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

Optimal. Leaf size=156 \[ -\frac{(6 A-3 B-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a^3 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^3 x (6 A+7 B+5 C)-\frac{(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

[Out]

(a^3*(6*A + 7*B + 5*C)*x)/2 + (a^3*(3*A + B)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(B + C)*Sin[c + d*x])/(2*d) - (
(3*A - C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(3*a*d) - ((6*A - 3*B - 5*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c
+ d*x])/(6*d) + (A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d

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Rubi [A]  time = 0.509521, antiderivative size = 156, normalized size of antiderivative = 1., 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, 2976, 2968, 3023, 2735, 3770} \[ -\frac{(6 A-3 B-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a^3 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^3 x (6 A+7 B+5 C)-\frac{(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac{5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

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]^2,x]

[Out]

(a^3*(6*A + 7*B + 5*C)*x)/2 + (a^3*(3*A + B)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(B + C)*Sin[c + d*x])/(2*d) - (
(3*A - C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(3*a*d) - ((6*A - 3*B - 5*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c
+ d*x])/(6*d) + (A*(a + a*Cos[c + d*x])^3*Tan[c + d*x])/d

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

Mathematica [A]  time = 0.958279, size = 227, normalized size = 1.46 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (6 (6 A+7 B+5 C) (c+d x)+3 (4 A+12 B+15 C) \sin (c+d x)-12 (3 A+B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 (3 A+B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 A \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 A \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+3 (B+3 C) \sin (2 (c+d x))+C \sin (3 (c+d x))\right )}{96 d} \]

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]^2,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(6*(6*A + 7*B + 5*C)*(c + d*x) - 12*(3*A + B)*Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] + 12*(3*A + B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*A*Sin[(c + d*x)/2])/(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]) + (12*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*(4*A + 12*
B + 15*C)*Sin[c + d*x] + 3*(B + 3*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.082, size = 221, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{5\,{a}^{3}Cx}{2}}+{\frac{5\,{a}^{3}Cc}{2\,d}}+3\,{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{7\,{a}^{3}Bx}{2}}+{\frac{7\,{a}^{3}Bc}{2\,d}}+{\frac{11\,{a}^{3}C\sin \left ( dx+c \right ) }{3\,d}}+3\,A{a}^{3}x+3\,{\frac{A{a}^{3}c}{d}}+3\,{\frac{{a}^{3}B\sin \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{3}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}} \end{align*}

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)^2,x)

[Out]

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

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Maxima [A]  time = 0.999814, size = 284, normalized size = 1.82 \begin{align*} \frac{36 \,{\left (d x + c\right )} A a^{3} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 36 \,{\left (d x + c\right )} B a^{3} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \,{\left (d x + c\right )} C a^{3} + 18 \, 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 )} + 6 \, 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 )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, B a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

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)^2,x, algorithm="maxima")

[Out]

1/12*(36*(d*x + c)*A*a^3 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 36*(d*x + c)*B*a^3 - 4*(sin(d*x + c)^3 -
3*sin(d*x + c))*C*a^3 + 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 12*(d*x + c)*C*a^3 + 18*A*a^3*(log(sin(d*x
+ c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d
*x + c) + 36*B*a^3*sin(d*x + c) + 36*C*a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d

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Fricas [A]  time = 2.12494, size = 398, normalized size = 2.55 \begin{align*} \frac{3 \,{\left (6 \, A + 7 \, B + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 3 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, A + 9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}

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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.26877, size = 379, normalized size = 2.43 \begin{align*} -\frac{\frac{12 \, A 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} - 3 \,{\left (6 \, A a^{3} + 7 \, B a^{3} + 5 \, C a^{3}\right )}{\left (d x + c\right )} - 6 \,{\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (3 \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, 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} + 15 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, 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} + 40 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, 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 ) + 33 \, 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} \end{align*}

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)^2,x, algorithm="giac")

[Out]

-1/6*(12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(6*A*a^3 + 7*B*a^3 + 5*C*a^3)*(d*x + c) -
6*(3*A*a^3 + B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(3*A*a^3 + B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) - 2*(6*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 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1
2*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 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*t
an(1/2*d*x + 1/2*c) + 21*B*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 +
1)^3)/d