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+1/105*(6*A-55*B)*sin(x)/a^4/(1+cos(x))^2+2/105*(3*A-80*B)*sin(x)/a^4/(1+cos(x))+1/7*(A-B
)*sin(x)/(a+a*cos(x))^4+1/35*(3*A-10*B)*sin(x)/a/(a+a*cos(x))^3

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Rubi [A]  time = 0.41, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 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))^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*}

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Mathematica [A]  time = 0.75, 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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fricas [A]  time = 0.57, size = 158, normalized size = 1.65 \[ \frac {105 \, {\left (B \cos \relax (x)^{4} + 4 \, B \cos \relax (x)^{3} + 6 \, B \cos \relax (x)^{2} + 4 \, B \cos \relax (x) + B\right )} \log \left (\sin \relax (x) + 1\right ) - 105 \, {\left (B \cos \relax (x)^{4} + 4 \, B \cos \relax (x)^{3} + 6 \, B \cos \relax (x)^{2} + 4 \, B \cos \relax (x) + B\right )} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B\right )} \cos \relax (x)^{3} + {\left (24 \, A - 535 \, B\right )} \cos \relax (x)^{2} + {\left (39 \, A - 620 \, B\right )} \cos \relax (x) + 36 \, A - 260 \, B\right )} \sin \relax (x)}{210 \, {\left (a^{4} \cos \relax (x)^{4} + 4 \, a^{4} \cos \relax (x)^{3} + 6 \, a^{4} \cos \relax (x)^{2} + 4 \, a^{4} \cos \relax (x) + a^{4}\right )}} \]

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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giac [A]  time = 0.15, size = 126, normalized size = 1.31 \[ \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a^{4}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, x\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, x\right )}{840 \, a^{28}} \]

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

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maple [A]  time = 0.10, size = 119, normalized size = 1.24 \[ \frac {\left (\tan ^{7}\left (\frac {x}{2}\right )\right ) A}{56 a^{4}}-\frac {\left (\tan ^{7}\left (\frac {x}{2}\right )\right ) B}{56 a^{4}}+\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) A}{8 a^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) B}{24 a^{4}}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a^{4}}+\frac {A \tan \left (\frac {x}{2}\right )}{8 a^{4}}-\frac {15 B \tan \left (\frac {x}{2}\right )}{8 a^{4}}+\frac {B \ln \left (1+\tan \left (\frac {x}{2}\right )\right )}{a^{4}}+\frac {3 \left (\tan ^{5}\left (\frac {x}{2}\right )\right ) A}{40 a^{4}}-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) B}{8 a^{4}} \]

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

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maxima [A]  time = 0.77, size = 143, normalized size = 1.49 \[ -\frac {1}{168} \, B {\left (\frac {\frac {315 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {77 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {21 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {3 \, \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} - 1\right )}{a^{4}}\right )} + \frac {A {\left (\frac {35 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {35 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {21 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {5 \, \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}}{280 \, a^{4}} \]

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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mupad [B]  time = 2.34, size = 140, normalized size = 1.46 \[ \mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {A-B}{8\,a^4}-\frac {3\,B}{4\,a^4}+\frac {2\,A-4\,B}{8\,a^4}-\frac {2\,A+4\,B}{8\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (\frac {A-B}{40\,a^4}+\frac {2\,A-4\,B}{40\,a^4}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {A-B}{24\,a^4}-\frac {B}{4\,a^4}+\frac {2\,A-4\,B}{24\,a^4}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}+\frac {2\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{\cos ^{4}{\relax (x )} + 4 \cos ^{3}{\relax (x )} + 6 \cos ^{2}{\relax (x )} + 4 \cos {\relax (x )} + 1}\, dx + \int \frac {B \sec {\relax (x )}}{\cos ^{4}{\relax (x )} + 4 \cos ^{3}{\relax (x )} + 6 \cos ^{2}{\relax (x )} + 4 \cos {\relax (x )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(cos(x)**4 + 4*cos(x)**3 + 6*cos(x)**2 + 4*cos(x) + 1), x) + Integral(B*sec(x)/(cos(x)**4 + 4*cos(
x)**3 + 6*cos(x)**2 + 4*cos(x) + 1), x))/a**4

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