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

Optimal. Leaf size=160 \[ \frac {a^2 (4 A+5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (7 A+8 B+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

[Out]

1/8*a^2*(7*A+8*B+12*C)*arctanh(sin(d*x+c))/d+1/3*a^2*(4*A+5*B+6*C)*tan(d*x+c)/d+1/24*a^2*(11*A+16*B+12*C)*sec(
d*x+c)*tan(d*x+c)/d+1/6*(A+2*B)*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a*cos(d*x+c))^2*sec(d*
x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.47, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3043, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^2 (4 A+5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (7 A+8 B+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A + 8*B + 12*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(4*A + 5*B + 6*C)*Tan[c + d*x])/(3*d) + (a^2*(11*A
+ 16*B + 12*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((A + 2*B)*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c +
d*x])/(6*d) + (A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[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))^2 \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))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^2 (2 a (A+2 B)+a (A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x)) \left (a^2 (11 A+16 B+12 C)+a^2 (5 A+4 B+12 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (a^3 (11 A+16 B+12 C)+\left (a^3 (5 A+4 B+12 C)+a^3 (11 A+16 B+12 C)\right ) \cos (c+d x)+a^3 (5 A+4 B+12 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {a^2 (11 A+16 B+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \left (8 a^3 (4 A+5 B+6 C)+3 a^3 (7 A+8 B+12 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {a^2 (11 A+16 B+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} \left (a^2 (4 A+5 B+6 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{8} \left (a^2 (7 A+8 B+12 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {a^2 (7 A+8 B+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (11 A+16 B+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (a^2 (4 A+5 B+6 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a^2 (7 A+8 B+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (4 A+5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (11 A+16 B+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [B]  time = 3.75, size = 404, normalized size = 2.52 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {16 (4 A+5 B+6 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {16 (4 A+5 B+6 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {29 A+28 B+12 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {-29 A-4 (7 B+3 C)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-6 (7 A+8 B+12 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 (7 A+8 B+12 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 (2 A+B) \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {8 (2 A+B) \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {3 A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-6*(7*A + 8*B + 12*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
6*(7*A + 8*B + 12*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4
+ (29*A + 28*B + 12*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (8*(2*A + B)*Sin[(c + d*x)/2])/(Cos[(c + d*x
)/2] - Sin[(c + d*x)/2])^3 + (16*(4*A + 5*B + 6*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (
3*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (8*(2*A + B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^3 + (-29*A - 4*(7*B + 3*C))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (16*(4*A + 5*B + 6*C)*Sin[(c + d*x
)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(192*d)

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fricas [A]  time = 0.51, size = 157, normalized size = 0.98 \[ \frac {3 \, {\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (4 \, A + 5 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 8 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(7*A + 8*B + 12*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(7*A + 8*B + 12*C)*a^2*cos(d*x + c)^4*
log(-sin(d*x + c) + 1) + 2*(8*(4*A + 5*B + 6*C)*a^2*cos(d*x + c)^3 + 3*(7*A + 8*B + 4*C)*a^2*cos(d*x + c)^2 +
8*(2*A + B)*a^2*cos(d*x + c) + 6*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.69, size = 290, normalized size = 1.81 \[ \frac {3 \, {\left (7 \, A a^{2} + 8 \, B a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, A a^{2} + 8 \, B a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 88 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 132 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*l
og(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 36*
C*a^2*tan(1/2*d*x + 1/2*c)^7 - 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 88*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 132*C*a^2*t
an(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 136*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C*a^2*tan(1/2
*d*x + 1/2*c)^3 - 75*A*a^2*tan(1/2*d*x + 1/2*c) - 72*B*a^2*tan(1/2*d*x + 1/2*c) - 60*C*a^2*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 = 0.44, size = 246, normalized size = 1.54 \[ \frac {7 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {3 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {4 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{d}+\frac {a^{2} A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 316, normalized size = 1.98 \[ \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{2} \tan \left (d x + c\right ) + 96 \, C a^{2} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 3*A*a^2*(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)) - 12*A*a^2*(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^2*(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^2
*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^2*(log(sin(d*x
+ c) + 1) - log(sin(d*x + c) - 1)) + 48*B*a^2*tan(d*x + c) + 96*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 4.66, size = 245, normalized size = 1.53 \[ \frac {2\,a^2\,\mathrm {atanh}\left (\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,A}{8}+B+\frac {3\,C}{2}\right )}{\frac {7\,A\,a^2}{2}+4\,B\,a^2+6\,C\,a^2}\right )\,\left (\frac {7\,A}{8}+B+\frac {3\,C}{2}\right )}{d}-\frac {\left (\frac {7\,A\,a^2}{4}+2\,B\,a^2+3\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {77\,A\,a^2}{12}-\frac {22\,B\,a^2}{3}-11\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {83\,A\,a^2}{12}+\frac {34\,B\,a^2}{3}+13\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {25\,A\,a^2}{4}-6\,B\,a^2-5\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*atanh((4*a^2*tan(c/2 + (d*x)/2)*((7*A)/8 + B + (3*C)/2))/((7*A*a^2)/2 + 4*B*a^2 + 6*C*a^2))*((7*A)/8 +
B + (3*C)/2))/d - (tan(c/2 + (d*x)/2)^7*((7*A*a^2)/4 + 2*B*a^2 + 3*C*a^2) - tan(c/2 + (d*x)/2)^5*((77*A*a^2)/1
2 + (22*B*a^2)/3 + 11*C*a^2) + tan(c/2 + (d*x)/2)^3*((83*A*a^2)/12 + (34*B*a^2)/3 + 13*C*a^2) - tan(c/2 + (d*x
)/2)*((25*A*a^2)/4 + 6*B*a^2 + 5*C*a^2))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*
x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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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))**2*(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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