3.251 \(\int (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=154 \[ \frac{a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac{5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

[Out]

(5*a^3*(3*B + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(9*B + 11*C)*Tan[c + d*x])/(3*d) + (a^3*(27*B + 28*C)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*B + 2*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) +
(a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.505596, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3029, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac{5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac{a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(5*a^3*(3*B + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(9*B + 11*C)*Tan[c + d*x])/(3*d) + (a^3*(27*B + 28*C)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*B + 2*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) +
(a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 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 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 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 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 8

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

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 (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\int (a+a \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+a \cos (c+d x))^2 (2 a (3 B+2 C)+a (B+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+a \cos (c+d x)) \left (a^2 (27 B+28 C)+a^2 (9 B+16 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int \left (a^3 (27 B+28 C)+\left (a^3 (9 B+16 C)+a^3 (27 B+28 C)\right ) \cos (c+d x)+a^3 (9 B+16 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (8 a^3 (9 B+11 C)+15 a^3 (3 B+4 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (5 a^3 (3 B+4 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{3} \left (a^3 (9 B+11 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (a^3 (9 B+11 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac{a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.21608, size = 273, normalized size = 1.77 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (120 (3 B+4 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-24 (9 B+11 C) \sin (c)+69 B \sin (2 c+d x)+264 B \sin (c+2 d x)-24 B \sin (3 c+2 d x)+45 B \sin (2 c+3 d x)+45 B \sin (4 c+3 d x)+72 B \sin (3 c+4 d x)+(69 B+36 C) \sin (d x)+36 C \sin (2 c+d x)+280 C \sin (c+2 d x)-72 C \sin (3 c+2 d x)+36 C \sin (2 c+3 d x)+36 C \sin (4 c+3 d x)+88 C \sin (3 c+4 d x))\right )}{1536 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

-(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(120*(3*B + 4*C)*Cos[c + d*x]^4*(Log[Cos[(c + d*x
)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-24*(9*B + 11*C)*Sin[c] + (69*B
+ 36*C)*Sin[d*x] + 69*B*Sin[2*c + d*x] + 36*C*Sin[2*c + d*x] + 264*B*Sin[c + 2*d*x] + 280*C*Sin[c + 2*d*x] -
24*B*Sin[3*c + 2*d*x] - 72*C*Sin[3*c + 2*d*x] + 45*B*Sin[2*c + 3*d*x] + 36*C*Sin[2*c + 3*d*x] + 45*B*Sin[4*c +
3*d*x] + 36*C*Sin[4*c + 3*d*x] + 72*B*Sin[3*c + 4*d*x] + 88*C*Sin[3*c + 4*d*x])))/(1536*d)

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Maple [A]  time = 0.076, size = 188, normalized size = 1.2 \begin{align*}{\frac{11\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

11/3/d*a^3*C*tan(d*x+c)+1/3/d*a^3*C*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^3*B*tan(d*x+c)*sec(d*x+c)^3+15/8/d*a^3*B*s
ec(d*x+c)*tan(d*x+c)+15/8/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*a^3*C*sec(d*x+c)*tan(d*x+c)+5/2/d*a^3*C*ln(s
ec(d*x+c)+tan(d*x+c))+3/d*a^3*B*tan(d*x+c)+1/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.27615, size = 363, normalized size = 2.36 \begin{align*} \frac{48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, B a^{3}{\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 )} - 36 \, 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 )} - 36 \, C 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 )} + 24 \, 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 )} + 48 \, B a^{3} \tan \left (d x + c\right ) + 144 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/48*(48*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*B*a^3*(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)) - 36*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
)) - 36*C*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)) + 24*C*a^3
*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*B*a^3*tan(d*x + c) + 144*C*a^3*tan(d*x + c))/d

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Fricas [A]  time = 1.75388, size = 366, normalized size = 2.38 \begin{align*} \frac{15 \,{\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \,{\left (5 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \,{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, B a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/48*(15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*(9*B + 11*C)*a^3*cos(d*x + c)^3 + 9*(5*B + 4*C)*a^3*cos(d*x + c)^2 + 8*(3*B + C)*a^3*cos(d*x
+ c) + 6*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.64964, size = 286, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (3 \, B a^{3} + 4 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (3 \, B a^{3} + 4 \, 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 (45 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 165 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 220 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 219 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 292 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 147 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 132 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/24*(15*(3*B*a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(45*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 165*B*a^3*tan(1/2*d*x +
1/2*c)^5 - 220*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*C*a^3*tan(1/2*d*x + 1/2*
c)^3 - 147*B*a^3*tan(1/2*d*x + 1/2*c) - 132*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d