\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [77]

Optimal result

Integrand size = 31, antiderivative size = 165 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a^4 (12 A+13 B) x+\frac {a^4 (A+4 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 165, 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} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 (A+4 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (12 A+13 B)+\frac {(2 A+B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(a^4*(12*A + 13*B)*x)/2 + (a^4*(A + 4*B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(2*A + B)*Sin[c + d*x])/(2*d) + (a*
A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((2*A + B)*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])
^2*Sin[c + d*x])/(2*d) - ((8*A - 3*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*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 ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (3 a (2 A+B)-a (A-3 B) \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (11 A+9 B)-a^2 (8 A-3 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (2 A+B)+6 a^3 (A+4 B) \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {1}{6} \int \left (-3 a^4 (12 A+13 B)-6 a^4 (A+4 B) \sec (c+d x)\right ) \, dx \\ & = \frac {1}{2} a^4 (12 A+13 B) x+\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^4 (A+4 B)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^4 (12 A+13 B) x+\frac {a^4 (A+4 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(2 A+B) \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(342\) vs. \(2(165)=330\).

Time = 4.95 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.07 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 \cos ^5(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 (A+B \sec (c+d x)) \left (72 A x+78 B x-\frac {12 (A+4 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 (A+4 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {3 (27 A+16 B) \cos (d x) \sin (c)}{d}+\frac {3 (4 A+B) \cos (2 d x) \sin (2 c)}{d}+\frac {A \cos (3 d x) \sin (3 c)}{d}+\frac {3 (27 A+16 B) \cos (c) \sin (d x)}{d}+\frac {3 (4 A+B) \cos (2 c) \sin (2 d x)}{d}+\frac {A \cos (3 c) \sin (3 d x)}{d}+\frac {12 B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{192 (B+A \cos (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^3*(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])*(72*A*x + 78*B*x - (12*(A + 4
*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (12*(A + 4*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d +
(3*(27*A + 16*B)*Cos[d*x]*Sin[c])/d + (3*(4*A + B)*Cos[2*d*x]*Sin[2*c])/d + (A*Cos[3*d*x]*Sin[3*c])/d + (3*(27
*A + 16*B)*Cos[c]*Sin[d*x])/d + (3*(4*A + B)*Cos[2*c]*Sin[2*d*x])/d + (A*Cos[3*c]*Sin[3*d*x])/d + (12*B*Sin[(d
*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (12*B*Sin[(d*x)/2])/(d*(Cos[c/2] + S
in[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(192*(B + A*Cos[c + d*x]))

Maple [A] (verified)

Time = 2.85 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {\left (-2 \cos \left (d x +c \right ) \left (A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \cos \left (d x +c \right ) \left (A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\frac {41 A}{6}+4 B \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {B}{4}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{12}+12 \left (A +\frac {13 B}{12}\right ) d x \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (A +\frac {9 B}{4}\right )\right ) a^{4}}{2 d \cos \left (d x +c \right )}\)	\(139\)
derivativedivides	\(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \sin \left (d x +c \right )+6 B \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(179\)
default	\(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \sin \left (d x +c \right )+6 B \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(179\)
risch	\(6 a^{4} A x +\frac {13 a^{4} x B}{2}-\frac {27 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{d}+\frac {27 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}+\frac {2 i B \,a^{4}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {a^{4} A \sin \left (3 d x +3 c \right )}{12 d}\)	\(296\)
norman	\(\frac {\left (6 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x +\left (-18 a^{4} A -\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-18 a^{4} A -\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-6 a^{4} A -\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-6 a^{4} A -\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (6 a^{4} A +\frac {13}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (18 a^{4} A +\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (18 a^{4} A +\frac {39}{2} B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a^{4} \left (18 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 a^{4} \left (2 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {8 a^{4} \left (5 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {20 a^{4} \left (7 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {7 a^{4} \left (10 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {4 a^{4} \left (11 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a^{4} \left (50 A -27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a^{4} \left (A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} \left (A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(457\)

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2*(-2*cos(d*x+c)*(A+4*B)*ln(tan(1/2*d*x+1/2*c)-1)+2*cos(d*x+c)*(A+4*B)*ln(tan(1/2*d*x+1/2*c)+1)+(41/6*A+4*B)
*sin(2*d*x+2*c)+(A+1/4*B)*sin(3*d*x+3*c)+1/12*A*sin(4*d*x+4*c)+12*(A+13/12*B)*d*x*cos(d*x+c)+sin(d*x+c)*(A+9/4
*B))*a^4/d/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.91 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (12 \, A + 13 \, B\right )} a^{4} d x \cos \left (d x + c\right ) + 3 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (5 \, A + 3 \, B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, B a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \, {\left (d x + c\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 72 \, {\left (d x + c\right )} B a^{4} - 6 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 48 \, B a^{4} \sin \left (d x + c\right ) - 12 \, B a^{4} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 48*(d*x + c)*A*
a^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 72*(d*x + c)*B*a^4 - 6*A*a^4*(log(sin(d*x + c) + 1) - log(sin
(d*x + c) - 1)) - 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 48*B*a^4*
sin(d*x + c) - 12*B*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.37 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=-\frac {\frac {12 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (12 \, A a^{4} + 13 \, B a^{4}\right )} {\left (d x + c\right )} - 6 \, {\left (A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (A a^{4} + 4 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (30 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(12*B*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(12*A*a^4 + 13*B*a^4)*(d*x + c) - 6*(A*a^
4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(A*a^4 + 4*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
30*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^4
*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)
^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.47 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4,x)

[Out]

(20*A*a^4*sin(c + d*x))/(3*d) + (4*B*a^4*sin(c + d*x))/d + (12*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/d + (2*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2)))/d + (8*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*cos(c + d*x)^2*sin(c + d*x))
/(3*d) + (B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (2*A*a^4*cos(c + d*x)*sin(c + d*x))/d + (B*a^4*cos(c + d*x)*s
in(c + d*x))/(2*d)