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

Optimal. Leaf size=147 \[ -\frac{2 \left (2 a^2 A b+a^2 b C+a^3 (-B)-A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

[Out]

(-2*(2*a^2*A*b - A*b^3 - a^3*B + a^2*b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/
2)*(a + b)^(3/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2
)*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.351364, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3055, 3001, 3770, 2659, 205} \[ -\frac{2 \left (2 a^2 A b+a^2 b C+a^3 (-B)-A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*a^2*A*b - A*b^3 - a^3*B + a^2*b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/
2)*(a + b)^(3/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(a*(a^2 - b^2
)*d*(a + b*Cos[c + d*x]))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + 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] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (A \left (a^2-b^2\right )-a (A b-a B+b C) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{A \int \sec (c+d x) \, dx}{a^2}+\frac{\left (A b^3+a^3 B-a^2 b (2 A+C)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 \left (A b^3+a^3 B-a^2 b (2 A+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 )}{a^2 \left (a^2-b^2\right ) d}\\ &=-\frac{2 \left (2 a^2 A b-A b^3-a^3 B+a^2 b C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 2.82591, size = 319, normalized size = 2.17 \[ \frac{2 \cos (c+d x) (A \sec (c+d x)+B+C \cos (c+d x)) \left (\frac{2 i (\cos (c)-i \sin (c))^3 \left (-a^2 b (2 A+C)+a^3 B+A b^3\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )}{\left (\left (b^2-a^2\right ) (\cos (c)-i \sin (c))^2\right )^{3/2}}+\frac{a \left (a (a C-b B)+A b^2\right ) (b \sin (d x)-a \sin (c))}{b (a-b) (a+b) \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) (a+b \cos (c+d x))}-A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a^2 d (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c + d*x]*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*(-(A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + A*Log[C
os[(c + d*x)/2] + Sin[(c + d*x)/2]] + ((2*I)*(A*b^3 + a^3*B - a^2*b*(2*A + 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])^3)/((-
a^2 + b^2)*(Cos[c] - I*Sin[c])^2)^(3/2) + (a*(A*b^2 + a*(-(b*B) + a*C))*(-(a*Sin[c]) + b*Sin[d*x]))/((a - b)*b
*(a + b)*(a + b*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(a^2*d*(2*A + C + 2*B*Cos[c + d*x
] + C*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.077, size = 458, normalized size = 3.1 \begin{align*} -{\frac{A}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ){b}^{2}}{da \left ({a}^{2}-{b}^{2} \right ) \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{\tan \left ( 1/2\,dx+c/2 \right ) bB}{d \left ({a}^{2}-{b}^{2} \right ) \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{\tan \left ( 1/2\,dx+c/2 \right ) aC}{d \left ({a}^{2}-{b}^{2} \right ) \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 ) }}-4\,{\frac{Ab}{d \left ( a+b \right ) \left ( a-b \right ) \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{A{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \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{Ba}{d \left ( a+b \right ) \left ( a-b \right ) \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{Cb}{d \left ( a+b \right ) \left ( a-b \right ) \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}{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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x)

[Out]

-1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)+2/d/a/(a^2-b^2)*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)*A*b^2-2/d/(a^2-b^2)*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)*b*B+2/
d/(a^2-b^2)*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)*a*C-4/d*b/(a+b)/(a-b)/((a+b
)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/d/a^2/(a+b)/(a-b)/((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^3+2/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*
tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-2/d/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/
((a+b)*(a-b))^(1/2))*C*b+1/d/a^2*A*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+B*cos(d*x+c)+C*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 = 61.1697, size = 1605, normalized size = 10.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((B*a^4 - (2*A + C)*a^3*b + A*a*b^3 + (B*a^3*b - (2*A + C)*a^2*b^2 + A*b^4)*cos(d*x + c))*sqrt(-a^2 + b^
2)*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^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b
- 2*A*a^2*b^3 + A*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + (A*a^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b - 2*A*
a^2*b^3 + A*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^5 - B*a^4*b + (A - C)*a^3*b^2 + B*a^2*b^3 - A*a
*b^4)*sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d), 1/2*(2*(B*
a^4 - (2*A + C)*a^3*b + A*a*b^3 + (B*a^3*b - (2*A + C)*a^2*b^2 + A*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-
(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (A*a^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b - 2*A*a^2*b^3
+ A*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - (A*a^5 - 2*A*a^3*b^2 + A*a*b^4 + (A*a^4*b - 2*A*a^2*b^3 + A*b^
5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C*a^5 - B*a^4*b + (A - C)*a^3*b^2 + B*a^2*b^3 - A*a*b^4)*sin(d*x
+ c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26244, size = 329, normalized size = 2.24 \begin{align*} \frac{\frac{2 \,{\left (B a^{3} - 2 \, A a^{2} b - C a^{2} b + A b^{3}\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 )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\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 )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(B*a^3 - 2*A*a^2*b - C*a^2*b + A*b^3)*(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)))/((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) + A*log(abs(tan(1/2
*d*x + 1/2*c) + 1))/a^2 - A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(C*a^2*tan(1/2*d*x + 1/2*c) - B*a*b*tan
(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2
*c)^2 + a + b)))/d