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

Optimal. Leaf size=299 \[ \frac{b \left (-15 a^2 A b^3+12 a^4 A b+5 a^3 b^2 B-6 a^5 B-2 a b^4 B+6 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (-11 a^2 A b^2+2 a^4 A+5 a^3 b B-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (6 a^2 A b-4 a^3 B+a b^2 B-3 A b^3\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

[Out]

(b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((3*A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((2*a
^4*A - 11*a^2*A*b^2 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b*(A*b - a*B)*
Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b*(6*a^2*A*b - 3*A*b^3 - 4*a^3*B + a*b^2*B)*Tan[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 = 1.76445, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3000, 3055, 3001, 3770, 2659, 205} \[ \frac{b \left (-15 a^2 A b^3+12 a^4 A b+5 a^3 b^2 B-6 a^5 B-2 a b^4 B+6 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (-11 a^2 A b^2+2 a^4 A+5 a^3 b B-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (6 a^2 A b-4 a^3 B+a b^2 B-3 A b^3\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((3*A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((2*a
^4*A - 11*a^2*A*b^2 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b*(A*b - a*B)*
Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b*(6*a^2*A*b - 3*A*b^3 - 4*a^3*B + a*b^2*B)*Tan[c
+ d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

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 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{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{b (A b-a B) \tan (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^2 A-3 A b^2+a b B-2 a (A b-a B) \cos (c+d x)+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (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^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \cos (c+d x)+b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (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 \left (a^2-b^2\right )^2 (3 A b-a B)+a b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{(3 A b-a B) \int \sec (c+d x) \, dx}{a^4}+\frac{\left (b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\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^4 \left (a^2-b^2\right )^2 d}\\ &=\frac{b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.5268, size = 352, normalized size = 1.18 \[ \frac{\frac{a b^2 \left (-7 a^2 A b+5 a^3 B-2 a b^2 B+4 A b^3\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{a^2 b^2 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}-\frac{2 b \left (-15 a^2 A b^3+12 a^4 A b+5 a^3 b^2 B-6 a^5 B-2 a b^4 B+6 A b^5\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+2 (3 A b-a B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 (a B-3 A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{2 a A \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{2 a A \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTanh[((a - b)*Tan[(c + d*x
)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*(3*A*b - a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(-3
*A*b + a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2]) + (2*a*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a^2*b^2*(-(A*b) + a*B)*Sin[c + d*x
])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b^2*(-7*a^2*A*b + 4*A*b^3 + 5*a^3*B - 2*a*b^2*B)*Sin[c + d*x]
)/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*a^4*d)

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Maple [B]  time = 0.187, size = 1358, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x)

[Out]

-1/d*A/a^3/(tan(1/2*d*x+1/2*c)-1)+3/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A*b-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d*
A/a^3/(tan(1/2*d*x+1/2*c)+1)-3/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*A*b+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B-8/d/a/(ta
n(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-1/d/a^2/(t
an(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^4/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+4/d*b^5/a
^3/(tan(1/2*d*x+1/2*c)^2*a-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+6/d/(tan
(1/2*d*x+1/2*c)^2*a-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+1/d*b^3/a/(
tan(1/2*d*x+1/2*c)^2*a-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/d*b^4/a^2/
(tan(1/2*d*x+1/2*c)^2*a-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-8/d/a/(tan(
1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+1/d/a^2/(tan(1/2*d*x+1
/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^4/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+4/d*b^5/a^3/(tan(1/2*d*x+1/2*c)
^2*a-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+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+
1/2*c)^2*b+a+b)^2*b^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-1/d*b^3/a/(tan(1/2*d*x+1/2*c)^2*a-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/d*b^4/a^2/(tan(1/2*d*x+1/2*c)^2*a-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+12/d*b^2/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c
)*(a-b)/((a-b)*(a+b))^(1/2))*A-15/d*b^4/a^2/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*
(a-b)/((a-b)*(a+b))^(1/2))*A+6/d*b^6/a^4/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-
b)/((a-b)*(a+b))^(1/2))*A-6/d*b/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)
*(a+b))^(1/2))*B*a+5/d*b^3/a/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a
+b))^(1/2))*B-2/d*b^5/a^3/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b)
)^(1/2))*B

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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))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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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 )}\right ) \sec ^{2}{\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+B*cos(d*x+c))*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 1.5991, size = 775, normalized size = 2.59 \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))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

((6*B*a^5*b - 12*A*a^4*b^2 - 5*B*a^3*b^3 + 15*A*a^2*b^4 + 2*B*a*b^5 - 6*A*b^6)*(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^8 - 2*a^6
*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + (6*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 -
5*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 +
5*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*A*b^6*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^
4*b^2*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 5*B*a^3*b^3*tan(1/2*d*x + 1/2*c) - 7*A*a^2*b^4
*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c) + 5*A*a*b^5*tan(1/2*d*x + 1/2*c) - 2*B*a*b^5*tan(1/2*
d*x + 1/2*c) + 4*A*b^6*tan(1/2*d*x + 1/2*c))/((a^7 - 2*a^5*b^2 + a^3*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) + (B*a - 3*A*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (B*a - 3*A*b)*log(abs(ta
n(1/2*d*x + 1/2*c) - 1))/a^4 - 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3))/d