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

Optimal. Leaf size=74 \[ -\frac{a^2 (B-C) \sin (c+d x)}{d}+\frac{a^2 (2 B+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{B \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+a^2 x (B+2 C) \]

[Out]

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

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Rubi [A]  time = 0.290653, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3029, 2975, 2968, 3023, 2735, 3770} \[ -\frac{a^2 (B-C) \sin (c+d x)}{d}+\frac{a^2 (2 B+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{B \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+a^2 x (B+2 C) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.31674, size = 143, normalized size = 1.93 \[ \frac{a^2 \left (B \tan (c+d x)-2 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+B c+B d x+C \sin (c+d x)-C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 c C+2 C d x\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(B*c + 2*c*C + B*d*x + 2*C*d*x - 2*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - C*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + 2*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
+ C*Sin[c + d*x] + B*Tan[c + d*x]))/d

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Maple [A]  time = 0.062, size = 107, normalized size = 1.5 \begin{align*}{a}^{2}Bx+2\,{a}^{2}Cx+2\,{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}B\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}c}{d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{C{a}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08597, size = 142, normalized size = 1.92 \begin{align*} \frac{2 \,{\left (d x + c\right )} B a^{2} + 4 \,{\left (d x + c\right )} C a^{2} + 2 \, B a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + C a^{2}{\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^{2} \sin \left (d x + c\right ) + 2 \, B a^{2} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72272, size = 278, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (B + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) +{\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (C a^{2} \cos \left (d x + c\right ) + B a^{2}\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))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/2*(2*(B + 2*C)*a^2*d*x*cos(d*x + c) + (2*B + C)*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) - (2*B + C)*a^2*cos(d
*x + c)*log(-sin(d*x + c) + 1) + 2*(C*a^2*cos(d*x + c) + B*a^2)*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))**2*(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 [B]  time = 1.56175, size = 209, normalized size = 2.82 \begin{align*} \frac{{\left (B a^{2} + 2 \, C a^{2}\right )}{\left (d x + c\right )} +{\left (2 \, B a^{2} + C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, B a^{2} + C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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