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{\tanh ^{-1}(\sin (c+d x))}{a d}-\frac{\tan (c+d x)}{d (a+a \cos (c+d x))}-\frac{2 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=-\frac{\tanh ^{-1}(\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]  time = 0.600276, size = 188, normalized size = 3.55 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+\cos \left (\frac{1}{2} (c+d x)\right ) \left (\frac{\sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[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]))

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Maple [A]  time = 0.057, size = 99, normalized size = 1.9 \begin{align*}{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16716, size = 161, normalized size = 3.04 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.6174, size = 266, normalized size = 5.02 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{2}{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.37635, size = 113, normalized size = 2.13 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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