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

Optimal. Leaf size=110 \[ \frac{a^3 (3 B+C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac{1}{2} a^3 x (6 B+7 C)+\frac{5 a^3 C \sin (c+d x)}{2 d}+\frac{a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.400305, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 2975, 2976, 2968, 3023, 2735, 3770} \[ \frac{a^3 (3 B+C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(2 B-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac{1}{2} a^3 x (6 B+7 C)+\frac{5 a^3 C \sin (c+d x)}{2 d}+\frac{a B \tan (c+d x) (a \cos (c+d x)+a)^2}{d} \]

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]^3,x]

[Out]

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

Mathematica [B]  time = 1.71076, size = 272, normalized size = 2.47 \[ \frac{1}{32} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 (B+3 C) \sin (c) \cos (d x)}{d}+\frac{4 (B+3 C) \cos (c) \sin (d x)}{d}-\frac{4 (3 B+C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{4 (3 B+C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{4 B \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 B \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+2 x (6 B+7 C)+\frac{C \sin (2 c) \cos (2 d x)}{d}+\frac{C \cos (2 c) \sin (2 d x)}{d}\right ) \]

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]^3,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*(6*B + 7*C)*x - (4*(3*B + C)*Log[Cos[(c + d*x)/2] - Sin[(c + d
*x)/2]])/d + (4*(3*B + C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(B + 3*C)*Cos[d*x]*Sin[c])/d + (C*C
os[2*d*x]*Sin[2*c])/d + (4*(B + 3*C)*Cos[c]*Sin[d*x])/d + (C*Cos[2*c]*Sin[2*d*x])/d + (4*B*Sin[(d*x)/2])/(d*(C
os[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*B*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]))))/32

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Maple [A]  time = 0.067, size = 145, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{7\,{a}^{3}Cx}{2}}+{\frac{7\,{a}^{3}Cc}{2\,d}}+3\,{\frac{{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+3\,{a}^{3}Bx+3\,{\frac{B{a}^{3}c}{d}}+{\frac{{a}^{3}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}B\sin \left ( dx+c \right ) }{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.20951, size = 189, normalized size = 1.72 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{3} +{\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} + 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 )} + 2 \, 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 )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, C a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right )}{4 \, 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.80824, size = 323, normalized size = 2.94 \begin{align*} \frac{{\left (6 \, B + 7 \, C\right )} a^{3} d x \cos \left (d x + c\right ) +{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (C a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{2 \, 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*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/2*((6*B + 7*C)*a^3*d*x*cos(d*x + c) + (3*B + C)*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - (3*B + C)*a^3*cos(d
*x + c)*log(-sin(d*x + c) + 1) + (C*a^3*cos(d*x + c)^2 + 2*(B + 3*C)*a^3*cos(d*x + c) + 2*B*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*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.41587, size = 259, normalized size = 2.35 \begin{align*} -\frac{\frac{4 \, B 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} -{\left (6 \, B a^{3} + 7 \, C a^{3}\right )}{\left (d x + c\right )} - 2 \,{\left (3 \, B a^{3} + C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (3 \, B a^{3} + 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 (2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, 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 )}^{2}}}{2 \, 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)^3,x, algorithm="giac")

[Out]

-1/2*(4*B*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (6*B*a^3 + 7*C*a^3)*(d*x + c) - 2*(3*B*a^3 +
C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*B*a^3 + C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(2*B*a
^3*tan(1/2*d*x + 1/2*c)^3 + 5*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1/2*d*x + 1/2*c) + 7*C*a^3*tan(1/2*d*
x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d