3.683 \(\int \frac{\cos ^4(c+d x) (A+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=232 \[ -\frac{b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{2 b^3 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right )}{8 a^5}-\frac{A b \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d}+\frac{A \sin (c+d x) \cos ^3(c+d x)}{4 a d} \]

[Out]

((8*A*b^4 + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*x)/(8*a^5) - (2*b^3*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Ta
n[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqrt[a - b]*Sqrt[a + b]*d) - (b*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[c + d*x])/(
3*a^4*d) + ((4*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) - (A*b*Cos[c + d*x]^2*Sin[c + d*x
])/(3*a^2*d) + (A*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.927148, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4105, 4104, 3919, 3831, 2659, 208} \[ -\frac{b \left (a^2 (2 A+3 C)+3 A b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{2 b^3 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right )}{8 a^5}-\frac{A b \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d}+\frac{A \sin (c+d x) \cos ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

((8*A*b^4 + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*x)/(8*a^5) - (2*b^3*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Ta
n[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqrt[a - b]*Sqrt[a + b]*d) - (b*(3*A*b^2 + a^2*(2*A + 3*C))*Sin[c + d*x])/(
3*a^4*d) + ((4*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) - (A*b*Cos[c + d*x]^2*Sin[c + d*x
])/(3*a^2*d) + (A*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d)

Rule 4105

Int[((A_.) + 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*(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, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]

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 ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\int \frac{\cos ^3(c+d x) \left (4 A b-a (3 A+4 C) \sec (c+d x)-3 A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{4 a}\\ &=-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (4 A b^2+a^2 (3 A+4 C)\right )+a A b \sec (c+d x)-8 A b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{12 a^2}\\ &=\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\int \frac{\cos (c+d x) \left (8 b \left (3 A b^2+a^2 (2 A+3 C)\right )+a \left (4 A b^2-3 a^2 (3 A+4 C)\right ) \sec (c+d x)-3 b \left (4 A b^2+a^2 (3 A+4 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{24 a^3}\\ &=-\frac{b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac{\int \frac{3 \left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )+3 a b \left (4 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{24 a^4}\\ &=\frac{\left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac{b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\left (b^3 \left (A b^2+a^2 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5}\\ &=\frac{\left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac{b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\left (b^2 \left (A b^2+a^2 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5}\\ &=\frac{\left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac{b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\left (2 b^2 \left (A b^2+a^2 C\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 d}\\ &=\frac{\left (8 A b^4+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) x}{8 a^5}-\frac{2 b^3 \left (A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 \sqrt{a-b} \sqrt{a+b} d}-\frac{b \left (3 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{A b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{A \cos ^3(c+d x) \sin (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.658355, size = 191, normalized size = 0.82 \[ \frac{12 (c+d x) \left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right )+24 a^2 \left (a^2 (A+C)+A b^2\right ) \sin (2 (c+d x))-24 a b \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)+\frac{192 b^3 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-8 a^3 A b \sin (3 (c+d x))+3 a^4 A \sin (4 (c+d x))}{96 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

(12*(8*A*b^4 + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*(c + d*x) + (192*b^3*(A*b^2 + a^2*C)*ArcTanh[((-a + b)*T
an[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 24*a*b*(4*A*b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] + 24*a^2*
(A*b^2 + a^2*(A + C))*Sin[2*(c + d*x)] - 8*a^3*A*b*Sin[3*(c + d*x)] + 3*a^4*A*Sin[4*(c + d*x)])/(96*a^5*d)

________________________________________________________________________________________

Maple [B]  time = 0.125, size = 1060, normalized size = 4.6 \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)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

-6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A*b^3+1/a/d*arctan(tan(1/2*d*x+1/2*c))*C+3/4/a/d*A*ar
ctan(tan(1/2*d*x+1/2*c))-2/d*b^5/a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))
*A-10/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A*b-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)*A*b-2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*A*b^3-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*t
an(1/2*d*x+1/2*c)^7*b*C-2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*A*b^3-10/3/d/a^2/(1+tan(1/2*d*
x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A*b-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*A*b-6/d/a^2/(1+
tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b*C-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b*C-6/d
/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A*b^3-6/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^5*b*C-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A*b^2+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(
1/2*d*x+1/2*c)*A*b^2+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*C-5/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*
tan(1/2*d*x+1/2*c)^7*A-3/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A+1/d/a/(1+tan(1/2*d*x+1/2*c)^2
)^4*tan(1/2*d*x+1/2*c)^3*C-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*C+2/d/a^3*arctan(tan(1/2*d*x+
1/2*c))*C*b^2+2/d/a^5*arctan(tan(1/2*d*x+1/2*c))*A*b^4+5/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*A
-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*C+1/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A*b^2+3/4/d/a/(1+t
an(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A*b^2-2/
d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/d/a^3/(1+tan(1/2*d*x+1
/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*A*b^2

________________________________________________________________________________________

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)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.662604, size = 1328, normalized size = 5.72 \begin{align*} \left [\frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{6} +{\left (A + 4 \, C\right )} a^{4} b^{2} + 4 \,{\left (A - 2 \, C\right )} a^{2} b^{4} - 8 \, A b^{6}\right )} d x + 12 \,{\left (C a^{2} b^{3} + A b^{5}\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) -{\left (8 \,{\left (2 \, A + 3 \, C\right )} a^{5} b + 8 \,{\left (A - 3 \, C\right )} a^{3} b^{3} - 24 \, A a b^{5} - 6 \,{\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (A a^{5} b - A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{6} +{\left (A - 4 \, C\right )} a^{4} b^{2} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{7} - a^{5} b^{2}\right )} d}, \frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{6} +{\left (A + 4 \, C\right )} a^{4} b^{2} + 4 \,{\left (A - 2 \, C\right )} a^{2} b^{4} - 8 \, A b^{6}\right )} d x - 24 \,{\left (C a^{2} b^{3} + A b^{5}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) -{\left (8 \,{\left (2 \, A + 3 \, C\right )} a^{5} b + 8 \,{\left (A - 3 \, C\right )} a^{3} b^{3} - 24 \, A a b^{5} - 6 \,{\left (A a^{6} - A a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (A a^{5} b - A a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{6} +{\left (A - 4 \, C\right )} a^{4} b^{2} - 4 \, A a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{7} - a^{5} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/24*(3*((3*A + 4*C)*a^6 + (A + 4*C)*a^4*b^2 + 4*(A - 2*C)*a^2*b^4 - 8*A*b^6)*d*x + 12*(C*a^2*b^3 + A*b^5)*sq
rt(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)) - (8*(2*A + 3*C)*a^5*b + 8*(A - 3
*C)*a^3*b^3 - 24*A*a*b^5 - 6*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^3 + 8*(A*a^5*b - A*a^3*b^3)*cos(d*x + c)^2 - 3*(
(3*A + 4*C)*a^6 + (A - 4*C)*a^4*b^2 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d), 1/24*(3*((
3*A + 4*C)*a^6 + (A + 4*C)*a^4*b^2 + 4*(A - 2*C)*a^2*b^4 - 8*A*b^6)*d*x - 24*(C*a^2*b^3 + A*b^5)*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))) - (8*(2*A + 3*C)*a^5*b + 8*(A -
3*C)*a^3*b^3 - 24*A*a*b^5 - 6*(A*a^6 - A*a^4*b^2)*cos(d*x + c)^3 + 8*(A*a^5*b - A*a^3*b^3)*cos(d*x + c)^2 - 3*
((3*A + 4*C)*a^6 + (A - 4*C)*a^4*b^2 - 4*A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*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)**4*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.38957, size = 775, normalized size = 3.34 \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)^4*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^4 + 4*C*a^4 + 4*A*a^2*b^2 + 8*C*a^2*b^2 + 8*A*b^4)*(d*x + c)/a^5 - 48*(C*a^2*b^3 + A*b^5)*(pi*f
loor(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)))/(sqrt(-a^2 + b^2)*a^5) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^7 +
24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24
*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^2*b*
tan(1/2*d*x + 1/2*c)^5 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*A*b^3*tan(
1/2*d*x + 1/2*c)^5 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 72*A*b^3*tan(1/2*d*x + 1
/2*c)^3 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^3*tan(1/2*d*x + 1/2*c) + 24*A*a^2*b*tan(1/2*d*x + 1/2*c) + 24
*C*a^2*b*tan(1/2*d*x + 1/2*c) - 12*A*a*b^2*tan(1/2*d*x + 1/2*c) + 24*A*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x
+ 1/2*c)^2 + 1)^4*a^4))/d