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

Optimal. Leaf size=346 \[ \frac{\left (7 a^2 A b^2+2 a^4 A-6 a^3 b B+9 a b^3 B-12 A b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}+\frac{\left (a^2 A+3 a b B-4 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{\left (2 a^2 A b+a^3 (-B)+3 a b^2 B-4 A b^3\right ) \sin (c+d x) \cos (c+d x)}{2 a^3 d \left (a^2-b^2\right )}+\frac{2 b^3 \left (5 a^2 A b-4 a^3 B+3 a b^2 B-4 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x \left (2 a^2 A b+a^3 (-B)-6 a b^2 B+8 A b^3\right )}{2 a^5} \]

[Out]

-((2*a^2*A*b + 8*A*b^3 - a^3*B - 6*a*b^2*B)*x)/(2*a^5) + (2*b^3*(5*a^2*A*b - 4*A*b^3 - 4*a^3*B + 3*a*b^2*B)*Ar
cTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a + b)^(3/2)*d) + ((2*a^4*A + 7*a^2*A*b
^2 - 12*A*b^4 - 6*a^3*b*B + 9*a*b^3*B)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*d) - ((2*a^2*A*b - 4*A*b^3 - a^3*B + 3
*a*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b^2)*d) + ((a^2*A - 4*A*b^2 + 3*a*b*B)*Cos[c + d*x]^2*Sin[c
+ d*x])/(3*a^2*(a^2 - b^2)*d) + (b*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d
*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.27456, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4030, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (7 a^2 A b^2+2 a^4 A-6 a^3 b B+9 a b^3 B-12 A b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}+\frac{\left (a^2 A+3 a b B-4 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{\left (2 a^2 A b+a^3 (-B)+3 a b^2 B-4 A b^3\right ) \sin (c+d x) \cos (c+d x)}{2 a^3 d \left (a^2-b^2\right )}+\frac{2 b^3 \left (5 a^2 A b-4 a^3 B+3 a b^2 B-4 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x \left (2 a^2 A b+a^3 (-B)-6 a b^2 B+8 A b^3\right )}{2 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a^2*A*b + 8*A*b^3 - a^3*B - 6*a*b^2*B)*x)/(2*a^5) + (2*b^3*(5*a^2*A*b - 4*A*b^3 - 4*a^3*B + 3*a*b^2*B)*Ar
cTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(3/2)*(a + b)^(3/2)*d) + ((2*a^4*A + 7*a^2*A*b
^2 - 12*A*b^4 - 6*a^3*b*B + 9*a*b^3*B)*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*d) - ((2*a^2*A*b - 4*A*b^3 - a^3*B + 3
*a*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*(a^2 - b^2)*d) + ((a^2*A - 4*A*b^2 + 3*a*b*B)*Cos[c + d*x]^2*Sin[c
+ d*x])/(3*a^2*(a^2 - b^2)*d) + (b*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d
*x]))

Rule 4030

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*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)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && 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 ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^3(c+d x) \left (-a^2 A+4 A b^2-3 a b B+a (A b-a B) \sec (c+d x)-3 b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos ^2(c+d x) \left (-3 \left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right )+a \left (2 a^2 A+A b^2-3 a b B\right ) \sec (c+d x)+2 b \left (a^2 A-4 A b^2+3 a b B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 \left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right )+a \left (2 a^2 A b+4 A b^3-3 a^3 B-3 a b^2 B\right ) \sec (c+d x)+3 b \left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-3 \left (a^2-b^2\right ) \left (2 a^2 A b+8 A b^3-a^3 B-6 a b^2 B\right )-3 a b \left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b+8 A b^3-a^3 B-6 a b^2 B\right ) x}{2 a^5}+\frac{\left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b^3 \left (5 a^2 A b-4 A b^3-4 a^3 B+3 a b^2 B\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b+8 A b^3-a^3 B-6 a b^2 B\right ) x}{2 a^5}+\frac{\left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b^2 \left (5 a^2 A b-4 A b^3-4 a^3 B+3 a b^2 B\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b+8 A b^3-a^3 B-6 a b^2 B\right ) x}{2 a^5}+\frac{\left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 b^2 \left (5 a^2 A b-4 A b^3-4 a^3 B+3 a b^2 B\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^5 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (2 a^2 A b+8 A b^3-a^3 B-6 a b^2 B\right ) x}{2 a^5}+\frac{2 b^3 \left (5 a^2 A b-4 A b^3-4 a^3 B+3 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (2 a^4 A+7 a^2 A b^2-12 A b^4-6 a^3 b B+9 a b^3 B\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}-\frac{\left (2 a^2 A b-4 A b^3-a^3 B+3 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-4 A b^2+3 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.37415, size = 224, normalized size = 0.65 \[ \frac{6 (c+d x) \left (-2 a^2 A b+a^3 B+6 a b^2 B-8 A b^3\right )+3 a \left (3 a^2 A-8 a b B+12 A b^2\right ) \sin (c+d x)+\frac{24 b^3 \left (-5 a^2 A b+4 a^3 B-3 a b^2 B+4 A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+3 a^2 (a B-2 A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))+\frac{12 a b^4 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}}{12 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(-2*a^2*A*b - 8*A*b^3 + a^3*B + 6*a*b^2*B)*(c + d*x) + (24*b^3*(-5*a^2*A*b + 4*A*b^3 + 4*a^3*B - 3*a*b^2*B)
*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 3*a*(3*a^2*A + 12*A*b^2 - 8*a*b*B)*
Sin[c + d*x] + (12*a*b^4*(-(A*b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + 3*a^2*(-2*A*b +
a*B)*Sin[2*(c + d*x)] + a^3*A*Sin[3*(c + d*x)])/(12*a^5*d)

________________________________________________________________________________________

Maple [B]  time = 0.12, size = 926, normalized size = 2.7 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)

[Out]

2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^5*A*b+6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A*b^2-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1
/2*d*x+1/2*c)^5*B-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B*b+4/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^
2)^3*tan(1/2*d*x+1/2*c)^3*A+12/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A*b^2-8/d/a^3/(1+tan(1/2*
d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*B*b+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A+6/d/a^4/(1+ta
n(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b^2-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B*b-2/d/a
^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B-2
/d/a^3*A*arctan(tan(1/2*d*x+1/2*c))*b-8/d/a^5*arctan(tan(1/2*d*x+1/2*c))*A*b^3+1/d/a^2*B*arctan(tan(1/2*d*x+1/
2*c))+6/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B*b^2+2/d*b^5/a^4/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*
a-tan(1/2*d*x+1/2*c)^2*b-a-b)*A-2/d*b^4/a^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1
/2*c)^2*b-a-b)*B+10/d/a^3*b^4/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(
1/2))*A-8/d*b^6/a^5/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A-8/
d/a^2*b^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+6/d*b^5/a^4/
(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.815434, size = 2592, normalized size = 7.49 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^8 - 2*A*a^7*b + 4*B*a^6*b^2 - 4*A*a^5*b^3 - 11*B*a^4*b^4 + 14*A*a^3*b^5 + 6*B*a^2*b^6 - 8*A*a*b^7
)*d*x*cos(d*x + c) + 3*(B*a^7*b - 2*A*a^6*b^2 + 4*B*a^5*b^3 - 4*A*a^4*b^4 - 11*B*a^3*b^5 + 14*A*a^2*b^6 + 6*B*
a*b^7 - 8*A*b^8)*d*x + 3*(4*B*a^3*b^4 - 5*A*a^2*b^5 - 3*B*a*b^6 + 4*A*b^7 + (4*B*a^4*b^3 - 5*A*a^3*b^4 - 3*B*a
^2*b^5 + 4*A*a*b^6)*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*s
qrt(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
)) + (4*A*a^7*b - 12*B*a^6*b^2 + 10*A*a^5*b^3 + 30*B*a^4*b^4 - 38*A*a^3*b^5 - 18*B*a^2*b^6 + 24*A*a*b^7 + 2*(A
*a^8 - 2*A*a^6*b^2 + A*a^4*b^4)*cos(d*x + c)^3 + (3*B*a^8 - 4*A*a^7*b - 6*B*a^6*b^2 + 8*A*a^5*b^3 + 3*B*a^4*b^
4 - 4*A*a^3*b^5)*cos(d*x + c)^2 + (4*A*a^8 - 9*B*a^7*b + 4*A*a^6*b^2 + 18*B*a^5*b^3 - 20*A*a^4*b^4 - 9*B*a^3*b
^5 + 12*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c) + (a^9*b - 2*a^7*b
^3 + a^5*b^5)*d), 1/6*(3*(B*a^8 - 2*A*a^7*b + 4*B*a^6*b^2 - 4*A*a^5*b^3 - 11*B*a^4*b^4 + 14*A*a^3*b^5 + 6*B*a^
2*b^6 - 8*A*a*b^7)*d*x*cos(d*x + c) + 3*(B*a^7*b - 2*A*a^6*b^2 + 4*B*a^5*b^3 - 4*A*a^4*b^4 - 11*B*a^3*b^5 + 14
*A*a^2*b^6 + 6*B*a*b^7 - 8*A*b^8)*d*x - 6*(4*B*a^3*b^4 - 5*A*a^2*b^5 - 3*B*a*b^6 + 4*A*b^7 + (4*B*a^4*b^3 - 5*
A*a^3*b^4 - 3*B*a^2*b^5 + 4*A*a*b^6)*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))) + (4*A*a^7*b - 12*B*a^6*b^2 + 10*A*a^5*b^3 + 30*B*a^4*b^4 - 38*A*a^3*b^5 - 18*
B*a^2*b^6 + 24*A*a*b^7 + 2*(A*a^8 - 2*A*a^6*b^2 + A*a^4*b^4)*cos(d*x + c)^3 + (3*B*a^8 - 4*A*a^7*b - 6*B*a^6*b
^2 + 8*A*a^5*b^3 + 3*B*a^4*b^4 - 4*A*a^3*b^5)*cos(d*x + c)^2 + (4*A*a^8 - 9*B*a^7*b + 4*A*a^6*b^2 + 18*B*a^5*b
^3 - 20*A*a^4*b^4 - 9*B*a^3*b^5 + 12*A*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*co
s(d*x + c) + (a^9*b - 2*a^7*b^3 + a^5*b^5)*d)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.49033, size = 639, normalized size = 1.85 \begin{align*} -\frac{\frac{12 \,{\left (4 \, B a^{3} b^{3} - 5 \, A a^{2} b^{4} - 3 \, B a b^{5} + 4 \, A 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^{7} - a^{5} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{12 \,{\left (B a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\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 )}} - \frac{3 \,{\left (B a^{3} - 2 \, A a^{2} b + 6 \, B a b^{2} - 8 \, A b^{3}\right )}{\left (d x + c\right )}}{a^{5}} - \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{4}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*(4*B*a^3*b^3 - 5*A*a^2*b^4 - 3*B*a*b^5 + 4*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^7 - a^5*b^2)*sqrt(-a^2 + b^2
)) + 12*(B*a*b^4*tan(1/2*d*x + 1/2*c) - A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - a^4*b^2)*(a*tan(1/2*d*x + 1/2*c)^2
- b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - 3*(B*a^3 - 2*A*a^2*b + 6*B*a*b^2 - 8*A*b^3)*(d*x + c)/a^5 - 2*(6*A*a^2
*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*b*tan(1/2*d
*x + 1/2*c)^5 + 18*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*
c)^3 + 36*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) - 6*A*a*b
*tan(1/2*d*x + 1/2*c) - 12*B*a*b*tan(1/2*d*x + 1/2*c) + 18*A*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^
2 + 1)^3*a^4))/d