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

Optimal. Leaf size=196 \[ \frac {5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac {(12 A-32 B-35 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{8} a^4 x (52 A+48 B+35 C)-\frac {(12 A-4 B-7 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}-\frac {a (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d} \]

[Out]

1/8*a^4*(52*A+48*B+35*C)*x+a^4*(4*A+B)*arctanh(sin(d*x+c))/d+5/8*a^4*(4*A+8*B+7*C)*sin(d*x+c)/d-1/4*a*(4*A-C)*
(a+a*cos(d*x+c))^3*sin(d*x+c)/d-1/12*(12*A-4*B-7*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d-1/24*(12*A-32*B-35*C)*
(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^4*tan(d*x+c)/d

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Rubi [A]  time = 0.68, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3043, 2976, 2968, 3023, 2735, 3770} \[ \frac {5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac {(12 A-4 B-7 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}-\frac {(12 A-32 B-35 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{8} a^4 x (52 A+48 B+35 C)-\frac {a (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(52*A + 48*B + 35*C)*x)/8 + (a^4*(4*A + B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(4*A + 8*B + 7*C)*Sin[c + d*
x])/(8*d) - (a*(4*A - C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) - ((12*A - 4*B - 7*C)*(a^2 + a^2*Cos[c + d
*x])^2*Sin[c + d*x])/(12*d) - ((12*A - 32*B - 35*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(24*d) + (A*(a + a*
Cos[c + d*x])^4*Tan[c + d*x])/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 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 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))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^4 (a (4 A+B)-a (4 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^3 \left (4 a^2 (4 A+B)-a^2 (12 A-4 B-7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{4 a}\\ &=-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^2 \left (12 a^3 (4 A+B)-a^3 (12 A-32 B-35 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x)) \left (24 a^4 (4 A+B)+15 a^4 (4 A+8 B+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int \left (24 a^5 (4 A+B)+\left (24 a^5 (4 A+B)+15 a^5 (4 A+8 B+7 C)\right ) \cos (c+d x)+15 a^5 (4 A+8 B+7 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac {\int \left (24 a^5 (4 A+B)+3 a^5 (52 A+48 B+35 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {1}{8} a^4 (52 A+48 B+35 C) x+\frac {5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\left (a^4 (4 A+B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a^4 (52 A+48 B+35 C) x+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 246, normalized size = 1.26 \[ \frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (12 (52 A+48 B+35 C) (c+d x)+24 (16 A+27 B+28 C) \sin (c+d x)+24 (A+4 B+7 C) \sin (2 (c+d x))-96 (4 A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 (4 A+B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {96 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {96 A \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+8 (B+4 C) \sin (3 (c+d x))+3 C \sin (4 (c+d x))\right )}{1536 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(12*(52*A + 48*B + 35*C)*(c + d*x) - 96*(4*A + B)*Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] + 96*(4*A + B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (96*A*Sin[(c + d*x)/2])/(C
os[(c + d*x)/2] - Sin[(c + d*x)/2]) + (96*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 24*(16*A
+ 27*B + 28*C)*Sin[c + d*x] + 24*(A + 4*B + 7*C)*Sin[2*(c + d*x)] + 8*(B + 4*C)*Sin[3*(c + d*x)] + 3*C*Sin[4*
(c + d*x)]))/(1536*d)

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fricas [A]  time = 0.44, size = 179, normalized size = 0.91 \[ \frac {3 \, {\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 12 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{4} \cos \left (d x + c\right )^{4} + 8 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 16 \, B + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(52*A + 48*B + 35*C)*a^4*d*x*cos(d*x + c) + 12*(4*A + B)*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 12*(
4*A + B)*a^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + (6*C*a^4*cos(d*x + c)^4 + 8*(B + 4*C)*a^4*cos(d*x + c)^3 +
3*(4*A + 16*B + 27*C)*a^4*cos(d*x + c)^2 + 32*(3*A + 5*B + 5*C)*a^4*cos(d*x + c) + 24*A*a^4)*sin(d*x + c))/(d*
cos(d*x + c))

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giac [A]  time = 2.55, size = 332, normalized size = 1.69 \[ -\frac {\frac {48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, C a^{4}\right )} {\left (d x + c\right )} - 24 \, {\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 24 \, {\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 276 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 108 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 279 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(52*A*a^4 + 48*B*a^4 + 35*C*a^4)*(d*x +
c) - 24*(4*A*a^4 + B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 24*(4*A*a^4 + B*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) - 1)) - 2*(84*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 105*C*a^4*tan(1/2*d*x + 1/2
*c)^7 + 276*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 385*C*a^4*tan(1/2*d*x + 1/2*c)^5
+ 300*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 511*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 10
8*A*a^4*tan(1/2*d*x + 1/2*c) + 216*B*a^4*tan(1/2*d*x + 1/2*c) + 279*C*a^4*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 = 0.43, size = 289, normalized size = 1.47 \[ \frac {35 a^{4} C x}{8}+\frac {B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 a^{4} B \sin \left (d x +c \right )}{3 d}+\frac {4 a^{4} C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3 d}+\frac {20 a^{4} C \sin \left (d x +c \right )}{3 d}+\frac {A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 A \,a^{4} \sin \left (d x +c \right )}{d}+\frac {4 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {6 a^{4} B c}{d}+\frac {13 A \,a^{4} c}{2 d}+\frac {2 a^{4} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+\frac {35 a^{4} C c}{8 d}+\frac {A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{4} C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {27 a^{4} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+6 a^{4} B x +\frac {13 A \,a^{4} x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*a^4*C*x+1/3/d*B*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*a^4*B*sin(d*x+c)+4/3/d*a^4*C*sin(d*x+c)*cos(d*x+c)^2+2
0/3/d*a^4*C*sin(d*x+c)+1/d*A*a^4*tan(d*x+c)+1/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+4/d*A*a^4*sin(d*x+c)+4/d*A*a^4
*ln(sec(d*x+c)+tan(d*x+c))+6/d*a^4*B*c+13/2/d*A*a^4*c+2/d*a^4*B*cos(d*x+c)*sin(d*x+c)+35/8/d*a^4*C*c+1/2/d*A*a
^4*cos(d*x+c)*sin(d*x+c)+1/4/d*a^4*C*sin(d*x+c)*cos(d*x+c)^3+27/8/d*a^4*C*cos(d*x+c)*sin(d*x+c)+6*a^4*B*x+13/2
*A*a^4*x

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maxima [A]  time = 0.35, size = 290, normalized size = 1.48 \[ \frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 576 \, {\left (d x + c\right )} A a^{4} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 384 \, {\left (d x + c\right )} B a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 96 \, {\left (d x + c\right )} C a^{4} + 192 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, A a^{4} \sin \left (d x + c\right ) + 576 \, B a^{4} \sin \left (d x + c\right ) + 384 \, C a^{4} \sin \left (d x + c\right ) + 96 \, A a^{4} \tan \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 576*(d*x + c)*A*a^4 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^4 + 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 384*(d*x + c)*B*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))
*C*a^4 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^4 + 144*(2*d*x + 2*c + sin(2*d*x + 2*c)
)*C*a^4 + 96*(d*x + c)*C*a^4 + 192*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*B*a^4*(log(sin(d
*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*A*a^4*sin(d*x + c) + 576*B*a^4*sin(d*x + c) + 384*C*a^4*sin(d*x +
c) + 96*A*a^4*tan(d*x + c))/d

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mupad [B]  time = 2.63, size = 1244, normalized size = 6.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (tan(c/2 + (d*x)/2)^5*(8*B*a^4 - 10*A*a^4 + (21*C*a^4)/2) + tan(c/2 + (d*x)/2)^9*(5*A*a^4 + 10*B*a^4 + (35*C
*a^4)/4) - tan(c/2 + (d*x)/2)^3*(24*A*a^4 + (76*B*a^4)/3 + (58*C*a^4)/3) + tan(c/2 + (d*x)/2)^7*(8*A*a^4 + (76
*B*a^4)/3 + (70*C*a^4)/3) - tan(c/2 + (d*x)/2)*(11*A*a^4 + 18*B*a^4 + (93*C*a^4)/4))/(d*(3*tan(c/2 + (d*x)/2)^
2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^10 + 1)) - (
a^4*atan((a^4*(tan(c/2 + (d*x)/2)*(1864*A^2*a^8 + 1184*B^2*a^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^
8 + 1680*B*C*a^8) + a^4*(4*A + B)*(336*A*a^4 + 224*B*a^4 + 140*C*a^4))*(4*A + B)*1i + a^4*(tan(c/2 + (d*x)/2)*
(1864*A^2*a^8 + 1184*B^2*a^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) - a^4*(4*A + B)*
(336*A*a^4 + 224*B*a^4 + 140*C*a^4))*(4*A + B)*1i)/(4160*A^3*a^12 + 1920*B^3*a^12 + 10720*A*B^2*a^12 + 13200*A
^2*B*a^12 + 4900*A*C^2*a^12 + 10080*A^2*C*a^12 + 1225*B*C^2*a^12 + 3080*B^2*C*a^12 + a^4*(tan(c/2 + (d*x)/2)*(
1864*A^2*a^8 + 1184*B^2*a^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) + a^4*(4*A + B)*(
336*A*a^4 + 224*B*a^4 + 140*C*a^4))*(4*A + B) - a^4*(tan(c/2 + (d*x)/2)*(1864*A^2*a^8 + 1184*B^2*a^8 + (1225*C
^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) - a^4*(4*A + B)*(336*A*a^4 + 224*B*a^4 + 140*C*a^4))*(
4*A + B) + 14840*A*B*C*a^12))*(4*A + B)*2i)/d - (a^4*atan(((a^4*(tan(c/2 + (d*x)/2)*(1864*A^2*a^8 + 1184*B^2*a
^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) - (a^4*(52*A + 48*B + 35*C)*(336*A*a^4 + 2
24*B*a^4 + 140*C*a^4)*1i)/8)*(52*A + 48*B + 35*C))/8 + (a^4*(tan(c/2 + (d*x)/2)*(1864*A^2*a^8 + 1184*B^2*a^8 +
(1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) + (a^4*(52*A + 48*B + 35*C)*(336*A*a^4 + 224*B
*a^4 + 140*C*a^4)*1i)/8)*(52*A + 48*B + 35*C))/8)/(4160*A^3*a^12 + 1920*B^3*a^12 + 10720*A*B^2*a^12 + 13200*A^
2*B*a^12 + 4900*A*C^2*a^12 + 10080*A^2*C*a^12 + 1225*B*C^2*a^12 + 3080*B^2*C*a^12 - (a^4*(tan(c/2 + (d*x)/2)*(
1864*A^2*a^8 + 1184*B^2*a^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) - (a^4*(52*A + 48
*B + 35*C)*(336*A*a^4 + 224*B*a^4 + 140*C*a^4)*1i)/8)*(52*A + 48*B + 35*C)*1i)/8 + (a^4*(tan(c/2 + (d*x)/2)*(1
864*A^2*a^8 + 1184*B^2*a^8 + (1225*C^2*a^8)/2 + 2752*A*B*a^8 + 1820*A*C*a^8 + 1680*B*C*a^8) + (a^4*(52*A + 48*
B + 35*C)*(336*A*a^4 + 224*B*a^4 + 140*C*a^4)*1i)/8)*(52*A + 48*B + 35*C)*1i)/8 + 14840*A*B*C*a^12))*(52*A + 4
8*B + 35*C))/(4*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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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