\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [68]

Optimal result

Integrand size = 31, antiderivative size = 125 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a^3 (5 A+7 B) x+\frac {a^3 B \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 A+3 B) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^3*(5*A+7*B)*x+a^3*B*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*sin(d*x+c)/d+1/3*a*A*cos(d*x+c)^2*(a+a*sec(d*x+c
))^2*sin(d*x+c)/d+1/6*(5*A+3*B)*cos(d*x+c)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4102, 4081, 3855} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(5 A+3 B) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {1}{2} a^3 x (5 A+7 B)+\frac {a^3 B \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[In]

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

[Out]

(a^3*(5*A + 7*B)*x)/2 + (a^3*B*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (a*A*Cos[c + d*
x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((5*A + 3*B)*Cos[c + d*x]*(a^3 + a^3*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]

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 A+3 B)+3 a B \sec (c+d x)) \, dx \\ & = \frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 A+3 B) \cos (c+d x) \left (a^3+a^3 \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^2 (A+B)+6 a^2 B \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 A+3 B) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {1}{6} \int \left (-3 a^3 (5 A+7 B)-6 a^3 B \sec (c+d x)\right ) \, dx \\ & = \frac {1}{2} a^3 (5 A+7 B) x+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 A+3 B) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 B\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (5 A+7 B) x+\frac {a^3 B \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 A+3 B) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (30 A d x+42 B d x-12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 (5 A+4 B) \sin (c+d x)+3 (3 A+B) \sin (2 (c+d x))+A \sin (3 (c+d x))\right )}{12 d} \]

[In]

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

[Out]

(a^3*(30*A*d*x + 42*B*d*x - 12*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*B*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 9*(5*A + 4*B)*Sin[c + d*x] + 3*(3*A + B)*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(12*d)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {5 \left (-\frac {2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{5}+\frac {2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{5}+\frac {\left (3 A +B \right ) \sin \left (2 d x +2 c \right )}{10}+\frac {A \sin \left (3 d x +3 c \right )}{30}+3 \left (\frac {A}{2}+\frac {2 B}{5}\right ) \sin \left (d x +c \right )+d x \left (A +\frac {7 B}{5}\right )\right ) a^{3}}{2 d}\)	\(91\)
derivativedivides	\(\frac {a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{3} \sin \left (d x +c \right )+\frac {a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(147\)
default	\(\frac {a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{3} \sin \left (d x +c \right )+\frac {a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(147\)
risch	\(\frac {5 a^{3} A x}{2}+\frac {7 a^{3} x B}{2}-\frac {15 i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {15 i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {a^{3} A \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{3} A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}\)	\(189\)
norman	\(\frac {\left (-\frac {5}{2} a^{3} A -\frac {7}{2} B \,a^{3}\right ) x +\left (-\frac {15}{2} a^{3} A -\frac {21}{2} B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {5}{2} a^{3} A +\frac {7}{2} B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {15}{2} a^{3} A +\frac {21}{2} B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {14 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {5 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a^{3} \left (A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (5 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a^{3} \left (11 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (59 A +27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(312\)

[In]

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

[Out]

5/2*(-2/5*B*ln(tan(1/2*d*x+1/2*c)-1)+2/5*B*ln(tan(1/2*d*x+1/2*c)+1)+1/10*(3*A+B)*sin(2*d*x+2*c)+1/30*A*sin(3*d
*x+3*c)+3*(1/2*A+2/5*B)*sin(d*x+c)+d*x*(A+7/5*B))*a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (5 \, A + 7 \, B\right )} a^{3} d x + 3 \, B a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, {\left (11 \, A + 9 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(5*A + 7*B)*a^3*d*x + 3*B*a^3*log(sin(d*x + c) + 1) - 3*B*a^3*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d*x
+ c)^2 + 3*(3*A + B)*a^3*cos(d*x + c) + 2*(11*A + 9*B)*a^3)*sin(d*x + c))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.18 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 12 \, {\left (d x + c\right )} A a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 36 \, {\left (d x + c\right )} B a^{3} - 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} \sin \left (d x + c\right ) - 36 \, B a^{3} \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(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^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 12*(d*x + c)*A*a
^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 36*(d*x + c)*B*a^3 - 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(
d*x + c) - 1)) - 36*A*a^3*sin(d*x + c) - 36*B*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.44 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(5*A*a^3 + 7*
B*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 36*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*A*a^3*tan(1/2*d*x + 1/2*c) + 21*B*a^3*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 = 13.92 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.42 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15\,A\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

(15*A*a^3*sin(c + d*x))/(4*d) + (3*B*a^3*sin(c + d*x))/d + (5*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (7*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (3*A*a^3*sin(2*c + 2*d*x))/(4*d) + (A*a^3*sin(3*c + 3*d*x))/(12*d) + (B*a^3*sin(2*c + 2*d*x))/(4
*d)