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

Optimal. Leaf size=169 \[ \frac {a^3 (5 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(5 A+6 B+3 C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}-\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+a^3 x (B+3 C)+\frac {(A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 a d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

a^3*(B+3*C)*x+1/2*a^3*(5*A+7*B+6*C)*arctanh(sin(d*x+c))/d-5/2*a^3*(A+B)*sin(d*x+c)/d+1/3*(5*A+6*B+3*C)*(a^3+a^
3*cos(d*x+c))*tan(d*x+c)/d+1/2*(A+B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)*tan(d*x+c)/a/d+1/3*A*(a+a*cos(d*x+c))^3
*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.57, antiderivative size = 169, 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, 3023, 2735, 3770} \[ \frac {a^3 (5 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(5 A+6 B+3 C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}-\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 a d}+a^3 x (B+3 C)+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

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]^4,x]

[Out]

a^3*(B + 3*C)*x + (a^3*(5*A + 7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (
(5*A + 6*B + 3*C)*(a^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/(3*d) + ((A + B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d
*x]*Tan[c + d*x])/(2*a*d) + (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^3 (3 a (A+B)-a (A-3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 \left (2 a^2 (5 A+6 B+3 C)-a^2 (5 A+3 B-6 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {(5 A+6 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^3 (5 A+7 B+6 C)-15 a^3 (A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {(5 A+6 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (3 a^4 (5 A+7 B+6 C)+\left (-15 a^4 (A+B)+3 a^4 (5 A+7 B+6 C)\right ) \cos (c+d x)-15 a^4 (A+B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(5 A+6 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (3 a^4 (5 A+7 B+6 C)+6 a^4 (B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=a^3 (B+3 C) x-\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(5 A+6 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (a^3 (5 A+7 B+6 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 (B+3 C) x+\frac {a^3 (5 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(5 A+6 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 4.12, size = 354, normalized size = 2.09 \[ \frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {4 (11 A+9 B+3 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 (11 A+9 B+3 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-6 (5 A+7 B+6 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 (5 A+7 B+6 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {-10 A-3 B}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {10 A+3 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+12 (B+3 C) (c+d x)+12 C \sin (c+d x)\right )}{96 d} \]

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]^4,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(12*(B + 3*C)*(c + d*x) - 6*(5*A + 7*B + 6*C)*Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] + 6*(5*A + 7*B + 6*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (10*A + 3*B)/(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])^2 + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*(11*A + 9*
B + 3*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^3 + (-10*A - 3*B)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*(11*A + 9*B + 3*C)*Sin[(c + d
*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*C*Sin[c + d*x]))/(96*d)

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fricas [A]  time = 0.48, size = 168, normalized size = 0.99 \[ \frac {12 \, {\left (B + 3 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 7 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A + 7 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A + 9 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

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)^4,x, algorithm="fricas")

[Out]

1/12*(12*(B + 3*C)*a^3*d*x*cos(d*x + c)^3 + 3*(5*A + 7*B + 6*C)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(
5*A + 7*B + 6*C)*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*C*a^3*cos(d*x + c)^3 + 2*(11*A + 9*B + 3*C)*
a^3*cos(d*x + c)^2 + 3*(3*A + B)*a^3*cos(d*x + c) + 2*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.55, size = 288, normalized size = 1.70 \[ \frac {\frac {12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 6 \, {\left (B a^{3} + 3 \, C a^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

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)^4,x, algorithm="giac")

[Out]

1/6*(12*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(B*a^3 + 3*C*a^3)*(d*x + c) + 3*(5*A*a^3 +
7*B*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(5*A*a^3 + 7*B*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*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 - 36*B*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) + 21*B*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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maple [A]  time = 0.42, size = 226, normalized size = 1.34 \[ \frac {5 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+a^{3} B x +\frac {a^{3} B c}{d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}+\frac {11 A \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {7 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+3 a^{3} C x +\frac {3 C \,a^{3} c}{d}+\frac {3 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \tan \left (d x +c \right )}{d} \]

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)^4,x)

[Out]

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

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maxima [A]  time = 0.36, size = 274, normalized size = 1.62 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{3} + 36 \, {\left (d x + c\right )} C a^{3} - 9 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{3} \sin \left (d x + c\right ) + 36 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 12*(d*x + c)*B*a^3 + 36*(d*x + c)*C*a^3 - 9*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)) - 3*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)) + 6*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x +
c) - 1)) + 18*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*C*a^3*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 12*C*a^3*sin(d*x + c) + 36*A*a^3*tan(d*x + c) + 36*B*a^3*tan(d*x + c) + 12*C*a^3*tan(d*x +
c))/d

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mupad [B]  time = 2.58, size = 541, normalized size = 3.20 \[ \frac {\frac {3\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {11\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {5\,A\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{4}-\frac {A\,a^3\,\cos \left (c+d\,x\right )\,\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}}{4}+\frac {3\,B\,a^3\,\cos \left (c+d\,x\right )\,\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 )}{2}-\frac {B\,a^3\,\cos \left (c+d\,x\right )\,\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}}{4}+\frac {9\,C\,a^3\,\cos \left (c+d\,x\right )\,\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 )}{2}-\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{2}-\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 (3\,c+3\,d\,x\right )\,5{}\mathrm {i}}{4}+\frac {B\,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 (3\,c+3\,d\,x\right )}{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 (3\,c+3\,d\,x\right )\,7{}\mathrm {i}}{4}+\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 )\,\cos \left (3\,c+3\,d\,x\right )}{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 (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{2}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\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)^4,x)

[Out]

((3*A*a^3*sin(2*c + 2*d*x))/4 + (11*A*a^3*sin(3*c + 3*d*x))/12 + (B*a^3*sin(2*c + 2*d*x))/4 + (3*B*a^3*sin(3*c
+ 3*d*x))/4 + (C*a^3*sin(2*c + 2*d*x))/4 + (C*a^3*sin(3*c + 3*d*x))/4 + (C*a^3*sin(4*c + 4*d*x))/8 + (5*A*a^3
*sin(c + d*x))/4 + (3*B*a^3*sin(c + d*x))/4 + (C*a^3*sin(c + d*x))/4 - (A*a^3*cos(c + d*x)*atan((sin(c/2 + (d*
x)/2)*1i)/cos(c/2 + (d*x)/2))*15i)/4 + (3*B*a^3*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 -
(B*a^3*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*21i)/4 + (9*C*a^3*cos(c + d*x)*atan(sin(c
/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (C*a^3*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i
)/2 - (A*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*5i)/4 + (B*a^3*atan(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(
3*c + 3*d*x)*7i)/4 + (3*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - (C*a^3*atan((s
in(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*3i)/2)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4)
)

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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)**4,x)

[Out]

Timed out

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