3.811 \(\int \frac{(B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=164 \[ \frac{\left (2 a^2 B-3 a b C+b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{(b B-a C) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

[Out]

((2*a^2*B + b^2*B - 3*a*b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*
d) - ((b*B - a*C)*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((3*a*b*B - a^2*C - 2*b^2*C)*Sin[c
+ d*x])/(2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.253487, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3029, 2754, 12, 2659, 205} \[ \frac{\left (2 a^2 B-3 a b C+b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-C)+3 a b B-2 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{(b B-a C) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]

[Out]

((2*a^2*B + b^2*B - 3*a*b*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*
d) - ((b*B - a*C)*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((3*a*b*B - a^2*C - 2*b^2*C)*Sin[c
+ d*x])/(2*(a^2 - b^2)^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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 (c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\int \frac{B+C \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac{(b B-a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{-2 (a B-b C)+(b B-a C) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{(b B-a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a^2 B+b^2 B-3 a b C}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{(b B-a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 a^2 B+b^2 B-3 a b C\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{(b B-a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 a^2 B+b^2 B-3 a b C\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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{\left (2 a^2 B+b^2 B-3 a b C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{(b B-a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 a b B-a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.653494, size = 157, normalized size = 0.96 \[ \frac{\frac{\left (a^2 C-3 a b B+2 b^2 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{2 \left (2 a^2 B-3 a b C+b^2 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{(a C-b B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]

[Out]

((-2*(2*a^2*B + b^2*B - 3*a*b*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((
-(b*B) + a*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + ((-3*a*b*B + a^2*C + 2*b^2*C)*Sin[c + d
*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*d)

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Maple [B]  time = 0.059, size = 886, normalized size = 5.4 \begin{align*} \text{result too large to display} \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)/(a+b*cos(d*x+c))^3,x)

[Out]

-4/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-1/
d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*B+2/d/(
a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*a^2*C+1/d/(a*t
an(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*a*b*C+2/d/(a*tan(
1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*C-4/d/(a*tan(1/2
*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*a*b*B+1/d/(a*tan(1/2*d*x+
1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*b^2*B+2/d/(a*tan(1/2*d*x+1/2*c
)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*a^2*C-1/d/(a*tan(1/2*d*x+1/2*c)^2-t
an(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*a*b*C+2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/
2*d*x+1/2*c)^2*b+a+b)^2*b^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+2/d*a^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b
))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+1/d*b^2/(a^4-2*a^2*b^2+b^4)/((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/d*b*a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arcta
n((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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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)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.00056, size = 1616, normalized size = 9.85 \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)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*((2*B*a^4 - 3*C*a^3*b + B*a^2*b^2 + (2*B*a^2*b^2 - 3*C*a*b^3 + B*b^4)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*
C*a^2*b^2 + B*a*b^3)*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)) - 2*(2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2*b^3 - C*a*b^4 - B*b^5 + (C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^
3 + 3*B*a*b^4 - 2*C*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2
+ 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d), 1/2*
((2*B*a^4 - 3*C*a^3*b + B*a^2*b^2 + (2*B*a^2*b^2 - 3*C*a*b^3 + B*b^4)*cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C*a^2*
b^2 + B*a*b^3)*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))) + (2
*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2*b^3 - C*a*b^4 - B*b^5 + (C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^
4 - 2*C*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b
- 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*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)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.35438, size = 527, normalized size = 3.21 \begin{align*} \frac{\frac{{\left (2 \, B a^{2} - 3 \, C a b + B 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 )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\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 )}^{2}}}{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)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

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