3.191 \(\int \frac {A+B \sec (x)}{(a+a \cos (x))^3} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 (A-11 B) \sin (x)}{15 \left (a^3 \cos (x)+a^3\right )}+\frac {B \tanh ^{-1}(\sin (x))}{a^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a \cos (x)+a)^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3} \]

[Out]

B*arctanh(sin(x))/a^3+1/5*(A-B)*sin(x)/(a+a*cos(x))^3+1/15*(2*A-7*B)*sin(x)/a/(a+a*cos(x))^2+2/15*(A-11*B)*sin
(x)/(a^3+a^3*cos(x))

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Rubi [A]  time = 0.31, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2828, 2978, 12, 3770} \[ \frac {2 (A-11 B) \sin (x)}{15 \left (a^3 \cos (x)+a^3\right )}+\frac {B \tanh ^{-1}(\sin (x))}{a^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a \cos (x)+a)^2}+\frac {(A-B) \sin (x)}{5 (a \cos (x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[x]])/a^3 + ((A - B)*Sin[x])/(5*(a + a*Cos[x])^3) + ((2*A - 7*B)*Sin[x])/(15*a*(a + a*Cos[x])^2)
+ (2*(A - 11*B)*Sin[x])/(15*(a^3 + a^3*Cos[x]))

Rule 12

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

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 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))^3} \, dx &=\int \frac {(B+A \cos (x)) \sec (x)}{(a+a \cos (x))^3} \, dx\\ &=\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {\int \frac {(5 a B+2 a (A-B) \cos (x)) \sec (x)}{(a+a \cos (x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a+a \cos (x))^2}+\frac {\int \frac {\left (15 a^2 B+a^2 (2 A-7 B) \cos (x)\right ) \sec (x)}{a+a \cos (x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a+a \cos (x))^2}+\frac {2 (A-11 B) \sin (x)}{15 \left (a^3+a^3 \cos (x)\right )}+\frac {\int 15 a^3 B \sec (x) \, dx}{15 a^6}\\ &=\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a+a \cos (x))^2}+\frac {2 (A-11 B) \sin (x)}{15 \left (a^3+a^3 \cos (x)\right )}+\frac {B \int \sec (x) \, dx}{a^3}\\ &=\frac {B \tanh ^{-1}(\sin (x))}{a^3}+\frac {(A-B) \sin (x)}{5 (a+a \cos (x))^3}+\frac {(2 A-7 B) \sin (x)}{15 a (a+a \cos (x))^2}+\frac {2 (A-11 B) \sin (x)}{15 \left (a^3+a^3 \cos (x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 88, normalized size = 1.17 \[ \frac {\sin (x) ((6 A-51 B) \cos (x)+(A-11 B) \cos (2 x)+8 A-43 B)-120 B \cos ^6\left (\frac {x}{2}\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right )}{15 a^3 (\cos (x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*B*Cos[x/2]^6*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]) + (8*A - 43*B + (6*A - 51*B)*Cos[x] +
(A - 11*B)*Cos[2*x])*Sin[x])/(15*a^3*(1 + Cos[x])^3)

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fricas [A]  time = 0.74, size = 122, normalized size = 1.63 \[ \frac {15 \, {\left (B \cos \relax (x)^{3} + 3 \, B \cos \relax (x)^{2} + 3 \, B \cos \relax (x) + B\right )} \log \left (\sin \relax (x) + 1\right ) - 15 \, {\left (B \cos \relax (x)^{3} + 3 \, B \cos \relax (x)^{2} + 3 \, B \cos \relax (x) + B\right )} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left (2 \, {\left (A - 11 \, B\right )} \cos \relax (x)^{2} + 3 \, {\left (2 \, A - 17 \, B\right )} \cos \relax (x) + 7 \, A - 32 \, B\right )} \sin \relax (x)}{30 \, {\left (a^{3} \cos \relax (x)^{3} + 3 \, a^{3} \cos \relax (x)^{2} + 3 \, a^{3} \cos \relax (x) + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*(B*cos(x)^3 + 3*B*cos(x)^2 + 3*B*cos(x) + B)*log(sin(x) + 1) - 15*(B*cos(x)^3 + 3*B*cos(x)^2 + 3*B*co
s(x) + B)*log(-sin(x) + 1) + 2*(2*(A - 11*B)*cos(x)^2 + 3*(2*A - 17*B)*cos(x) + 7*A - 32*B)*sin(x))/(a^3*cos(x
)^3 + 3*a^3*cos(x)^2 + 3*a^3*cos(x) + a^3)

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giac [A]  time = 0.16, size = 102, normalized size = 1.36 \[ \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a^{3}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a^{3}} + \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, x\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, x\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, x\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, x\right )}{60 \, a^{15}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(abs(tan(1/2*x) + 1))/a^3 - B*log(abs(tan(1/2*x) - 1))/a^3 + 1/60*(3*A*a^12*tan(1/2*x)^5 - 3*B*a^12*tan(1
/2*x)^5 + 10*A*a^12*tan(1/2*x)^3 - 20*B*a^12*tan(1/2*x)^3 + 15*A*a^12*tan(1/2*x) - 105*B*a^12*tan(1/2*x))/a^15

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maple [A]  time = 0.10, size = 95, normalized size = 1.27 \[ \frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) A}{6 a^{3}}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) B}{3 a^{3}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{3}}+\frac {A \tan \left (\frac {x}{2}\right )}{4 a^{3}}-\frac {7 B \tan \left (\frac {x}{2}\right )}{4 a^{3}}+\frac {B \ln \left (1+\tan \left (\frac {x}{2}\right )\right )}{a^{3}}+\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) A}{20 a^{3}}-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) B}{20 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^3*tan(1/2*x)^3*A-1/3/a^3*tan(1/2*x)^3*B-1/a^3*B*ln(tan(1/2*x)-1)+1/4/a^3*A*tan(1/2*x)-7/4/a^3*B*tan(1/2*
x)+1/a^3*B*ln(1+tan(1/2*x))+1/20/a^3*tan(1/2*x)^5*A-1/20/a^3*tan(1/2*x)^5*B

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maxima [A]  time = 0.64, size = 119, normalized size = 1.59 \[ -\frac {1}{60} \, B {\left (\frac {\frac {105 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {20 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {10 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}\right )}}{60 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*B*((105*sin(x)/(cos(x) + 1) + 20*sin(x)^3/(cos(x) + 1)^3 + 3*sin(x)^5/(cos(x) + 1)^5)/a^3 - 60*log(sin(x
)/(cos(x) + 1) + 1)/a^3 + 60*log(sin(x)/(cos(x) + 1) - 1)/a^3) + 1/60*A*(15*sin(x)/(cos(x) + 1) + 10*sin(x)^3/
(cos(x) + 1)^3 + 3*sin(x)^5/(cos(x) + 1)^5)/a^3

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mupad [B]  time = 2.37, size = 92, normalized size = 1.23 \[ {\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {A-B}{12\,a^3}+\frac {A-3\,B}{12\,a^3}\right )+\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {A-B}{4\,a^3}+\frac {A-3\,B}{4\,a^3}-\frac {A+3\,B}{4\,a^3}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(x/2)^3*((A - B)/(12*a^3) + (A - 3*B)/(12*a^3)) + tan(x/2)*((A - B)/(4*a^3) + (A - 3*B)/(4*a^3) - (A + 3*B)
/(4*a^3)) + (tan(x/2)^5*(A - B))/(20*a^3) + (2*B*atanh(tan(x/2)))/a^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{\cos ^{3}{\relax (x )} + 3 \cos ^{2}{\relax (x )} + 3 \cos {\relax (x )} + 1}\, dx + \int \frac {B \sec {\relax (x )}}{\cos ^{3}{\relax (x )} + 3 \cos ^{2}{\relax (x )} + 3 \cos {\relax (x )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(cos(x)**3 + 3*cos(x)**2 + 3*cos(x) + 1), x) + Integral(B*sec(x)/(cos(x)**3 + 3*cos(x)**2 + 3*cos(
x) + 1), x))/a**3

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