3.2.7 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [107]

Optimal. Leaf size=156 \[ \frac {1}{2} a^3 (5 A+6 C) x+\frac {3 a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4172, 4102, 4103, 4081, 3855} \begin {gather*} -\frac {(5 A-6 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {1}{2} a^3 x (5 A+6 C)+\frac {3 a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(5*A + 6*C)*x)/2 + (3*a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*A*Sin[c + d*x])/(2*d) + (A*Cos[c + d*x]^2*(
a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*a*d) -
((5*A - 6*C)*(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]

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]

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 ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (3 a A-a (A-3 C) \sec (c+d x)) \, dx}{3 a}\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (5 A+3 C)-a^2 (5 A-6 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 A+18 a^3 C \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {\int \left (-3 a^4 (5 A+6 C)-18 a^4 C \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {1}{2} a^3 (5 A+6 C) x+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (3 a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (5 A+6 C) x+\frac {3 a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1014\) vs. \(2(156)=312\).
time = 6.20, size = 1014, normalized size = 6.50 \begin {gather*} a^3 \left (\frac {(5 A+6 C) x \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{8 (A+2 C+A \cos (2 c+2 d x))}-\frac {3 C \cos ^2(c+d x) (1+\cos (c+d x))^3 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{4 d (A+2 C+A \cos (2 c+2 d x))}+\frac {3 C \cos ^2(c+d x) (1+\cos (c+d x))^3 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{4 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(15 A+4 C) \cos (d x) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {3 A \cos (2 d x) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (2 c)}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (3 d x) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (3 c)}{48 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(15 A+4 C) \cos (c) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (d x)}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {3 A \cos (2 c) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (2 d x)}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (3 c) \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (3 d x)}{48 d (A+2 C+A \cos (2 c+2 d x))}+\frac {C \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{4 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {C \cos ^2(c+d x) (1+\cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{4 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

a^3*(((5*A + 6*C)*x*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2))/(8*(A + 2
*C + A*Cos[2*c + 2*d*x])) - (3*C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/
2]]*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2))/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (3*C*Cos[c + d*x]^2*(1
+ Cos[c + d*x])^3*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2))/(
4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((15*A + 4*C)*Cos[d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*
x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[c])/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (3*A*Cos[2*d*x]*Cos[c + d*x]^2*
(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[2*c])/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x]
)) + (A*Cos[3*d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[3*c])/(
48*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((15*A + 4*C)*Cos[c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x
)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[d*x])/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (3*A*Cos[2*c]*Cos[c + d*x]^2*(
1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x
])) + (A*Cos[3*c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[3*d*x])/
(48*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec
[c + d*x]^2)*Sin[(d*x)/2])/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin
[c/2 + (d*x)/2])) + (C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[(d*
x)/2])/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])))

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Maple [A]
time = 0.67, size = 131, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {A \,a^{3} \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+\frac {A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{3} C \sin \left (d x +c \right )}{d}\)	\(131\)
default	\(\frac {A \,a^{3} \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+\frac {A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{3} C \sin \left (d x +c \right )}{d}\)	\(131\)
risch	\(\frac {5 a^{3} A x}{2}+3 a^{3} x C -\frac {3 i A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {15 i A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {15 i A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {3 i A \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3} C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {A \,a^{3} \sin \left (3 d x +3 c \right )}{12 d}\)	\(215\)
norman	\(\frac {\left (\frac {5}{2} A \,a^{3}+3 a^{3} C \right ) x +\left (-\frac {15}{2} A \,a^{3}-9 a^{3} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {15}{2} A \,a^{3}-9 a^{3} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {5}{2} A \,a^{3}-3 a^{3} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {5}{2} A \,a^{3}-3 a^{3} C \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} A \,a^{3}+3 a^{3} C \right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{2} A \,a^{3}+9 a^{3} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{2} A \,a^{3}+9 a^{3} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{3} \left (11 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 A \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (5 A +3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{3} \left (23 A +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (37 A -12 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{3} \left (53 A -24 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {3 a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {3 a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^3*(d*x+c)+a^3*C*tan(d*x+c)+3*A*a^3*sin(d*x+c)+3*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3*A*a^3*(1/2*cos(d*x+
c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a^3*C*(d*x+c)+1/3*A*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+a^3*C*sin(d*x+c))

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Maxima [A]
time = 0.28, size = 137, normalized size = 0.88 \begin {gather*} -\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} - 36 \, {\left (d x + c\right )} C a^{3} - 18 \, C 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 ) - 12 \, C a^{3} \sin \left (d x + c\right ) - 12 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),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 - 36*(d*x + c)*C*a^3 - 18*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*A*a^3*sin(d*x + c) - 1
2*C*a^3*sin(d*x + c) - 12*C*a^3*tan(d*x + c))/d

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Fricas [A]
time = 4.70, size = 138, normalized size = 0.88 \begin {gather*} \frac {3 \, {\left (5 \, A + 6 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 9 \, C a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.51, size = 210, normalized size = 1.35 \begin {gather*} \frac {18 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, C a^{3} \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 (5 \, A a^{3} + 6 \, C 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} + 6 \, C 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} + 12 \, C 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 ) + 6 \, C 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(18*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 12*C*a^3*tan(
1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(5*A*a^3 + 6*C*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/
2*c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 +
33*A*a^3*tan(1/2*d*x + 1/2*c) + 6*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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Mupad [B]
time = 2.72, size = 189, normalized size = 1.21 \begin {gather*} \frac {11\,A\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,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 {6\,C\,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 {6\,C\,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 {A\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {3\,A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(11*A*a^3*sin(c + d*x))/(3*d) + (C*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 + (6*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/d + (A*a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (C*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (3*A*a^3*cos(c
+ d*x)*sin(c + d*x))/(2*d)

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