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

Optimal. Leaf size=74 \[ \frac{2 \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{x}{2}\right )}{\sqrt{c+d}}\right )}{c^2 \sqrt{c-d} \sqrt{c+d}}-\frac{b d \tanh ^{-1}(\sin (x))}{c^2}+\frac{b \tan (x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.245693, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {4234, 3056, 3001, 3770, 2659, 205} \[ \frac{2 \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{x}{2}\right )}{\sqrt{c+d}}\right )}{c^2 \sqrt{c-d} \sqrt{c+d}}-\frac{b d \tanh ^{-1}(\sin (x))}{c^2}+\frac{b \tan (x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4234

Int[(u_)*((A_) + (C_.)*sec[(a_.) + (b_.)*(x_)]^2), x_Symbol] :> Int[(ActivateTrig[u]*(C + A*Cos[a + b*x]^2))/C
os[a + b*x]^2, x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, 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 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 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{a+b \sec ^2(x)}{c+d \cos (x)} \, dx &=\int \frac{\left (b+a \cos ^2(x)\right ) \sec ^2(x)}{c+d \cos (x)} \, dx\\ &=\frac{b \tan (x)}{c}+\frac{\int \frac{(-b d+a c \cos (x)) \sec (x)}{c+d \cos (x)} \, dx}{c}\\ &=\frac{b \tan (x)}{c}-\frac{(b d) \int \sec (x) \, dx}{c^2}+\left (a+\frac{b d^2}{c^2}\right ) \int \frac{1}{c+d \cos (x)} \, dx\\ &=-\frac{b d \tanh ^{-1}(\sin (x))}{c^2}+\frac{b \tan (x)}{c}+\left (2 \left (a+\frac{b d^2}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+d+(c-d) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=\frac{2 \left (a+\frac{b d^2}{c^2}\right ) \tan ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{x}{2}\right )}{\sqrt{c+d}}\right )}{\sqrt{c-d} \sqrt{c+d}}-\frac{b d \tanh ^{-1}(\sin (x))}{c^2}+\frac{b \tan (x)}{c}\\ \end{align*}

Mathematica [A]  time = 0.442426, size = 98, normalized size = 1.32 \[ \frac{-\frac{2 \left (a c^2+b d^2\right ) \tanh ^{-1}\left (\frac{(c-d) \tan \left (\frac{x}{2}\right )}{\sqrt{d^2-c^2}}\right )}{\sqrt{d^2-c^2}}+b c \tan (x)+b d \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.036, size = 135, normalized size = 1.8 \begin{align*} 2\,{\frac{a}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}\arctan \left ({\frac{ \left ( c-d \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }+2\,{\frac{b{d}^{2}}{{c}^{2}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}\arctan \left ({\frac{ \left ( c-d \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }-{\frac{b}{c} \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{db}{{c}^{2}}\ln \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{b}{c} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{db}{{c}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/((c+d)*(c-d))^(1/2)*arctan((c-d)*tan(1/2*x)/((c+d)*(c-d))^(1/2))*a+2/c^2/((c+d)*(c-d))^(1/2)*arctan((c-d)*ta
n(1/2*x)/((c+d)*(c-d))^(1/2))*b*d^2-b/c/(1+tan(1/2*x))-d*b/c^2*ln(1+tan(1/2*x))-b/c/(tan(1/2*x)-1)+d*b/c^2*ln(
tan(1/2*x)-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((a+b*sec(x)^2)/(c+d*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.93377, size = 768, normalized size = 10.38 \begin{align*} \left [-\frac{{\left (a c^{2} + b d^{2}\right )} \sqrt{-c^{2} + d^{2}} \cos \left (x\right ) \log \left (\frac{2 \, c d \cos \left (x\right ) +{\left (2 \, c^{2} - d^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-c^{2} + d^{2}}{\left (c \cos \left (x\right ) + d\right )} \sin \left (x\right ) - c^{2} + 2 \, d^{2}}{d^{2} \cos \left (x\right )^{2} + 2 \, c d \cos \left (x\right ) + c^{2}}\right ) +{\left (b c^{2} d - b d^{3}\right )} \cos \left (x\right ) \log \left (\sin \left (x\right ) + 1\right ) -{\left (b c^{2} d - b d^{3}\right )} \cos \left (x\right ) \log \left (-\sin \left (x\right ) + 1\right ) - 2 \,{\left (b c^{3} - b c d^{2}\right )} \sin \left (x\right )}{2 \,{\left (c^{4} - c^{2} d^{2}\right )} \cos \left (x\right )}, \frac{2 \,{\left (a c^{2} + b d^{2}\right )} \sqrt{c^{2} - d^{2}} \arctan \left (-\frac{c \cos \left (x\right ) + d}{\sqrt{c^{2} - d^{2}} \sin \left (x\right )}\right ) \cos \left (x\right ) -{\left (b c^{2} d - b d^{3}\right )} \cos \left (x\right ) \log \left (\sin \left (x\right ) + 1\right ) +{\left (b c^{2} d - b d^{3}\right )} \cos \left (x\right ) \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (b c^{3} - b c d^{2}\right )} \sin \left (x\right )}{2 \,{\left (c^{4} - c^{2} d^{2}\right )} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18005, size = 169, normalized size = 2.28 \begin{align*} -\frac{b d \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{c^{2}} + \frac{b d \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{c^{2}} - \frac{2 \, b \tan \left (\frac{1}{2} \, x\right )}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )} c} - \frac{2 \,{\left (a c^{2} + b d^{2}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac{c \tan \left (\frac{1}{2} \, x\right ) - d \tan \left (\frac{1}{2} \, x\right )}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{\sqrt{c^{2} - d^{2}} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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