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

Optimal. Leaf size=133 \[ -\frac{2 \left (2 a^2 b B+a^3 (-C)-b^3 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{B \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

[Out]

(-2*(2*a^2*b*B - b^3*B - a^3*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/2)*(a + b)
^(3/2)*d) + (B*ArcTanh[Sin[c + d*x]])/(a^2*d) + (b*(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 = 0.385672, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3029, 3000, 3001, 3770, 2659, 205} \[ -\frac{2 \left (2 a^2 b B+a^3 (-C)-b^3 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{B \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3000

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

Mathematica [A]  time = 0.588287, size = 191, normalized size = 1.44 \[ \frac{\cos (c+d x) (B \sec (c+d x)+C) \left (\frac{2 \left (-2 a^2 b B+a^3 C+b^3 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a^2 d (B+C \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.074, size = 342, normalized size = 2.6 \begin{align*} -{\frac{B}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) B}{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{b\tan \left ( 1/2\,dx+c/2 \right ) C}{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{bB}{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{{b}^{3}B}{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{aC}{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{B}{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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x)

[Out]

-1/d/a^2*B*ln(tan(1/2*d*x+1/2*c)-1)+2/d/a*b^2/(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-2/d*b/(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)*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))*B+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))*b^3*B+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))*C+1/d/a^2*B*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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 29.9143, size = 1534, normalized size = 11.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((C*a^4 - 2*B*a^3*b + B*a*b^3 + (C*a^3*b - 2*B*a^2*b^2 + B*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)) + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b - 2*B*a^2*b^
3 + B*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - (B*a^5 - 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b - 2*B*a^2*b^3 + B*b
^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^4*b - B*a^3*b^2 - C*a^2*b^3 + B*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*(C*a^4 - 2*B*a^3*b + B*a*b^3 +
(C*a^3*b - 2*B*a^2*b^2 + B*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*s
in(d*x + c))) + (B*a^5 - 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b - 2*B*a^2*b^3 + B*b^5)*cos(d*x + c))*log(sin(d*x + c
) + 1) - (B*a^5 - 2*B*a^3*b^2 + B*a*b^4 + (B*a^4*b - 2*B*a^2*b^3 + B*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1)
- 2*(C*a^4*b - B*a^3*b^2 - C*a^2*b^3 + B*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.34965, size = 304, normalized size = 2.29 \begin{align*} -\frac{\frac{2 \,{\left (C a^{3} - 2 \, B a^{2} b + B 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{B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac{B \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 b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B 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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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