3.2.18 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [118]

Optimal. Leaf size=207 \[ \frac {1}{2} a^4 (7 A+12 C) x+\frac {a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^4*(7*A+12*C)*x+a^4*C*arctanh(sin(d*x+c))/d+1/2*a^4*(7*A+10*C)*sin(d*x+c)/d+1/5*a*A*cos(d*x+c)^3*(a+a*sec
(d*x+c))^3*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+1/15*(7*A+5*C)*cos(d*x+c)^2*(a^2+a^
2*sec(d*x+c))^2*sin(d*x+c)/d+1/6*(7*A+8*C)*cos(d*x+c)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]
time = 0.37, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4172, 4102, 4081, 3855} \begin {gather*} \frac {a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac {(7 A+8 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (7 A+12 C)+\frac {a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(7 A+5 C) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(7*A + 12*C)*x)/2 + (a^4*C*ArcTanh[Sin[c + d*x]])/d + (a^4*(7*A + 10*C)*Sin[c + d*x])/(2*d) + (a*A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(
5*d) + ((7*A + 5*C)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + ((7*A + 8*C)*Cos[c + d*x]
*(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 4172

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (4 a A+5 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (4 a^2 (7 A+5 C)+20 a^2 C \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (20 a^3 (7 A+8 C)+60 a^3 C \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (60 a^4 (7 A+10 C)+120 a^4 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {\int \left (-60 a^5 (7 A+12 C)-120 a^5 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {1}{2} a^4 (7 A+12 C) x+\frac {a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^4 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (7 A+12 C) x+\frac {a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac {a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 147, normalized size = 0.71 \begin {gather*} \frac {a^4 \left (840 A d x+1440 C d x-240 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 (49 A+54 C) \sin (c+d x)+240 (2 A+C) \sin (2 (c+d x))+145 A \sin (3 (c+d x))+20 C \sin (3 (c+d x))+30 A \sin (4 (c+d x))+3 A \sin (5 (c+d x))\right )}{240 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(840*A*d*x + 1440*C*d*x - 240*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 240*C*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] + 30*(49*A + 54*C)*Sin[c + d*x] + 240*(2*A + C)*Sin[2*(c + d*x)] + 145*A*Sin[3*(c + d*x)] + 2
0*C*Sin[3*(c + d*x)] + 30*A*Sin[4*(c + d*x)] + 3*A*Sin[5*(c + d*x)]))/(240*d)

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Maple [A]
time = 0.73, size = 230, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {A \,a^{4} \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (d x +c \right )+2 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 a^{4} C \sin \left (d x +c \right )+4 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{4} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(230\)
default	\(\frac {A \,a^{4} \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (d x +c \right )+2 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 a^{4} C \sin \left (d x +c \right )+4 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{4} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(230\)
risch	\(\frac {7 a^{4} A x}{2}+6 a^{4} x C -\frac {49 i A \,a^{4} {\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{8 d}+\frac {49 i A \,a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{8 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {A \,a^{4} \sin \left (5 d x +5 c \right )}{80 d}+\frac {A \,a^{4} \sin \left (4 d x +4 c \right )}{8 d}+\frac {29 A \,a^{4} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} C}{12 d}+\frac {2 \sin \left (2 d x +2 c \right ) A \,a^{4}}{d}+\frac {\sin \left (2 d x +2 c \right ) a^{4} C}{d}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^4*sin(d*x+c)+a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4*A*a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a^4*
C*(d*x+c)+2*A*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+6*a^4*C*sin(d*x+c)+4*A*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*si
n(d*x+c)+3/8*d*x+3/8*c)+4*a^4*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/5*A*a^4*(8/3+cos(d*x+c)^4+4/3*cos(
d*x+c)^2)*sin(d*x+c)+1/3*a^4*C*(2+cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.30, size = 229, normalized size = 1.11 \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 a^{4} - 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A 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 a^{4} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 40 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 480 \, {\left (d x + c\right )} C a^{4} + 60 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, A a^{4} \sin \left (d x + c\right ) + 720 \, C a^{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+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),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*a^4 - 240*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*A*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 120*(2*d*x + 2*c + sin(2*d*x + 2*
c))*A*a^4 - 40*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 480*(d*x
+ c)*C*a^4 + 60*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 120*A*a^4*sin(d*x + c) + 720*C*a^4*si
n(d*x + c))/d

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Fricas [A]
time = 3.32, size = 138, normalized size = 0.67 \begin {gather*} \frac {15 \, {\left (7 \, A + 12 \, C\right )} a^{4} d x + 15 \, C a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, C a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (34 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \, {\left (83 \, A + 100 \, C\right )} a^{4}\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+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/30*(15*(7*A + 12*C)*a^4*d*x + 15*C*a^4*log(sin(d*x + c) + 1) - 15*C*a^4*log(-sin(d*x + c) + 1) + (6*A*a^4*co
s(d*x + c)^4 + 30*A*a^4*cos(d*x + c)^3 + 2*(34*A + 5*C)*a^4*cos(d*x + c)^2 + 15*(7*A + 4*C)*a^4*cos(d*x + c) +
2*(83*A + 100*C)*a^4)*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+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [A]
time = 0.51, size = 248, normalized size = 1.20 \begin {gather*} \frac {30 \, C a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 30 \, C a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 15 \, {\left (7 \, A a^{4} + 12 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 490 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 680 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 896 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1180 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 790 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 920 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 375 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 270 \, C 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 )}^{5}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/30*(30*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(7*A*a^4
+ 12*C*a^4)*(d*x + c) + 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 490*A*a^4*tan
(1/2*d*x + 1/2*c)^7 + 680*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 896*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1180*C*a^4*tan(1/2
*d*x + 1/2*c)^5 + 790*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 920*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 375*A*a^4*tan(1/2*d*x
+ 1/2*c) + 270*C*a^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 = 3.31, size = 202, normalized size = 0.98 \begin {gather*} \frac {7\,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 )+12\,C\,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 )+2\,C\,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 )+2\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {29\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {A\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{80}+C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{12}+\frac {49\,A\,a^4\,\sin \left (c+d\,x\right )}{8}+\frac {27\,C\,a^4\,\sin \left (c+d\,x\right )}{4}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^4,x)

[Out]

(7*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 12*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) +
2*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*A*a^4*sin(2*c + 2*d*x) + (29*A*a^4*sin(3*c + 3*d*x))/
48 + (A*a^4*sin(4*c + 4*d*x))/8 + (A*a^4*sin(5*c + 5*d*x))/80 + C*a^4*sin(2*c + 2*d*x) + (C*a^4*sin(3*c + 3*d*
x))/12 + (49*A*a^4*sin(c + d*x))/8 + (27*C*a^4*sin(c + d*x))/4)/d

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