\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [76]

Optimal result

Integrand size = 31, antiderivative size = 160 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a^4 (13 A+8 B) x+\frac {a^4 (8 A+13 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4102, 4103, 4081, 3855} \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 (8 A+13 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac {1}{2} a^4 x (13 A+8 B)-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

[In]

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

[Out]

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

Rule 3855

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

Rule 4081

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

Rule 4102

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

Rule 4103

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

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^3 (a (5 A+2 B)-2 a (A-B) \sec (c+d x)) \, dx \\ & = \frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {1}{4} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (6 A+B)+2 a^2 (A+6 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{4} \int \cos (c+d x) (a+a \sec (c+d x)) \left (10 a^3 (A-B)+2 a^3 (8 A+13 B) \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac {1}{4} \int \left (-2 a^4 (13 A+8 B)-2 a^4 (8 A+13 B) \sec (c+d x)\right ) \, dx \\ & = \frac {1}{2} a^4 (13 A+8 B) x+\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{2} \left (a^4 (8 A+13 B)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^4 (13 A+8 B) x+\frac {a^4 (8 A+13 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(373\) vs. \(2(160)=320\).

Time = 8.93 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.33 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 \cos ^5(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 (A+B \sec (c+d x)) \left (2 (13 A+8 B) x-\frac {2 (8 A+13 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (8 A+13 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (4 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (4 A+B) \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {B}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (A+4 B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {B}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (A+4 B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{64 (B+A \cos (c+d x))} \]

[In]

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

[Out]

(a^4*Cos[c + d*x]^5*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(2*(13*A + 8*B)*x - (2*(8*A +
13*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(8*A + 13*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])
/d + (4*(4*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(4*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*
c]*Sin[2*d*x])/d + B/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(A + 4*B)*Sin[(d*x)/2])/(d*(Cos[c/2] - S
in[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - B/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(A + 4*B)
*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(64*(B + A*Cos[c + d*x]))

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {2 \left (-2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {13 B}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {13 B}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {13 \left (A +\frac {8 B}{13}\right ) d x \cos \left (2 d x +2 c \right )}{4}+\left (\frac {5 A}{8}+2 B \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {B}{4}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{16}+\left (A +\frac {3 B}{4}\right ) \sin \left (d x +c \right )+\frac {13 \left (A +\frac {8 B}{13}\right ) d x}{4}\right ) a^{4}}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(166\)
derivativedivides	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} A \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )}{d}\)	\(177\)
default	\(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} A \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )}{d}\)	\(177\)
risch	\(\frac {13 a^{4} A x}{2}+4 a^{4} x B -\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {2 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{4} \left (B \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-8 B \,{\mathrm e}^{2 i \left (d x +c \right )}-B \,{\mathrm e}^{i \left (d x +c \right )}-2 A -8 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(294\)
norman	\(\frac {\left (\frac {13}{2} a^{4} A +4 B \,a^{4}\right ) x +\left (-\frac {13}{2} a^{4} A -4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {13}{2} a^{4} A -4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {13}{2} a^{4} A +4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-13 a^{4} A -8 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-13 a^{4} A -8 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (26 a^{4} A +16 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {5 a^{4} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {11 a^{4} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{4} \left (5 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{4} \left (11 A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{4} \left (17 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {a^{4} \left (31 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {a^{4} \left (8 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (8 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(407\)

[In]

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

[Out]

2*(-2*(1+cos(2*d*x+2*c))*(A+13/8*B)*ln(tan(1/2*d*x+1/2*c)-1)+2*(1+cos(2*d*x+2*c))*(A+13/8*B)*ln(tan(1/2*d*x+1/
2*c)+1)+13/4*(A+8/13*B)*d*x*cos(2*d*x+2*c)+(5/8*A+2*B)*sin(2*d*x+2*c)+(A+1/4*B)*sin(3*d*x+3*c)+1/16*A*sin(4*d*
x+4*c)+(A+3/4*B)*sin(d*x+c)+13/4*(A+8/13*B)*d*x)*a^4/d/(1+cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.98 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (13 \, A + 8 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{2} + {\left (8 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + B a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*(2*(13*A + 8*B)*a^4*d*x*cos(d*x + c)^2 + (8*A + 13*B)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (8*A + 13
*B)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4*cos(d*x + c)^3 + 2*(4*A + B)*a^4*cos(d*x + c)^2 + 2*(
A + 4*B)*a^4*cos(d*x + c) + B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

\[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^{4} \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**4*(Integral(A*cos(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(6*A*cos(c + d*
x)**2*sec(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)**2*sec(
c + d*x)**4, x) + Integral(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)**2*sec(c + d*x)**2,
x) + Integral(6*B*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(4*B*cos(c + d*x)**2*sec(c + d*x)**4, x) + Int
egral(B*cos(c + d*x)**2*sec(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 24 \, {\left (d x + c\right )} A a^{4} + 16 \, {\left (d x + c\right )} B a^{4} - B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{4} \sin \left (d x + c\right ) + 4 \, B a^{4} \sin \left (d x + c\right ) + 4 \, A a^{4} \tan \left (d x + c\right ) + 16 \, B a^{4} \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)^2*(a+a*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 + 24*(d*x + c)*A*a^4 + 16*(d*x + c)*B*a^4 - B*a^4*(2*sin(d*x + c)/
(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 8*A*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 12*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 16*A*a^4*sin(d*x + c) + 4*B*a^4
*sin(d*x + c) + 4*A*a^4*tan(d*x + c) + 16*B*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.44 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {{\left (13 \, A a^{4} + 8 \, B a^{4}\right )} {\left (d x + c\right )} + {\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 7 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*((13*A*a^4 + 8*B*a^4)*(d*x + c) + (8*A*a^4 + 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (8*A*a^4 + 13*
B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 5*B*a^4*tan(1/2*d*x + 1/2*c)^7
- 7*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 7*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*B*a^4*
tan(1/2*d*x + 1/2*c)^3 + 11*A*a^4*tan(1/2*d*x + 1/2*c) + 11*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
4 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 13.55 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.52 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

[In]

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

[Out]

(4*A*a^4*sin(c + d*x))/d + (B*a^4*sin(c + d*x))/d + (13*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(8*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/d + (13*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (4
*B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (B*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (A*a^4*cos(c + d*x)*sin(c
+ d*x))/(2*d)