\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [69]

Optimal result

Integrand size = 31, antiderivative size = 124 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {5}{8} a^3 (3 A+4 B) x+\frac {a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac {3 a^3 (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4098, 3876, 2717, 2715, 8, 2713} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d}+\frac {a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac {3 a^3 (3 A+4 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5}{8} a^3 x (3 A+4 B)+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2713

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4098

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

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} (3 A+4 B) \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx \\ & = \frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} (3 A+4 B) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx \\ & = \frac {1}{4} a^3 (3 A+4 B) x+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \left (a^3 (3 A+4 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (3 A+4 B)\right ) \int \cos (c+d x) \, dx+\frac {1}{4} \left (3 a^3 (3 A+4 B)\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {1}{4} a^3 (3 A+4 B) x+\frac {3 a^3 (3 A+4 B) \sin (c+d x)}{4 d}+\frac {3 a^3 (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 (3 A+4 B)\right ) \int 1 \, dx-\frac {\left (a^3 (3 A+4 B)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d} \\ & = \frac {5}{8} a^3 (3 A+4 B) x+\frac {a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac {3 a^3 (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \sin (c+d x) \left (72 A+88 B+9 (5 A+4 B) \cos (c+d x)+8 (3 A+B) \cos ^2(c+d x)+6 A \cos ^3(c+d x)+\frac {30 (3 A+4 B) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{24 d} \]

[In]

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

[Out]

(a^3*Sin[c + d*x]*(72*A + 88*B + 9*(5*A + 4*B)*Cos[c + d*x] + 8*(3*A + B)*Cos[c + d*x]^2 + 6*A*Cos[c + d*x]^3
+ (30*(3*A + 4*B)*ArcSin[Sqrt[Sin[(c + d*x)/2]^2]])/Sqrt[Sin[c + d*x]^2]))/(24*d)

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63

method	result	size
parallelrisch	\(\frac {\left (\left (32 A +24 B \right ) \sin \left (2 d x +2 c \right )+\left (8 A +\frac {8 B}{3}\right ) \sin \left (3 d x +3 c \right )+A \sin \left (4 d x +4 c \right )+\left (104 A +120 B \right ) \sin \left (d x +c \right )+60 d \left (A +\frac {4 B}{3}\right ) x \right ) a^{3}}{32 d}\)	\(78\)
risch	\(\frac {15 a^{3} A x}{8}+\frac {5 a^{3} x B}{2}+\frac {13 a^{3} A \sin \left (d x +c \right )}{4 d}+\frac {15 a^{3} B \sin \left (d x +c \right )}{4 d}+\frac {a^{3} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} A \sin \left (3 d x +3 c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} A}{d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}\)	\(135\)
derivativedivides	\(\frac {a^{3} A \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{3} \sin \left (d x +c \right )+a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(176\)
default	\(\frac {a^{3} A \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \,a^{3} \sin \left (d x +c \right )+a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(176\)
norman	\(\frac {-\frac {5 a^{3} \left (3 A +4 B \right ) x}{8}-\frac {47 a^{3} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {5 a^{3} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {5 a^{3} \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}-\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {15 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {5 a^{3} \left (3 A +4 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {a^{3} \left (5 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{3} \left (49 A +44 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (51 A +260 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a^{3} \left (111 A +52 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(394\)

[In]

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

[Out]

1/32*((32*A+24*B)*sin(2*d*x+2*c)+(8*A+8/3*B)*sin(3*d*x+3*c)+A*sin(4*d*x+4*c)+(104*A+120*B)*sin(d*x+c)+60*d*(A+
4/3*B)*x)*a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (3 \, A + 4 \, B\right )} a^{3} d x + {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \, {\left (5 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (9 \, A + 11 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

1/24*(15*(3*A + 4*B)*a^3*d*x + (6*A*a^3*cos(d*x + c)^3 + 8*(3*A + B)*a^3*cos(d*x + c)^2 + 9*(5*A + 4*B)*a^3*co
s(d*x + c) + 8*(9*A + 11*B)*a^3)*sin(d*x + c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.35 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=-\frac {96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 96 \, {\left (d x + c\right )} B a^{3} - 96 \, A a^{3} \sin \left (d x + c\right ) - 288 \, B a^{3} \sin \left (d x + c\right )}{96 \, d} \]

[In]

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

[Out]

-1/96*(96*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
A*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 72*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a^3 - 96*(d*x + c)*B*a^3 - 96*A*a^3*sin(d*x + c) - 288*B*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.42 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (3 \, A a^{3} + 4 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 147 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(15*(3*A*a^3 + 4*B*a^3)*(d*x + c) + 2*(45*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1/2*d*x + 1/2*c)^7
+ 165*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292
*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 147*A*a^3*tan(1/2*d*x + 1/2*c) + 132*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 + 1)^4)/d

Mupad [B] (verification not implemented)

Time = 13.82 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15\,A\,a^3\,x}{8}+\frac {5\,B\,a^3\,x}{2}+\frac {13\,A\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

(15*A*a^3*x)/8 + (5*B*a^3*x)/2 + (13*A*a^3*sin(c + d*x))/(4*d) + (15*B*a^3*sin(c + d*x))/(4*d) + (A*a^3*sin(2*
c + 2*d*x))/d + (A*a^3*sin(3*c + 3*d*x))/(4*d) + (A*a^3*sin(4*c + 4*d*x))/(32*d) + (3*B*a^3*sin(2*c + 2*d*x))/
(4*d) + (B*a^3*sin(3*c + 3*d*x))/(12*d)