3.512 \(\int \frac{\cos (c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=223 \[ \frac{\left (-11 a^2 b^2+2 a^4+6 b^4\right ) \sin (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{3 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{3 b x}{a^4} \]

[Out]

(-3*b*x)/a^4 + (3*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a
- b)^(5/2)*(a + b)^(5/2)*d) + ((2*a^4 - 11*a^2*b^2 + 6*b^4)*Sin[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b^2*Sin[
c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (3*b^2*(2*a^2 - b^2)*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2
*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.623448, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3847, 4100, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (-11 a^2 b^2+2 a^4+6 b^4\right ) \sin (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{3 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{3 b x}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*x)/a^4 + (3*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a
- b)^(5/2)*(a + b)^(5/2)*d) + ((2*a^4 - 11*a^2*b^2 + 6*b^4)*Sin[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b^2*Sin[
c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + (3*b^2*(2*a^2 - b^2)*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2
*d*(a + b*Sec[c + d*x]))

Rule 3847

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)
*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 4100

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4104

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+b \sec (c+d x))^3} \, dx &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (-2 a^2+3 b^2+2 a b \sec (c+d x)-2 b^2 \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (2 a^4-11 a^2 b^2+6 b^4-a b \left (4 a^2-b^2\right ) \sec (c+d x)+3 b^2 \left (2 a^2-b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 a^4-11 a^2 b^2+6 b^4\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{6 b \left (a^2-b^2\right )^2-3 a b^2 \left (2 a^2-b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{3 b x}{a^4}+\frac{\left (2 a^4-11 a^2 b^2+6 b^4\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 b x}{a^4}+\frac{\left (2 a^4-11 a^2 b^2+6 b^4\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (3 b \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 b x}{a^4}+\frac{\left (2 a^4-11 a^2 b^2+6 b^4\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (3 b \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) 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{3 b x}{a^4}+\frac{3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \tanh ^{-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{\left (2 a^4-11 a^2 b^2+6 b^4\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac{3 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.881031, size = 229, normalized size = 1.03 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b) \left (-\frac{a b^3 \left (8 a^2-5 b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}-\frac{6 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^4 \sin (c+d x)}{(a-b) (a+b)}-6 b (c+d x) (a \cos (c+d x)+b)^2+2 a \sin (c+d x) (a \cos (c+d x)+b)^2\right )}{2 a^4 d (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^3*(-6*b*(c + d*x)*(b + a*Cos[c + d*x])^2 - (6*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^
4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + (a*b^4*Sin
[c + d*x])/((a - b)*(a + b)) - (a*b^3*(8*a^2 - 5*b^2)*(b + a*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2)
+ 2*a*(b + a*Cos[c + d*x])^2*Sin[c + d*x]))/(2*a^4*d*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.086, size = 702, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^3*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^4*b*arctan(tan(1/2*d*x+1/2*c))+8/d/a*b^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+1/d/a^2*b^4/(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)^3-4/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)^3-8/d/a*b^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)+1/d/a^2*b^4/(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)+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)+12/d*b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(
a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-15/d*b^4/a^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b
))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+6/d*b^6/a^4/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(
1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.34654, size = 2225, normalized size = 9.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(12*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*x*cos(d*x + c)^2 + 24*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 -
a*b^8)*d*x*cos(d*x + c) + 12*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*x - 3*(4*a^4*b^4 - 5*a^2*b^6 + 2*b^8 +
(4*a^6*b^2 - 5*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c)^2 + 2*(4*a^5*b^3 - 5*a^3*b^5 + 2*a*b^7)*cos(d*x + c))*sqrt(a
^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(
d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(2*a^7*b^2 - 13*a^5*b^4 + 17*a^3*
b^6 - 6*a*b^8 + 2*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^2 + (4*a^8*b - 20*a^6*b^3 + 25*a^4*b^5
- 9*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^11
*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) + (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d), -1/2*(
6*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*x*cos(d*x + c)^2 + 12*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*
d*x*cos(d*x + c) + 6*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*x - 3*(4*a^4*b^4 - 5*a^2*b^6 + 2*b^8 + (4*a^6*b
^2 - 5*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c)^2 + 2*(4*a^5*b^3 - 5*a^3*b^5 + 2*a*b^7)*cos(d*x + c))*sqrt(-a^2 + b^2
)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (2*a^7*b^2 - 13*a^5*b^4 + 17*a^3
*b^6 - 6*a*b^8 + 2*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^2 + (4*a^8*b - 20*a^6*b^3 + 25*a^4*b^5
- 9*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^1
1*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) + (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+b*sec(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.35167, size = 482, normalized size = 2.16 \begin{align*} -\frac{\frac{3 \,{\left (4 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\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^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{8 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} 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}} + \frac{3 \,{\left (d x + c\right )} b}{a^{4}} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(4*a^4*b^2 - 5*a^2*b^4 + 2*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)) - (8*a^3*b
^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 5*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*b^6*tan(1/2*
d*x + 1/2*c)^3 - 8*a^3*b^3*tan(1/2*d*x + 1/2*c) - 7*a^2*b^4*tan(1/2*d*x + 1/2*c) + 5*a*b^5*tan(1/2*d*x + 1/2*c
) + 4*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) + 3*(d*x + c)*b/a^4 - 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3))/d