3.5.84 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \, dx\) [484]

Optimal. Leaf size=173 \[ \frac {1}{2} a b \left (3 a^2+4 b^2\right ) x+\frac {\left (4 a^4+29 a^2 b^2+5 b^4\right ) \sin (c+d x)}{5 d}+\frac {a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d} \]

[Out]

1/2*a*b*(3*a^2+4*b^2)*x+1/5*(4*a^4+29*a^2*b^2+5*b^4)*sin(d*x+c)/d+1/2*a*b*(3*a^2+4*b^2)*cos(d*x+c)*sin(d*x+c)/
d+3/5*a^3*b*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a^2*cos(d*x+c)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*a^2*(4*a^2+27*
b^2)*sin(d*x+c)^3/d

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Rubi [A]
time = 0.25, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3926, 4159, 4132, 2715, 8, 4129, 3092} \begin {gather*} \frac {3 a^3 b \sin (c+d x) \cos ^3(c+d x)}{5 d}-\frac {a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {a b \left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a b x \left (3 a^2+4 b^2\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}+\frac {\left (4 a^4+29 a^2 b^2+5 b^4\right ) \sin (c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4,x]

[Out]

(a*b*(3*a^2 + 4*b^2)*x)/2 + ((4*a^4 + 29*a^2*b^2 + 5*b^4)*Sin[c + d*x])/(5*d) + (a*b*(3*a^2 + 4*b^2)*Cos[c + d
*x]*Sin[c + d*x])/(2*d) + (3*a^3*b*Cos[c + d*x]^3*Sin[c + d*x])/(5*d) + (a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x
])^2*Sin[c + d*x])/(5*d) - (a^2*(4*a^2 + 27*b^2)*Sin[c + d*x]^3)/(15*d)

Rule 8

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

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 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 3926

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^2*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 4132

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

Rule 4159

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_)), 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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (12 a^2 b+a \left (4 a^2+15 b^2\right ) \sec (c+d x)+b \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos ^3(c+d x) \left (-4 a^2 \left (4 a^2+27 b^2\right )-20 a b \left (3 a^2+4 b^2\right ) \sec (c+d x)-4 b^2 \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos ^3(c+d x) \left (-4 a^2 \left (4 a^2+27 b^2\right )-4 b^2 \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\left (a b \left (3 a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos (c+d x) \left (-4 b^2 \left (2 a^2+5 b^2\right )-4 a^2 \left (4 a^2+27 b^2\right ) \cos ^2(c+d x)\right ) \, dx+\frac {1}{2} \left (a b \left (3 a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac {1}{2} a b \left (3 a^2+4 b^2\right ) x+\frac {a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {\text {Subst}\left (\int \left (-4 b^2 \left (2 a^2+5 b^2\right )-4 a^2 \left (4 a^2+27 b^2\right )+4 a^2 \left (4 a^2+27 b^2\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{2} a b \left (3 a^2+4 b^2\right ) x+\frac {\left (4 a^4+29 a^2 b^2+5 b^4\right ) \sin (c+d x)}{5 d}+\frac {a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 0.52, size = 133, normalized size = 0.77 \begin {gather*} \frac {30 \left (5 a^4+36 a^2 b^2+8 b^4\right ) \sin (c+d x)+a \left (360 a^2 b c+480 b^3 c+360 a^2 b d x+480 b^3 d x+240 b \left (a^2+b^2\right ) \sin (2 (c+d x))+5 \left (5 a^3+24 a b^2\right ) \sin (3 (c+d x))+30 a^2 b \sin (4 (c+d x))+3 a^3 \sin (5 (c+d x))\right )}{240 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4,x]

[Out]

(30*(5*a^4 + 36*a^2*b^2 + 8*b^4)*Sin[c + d*x] + a*(360*a^2*b*c + 480*b^3*c + 360*a^2*b*d*x + 480*b^3*d*x + 240
*b*(a^2 + b^2)*Sin[2*(c + d*x)] + 5*(5*a^3 + 24*a*b^2)*Sin[3*(c + d*x)] + 30*a^2*b*Sin[4*(c + d*x)] + 3*a^3*Si
n[5*(c + d*x)]))/(240*d)

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Maple [A]
time = 0.14, size = 138, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 b \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 b^{2} a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 b^{3} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{4} \sin \left (d x +c \right )}{d}\)	\(138\)
default	\(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 b \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 b^{2} a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 b^{3} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{4} \sin \left (d x +c \right )}{d}\)	\(138\)
risch	\(\frac {3 a^{3} b x}{2}+2 a \,b^{3} x +\frac {5 a^{4} \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) b^{2} a^{2}}{2 d}+\frac {\sin \left (d x +c \right ) b^{4}}{d}+\frac {a^{4} \sin \left (5 d x +5 c \right )}{80 d}+\frac {b \,a^{3} \sin \left (4 d x +4 c \right )}{8 d}+\frac {5 a^{4} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) b^{2} a^{2}}{2 d}+\frac {b \,a^{3} \sin \left (2 d x +2 c \right )}{d}+\frac {\sin \left (2 d x +2 c \right ) b^{3} a}{d}\)	\(166\)
norman	\(\frac {\left (-\frac {3}{2} b \,a^{3}-2 b^{3} a \right ) x +\left (\frac {3}{2} b \,a^{3}+2 b^{3} a \right ) x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-9 b \,a^{3}-12 b^{3} a \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 b \,a^{3}-4 b^{3} a \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 b \,a^{3}-4 b^{3} a \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 b \,a^{3}+4 b^{3} a \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 b \,a^{3}+4 b^{3} a \right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (9 b \,a^{3}+12 b^{3} a \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 a^{4}-5 b \,a^{3}+12 b^{2} a^{2}-4 b^{3} a +2 b^{4}\right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a^{4}+5 b \,a^{3}+12 b^{2} a^{2}+4 b^{3} a +2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (10 a^{4}-39 b \,a^{3}+12 b^{2} a^{2}-12 b^{3} a -6 b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (10 a^{4}+39 b \,a^{3}+12 b^{2} a^{2}+12 b^{3} a -6 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (86 a^{4}-135 b \,a^{3}-300 b^{2} a^{2}+180 b^{3} a -90 b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (86 a^{4}+135 b \,a^{3}-300 b^{2} a^{2}-180 b^{3} a -90 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {\left (218 a^{4}-15 b \,a^{3}+60 b^{2} a^{2}+180 b^{3} a +90 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {\left (218 a^{4}+15 b \,a^{3}+60 b^{2} a^{2}-180 b^{3} a +90 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(603\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/5*a^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+4*b*a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+2*b^2*a^2*(2+cos(d*x+c)^2)*sin(d*x+c)+4*b^3*a*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+b^4
*sin(d*x+c))

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Maxima [A]
time = 0.26, size = 133, normalized size = 0.77 \begin {gather*} \frac {8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} + 15 \, {\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^{3} b - 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} + 120 \, b^{4} \sin \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/120*(8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*
c))*a*b^3 + 120*b^4*sin(d*x + c))/d

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Fricas [A]
time = 2.81, size = 121, normalized size = 0.70 \begin {gather*} \frac {15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} d x + {\left (6 \, a^{4} \cos \left (d x + c\right )^{4} + 30 \, a^{3} b \cos \left (d x + c\right )^{3} + 16 \, a^{4} + 120 \, a^{2} b^{2} + 30 \, b^{4} + 4 \, {\left (2 \, a^{4} + 15 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/30*(15*(3*a^3*b + 4*a*b^3)*d*x + (6*a^4*cos(d*x + c)^4 + 30*a^3*b*cos(d*x + c)^3 + 16*a^4 + 120*a^2*b^2 + 30
*b^4 + 4*(2*a^4 + 15*a^2*b^2)*cos(d*x + c)^2 + 15*(3*a^3*b + 4*a*b^3)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))**4,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (161) = 322\).
time = 0.46, size = 425, normalized size = 2.46 \begin {gather*} \frac {15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (30 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 180 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 30 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 116 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 600 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 180 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, 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 )^{2} + 1\right )}^{5}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/30*(15*(3*a^3*b + 4*a*b^3)*(d*x + c) + 2*(30*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*a^3*b*tan(1/2*d*x + 1/2*c)^9 +
180*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*b^4*tan(1/2*d*x + 1/2*c)^9 + 40*a^4*
tan(1/2*d*x + 1/2*c)^7 - 30*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*a*b^3*tan(
1/2*d*x + 1/2*c)^7 + 120*b^4*tan(1/2*d*x + 1/2*c)^7 + 116*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^5 + 180*b^4*tan(1/2*d*x + 1/2*c)^5 + 40*a^4*tan(1/2*d*x + 1/2*c)^3 + 30*a^3*b*tan(1/2*d*x + 1/2*c)^3
+ 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*b^4*tan(1/2*d*x + 1/2*c)^3 + 30
*a^4*tan(1/2*d*x + 1/2*c) + 75*a^3*b*tan(1/2*d*x + 1/2*c) + 180*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*a*b^3*tan(1/
2*d*x + 1/2*c) + 30*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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Mupad [B]
time = 3.82, size = 330, normalized size = 1.91 \begin {gather*} \frac {\left (2\,a^4-5\,a^3\,b+12\,a^2\,b^2-4\,a\,b^3+2\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a^4}{3}-2\,a^3\,b+32\,a^2\,b^2-8\,a\,b^3+8\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^4}{15}+40\,a^2\,b^2+12\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,a^4}{3}+2\,a^3\,b+32\,a^2\,b^2+8\,a\,b^3+8\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^4+5\,a^3\,b+12\,a^2\,b^2+4\,a\,b^3+2\,b^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,b\,\mathrm {atan}\left (\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+4\,b^2\right )}{3\,a^3\,b+4\,a\,b^3}\right )\,\left (3\,a^2+4\,b^2\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + b/cos(c + d*x))^4,x)

[Out]

(tan(c/2 + (d*x)/2)^5*((116*a^4)/15 + 12*b^4 + 40*a^2*b^2) + tan(c/2 + (d*x)/2)^9*(2*a^4 - 5*a^3*b - 4*a*b^3 +
2*b^4 + 12*a^2*b^2) + tan(c/2 + (d*x)/2)^3*(8*a*b^3 + 2*a^3*b + (8*a^4)/3 + 8*b^4 + 32*a^2*b^2) + tan(c/2 + (
d*x)/2)^7*((8*a^4)/3 - 2*a^3*b - 8*a*b^3 + 8*b^4 + 32*a^2*b^2) + tan(c/2 + (d*x)/2)*(4*a*b^3 + 5*a^3*b + 2*a^4
+ 2*b^4 + 12*a^2*b^2))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan
(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (a*b*atan((a*b*tan(c/2 + (d*x)/2)*(3*a^2 + 4*b^2))/(4*a*b^3
+ 3*a^3*b))*(3*a^2 + 4*b^2))/d

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