3.33 \(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=186 \[ -\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac {a^4 (13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(9 A-4 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+2 a^4 x (2 A+3 C)-\frac {(15 A-2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}+\frac {2 a A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d} \]

[Out]

2*a^4*(2*A+3*C)*x+1/2*a^4*(13*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^4*(A-2*C)*sin(d*x+c)/d-1/6*(15*A-2*C)*(a^2+a^
2*cos(d*x+c))^2*sin(d*x+c)/d-1/3*(9*A-4*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+2*a*A*(a+a*cos(d*x+c))^3*tan(d*x+
c)/d+1/2*A*(a+a*cos(d*x+c))^4*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.61, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3044, 2975, 2976, 2968, 3023, 2735, 3770} \[ -\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac {a^4 (13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(15 A-2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{6 d}-\frac {(9 A-4 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+2 a^4 x (2 A+3 C)+\frac {2 a A \tan (c+d x) (a \cos (c+d x)+a)^3}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

2*a^4*(2*A + 3*C)*x + (a^4*(13*A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A - 2*C)*Sin[c + d*x])/(2*d) -
((15*A - 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) - ((9*A - 4*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d
*x])/(3*d) + (2*a*A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d + (A*(a + a*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*
x])/(2*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 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3023

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

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a A-a (3 A-2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^3 \left (a^2 (13 A+2 C)-a^2 (15 A-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^2 \left (3 a^3 (13 A+2 C)-4 a^3 (9 A-4 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x)) \left (6 a^4 (13 A+2 C)-30 a^4 (A-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (6 a^5 (13 A+2 C)+\left (-30 a^5 (A-2 C)+6 a^5 (13 A+2 C)\right ) \cos (c+d x)-30 a^5 (A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (6 a^5 (13 A+2 C)+24 a^5 (2 A+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=2 a^4 (2 A+3 C) x-\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^4 (13 A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=2 a^4 (2 A+3 C) x+\frac {a^4 (13 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {(15 A-2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{6 d}-\frac {(9 A-4 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 6.24, size = 756, normalized size = 4.06 \[ \frac {1}{8} x (2 A+3 C) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4+\frac {(4 A+27 C) \sin (c) \cos (d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{64 d}+\frac {(4 A+27 C) \cos (c) \sin (d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{64 d}+\frac {(-13 A-2 C) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{32 d}+\frac {(13 A+2 C) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4 \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{32 d}+\frac {A \sin \left (\frac {d x}{2}\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{4 d \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 {A \sin \left (\frac {d x}{2}\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{4 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {A \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{64 d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}-\frac {A \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{64 d \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {C \sin (2 c) \cos (2 d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{16 d}+\frac {C \sin (3 c) \cos (3 d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{192 d}+\frac {C \cos (2 c) \sin (2 d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{16 d}+\frac {C \cos (3 c) \sin (3 d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^4}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

((2*A + 3*C)*x*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/8 + ((-13*A - 2*C)*(a + a*Cos[c + d*x])^4*Log[Cos[
c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(32*d) + ((13*A + 2*C)*(a + a*Cos[c + d*x])^4*Log[C
os[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8)/(32*d) + ((4*A + 27*C)*Cos[d*x]*(a + a*Cos[c + d
*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[c])/(64*d) + (C*Cos[2*d*x]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[2*c
])/(16*d) + (C*Cos[3*d*x]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[3*c])/(192*d) + ((4*A + 27*C)*Cos[c]
*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[d*x])/(64*d) + (C*Cos[2*c]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (
d*x)/2]^8*Sin[2*d*x])/(16*d) + (C*Cos[3*c]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[3*d*x])/(192*d) + (
A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(64*d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (A*(a + a*
Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(4*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 +
(d*x)/2])) - (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8)/(64*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2
) + (A*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*Sin[(d*x)/2])/(4*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2
] + Sin[c/2 + (d*x)/2]))

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fricas [A]  time = 0.60, size = 171, normalized size = 0.92 \[ \frac {24 \, {\left (2 \, A + 3 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (13 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{4} \cos \left (d x + c\right )^{4} + 12 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 24 \, A a^{4} \cos \left (d x + c\right ) + 3 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/12*(24*(2*A + 3*C)*a^4*d*x*cos(d*x + c)^2 + 3*(13*A + 2*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(13*
A + 2*C)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*a^4*cos(d*x + c)^4 + 12*C*a^4*cos(d*x + c)^3 + 2*(
3*A + 20*C)*a^4*cos(d*x + c)^2 + 24*A*a^4*cos(d*x + c) + 3*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.64, size = 248, normalized size = 1.33 \[ \frac {12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 38 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/6*(12*(2*A*a^4 + 3*C*a^4)*(d*x + c) + 3*(13*A*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(13*A*a^
4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(7*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 6
*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 38*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(
1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.40, size = 190, normalized size = 1.02 \[ \frac {A \,a^{4} \sin \left (d x +c \right )}{d}+\frac {a^{4} C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}+\frac {20 a^{4} C \sin \left (d x +c \right )}{3 d}+4 A \,a^{4} x +\frac {4 A \,a^{4} c}{d}+\frac {2 a^{4} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+6 a^{4} C x +\frac {6 a^{4} C c}{d}+\frac {13 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

[Out]

1/d*A*a^4*sin(d*x+c)+1/3/d*a^4*C*sin(d*x+c)*cos(d*x+c)^2+20/3/d*a^4*C*sin(d*x+c)+4*A*a^4*x+4/d*A*a^4*c+2/d*a^4
*C*cos(d*x+c)*sin(d*x+c)+6*a^4*C*x+6/d*a^4*C*c+13/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a^4*tan(d*x+c)+1/2
/d*A*a^4*sec(d*x+c)*tan(d*x+c)+1/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.33, size = 211, normalized size = 1.13 \[ \frac {48 \, {\left (d x + c\right )} A a^{4} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 48 \, {\left (d x + c\right )} C a^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 72 \, C a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*A*a^4 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a
^4 + 48*(d*x + c)*C*a^4 - 3*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x +
c) - 1)) + 36*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^4*(log(sin(d*x + c) + 1) - log(si
n(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 72*C*a^4*sin(d*x + c) + 48*A*a^4*tan(d*x + c))/d

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mupad [B]  time = 1.14, size = 244, normalized size = 1.31 \[ \frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^3,x)

[Out]

(A*a^4*sin(c + d*x))/d + (20*C*a^4*sin(c + d*x))/(3*d) + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))
/d + (13*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (2*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^4*sin(c + d*x))/(d*cos(c + d*x
)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (C*a^4*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (2*C*a^4*cos(c +
d*x)*sin(c + d*x))/d

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

[Out]

Timed out

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