3.306 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=209 \[ \frac {b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )-\frac {b \left (4 a^3 B+13 a^2 A b-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac {a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

[Out]

1/2*a^2*(A*a^2+12*A*b^2+8*B*a*b)*x+1/2*b^2*(8*A*a*b+12*B*a^2+B*b^2)*arctanh(sin(d*x+c))/d+1/2*a*(5*A*b+2*B*a)*
(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/2*b*(13*A*a^2*b-2*A*b^3+4
*B*a^3-8*B*a*b^2)*tan(d*x+c)/d-1/2*b^2*(6*A*a*b+2*B*a^2-B*b^2)*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.46, antiderivative size = 209, normalized size of antiderivative = 1.00, 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 = {4025, 4094, 4048, 3770, 3767, 8} \[ -\frac {b \left (13 a^2 A b+4 a^3 B-8 a b^2 B-2 A b^3\right ) \tan (c+d x)}{2 d}+\frac {b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^2 \left (2 a^2 B+6 a A b-b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} a^2 x \left (a^2 A+8 a b B+12 A b^2\right )+\frac {a (2 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*x)/2 + (b^2*(8*a*A*b + 12*a^2*B + b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(
5*A*b + 2*a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (a*A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d
*x])/(2*d) - (b*(13*a^2*A*b - 2*A*b^3 + 4*a^3*B - 8*a*b^2*B)*Tan[c + d*x])/(2*d) - (b^2*(6*a*A*b + 2*a^2*B - b
^2*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4025

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

Rule 4048

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

Rule 4094

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*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (-a (5 A b+2 a B)-\left (a^2 A+2 A b^2+4 a b B\right ) \sec (c+d x)+2 b (a A-b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {1}{2} \int (a+b \sec (c+d x)) \left (-a \left (a^2 A+12 A b^2+8 a b B\right )+b \left (a^2 A-2 A b^2-6 a b B\right ) \sec (c+d x)+2 b \left (6 a A b+2 a^2 B-b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{4} \int \left (-2 a^2 \left (a^2 A+12 A b^2+8 a b B\right )-2 b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \sec (c+d x)+2 b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac {a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (b^2 \left (8 a A b+12 a^2 B+b^2 B\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{2} \left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac {b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\left (b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac {1}{2} a^2 \left (a^2 A+12 A b^2+8 a b B\right ) x+\frac {b^2 \left (8 a A b+12 a^2 B+b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (5 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b \left (13 a^2 A b-2 A b^3+4 a^3 B-8 a b^2 B\right ) \tan (c+d x)}{2 d}-\frac {b^2 \left (6 a A b+2 a^2 B-b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 2.01, size = 310, normalized size = 1.48 \[ \frac {a^4 A \sin (2 (c+d x))+4 a^3 (a B+4 A b) \sin (c+d x)+2 a^2 (c+d x) \left (a^2 A+8 a b B+12 A b^2\right )-2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b^2 \left (12 a^2 B+8 a A b+b^2 B\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 b^3 (4 a B+A b) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b^3 (4 a B+A b) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {b^4 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^4 B}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(a^2*A + 12*A*b^2 + 8*a*b*B)*(c + d*x) - 2*b^2*(8*a*A*b + 12*a^2*B + b^2*B)*Log[Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]] + 2*b^2*(8*a*A*b + 12*a^2*B + b^2*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^4*B)/(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b^3*(A*b + 4*a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
- (b^4*B)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b^3*(A*b + 4*a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]) + 4*a^3*(4*A*b + a*B)*Sin[c + d*x] + a^4*A*Sin[2*(c + d*x)])/(4*d)

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fricas [A]  time = 0.49, size = 202, normalized size = 0.97 \[ \frac {2 \, {\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{2} + {\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{4} \cos \left (d x + c\right )^{3} + B b^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(A*a^4 + 8*B*a^3*b + 12*A*a^2*b^2)*d*x*cos(d*x + c)^2 + (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*cos(d*x + c)
^2*log(sin(d*x + c) + 1) - (12*B*a^2*b^2 + 8*A*a*b^3 + B*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4
*cos(d*x + c)^3 + B*b^4 + 2*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 2*(4*B*a*b^3 + A*b^4)*cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^2)

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giac [B]  time = 2.11, size = 528, normalized size = 2.53 \[ \frac {{\left (A a^{4} + 8 \, B a^{3} b + 12 \, A a^{2} b^{2}\right )} {\left (d x + c\right )} + {\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (12 \, B a^{2} b^{2} + 8 \, A a b^{3} + B b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{4} \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 )^{4} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.07, size = 236, normalized size = 1.13 \[ \frac {A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {A \,a^{4} x}{2}+\frac {A \,a^{4} c}{2 d}+\frac {a^{4} B \sin \left (d x +c \right )}{d}+\frac {4 A \,a^{3} b \sin \left (d x +c \right )}{d}+4 B x \,a^{3} b +\frac {4 B \,a^{3} b c}{d}+6 A x \,a^{2} b^{2}+\frac {6 A \,a^{2} b^{2} c}{d}+\frac {6 a^{2} b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 B a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right )}{d}+\frac {B \,b^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+1/2*A*a^4*x+1/2/d*A*a^4*c+1/d*a^4*B*sin(d*x+c)+4/d*A*a^3*b*sin(d*x+c)+4*B*x*
a^3*b+4/d*B*a^3*b*c+6*A*x*a^2*b^2+6/d*A*a^2*b^2*c+6/d*a^2*b^2*B*ln(sec(d*x+c)+tan(d*x+c))+4/d*a*A*b^3*ln(sec(d
*x+c)+tan(d*x+c))+4/d*B*a*b^3*tan(d*x+c)+1/d*A*b^4*tan(d*x+c)+1/2/d*B*b^4*sec(d*x+c)*tan(d*x+c)+1/2/d*B*b^4*ln
(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.72, size = 209, normalized size = 1.00 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 16 \, {\left (d x + c\right )} B a^{3} b + 24 \, {\left (d x + c\right )} A a^{2} b^{2} - B b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{4} \sin \left (d x + c\right ) + 16 \, A a^{3} b \sin \left (d x + c\right ) + 16 \, B a b^{3} \tan \left (d x + c\right ) + 4 \, A b^{4} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 16*(d*x + c)*B*a^3*b + 24*(d*x + c)*A*a^2*b^2 - B*b^4*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*B*a^2*b^2*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 8*A*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^4*sin(d*x + c
) + 16*A*a^3*b*sin(d*x + c) + 16*B*a*b^3*tan(d*x + c) + 4*A*b^4*tan(d*x + c))/d

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mupad [B]  time = 4.39, size = 330, normalized size = 1.58 \[ \frac {2\,\left (\frac {A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {B\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+4\,A\,a\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+4\,B\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,A\,a^2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a^2\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {A\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{4}+\frac {B\,b^4\,\sin \left (c+d\,x\right )}{2}+A\,a^3\,b\,\sin \left (c+d\,x\right )+A\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )+2\,B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4,x)

[Out]

(2*((A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/2 + 4*A*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 4*B*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d
*x)/2)) + 6*A*a^2*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*B*a^2*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c
/2 + (d*x)/2))))/d + ((A*a^4*sin(2*c + 2*d*x))/8 + (A*a^4*sin(4*c + 4*d*x))/16 + (A*b^4*sin(2*c + 2*d*x))/2 +
(B*a^4*sin(3*c + 3*d*x))/4 + (B*a^4*sin(c + d*x))/4 + (B*b^4*sin(c + d*x))/2 + A*a^3*b*sin(c + d*x) + A*a^3*b*
sin(3*c + 3*d*x) + 2*B*a*b^3*sin(2*c + 2*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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