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

Optimal. Leaf size=95 \[ \frac{2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{A b \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{A \tan (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.22504, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3056, 3001, 3770, 2659, 205} \[ \frac{2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{A b \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{A \tan (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.18278, size = 306, normalized size = 3.22 \[ \frac{2 \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-\frac{2 i (\cos (c)-i \sin (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )}{\sqrt{\left (b^2-a^2\right ) (\cos (c)-i \sin (c))^2}}+\frac{a A \sin \left (\frac{d x}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{a A \sin \left (\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a^2 d (2 A+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(A*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - A*b*Log[Cos[(c + d*x)
/2] + Sin[(c + d*x)/2]] - ((2*I)*(A*b^2 + a^2*C)*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(
d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(Cos[c] - I*Sin[c]))/Sqrt[(-a^2 + b^2)*(Cos[c] - I*Sin[c
])^2] + (a*A*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (a*A*Sin[(d*x)/2])/
((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(a^2*d*(2*A + C + C*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.064, size = 183, normalized size = 1.9 \begin{align*} 2\,{\frac{A{b}^{2}}{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 ) }+2\,{\frac{C}{d\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{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{Ab}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{Ab}{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((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.59067, size = 952, normalized size = 10.02 \begin{align*} \left [-\frac{{\left (C a^{2} + A b^{2}\right )} \sqrt{-a^{2} + b^{2}} \cos \left (d x + c\right ) \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 a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac{2 \,{\left (C a^{2} + A b^{2}\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 ) \cos \left (d x + c\right ) -{\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (A a^{2} b - A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.4468, size = 221, normalized size = 2.33 \begin{align*} -\frac{\frac{A b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{A b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \, A \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} + \frac{2 \,{\left (C a^{2} + A b^{2}\right )}{\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 )}}{\sqrt{a^{2} - b^{2}} a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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