3.278 \(\int \frac{(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=301 \[ -\frac{\left (-8 a^4 A b^3+7 a^2 A b^5+8 a^6 A b-3 a^5 b^2 B-2 a^7 B-2 A b^7\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{7/2} (a+b)^{7/2}}+\frac{b \left (-17 a^2 A b^3+26 a^4 A b-4 a^3 b^2 B-11 a^5 B+6 A b^5\right ) \sin (c+d x)}{6 a^3 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac{b \left (8 a^2 A b-5 a^3 B-3 A b^3\right ) \sin (c+d x)}{6 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

[Out]

-(((8*a^6*A*b - 8*a^4*A*b^3 + 7*a^2*A*b^5 - 2*A*b^7 - 2*a^7*B - 3*a^5*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^4*(a - b)^(7/2)*(a + b)^(7/2)*d)) + (A*ArcTanh[Sin[c + d*x]])/(a^4*d) + (b*(A*b - a*B)*S
in[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + (b*(8*a^2*A*b - 3*A*b^3 - 5*a^3*B)*Sin[c + d*x])/(6*
a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) + (b*(26*a^4*A*b - 17*a^2*A*b^3 + 6*A*b^5 - 11*a^5*B - 4*a^3*b^2*B
)*Sin[c + d*x])/(6*a^3*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 1.50886, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {3000, 3055, 3001, 3770, 2659, 205} \[ -\frac{\left (-8 a^4 A b^3+7 a^2 A b^5+8 a^6 A b-3 a^5 b^2 B-2 a^7 B-2 A b^7\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{7/2} (a+b)^{7/2}}+\frac{b \left (-17 a^2 A b^3+26 a^4 A b-4 a^3 b^2 B-11 a^5 B+6 A b^5\right ) \sin (c+d x)}{6 a^3 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac{b \left (8 a^2 A b-5 a^3 B-3 A b^3\right ) \sin (c+d x)}{6 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac{b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]

[Out]

-(((8*a^6*A*b - 8*a^4*A*b^3 + 7*a^2*A*b^5 - 2*A*b^7 - 2*a^7*B - 3*a^5*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^4*(a - b)^(7/2)*(a + b)^(7/2)*d)) + (A*ArcTanh[Sin[c + d*x]])/(a^4*d) + (b*(A*b - a*B)*S
in[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + (b*(8*a^2*A*b - 3*A*b^3 - 5*a^3*B)*Sin[c + d*x])/(6*
a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) + (b*(26*a^4*A*b - 17*a^2*A*b^3 + 6*A*b^5 - 11*a^5*B - 4*a^3*b^2*B
)*Sin[c + d*x])/(6*a^3*(a^2 - b^2)^3*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 (c+d x)}{(a+b \cos (c+d x))^4} \, dx &=\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{\int \frac{\left (3 A \left (a^2-b^2\right )-3 a (A b-a B) \cos (c+d x)+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (6 A \left (a^2-b^2\right )^2-2 a \left (6 a^2 A b-A b^3-3 a^3 B-2 a b^2 B\right ) \cos (c+d x)+b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{b \left (26 a^4 A b-17 a^2 A b^3+6 A b^5-11 a^5 B-4 a^3 b^2 B\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (6 A \left (a^2-b^2\right )^3-3 a \left (6 a^4 A b-2 a^2 A b^3+A b^5-2 a^5 B-3 a^3 b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )^3}\\ &=\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{b \left (26 a^4 A b-17 a^2 A b^3+6 A b^5-11 a^5 B-4 a^3 b^2 B\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac{A \int \sec (c+d x) \, dx}{a^4}-\frac{\left (8 a^6 A b-8 a^4 A b^3+7 a^2 A b^5-2 A b^7-2 a^7 B-3 a^5 b^2 B\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^3}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{b \left (26 a^4 A b-17 a^2 A b^3+6 A b^5-11 a^5 B-4 a^3 b^2 B\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (8 a^6 A b-8 a^4 A b^3+7 a^2 A b^5-2 A b^7-2 a^7 B-3 a^5 b^2 B\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 )^3 d}\\ &=-\frac{\left (8 a^6 A b-8 a^4 A b^3+7 a^2 A b^5-2 A b^7-2 a^7 B-3 a^5 b^2 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)^{7/2} (a+b)^{7/2} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac{b \left (8 a^2 A b-3 A b^3-5 a^3 B\right ) \sin (c+d x)}{6 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{b \left (26 a^4 A b-17 a^2 A b^3+6 A b^5-11 a^5 B-4 a^3 b^2 B\right ) \sin (c+d x)}{6 a^3 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.60375, size = 368, normalized size = 1.22 \[ \frac{\cos (c+d x) (A \sec (c+d x)+B) \left (-\frac{2 a b \sin (c+d x) \left (6 a b \left (15 a^2 A b^3-20 a^4 A b+a^3 b^2 B+9 a^5 B-5 A b^5\right ) \cos (c+d x)+b^2 \left (17 a^2 A b^3-26 a^4 A b+4 a^3 b^2 B+11 a^5 B-6 A b^5\right ) \cos (2 (c+d x))+38 a^4 A b^3-5 a^2 A b^5-72 a^6 A b+a^5 b^2 B+8 a^3 b^4 B+36 a^7 B-6 A b^7\right )}{\left (a^2-b^2\right )^3 (a+b \cos (c+d x))^3}+\frac{24 \left (8 a^4 A b^3-7 a^2 A b^5-8 a^6 A b+3 a^5 b^2 B+2 a^7 B+2 A b^7\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}-24 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{24 a^4 d (A+B \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^4,x]

[Out]

(Cos[c + d*x]*(B + A*Sec[c + d*x])*((24*(-8*a^6*A*b + 8*a^4*A*b^3 - 7*a^2*A*b^5 + 2*A*b^7 + 2*a^7*B + 3*a^5*b^
2*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(7/2) - 24*A*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] + 24*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (2*a*b*(-72*a^6*A*b + 38*a^4*A*b^3 - 5*a^2*A
*b^5 - 6*A*b^7 + 36*a^7*B + a^5*b^2*B + 8*a^3*b^4*B + 6*a*b*(-20*a^4*A*b + 15*a^2*A*b^3 - 5*A*b^5 + 9*a^5*B +
a^3*b^2*B)*Cos[c + d*x] + b^2*(-26*a^4*A*b + 17*a^2*A*b^3 - 6*A*b^5 + 11*a^5*B + 4*a^3*b^2*B)*Cos[2*(c + d*x)]
)*Sin[c + d*x])/((a^2 - b^2)^3*(a + b*Cos[c + d*x])^3)))/(24*a^4*d*(A + B*Cos[c + d*x]))

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Maple [B]  time = 0.187, size = 2180, normalized size = 7.2 \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)/(a+b*cos(d*x+c))^4,x)

[Out]

3/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*B
*a*b^2-3/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1
/2*c)^5*B*a*b^2+4/d/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*b^6/(a^2+2*a*b+b^2)/(a^2-2*a*b+b
^2)*tan(1/2*d*x+1/2*c)^3*A-6/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*b^4/(a-b)/(a^3+3*a^2*b+
3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*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)^3*b^5/(a-b)/(a
^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A+2/d/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*b
^6/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A+2/d/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b
+a+b)^3*b^6/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*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)^3*b^5/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A-44/3/d/a/(tan(1/2*d*x+1/2*c)^2*a-t
an(1/2*d*x+1/2*c)^2*b+a+b)^3*b^4/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-6/d/a/(tan(1/2*d*x+1/2
*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*b^4/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A+24/d*b^2/(tan(1
/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-6/d*b
/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^2/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*
B-12/d*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^2/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x
+1/2*c)^3*B-6/d*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^2/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*ta
n(1/2*d*x+1/2*c)^5*B+12/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a-b)/(a^3+3*a^2*b+3*a*b
^2+b^3)*tan(1/2*d*x+1/2*c)^5*A+12/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a/(a+b)/(a^3-3*a
^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A-7/d/a^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/
2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*b^5+2/d/a^4/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan
(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*b^7+1/d/a^4*A*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^4*A*ln(tan(1/2*d
*x+1/2*c)-1)-4/3/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1
/2*d*x+1/2*c)^3*B*b^3-2/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3
)*tan(1/2*d*x+1/2*c)*B*b^3+3/d*b^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((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-8/d*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2
*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*a^2+4/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^
2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A*b^3-2/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a
^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*B*b^3-4/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(
a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A*b^3+8/d*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1
/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+2/d/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((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^3

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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)/(a+b*cos(d*x+c))^4,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)/(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

Timed out

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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((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.89333, size = 1130, normalized size = 3.75 \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)/(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(2*B*a^7 - 8*A*a^6*b + 3*B*a^5*b^2 + 8*A*a^4*b^3 - 7*A*a^2*b^5 + 2*A*b^7)*(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^10 - 3*a^
8*b^2 + 3*a^6*b^4 - a^4*b^6)*sqrt(a^2 - b^2)) + 3*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3*A*log(abs(tan(1
/2*d*x + 1/2*c) - 1))/a^4 - (18*B*a^7*b*tan(1/2*d*x + 1/2*c)^5 - 36*A*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 - 27*B*a^
6*b^2*tan(1/2*d*x + 1/2*c)^5 + 60*A*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*B*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*A*
a^4*b^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 - 45*A*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 + 6*
B*a^3*b^5*tan(1/2*d*x + 1/2*c)^5 + 6*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 15*A*a*b^7*tan(1/2*d*x + 1/2*c)^5 - 6*
A*b^8*tan(1/2*d*x + 1/2*c)^5 + 36*B*a^7*b*tan(1/2*d*x + 1/2*c)^3 - 72*A*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 - 32*B*
a^5*b^3*tan(1/2*d*x + 1/2*c)^3 + 116*A*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*B*a^3*b^5*tan(1/2*d*x + 1/2*c)^3 - 5
6*A*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^8*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^7*b*tan(1/2*d*x + 1/2*c) - 36*A*
a^6*b^2*tan(1/2*d*x + 1/2*c) + 27*B*a^6*b^2*tan(1/2*d*x + 1/2*c) - 60*A*a^5*b^3*tan(1/2*d*x + 1/2*c) + 6*B*a^5
*b^3*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*b^4*tan(1/2*d*x + 1/2*c) + 45*A*a^3*b^5
*tan(1/2*d*x + 1/2*c) + 6*B*a^3*b^5*tan(1/2*d*x + 1/2*c) + 6*A*a^2*b^6*tan(1/2*d*x + 1/2*c) - 15*A*a*b^7*tan(1
/2*d*x + 1/2*c) - 6*A*b^8*tan(1/2*d*x + 1/2*c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*(a*tan(1/2*d*x + 1/2*
c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3))/d