3.187 \(\int (a+a \cos (x))^3 (A+B \sec (x)) \, dx\)

Optimal. Leaf size=75 \[ \frac {1}{2} a^3 x (5 A+7 B)+\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{6} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )+a^3 B \tanh ^{-1}(\sin (x))+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2 \]

[Out]

1/2*a^3*(5*A+7*B)*x+a^3*B*arctanh(sin(x))+5/2*a^3*(A+B)*sin(x)+1/3*a*A*(a+a*cos(x))^2*sin(x)+1/6*(5*A+3*B)*(a^
3+a^3*cos(x))*sin(x)

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Rubi [A]  time = 0.30, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2828, 2976, 2968, 3023, 2735, 3770} \[ \frac {1}{2} a^3 x (5 A+7 B)+\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{6} (5 A+3 B) \sin (x) \left (a^3 \cos (x)+a^3\right )+a^3 B \tanh ^{-1}(\sin (x))+\frac {1}{3} a A \sin (x) (a \cos (x)+a)^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[x])^3*(A + B*Sec[x]),x]

[Out]

(a^3*(5*A + 7*B)*x)/2 + a^3*B*ArcTanh[Sin[x]] + (5*a^3*(A + B)*Sin[x])/2 + (a*A*(a + a*Cos[x])^2*Sin[x])/3 + (
(5*A + 3*B)*(a^3 + a^3*Cos[x])*Sin[x])/6

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 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 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 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 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 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))^3 (A+B \sec (x)) \, dx &=\int (a+a \cos (x))^3 (B+A \cos (x)) \sec (x) \, dx\\ &=\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{3} \int (a+a \cos (x))^2 (3 a B+a (5 A+3 B) \cos (x)) \sec (x) \, dx\\ &=\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x)+\frac {1}{6} \int (a+a \cos (x)) \left (6 a^2 B+15 a^2 (A+B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x)+\frac {1}{6} \int \left (6 a^3 B+\left (6 a^3 B+15 a^3 (A+B)\right ) \cos (x)+15 a^3 (A+B) \cos ^2(x)\right ) \sec (x) \, dx\\ &=\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x)+\frac {1}{6} \int \left (6 a^3 B+3 a^3 (5 A+7 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac {1}{2} a^3 (5 A+7 B) x+\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x)+\left (a^3 B\right ) \int \sec (x) \, dx\\ &=\frac {1}{2} a^3 (5 A+7 B) x+a^3 B \tanh ^{-1}(\sin (x))+\frac {5}{2} a^3 (A+B) \sin (x)+\frac {1}{3} a A (a+a \cos (x))^2 \sin (x)+\frac {1}{6} (5 A+3 B) \left (a^3+a^3 \cos (x)\right ) \sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 80, normalized size = 1.07 \[ \frac {1}{12} a^3 \left (9 (5 A+4 B) \sin (x)+3 (3 A+B) \sin (2 x)+30 A x+A \sin (3 x)+42 B x-12 B \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+12 B \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[x])^3*(A + B*Sec[x]),x]

[Out]

(a^3*(30*A*x + 42*B*x - 12*B*Log[Cos[x/2] - Sin[x/2]] + 12*B*Log[Cos[x/2] + Sin[x/2]] + 9*(5*A + 4*B)*Sin[x] +
3*(3*A + B)*Sin[2*x] + A*Sin[3*x]))/12

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fricas [A]  time = 1.19, size = 77, normalized size = 1.03 \[ \frac {1}{2} \, {\left (5 \, A + 7 \, B\right )} a^{3} x + \frac {1}{2} \, B a^{3} \log \left (\sin \relax (x) + 1\right ) - \frac {1}{2} \, B a^{3} \log \left (-\sin \relax (x) + 1\right ) + \frac {1}{6} \, {\left (2 \, A a^{3} \cos \relax (x)^{2} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \relax (x) + 2 \, {\left (11 \, A + 9 \, B\right )} a^{3}\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^3*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/2*(5*A + 7*B)*a^3*x + 1/2*B*a^3*log(sin(x) + 1) - 1/2*B*a^3*log(-sin(x) + 1) + 1/6*(2*A*a^3*cos(x)^2 + 3*(3*
A + B)*a^3*cos(x) + 2*(11*A + 9*B)*a^3)*sin(x)

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giac [A]  time = 0.18, size = 125, normalized size = 1.67 \[ B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) + \frac {1}{2} \, {\left (5 \, A a^{3} + 7 \, B a^{3}\right )} x + \frac {15 \, A a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, x\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^3*(A+B*sec(x)),x, algorithm="giac")

[Out]

B*a^3*log(abs(tan(1/2*x) + 1)) - B*a^3*log(abs(tan(1/2*x) - 1)) + 1/2*(5*A*a^3 + 7*B*a^3)*x + 1/3*(15*A*a^3*ta
n(1/2*x)^5 + 15*B*a^3*tan(1/2*x)^5 + 40*A*a^3*tan(1/2*x)^3 + 36*B*a^3*tan(1/2*x)^3 + 33*A*a^3*tan(1/2*x) + 21*
B*a^3*tan(1/2*x))/(tan(1/2*x)^2 + 1)^3

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maple [A]  time = 0.11, size = 77, normalized size = 1.03 \[ \frac {A \,a^{3} \left (2+\cos ^{2}\relax (x )\right ) \sin \relax (x )}{3}+\frac {B \,a^{3} \sin \relax (x ) \cos \relax (x )}{2}+\frac {7 B \,a^{3} x}{2}+\frac {3 A \,a^{3} \sin \relax (x ) \cos \relax (x )}{2}+\frac {5 A \,a^{3} x}{2}+3 B \,a^{3} \sin \relax (x )+3 A \,a^{3} \sin \relax (x )+B \,a^{3} \ln \left (\sec \relax (x )+\tan \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(x))^3*(A+B*sec(x)),x)

[Out]

1/3*A*a^3*(2+cos(x)^2)*sin(x)+1/2*B*a^3*sin(x)*cos(x)+7/2*B*a^3*x+3/2*A*a^3*sin(x)*cos(x)+5/2*A*a^3*x+3*B*a^3*
sin(x)+3*A*a^3*sin(x)+B*a^3*ln(sec(x)+tan(x))

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maxima [A]  time = 0.31, size = 84, normalized size = 1.12 \[ -\frac {1}{3} \, {\left (\sin \relax (x)^{3} - 3 \, \sin \relax (x)\right )} A a^{3} + \frac {3}{4} \, A a^{3} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + \frac {1}{4} \, B a^{3} {\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{3} x + 3 \, B a^{3} x + B a^{3} \log \left (\sec \relax (x) + \tan \relax (x)\right ) + 3 \, A a^{3} \sin \relax (x) + 3 \, B a^{3} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^3*(A+B*sec(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.48, size = 431, normalized size = 5.75 \[ \frac {\left (5\,A\,a^3+5\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+\left (\frac {40\,A\,a^3}{3}+12\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (11\,A\,a^3+7\,B\,a^3\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+a^3\,\mathrm {atan}\left (\frac {1000\,A^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {2968\,B^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {6040\,A\,B^2\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}+\frac {4200\,A^2\,B\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{1000\,A^3\,a^9+4200\,A^2\,B\,a^9+6040\,A\,B^2\,a^9+2968\,B^3\,a^9}\right )\,\left (5\,A+7\,B\right )+2\,B\,a^3\,\mathrm {atanh}\left (\frac {848\,B^3\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}+\frac {1120\,A\,B^2\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}+\frac {400\,A^2\,B\,a^9\,\mathrm {tan}\left (\frac {x}{2}\right )}{400\,A^2\,B\,a^9+1120\,A\,B^2\,a^9+848\,B^3\,a^9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(x))^3*(A + B/cos(x)),x)

[Out]

(tan(x/2)^5*(5*A*a^3 + 5*B*a^3) + tan(x/2)^3*((40*A*a^3)/3 + 12*B*a^3) + tan(x/2)*(11*A*a^3 + 7*B*a^3))/(3*tan
(x/2)^2 + 3*tan(x/2)^4 + tan(x/2)^6 + 1) + a^3*atan((1000*A^3*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 604
0*A*B^2*a^9 + 4200*A^2*B*a^9) + (2968*B^3*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 6040*A*B^2*a^9 + 4200*A
^2*B*a^9) + (6040*A*B^2*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 6040*A*B^2*a^9 + 4200*A^2*B*a^9) + (4200*
A^2*B*a^9*tan(x/2))/(1000*A^3*a^9 + 2968*B^3*a^9 + 6040*A*B^2*a^9 + 4200*A^2*B*a^9))*(5*A + 7*B) + 2*B*a^3*ata
nh((848*B^3*a^9*tan(x/2))/(848*B^3*a^9 + 1120*A*B^2*a^9 + 400*A^2*B*a^9) + (1120*A*B^2*a^9*tan(x/2))/(848*B^3*
a^9 + 1120*A*B^2*a^9 + 400*A^2*B*a^9) + (400*A^2*B*a^9*tan(x/2))/(848*B^3*a^9 + 1120*A*B^2*a^9 + 400*A^2*B*a^9
))

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sympy [A]  time = 7.63, size = 92, normalized size = 1.23 \[ \frac {5 A a^{3} x}{2} - \frac {A a^{3} \sin ^{3}{\relax (x )}}{3} + 4 A a^{3} \sin {\relax (x )} + \frac {3 A a^{3} \sin {\left (2 x \right )}}{4} + \frac {7 B a^{3} x}{2} + B a^{3} \log {\left (\tan {\relax (x )} + \sec {\relax (x )} \right )} + \frac {B a^{3} \sin {\relax (x )} \cos {\relax (x )}}{2} + 3 B a^{3} \sin {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))**3*(A+B*sec(x)),x)

[Out]

5*A*a**3*x/2 - A*a**3*sin(x)**3/3 + 4*A*a**3*sin(x) + 3*A*a**3*sin(2*x)/4 + 7*B*a**3*x/2 + B*a**3*log(tan(x) +
sec(x)) + B*a**3*sin(x)*cos(x)/2 + 3*B*a**3*sin(x)

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