3.325 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=183 \[ \frac {5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac {a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(15 A+20 B+12 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+a^3 C x+\frac {(3 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

a^3*C*x+1/8*a^3*(15*A+20*B+28*C)*arctanh(sin(d*x+c))/d+5/8*a^3*(3*A+4*B+4*C)*tan(d*x+c)/d+1/24*(15*A+20*B+12*C
)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)*tan(d*x+c)/d+1/12*(3*A+4*B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/a
/d+1/4*A*(a+a*cos(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.55, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3043, 2975, 2968, 3021, 2735, 3770} \[ \frac {5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac {a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(15 A+20 B+12 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac {(3 A+4 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+a^3 C x+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

a^3*C*x + (a^3*(15*A + 20*B + 28*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*(3*A + 4*(B + C))*Tan[c + d*x])/(8*d
) + ((15*A + 20*B + 12*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*A + 4*B)*(a^2 + a^2
*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(12*a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x]
)/(4*d)

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^3 (a (3 A+4 B)+4 a C \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^2 \left (a^2 (15 A+20 B+12 C)+12 a^2 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x)) \left (15 a^3 (3 A+4 (B+C))+24 a^3 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (15 a^4 (3 A+4 (B+C))+\left (24 a^4 C+15 a^4 (3 A+4 (B+C))\right ) \cos (c+d x)+24 a^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac {(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (3 a^4 (15 A+20 B+28 C)+24 a^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=a^3 C x+\frac {5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac {(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a^3 (15 A+20 B+28 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 C x+\frac {a^3 (15 A+20 B+28 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 (3 A+4 (B+C)) \tan (c+d x)}{8 d}+\frac {(15 A+20 B+12 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+4 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{12 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 793, normalized size = 4.33 \[ \frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (9 A \sin \left (\frac {1}{2} (c+d x)\right )+11 B \sin \left (\frac {1}{2} (c+d x)\right )+9 C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (9 A \sin \left (\frac {1}{2} (c+d x)\right )+11 B \sin \left (\frac {1}{2} (c+d x)\right )+9 C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(57 A+40 B+12 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{384 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(-57 A-40 B-12 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{384 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(-15 A-20 B-28 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {(15 A+20 B+28 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{64 d}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (3 A \sin \left (\frac {1}{2} (c+d x)\right )+B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (3 A \sin \left (\frac {1}{2} (c+d x)\right )+B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{128 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {A \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{128 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {C (c+d x) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(C*(c + d*x)*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/(8*d) + ((-15*A - 20*B - 28*C)*(a + a*Cos[c + d*x])^
3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(64*d) + ((15*A + 20*B + 28*C)*(a + a*Cos[c +
d*x])^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(64*d) + (A*(a + a*Cos[c + d*x])^3*Sec
[c/2 + (d*x)/2]^6)/(128*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((57*A + 40*B + 12*C)*(a + a*Cos[c + d*x]
)^3*Sec[c/2 + (d*x)/2]^6)/(384*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (A*(a + a*Cos[c + d*x])^3*Sec[c/2
+ (d*x)/2]^6)/(128*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + ((-57*A - 40*B - 12*C)*(a + a*Cos[c + d*x])^3*
Sec[c/2 + (d*x)/2]^6)/(384*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x
)/2]^6*(3*A*Sin[(c + d*x)/2] + B*Sin[(c + d*x)/2]))/(48*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + ((a + a*C
os[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(3*A*Sin[(c + d*x)/2] + B*Sin[(c + d*x)/2]))/(48*d*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^3) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(9*A*Sin[(c + d*x)/2] + 11*B*Sin[(c + d*x)/2]
+ 9*C*Sin[(c + d*x)/2]))/(24*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*
x)/2]^6*(9*A*Sin[(c + d*x)/2] + 11*B*Sin[(c + d*x)/2] + 9*C*Sin[(c + d*x)/2]))/(24*d*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]))

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fricas [A]  time = 0.50, size = 173, normalized size = 0.95 \[ \frac {48 \, C a^{3} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (15 \, A + 20 \, B + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (15 \, A + 20 \, B + 28 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A + 12 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(48*C*a^3*d*x*cos(d*x + c)^4 + 3*(15*A + 20*B + 28*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(15*A
+ 20*B + 28*C)*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(9*A + 11*B + 9*C)*a^3*cos(d*x + c)^3 + 3*(15*
A + 12*B + 4*C)*a^3*cos(d*x + c)^2 + 8*(3*A + B)*a^3*cos(d*x + c) + 6*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.59, size = 301, normalized size = 1.64 \[ \frac {24 \, {\left (d x + c\right )} C a^{3} + 3 \, {\left (15 \, A a^{3} + 20 \, B a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (15 \, A a^{3} + 20 \, B a^{3} + 28 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 204 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 228 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 147 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 84 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*C*a^3 + 3*(15*A*a^3 + 20*B*a^3 + 28*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(15*A*a^3
+ 20*B*a^3 + 28*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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 + 60*C*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 - 204*C*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 + 228*C*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)
- 84*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.48, size = 262, normalized size = 1.43 \[ \frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {5 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {15 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {11 a^{3} B \tan \left (d x +c \right )}{3 d}+\frac {7 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 C \,a^{3} \tan \left (d x +c \right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

3/d*A*a^3*tan(d*x+c)+5/2/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+a^3*C*x+1/d*C*a^3*c+15/8/d*A*a^3*sec(d*x+c)*tan(d*x
+c)+15/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+11/3/d*a^3*B*tan(d*x+c)+7/2/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*A
*a^3*tan(d*x+c)*sec(d*x+c)^2+3/2/d*a^3*B*sec(d*x+c)*tan(d*x+c)+3/d*C*a^3*tan(d*x+c)+1/4/d*A*a^3*tan(d*x+c)*sec
(d*x+c)^3+1/3/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+1/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)

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maxima [B]  time = 0.37, size = 366, normalized size = 2.00 \[ \frac {48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (d x + c\right )} C a^{3} - 3 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 144 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(48*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(d*x + c)*C
*a^3 - 3*A*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x
+ c) + 1) + 3*log(sin(d*x + c) - 1)) - 36*A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) +
log(sin(d*x + c) - 1)) - 36*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x
+ c) - 1)) - 12*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) +
24*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 72*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c)
- 1)) + 48*A*a^3*tan(d*x + c) + 144*B*a^3*tan(d*x + c) + 144*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.02, size = 636, normalized size = 3.48 \[ \frac {\frac {3\,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 )}{4}-\frac {B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,15{}\mathrm {i}}{8}-\frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,45{}\mathrm {i}}{32}-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,21{}\mathrm {i}}{8}+\frac {5\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {15\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{32}+\frac {3\,A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {13\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {3\,B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {11\,B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+\frac {3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {3\,C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {23\,A\,a^3\,\sin \left (c+d\,x\right )}{32}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{8}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{8}-\frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )\,15{}\mathrm {i}}{8}-\frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )\,15{}\mathrm {i}}{32}-\frac {B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )\,5{}\mathrm {i}}{2}-\frac {B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}}{8}+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 )\,\cos \left (2\,c+2\,d\,x\right )+\frac {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 )\,\cos \left (4\,c+4\,d\,x\right )}{4}-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )\,7{}\mathrm {i}}{2}-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )\,7{}\mathrm {i}}{8}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {\cos \left (4\,c+4\,d\,x\right )}{8}+\frac {3}{8}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^5,x)

[Out]

((3*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - (B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)
/2))*15i)/8 - (A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*45i)/32 - (C*a^3*atan((sin(c/2 + (d*x)/2
)*1i)/cos(c/2 + (d*x)/2))*21i)/8 + (5*A*a^3*sin(2*c + 2*d*x))/4 + (15*A*a^3*sin(3*c + 3*d*x))/32 + (3*A*a^3*si
n(4*c + 4*d*x))/8 + (13*B*a^3*sin(2*c + 2*d*x))/12 + (3*B*a^3*sin(3*c + 3*d*x))/8 + (11*B*a^3*sin(4*c + 4*d*x)
)/24 + (3*C*a^3*sin(2*c + 2*d*x))/4 + (C*a^3*sin(3*c + 3*d*x))/8 + (3*C*a^3*sin(4*c + 4*d*x))/8 + (23*A*a^3*si
n(c + d*x))/32 + (3*B*a^3*sin(c + d*x))/8 + (C*a^3*sin(c + d*x))/8 - (A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c
/2 + (d*x)/2))*cos(2*c + 2*d*x)*15i)/8 - (A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d
*x)*15i)/32 - (B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*5i)/2 - (B*a^3*atan((si
n(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x)*5i)/8 + C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2))*cos(2*c + 2*d*x) + (C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/4 - (C*a^3*atan(
(sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*7i)/2 - (C*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c
/2 + (d*x)/2))*cos(4*c + 4*d*x)*7i)/8)/(d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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