3.606 \(\int \frac{(1-\cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 b \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}+\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{\sin (c+d x)}{a d (a+b \cos (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.170311, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3056, 12, 2747, 3770, 2659, 205} \[ -\frac{2 b \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}+\frac{\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{\sin (c+d x)}{a d (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^2,x]

[Out]

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3770

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.18973, size = 123, normalized size = 1.31 \[ \frac{\frac{2 b \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-\frac{a \sin (c+d x)}{a+b \cos (c+d x)}-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.047, size = 137, normalized size = 1.5 \begin{align*} -2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-2\,{\frac{b}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d{a}^{2}}\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((1-cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x)

[Out]

-2/d/a*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)-2/d/a^2*b/((a+b)*(a-b))^(1/2)*ar
ctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^2*ln(tan(1/2*d*x+1/2
*c)+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94589, size = 1056, normalized size = 11.23 \begin{align*} \left [-\frac{{\left (b^{2} \cos \left (d x + c\right ) + a b\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} - a^{3} b^{2}\right )} d\right )}}, -\frac{2 \,{\left (b^{2} \cos \left (d x + c\right ) + a b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b - a^{2} b^{3}\right )} d \cos \left (d x + c\right ) +{\left (a^{5} - a^{3} b^{2}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*((b^2*cos(d*x + c) + a*b)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sq
rt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2
)) - (a^3 - a*b^2 + (a^2*b - b^3)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^3 - a*b^2 + (a^2*b - b^3)*cos(d*x +
c))*log(-sin(d*x + c) + 1) + 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4*b - a^2*b^3)*d*cos(d*x + c) + (a^5 - a^3*b^2
)*d), -1/2*(2*(b^2*cos(d*x + c) + a*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x +
c))) - (a^3 - a*b^2 + (a^2*b - b^3)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^3 - a*b^2 + (a^2*b - b^3)*cos(d*
x + c))*log(-sin(d*x + c) + 1) + 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4*b - a^2*b^3)*d*cos(d*x + c) + (a^5 - a^3*
b^2)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.6128, size = 223, normalized size = 2.37 \begin{align*} -\frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{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 )}{\sqrt{a^{2} - b^{2}}}\right )\right )} b}{\sqrt{a^{2} - b^{2}} a^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )} a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c)
)/sqrt(a^2 - b^2)))*b/(sqrt(a^2 - b^2)*a^2) - log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + log(abs(tan(1/2*d*x + 1
/2*c) - 1))/a^2 + 2*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)*a))/d