3.186 \(\int (a+a \cos (x))^2 (A+B \sec (x)) \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{2} a^2 x (3 A+4 B)+\frac{1}{2} a^2 (3 A+2 B) \sin (x)+\frac{1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )+a^2 B \tanh ^{-1}(\sin (x)) \]

[Out]

(a^2*(3*A + 4*B)*x)/2 + a^2*B*ArcTanh[Sin[x]] + (a^2*(3*A + 2*B)*Sin[x])/2 + (A*(a^2 + a^2*Cos[x])*Sin[x])/2

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Rubi [A]  time = 0.201573, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2828, 2976, 2968, 3023, 2735, 3770} \[ \frac{1}{2} a^2 x (3 A+4 B)+\frac{1}{2} a^2 (3 A+2 B) \sin (x)+\frac{1}{2} A \sin (x) \left (a^2 \cos (x)+a^2\right )+a^2 B \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[x])^2*(A + B*Sec[x]),x]

[Out]

(a^2*(3*A + 4*B)*x)/2 + a^2*B*ArcTanh[Sin[x]] + (a^2*(3*A + 2*B)*Sin[x])/2 + (A*(a^2 + a^2*Cos[x])*Sin[x])/2

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

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 (x))^2 (A+B \sec (x)) \, dx &=\int (a+a \cos (x))^2 (B+A \cos (x)) \sec (x) \, dx\\ &=\frac{1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x)+\frac{1}{2} \int (a+a \cos (x)) (2 a B+a (3 A+2 B) \cos (x)) \sec (x) \, dx\\ &=\frac{1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x)+\frac{1}{2} \int \left (2 a^2 B+\left (2 a^2 B+a^2 (3 A+2 B)\right ) \cos (x)+a^2 (3 A+2 B) \cos ^2(x)\right ) \sec (x) \, dx\\ &=\frac{1}{2} a^2 (3 A+2 B) \sin (x)+\frac{1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x)+\frac{1}{2} \int \left (2 a^2 B+a^2 (3 A+4 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{1}{2} a^2 (3 A+4 B) x+\frac{1}{2} a^2 (3 A+2 B) \sin (x)+\frac{1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x)+\left (a^2 B\right ) \int \sec (x) \, dx\\ &=\frac{1}{2} a^2 (3 A+4 B) x+a^2 B \tanh ^{-1}(\sin (x))+\frac{1}{2} a^2 (3 A+2 B) \sin (x)+\frac{1}{2} A \left (a^2+a^2 \cos (x)\right ) \sin (x)\\ \end{align*}

Mathematica [A]  time = 0.0823791, size = 67, normalized size = 1.18 \[ \frac{1}{4} a^2 \left (4 (2 A+B) \sin (x)+6 A x+A \sin (2 x)+8 B x-4 B \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+4 B \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[x])^2*(A + B*Sec[x]),x]

[Out]

(a^2*(6*A*x + 8*B*x - 4*B*Log[Cos[x/2] - Sin[x/2]] + 4*B*Log[Cos[x/2] + Sin[x/2]] + 4*(2*A + B)*Sin[x] + A*Sin
[2*x]))/4

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Maple [A]  time = 0.046, size = 52, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}A\sin \left ( x \right ) \cos \left ( x \right ) }{2}}+{\frac{3\,{a}^{2}Ax}{2}}+{a}^{2}B\sin \left ( x \right ) +2\,{a}^{2}A\sin \left ( x \right ) +2\,{a}^{2}Bx+{a}^{2}B\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(x))^2*(A+B*sec(x)),x)

[Out]

1/2*a^2*A*sin(x)*cos(x)+3/2*a^2*A*x+a^2*B*sin(x)+2*a^2*A*sin(x)+2*a^2*B*x+a^2*B*ln(sec(x)+tan(x))

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Maxima [A]  time = 1.01363, size = 73, normalized size = 1.28 \begin{align*} \frac{1}{4} \, A a^{2}{\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{2} x + 2 \, B a^{2} x + B a^{2} \log \left (\sec \left (x\right ) + \tan \left (x\right )\right ) + 2 \, A a^{2} \sin \left (x\right ) + B a^{2} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^2*(A+B*sec(x)),x, algorithm="maxima")

[Out]

1/4*A*a^2*(2*x + sin(2*x)) + A*a^2*x + 2*B*a^2*x + B*a^2*log(sec(x) + tan(x)) + 2*A*a^2*sin(x) + B*a^2*sin(x)

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Fricas [A]  time = 2.57262, size = 170, normalized size = 2.98 \begin{align*} \frac{1}{2} \,{\left (3 \, A + 4 \, B\right )} a^{2} x + \frac{1}{2} \, B a^{2} \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{2} \, B a^{2} \log \left (-\sin \left (x\right ) + 1\right ) + \frac{1}{2} \,{\left (A a^{2} \cos \left (x\right ) + 2 \,{\left (2 \, A + B\right )} a^{2}\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^2*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/2*(3*A + 4*B)*a^2*x + 1/2*B*a^2*log(sin(x) + 1) - 1/2*B*a^2*log(-sin(x) + 1) + 1/2*(A*a^2*cos(x) + 2*(2*A +
B)*a^2)*sin(x)

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Sympy [A]  time = 8.7605, size = 61, normalized size = 1.07 \begin{align*} \frac{3 A a^{2} x}{2} + 2 A a^{2} \sin{\left (x \right )} + \frac{A a^{2} \sin{\left (2 x \right )}}{4} + 2 B a^{2} x + B a^{2} \log{\left (\tan{\left (x \right )} + \sec{\left (x \right )} \right )} + B a^{2} \sin{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))**2*(A+B*sec(x)),x)

[Out]

3*A*a**2*x/2 + 2*A*a**2*sin(x) + A*a**2*sin(2*x)/4 + 2*B*a**2*x + B*a**2*log(tan(x) + sec(x)) + B*a**2*sin(x)

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Giac [A]  time = 1.18303, size = 135, normalized size = 2.37 \begin{align*} B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right ) - B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right ) + \frac{1}{2} \,{\left (3 \, A a^{2} + 4 \, B a^{2}\right )} x + \frac{3 \, A a^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, B a^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 5 \, A a^{2} \tan \left (\frac{1}{2} \, x\right ) + 2 \, B a^{2} \tan \left (\frac{1}{2} \, x\right )}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^2*(A+B*sec(x)),x, algorithm="giac")

[Out]

B*a^2*log(abs(tan(1/2*x) + 1)) - B*a^2*log(abs(tan(1/2*x) - 1)) + 1/2*(3*A*a^2 + 4*B*a^2)*x + (3*A*a^2*tan(1/2
*x)^3 + 2*B*a^2*tan(1/2*x)^3 + 5*A*a^2*tan(1/2*x) + 2*B*a^2*tan(1/2*x))/(tan(1/2*x)^2 + 1)^2