3.192 \(\int \frac{A+B \sec (x)}{(a+a \cos (x))^4} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 (3 A-80 B) \sin (x)}{105 a^4 (\cos (x)+1)}+\frac{(6 A-55 B) \sin (x)}{105 a^4 (\cos (x)+1)^2}+\frac{B \tanh ^{-1}(\sin (x))}{a^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a \cos (x)+a)^3}+\frac{(A-B) \sin (x)}{7 (a \cos (x)+a)^4} \]

[Out]

(B*ArcTanh[Sin[x]])/a^4 + ((6*A - 55*B)*Sin[x])/(105*a^4*(1 + Cos[x])^2) + (2*(3*A - 80*B)*Sin[x])/(105*a^4*(1
+ Cos[x])) + ((A - B)*Sin[x])/(7*(a + a*Cos[x])^4) + ((3*A - 10*B)*Sin[x])/(35*a*(a + a*Cos[x])^3)

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Rubi [A]  time = 0.414411, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2828, 2978, 12, 3770} \[ \frac{2 (3 A-80 B) \sin (x)}{105 a^4 (\cos (x)+1)}+\frac{(6 A-55 B) \sin (x)}{105 a^4 (\cos (x)+1)^2}+\frac{B \tanh ^{-1}(\sin (x))}{a^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a \cos (x)+a)^3}+\frac{(A-B) \sin (x)}{7 (a \cos (x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[x]])/a^4 + ((6*A - 55*B)*Sin[x])/(105*a^4*(1 + Cos[x])^2) + (2*(3*A - 80*B)*Sin[x])/(105*a^4*(1
+ Cos[x])) + ((A - B)*Sin[x])/(7*(a + a*Cos[x])^4) + ((3*A - 10*B)*Sin[x])/(35*a*(a + a*Cos[x])^3)

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 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \frac{A+B \sec (x)}{(a+a \cos (x))^4} \, dx &=\int \frac{(B+A \cos (x)) \sec (x)}{(a+a \cos (x))^4} \, dx\\ &=\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{\int \frac{(7 a B+3 a (A-B) \cos (x)) \sec (x)}{(a+a \cos (x))^3} \, dx}{7 a^2}\\ &=\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac{\int \frac{\left (35 a^2 B+2 a^2 (3 A-10 B) \cos (x)\right ) \sec (x)}{(a+a \cos (x))^2} \, dx}{35 a^4}\\ &=\frac{(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac{\int \frac{\left (105 a^3 B+a^3 (6 A-55 B) \cos (x)\right ) \sec (x)}{a+a \cos (x)} \, dx}{105 a^6}\\ &=\frac{(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac{2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )}+\frac{\int 105 a^4 B \sec (x) \, dx}{105 a^8}\\ &=\frac{(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac{2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )}+\frac{B \int \sec (x) \, dx}{a^4}\\ &=\frac{B \tanh ^{-1}(\sin (x))}{a^4}+\frac{(6 A-55 B) \sin (x)}{105 a^4 (1+\cos (x))^2}+\frac{(A-B) \sin (x)}{7 (a+a \cos (x))^4}+\frac{(3 A-10 B) \sin (x)}{35 a (a+a \cos (x))^3}+\frac{2 (3 A-80 B) \sin (x)}{105 \left (a^4+a^4 \cos (x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.805181, size = 104, normalized size = 1.08 \[ \frac{\sin (x) ((87 A-1480 B) \cos (x)+(24 A-535 B) \cos (2 x)+3 A \cos (3 x)+96 A-80 B \cos (3 x)-1055 B)-3360 B \cos ^8\left (\frac{x}{2}\right ) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{210 a^4 (\cos (x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3360*B*Cos[x/2]^8*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (96*A - 1055*B + (87*A - 1480*B)*C
os[x] + (24*A - 535*B)*Cos[2*x] + 3*A*Cos[3*x] - 80*B*Cos[3*x])*Sin[x])/(210*a^4*(1 + Cos[x])^4)

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Maple [A]  time = 0.037, size = 119, normalized size = 1.2 \begin{align*}{\frac{A}{56\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{7}}-{\frac{B}{56\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{7}}+{\frac{3\,A}{40\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{5}}-{\frac{B}{8\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{5}}+{\frac{A}{8\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{11\,B}{24\,{a}^{4}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{B}{{a}^{4}}\ln \left ( 1+\tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{B}{{a}^{4}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{A}{8\,{a}^{4}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{15\,B}{8\,{a}^{4}}\tan \left ({\frac{x}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(x))/(a+a*cos(x))^4,x)

[Out]

1/56/a^4*tan(1/2*x)^7*A-1/56/a^4*tan(1/2*x)^7*B+3/40/a^4*tan(1/2*x)^5*A-1/8/a^4*tan(1/2*x)^5*B+1/8/a^4*tan(1/2
*x)^3*A-11/24/a^4*tan(1/2*x)^3*B+1/a^4*B*ln(1+tan(1/2*x))-1/a^4*B*ln(tan(1/2*x)-1)+1/8/a^4*A*tan(1/2*x)-15/8/a
^4*B*tan(1/2*x)

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Maxima [A]  time = 1.09966, size = 193, normalized size = 2.01 \begin{align*} -\frac{1}{168} \, B{\left (\frac{\frac{315 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{77 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{3 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}}{a^{4}} - \frac{168 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{a^{4}} + \frac{168 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac{A{\left (\frac{35 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{35 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}\right )}}{280 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*B*((315*sin(x)/(cos(x) + 1) + 77*sin(x)^3/(cos(x) + 1)^3 + 21*sin(x)^5/(cos(x) + 1)^5 + 3*sin(x)^7/(cos
(x) + 1)^7)/a^4 - 168*log(sin(x)/(cos(x) + 1) + 1)/a^4 + 168*log(sin(x)/(cos(x) + 1) - 1)/a^4) + 1/280*A*(35*s
in(x)/(cos(x) + 1) + 35*sin(x)^3/(cos(x) + 1)^3 + 21*sin(x)^5/(cos(x) + 1)^5 + 5*sin(x)^7/(cos(x) + 1)^7)/a^4

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Fricas [A]  time = 2.53879, size = 464, normalized size = 4.83 \begin{align*} \frac{105 \,{\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - 105 \,{\left (B \cos \left (x\right )^{4} + 4 \, B \cos \left (x\right )^{3} + 6 \, B \cos \left (x\right )^{2} + 4 \, B \cos \left (x\right ) + B\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, A - 80 \, B\right )} \cos \left (x\right )^{3} +{\left (24 \, A - 535 \, B\right )} \cos \left (x\right )^{2} +{\left (39 \, A - 620 \, B\right )} \cos \left (x\right ) + 36 \, A - 260 \, B\right )} \sin \left (x\right )}{210 \,{\left (a^{4} \cos \left (x\right )^{4} + 4 \, a^{4} \cos \left (x\right )^{3} + 6 \, a^{4} \cos \left (x\right )^{2} + 4 \, a^{4} \cos \left (x\right ) + a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(B*cos(x)^4 + 4*B*cos(x)^3 + 6*B*cos(x)^2 + 4*B*cos(x) + B)*log(sin(x) + 1) - 105*(B*cos(x)^4 + 4*B
*cos(x)^3 + 6*B*cos(x)^2 + 4*B*cos(x) + B)*log(-sin(x) + 1) + 2*(2*(3*A - 80*B)*cos(x)^3 + (24*A - 535*B)*cos(
x)^2 + (39*A - 620*B)*cos(x) + 36*A - 260*B)*sin(x))/(a^4*cos(x)^4 + 4*a^4*cos(x)^3 + 6*a^4*cos(x)^2 + 4*a^4*c
os(x) + a^4)

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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+B*sec(x))/(a+a*cos(x))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.15574, size = 170, normalized size = 1.77 \begin{align*} \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{a^{4}} - \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{a^{4}} + \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, x\right )^{7} - 15 \, B a^{24} \tan \left (\frac{1}{2} \, x\right )^{7} + 63 \, A a^{24} \tan \left (\frac{1}{2} \, x\right )^{5} - 105 \, B a^{24} \tan \left (\frac{1}{2} \, x\right )^{5} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, x\right )^{3} - 385 \, B a^{24} \tan \left (\frac{1}{2} \, x\right )^{3} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, x\right ) - 1575 \, B a^{24} \tan \left (\frac{1}{2} \, x\right )}{840 \, a^{28}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(abs(tan(1/2*x) + 1))/a^4 - B*log(abs(tan(1/2*x) - 1))/a^4 + 1/840*(15*A*a^24*tan(1/2*x)^7 - 15*B*a^24*ta
n(1/2*x)^7 + 63*A*a^24*tan(1/2*x)^5 - 105*B*a^24*tan(1/2*x)^5 + 105*A*a^24*tan(1/2*x)^3 - 385*B*a^24*tan(1/2*x
)^3 + 105*A*a^24*tan(1/2*x) - 1575*B*a^24*tan(1/2*x))/a^28