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

Optimal. Leaf size=211 \[ \frac{b \left (5 a^2 A b^2-3 a^4 (2 A+C)-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (5 A+2 C)+a^4 (-C)+2 A b^4\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

[Out]

(b*(5*a^2*A*b^2 - 2*A*b^4 - 3*a^4*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^
(5/2)*(a + b)^(5/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d
*(a + b*Cos[c + d*x])^2) - ((2*A*b^4 - a^4*C - a^2*b^2*(5*A + 2*C))*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a +
b*Cos[c + d*x]))

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Rubi [A]  time = 0.648892, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3056, 3055, 3001, 3770, 2659, 205} \[ \frac{b \left (5 a^2 A b^2-3 a^4 (2 A+C)-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (-a^2 b^2 (5 A+2 C)+a^4 (-C)+2 A b^4\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*a^2*A*b^2 - 2*A*b^4 - 3*a^4*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^
(5/2)*(a + b)^(5/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d
*(a + b*Cos[c + d*x])^2) - ((2*A*b^4 - a^4*C - a^2*b^2*(5*A + 2*C))*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a +
b*Cos[c + d*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 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+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (2 A \left (a^2-b^2\right )-2 a b (A+C) \cos (c+d x)+\left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 A \left (a^2-b^2\right )^2+a b \left (A b^2-a^2 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{A \int \sec (c+d x) \, dx}{a^3}+\frac{\left (b \left (5 a^2 A b^2-2 A b^4-3 a^4 (2 A+C)\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (b \left (5 a^2 A b^2-2 A b^4-3 a^4 (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^3 \left (a^2-b^2\right )^2 d}\\ &=-\frac{b \left (6 a^4 A-5 a^2 A b^2+2 A b^4+3 a^4 C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.58006, size = 409, normalized size = 1.94 \[ \frac{\cos (c+d x) (A \sec (c+d x)+C \cos (c+d x)) \left (\frac{4 b (\sin (c)+i \cos (c)) \left (-5 a^2 A b^2+3 a^4 (2 A+C)+2 A b^4\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^2 \sqrt{\left (b^2-a^2\right ) (\cos (c)-i \sin (c))^2}}-\frac{a \sec (c) \left (b \left (b \left (a b \left (a^2 (4 A+3 C)-A b^2\right ) \sin (2 c+d x)-\left (a^2 b^2 (5 A+2 C)+a^4 C-2 A b^4\right ) \sin (c+2 d x)\right )-a \sin (d x) \left (a^2 b^2 (16 A+5 C)+4 a^4 C-7 A b^4\right )\right )+\left (2 a^2+b^2\right ) \sin (c) \left (a^2 b^2 (5 A+2 C)+a^4 C-2 A b^4\right )\right )}{b \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-4 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 a^3 d (2 A+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*(-4*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*A*Log[Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]] + (4*b*(-5*a^2*A*b^2 + 2*A*b^4 + 3*a^4*(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)]]*(I*Cos[c] + Sin[c]))/((
a^2 - b^2)^2*Sqrt[(-a^2 + b^2)*(Cos[c] - I*Sin[c])^2]) - (a*Sec[c]*((2*a^2 + b^2)*(-2*A*b^4 + a^4*C + a^2*b^2*
(5*A + 2*C))*Sin[c] + b*(-(a*(-7*A*b^4 + 4*a^4*C + a^2*b^2*(16*A + 5*C))*Sin[d*x]) + b*(a*b*(-(A*b^2) + a^2*(4
*A + 3*C))*Sin[2*c + d*x] - (-2*A*b^4 + a^4*C + a^2*b^2*(5*A + 2*C))*Sin[c + 2*d*x]))))/(b*(a^2 - b^2)^2*(a +
b*Cos[c + d*x])^2)))/(2*a^3*d*(2*A + C + C*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.073, size = 1115, normalized size = 5.3 \begin{align*} \text{result too large to display} \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)/(a+b*cos(d*x+c))^3,x)

[Out]

6/d*b^2/(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+1/d
/a/(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*b^3-2/d/
a^2/(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*b^4+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
*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*b*C+2/d/(a*ta
n(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+6/d*b^2/(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-b)^2*tan(1/2*d*x+1/2*c)*A-1/d/a/(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-b)^2*tan(1/2*d*x+1/2*c)*A*b^3-2/d/a^2/(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-b)^2*tan(1/2*d*x+1/2*c)*A*b^4+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-b)^2*tan(1/2*d*x+1/2*c)*a^2*C-1/d*a/(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-b)^2*tan(1/2*d*x+1/2*c)*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-b)^2
*tan(1/2*d*x+1/2*c)*b^2*C-6/d*a*b/(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))*A+5/d/a*b^3/(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))*A-2/d/a^3*b^5/(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))*A-3/d*b*a/(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)
)*C-1/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^3*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+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 = 46.9688, size = 2851, normalized size = 13.51 \begin{align*} \text{result too large to display} \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)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*((3*(2*A + C)*a^6*b - 5*A*a^4*b^3 + 2*A*a^2*b^5 + (3*(2*A + C)*a^4*b^3 - 5*A*a^2*b^5 + 2*A*b^7)*cos(d*x
+ c)^2 + 2*(3*(2*A + C)*a^5*b^2 - 5*A*a^3*b^4 + 2*A*a*b^6)*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*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 -
3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*cos(d*x
+ c))*log(sin(d*x + c) + 1) + 2*(A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4*b^4 + 3*
A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*cos(d*x + c))*log(-sin(d
*x + c) + 1) - 2*(2*C*a^8 + (6*A - C)*a^6*b^2 - (9*A + C)*a^4*b^4 + 3*A*a^2*b^6 + (C*a^7*b + (5*A + C)*a^5*b^3
- (7*A + 2*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*d
*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*b^4
- a^5*b^6)*d), -1/2*((3*(2*A + C)*a^6*b - 5*A*a^4*b^3 + 2*A*a^2*b^5 + (3*(2*A + C)*a^4*b^3 - 5*A*a^2*b^5 + 2*A
*b^7)*cos(d*x + c)^2 + 2*(3*(2*A + C)*a^5*b^2 - 5*A*a^3*b^4 + 2*A*a*b^6)*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^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^
6*b^2 - 3*A*a^4*b^4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*
cos(d*x + c))*log(sin(d*x + c) + 1) + (A*a^8 - 3*A*a^6*b^2 + 3*A*a^4*b^4 - A*a^2*b^6 + (A*a^6*b^2 - 3*A*a^4*b^
4 + 3*A*a^2*b^6 - A*b^8)*cos(d*x + c)^2 + 2*(A*a^7*b - 3*A*a^5*b^3 + 3*A*a^3*b^5 - A*a*b^7)*cos(d*x + c))*log(
-sin(d*x + c) + 1) - (2*C*a^8 + (6*A - C)*a^6*b^2 - (9*A + C)*a^4*b^4 + 3*A*a^2*b^6 + (C*a^7*b + (5*A + C)*a^5
*b^3 - (7*A + 2*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^
8)*d*cos(d*x + c)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c) + (a^11 - 3*a^9*b^2 + 3*a^7*
b^4 - a^5*b^6)*d)]

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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{\left (c + d x \right )}}{\left (a + b \cos{\left (c + d x \right )}\right )^{3}}\, 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)/(a+b*cos(d*x+c))**3,x)

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x))**3, x)

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Giac [B]  time = 1.68566, size = 680, normalized size = 3.22 \begin{align*} \text{result too large to display} \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)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-((6*A*a^4*b + 3*C*a^4*b - 5*A*a^2*b^3 + 2*A*b^5)*(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^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)
) - A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 + A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - (2*C*a^5*tan(1/2*d*x
+ 1/2*c)^3 - C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + C*a^3*b^2*tan(1/2*d*x + 1/
2*c)^3 - 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a*b^4*tan(1/2*d*x + 1/2
*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^5*tan(1/2*d*x + 1/2*c) + C*a^4*b*tan(1/2*d*x + 1/2*c) + 6*A*a^3
*b^2*tan(1/2*d*x + 1/2*c) + C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 2*C*a^2*b^3*ta
n(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/2*d*x + 1/2*c) - 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a^4*b^2 + a^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