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

Optimal. Leaf size=156 \[ \frac{\left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{3 d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

((3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a*b*B + 2*a^2*C + 3*b^2*C)*Tan[c + d*x])/(3*
d) + ((3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*(2*b*B + a*C)*Sec[c + d*x]^2*Tan[c +
d*x])/(3*d) + (a^2*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.376542, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3029, 2988, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{3 d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (3 a^2 B+8 a b C+4 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a*b*B + 2*a^2*C + 3*b^2*C)*Tan[c + d*x])/(3*
d) + ((3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*(2*b*B + a*C)*Sec[c + d*x]^2*Tan[c +
d*x])/(3*d) + (a^2*B*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 2988

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

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

Mathematica [A]  time = 0.719009, size = 120, normalized size = 0.77 \[ \frac{3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \sec (c+d x)+24 \left (a^2 C+2 a b B+b^2 C\right )+6 a^2 B \sec ^3(c+d x)+8 a (a C+2 b B) \tan ^2(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(2*a*b*B + a^2*C + b^2*C) + 3*(3*a^2
*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x] + 6*a^2*B*Sec[c + d*x]^3 + 8*a*(2*b*B + a*C)*Tan[c + d*x]^2))/(24*d)

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Maple [A]  time = 0.054, size = 241, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{abC\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,abB\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,abB\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04523, size = 308, normalized size = 1.97 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b - 3 \, B 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 )} - 24 \, C a b{\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 \, B b^{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 )} + 48 \, C b^{2} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.45799, size = 443, normalized size = 2.84 \begin{align*} \frac{3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (2 \, C a^{2} + 4 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, B a^{2} + 3 \,{\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\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+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/48*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*c
os(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(2*C*a^2 + 4*B*a*b + 3*C*b^2)*cos(d*x + c)^3 + 6*B*a^2 + 3*(3*B*a^
2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c)^2 + 8*(C*a^2 + 2*B*a*b)*cos(d*x + c))*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+b*cos(d*x+c))**2*(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 [B]  time = 1.75789, size = 645, normalized size = 4.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*B*
a*b*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*C*b^2*tan(
1/2*d*x + 1/2*c)^7 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 40*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*B*a*b*tan(1/2*d*x +
1/2*c)^5 - 24*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*C*b^2*tan(1/2*d*x + 1/2*c)^
5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*C*
a*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^2*tan(
1/2*d*x + 1/2*c) + 24*C*a^2*tan(1/2*d*x + 1/2*c) + 48*B*a*b*tan(1/2*d*x + 1/2*c) + 24*C*a*b*tan(1/2*d*x + 1/2*
c) + 12*B*b^2*tan(1/2*d*x + 1/2*c) + 24*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d