3.188 \(\int (a+a \cos (x))^4 (A+B \sec (x)) \, dx\)

Optimal. Leaf size=104 \[ \frac{1}{8} a^4 x (35 A+48 B)+\frac{5}{8} a^4 (7 A+8 B) \sin (x)+\frac{1}{12} (7 A+4 B) \sin (x) \left (a^2 \cos (x)+a^2\right )^2+\frac{1}{24} (35 A+32 B) \sin (x) \left (a^4 \cos (x)+a^4\right )+a^4 B \tanh ^{-1}(\sin (x))+\frac{1}{4} a A \sin (x) (a \cos (x)+a)^3 \]

[Out]

(a^4*(35*A + 48*B)*x)/8 + a^4*B*ArcTanh[Sin[x]] + (5*a^4*(7*A + 8*B)*Sin[x])/8 + (a*A*(a + a*Cos[x])^3*Sin[x])
/4 + ((7*A + 4*B)*(a^2 + a^2*Cos[x])^2*Sin[x])/12 + ((35*A + 32*B)*(a^4 + a^4*Cos[x])*Sin[x])/24

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Rubi [A]  time = 0.402811, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{1}{8} a^4 x (35 A+48 B)+\frac{5}{8} a^4 (7 A+8 B) \sin (x)+\frac{1}{12} (7 A+4 B) \sin (x) \left (a^2 \cos (x)+a^2\right )^2+\frac{1}{24} (35 A+32 B) \sin (x) \left (a^4 \cos (x)+a^4\right )+a^4 B \tanh ^{-1}(\sin (x))+\frac{1}{4} a A \sin (x) (a \cos (x)+a)^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[x])^4*(A + B*Sec[x]),x]

[Out]

(a^4*(35*A + 48*B)*x)/8 + a^4*B*ArcTanh[Sin[x]] + (5*a^4*(7*A + 8*B)*Sin[x])/8 + (a*A*(a + a*Cos[x])^3*Sin[x])
/4 + ((7*A + 4*B)*(a^2 + a^2*Cos[x])^2*Sin[x])/12 + ((35*A + 32*B)*(a^4 + a^4*Cos[x])*Sin[x])/24

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))^4 (A+B \sec (x)) \, dx &=\int (a+a \cos (x))^4 (B+A \cos (x)) \sec (x) \, dx\\ &=\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{4} \int (a+a \cos (x))^3 (4 a B+a (7 A+4 B) \cos (x)) \sec (x) \, dx\\ &=\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{12} \int (a+a \cos (x))^2 \left (12 a^2 B+a^2 (35 A+32 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac{1}{24} \int (a+a \cos (x)) \left (24 a^3 B+15 a^3 (7 A+8 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac{1}{24} \int \left (24 a^4 B+\left (24 a^4 B+15 a^4 (7 A+8 B)\right ) \cos (x)+15 a^4 (7 A+8 B) \cos ^2(x)\right ) \sec (x) \, dx\\ &=\frac{5}{8} a^4 (7 A+8 B) \sin (x)+\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\frac{1}{24} \int \left (24 a^4 B+3 a^4 (35 A+48 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{1}{8} a^4 (35 A+48 B) x+\frac{5}{8} a^4 (7 A+8 B) \sin (x)+\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)+\left (a^4 B\right ) \int \sec (x) \, dx\\ &=\frac{1}{8} a^4 (35 A+48 B) x+a^4 B \tanh ^{-1}(\sin (x))+\frac{5}{8} a^4 (7 A+8 B) \sin (x)+\frac{1}{4} a A (a+a \cos (x))^3 \sin (x)+\frac{1}{12} (7 A+4 B) \left (a^2+a^2 \cos (x)\right )^2 \sin (x)+\frac{1}{24} (35 A+32 B) \left (a^4+a^4 \cos (x)\right ) \sin (x)\\ \end{align*}

Mathematica [A]  time = 0.126385, size = 97, normalized size = 0.93 \[ \frac{1}{96} a^4 \left (24 (28 A+27 B) \sin (x)+24 (7 A+4 B) \sin (2 x)+420 A x+32 A \sin (3 x)+3 A \sin (4 x)+576 B x+8 B \sin (3 x)-96 B \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+96 B \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[x])^4*(A + B*Sec[x]),x]

[Out]

(a^4*(420*A*x + 576*B*x - 96*B*Log[Cos[x/2] - Sin[x/2]] + 96*B*Log[Cos[x/2] + Sin[x/2]] + 24*(28*A + 27*B)*Sin
[x] + 24*(7*A + 4*B)*Sin[2*x] + 32*A*Sin[3*x] + 8*B*Sin[3*x] + 3*A*Sin[4*x]))/96

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Maple [A]  time = 0.088, size = 103, normalized size = 1. \begin{align*}{\frac{A{a}^{4}\sin \left ( x \right ) \left ( \cos \left ( x \right ) \right ) ^{3}}{4}}+{\frac{27\,A{a}^{4}\sin \left ( x \right ) \cos \left ( x \right ) }{8}}+{\frac{35\,A{a}^{4}x}{8}}+{\frac{B{a}^{4} \left ( 2+ \left ( \cos \left ( x \right ) \right ) ^{2} \right ) \sin \left ( x \right ) }{3}}+{\frac{4\,A{a}^{4} \left ( 2+ \left ( \cos \left ( x \right ) \right ) ^{2} \right ) \sin \left ( x \right ) }{3}}+2\,B{a}^{4}\sin \left ( x \right ) \cos \left ( x \right ) +6\,B{a}^{4}x+6\,B{a}^{4}\sin \left ( x \right ) +4\,A{a}^{4}\sin \left ( x \right ) +B{a}^{4}\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))^4*(A+B*sec(x)),x)

[Out]

1/4*A*a^4*sin(x)*cos(x)^3+27/8*A*a^4*sin(x)*cos(x)+35/8*A*a^4*x+1/3*B*a^4*(2+cos(x)^2)*sin(x)+4/3*A*a^4*(2+cos
(x)^2)*sin(x)+2*B*a^4*sin(x)*cos(x)+6*B*a^4*x+6*B*a^4*sin(x)+4*A*a^4*sin(x)+B*a^4*ln(sec(x)+tan(x))

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Maxima [A]  time = 1.00779, size = 159, normalized size = 1.53 \begin{align*} -\frac{4}{3} \,{\left (\sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )} A a^{4} - \frac{1}{3} \,{\left (\sin \left (x\right )^{3} - 3 \, \sin \left (x\right )\right )} B a^{4} + \frac{1}{32} \, A a^{4}{\left (12 \, x + \sin \left (4 \, x\right ) + 8 \, \sin \left (2 \, x\right )\right )} + \frac{3}{2} \, A a^{4}{\left (2 \, x + \sin \left (2 \, x\right )\right )} + B a^{4}{\left (2 \, x + \sin \left (2 \, x\right )\right )} + A a^{4} x + 4 \, B a^{4} x + B a^{4} \log \left (\sec \left (x\right ) + \tan \left (x\right )\right ) + 4 \, A a^{4} \sin \left (x\right ) + 6 \, B a^{4} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="maxima")

[Out]

-4/3*(sin(x)^3 - 3*sin(x))*A*a^4 - 1/3*(sin(x)^3 - 3*sin(x))*B*a^4 + 1/32*A*a^4*(12*x + sin(4*x) + 8*sin(2*x))
+ 3/2*A*a^4*(2*x + sin(2*x)) + B*a^4*(2*x + sin(2*x)) + A*a^4*x + 4*B*a^4*x + B*a^4*log(sec(x) + tan(x)) + 4*
A*a^4*sin(x) + 6*B*a^4*sin(x)

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Fricas [A]  time = 2.54186, size = 255, normalized size = 2.45 \begin{align*} \frac{1}{8} \,{\left (35 \, A + 48 \, B\right )} a^{4} x + \frac{1}{2} \, B a^{4} \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{2} \, B a^{4} \log \left (-\sin \left (x\right ) + 1\right ) + \frac{1}{24} \,{\left (6 \, A a^{4} \cos \left (x\right )^{3} + 8 \,{\left (4 \, A + B\right )} a^{4} \cos \left (x\right )^{2} + 3 \,{\left (27 \, A + 16 \, B\right )} a^{4} \cos \left (x\right ) + 160 \,{\left (A + B\right )} a^{4}\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/8*(35*A + 48*B)*a^4*x + 1/2*B*a^4*log(sin(x) + 1) - 1/2*B*a^4*log(-sin(x) + 1) + 1/24*(6*A*a^4*cos(x)^3 + 8*
(4*A + B)*a^4*cos(x)^2 + 3*(27*A + 16*B)*a^4*cos(x) + 160*(A + B)*a^4)*sin(x)

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Sympy [A]  time = 174.516, size = 116, normalized size = 1.12 \begin{align*} \frac{35 A a^{4} x}{8} - \frac{4 A a^{4} \sin ^{3}{\left (x \right )}}{3} + 8 A a^{4} \sin{\left (x \right )} + \frac{7 A a^{4} \sin{\left (2 x \right )}}{4} + \frac{A a^{4} \sin{\left (4 x \right )}}{32} + 6 B a^{4} x + B a^{4} \log{\left (\tan{\left (x \right )} + \sec{\left (x \right )} \right )} - \frac{B a^{4} \sin ^{3}{\left (x \right )}}{3} + 2 B a^{4} \sin{\left (x \right )} \cos{\left (x \right )} + 7 B a^{4} \sin{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))**4*(A+B*sec(x)),x)

[Out]

35*A*a**4*x/8 - 4*A*a**4*sin(x)**3/3 + 8*A*a**4*sin(x) + 7*A*a**4*sin(2*x)/4 + A*a**4*sin(4*x)/32 + 6*B*a**4*x
+ B*a**4*log(tan(x) + sec(x)) - B*a**4*sin(x)**3/3 + 2*B*a**4*sin(x)*cos(x) + 7*B*a**4*sin(x)

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Giac [A]  time = 1.16871, size = 201, normalized size = 1.93 \begin{align*} B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right ) - B a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right ) + \frac{1}{8} \,{\left (35 \, A a^{4} + 48 \, B a^{4}\right )} x + \frac{105 \, A a^{4} \tan \left (\frac{1}{2} \, x\right )^{7} + 120 \, B a^{4} \tan \left (\frac{1}{2} \, x\right )^{7} + 385 \, A a^{4} \tan \left (\frac{1}{2} \, x\right )^{5} + 424 \, B a^{4} \tan \left (\frac{1}{2} \, x\right )^{5} + 511 \, A a^{4} \tan \left (\frac{1}{2} \, x\right )^{3} + 520 \, B a^{4} \tan \left (\frac{1}{2} \, x\right )^{3} + 279 \, A a^{4} \tan \left (\frac{1}{2} \, x\right ) + 216 \, B a^{4} \tan \left (\frac{1}{2} \, x\right )}{12 \,{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(x))^4*(A+B*sec(x)),x, algorithm="giac")

[Out]

B*a^4*log(abs(tan(1/2*x) + 1)) - B*a^4*log(abs(tan(1/2*x) - 1)) + 1/8*(35*A*a^4 + 48*B*a^4)*x + 1/12*(105*A*a^
4*tan(1/2*x)^7 + 120*B*a^4*tan(1/2*x)^7 + 385*A*a^4*tan(1/2*x)^5 + 424*B*a^4*tan(1/2*x)^5 + 511*A*a^4*tan(1/2*
x)^3 + 520*B*a^4*tan(1/2*x)^3 + 279*A*a^4*tan(1/2*x) + 216*B*a^4*tan(1/2*x))/(tan(1/2*x)^2 + 1)^4