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

Optimal. Leaf size=147 \[ -\frac{2 (80 A-3 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(55 A-6 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(10 A-3 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

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

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Rubi [A]  time = 0.465687, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2978, 12, 3770} \[ -\frac{2 (80 A-3 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac{(55 A-6 B) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(10 A-3 B) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{(A-B) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.4191, size = 239, normalized size = 1.63 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-70 (49 A-3 B) \sin \left (\frac{d x}{2}\right )+2170 A \sin \left (c+\frac{d x}{2}\right )-2625 A \sin \left (c+\frac{3 d x}{2}\right )+735 A \sin \left (2 c+\frac{3 d x}{2}\right )-1015 A \sin \left (2 c+\frac{5 d x}{2}\right )+105 A \sin \left (3 c+\frac{5 d x}{2}\right )-160 A \sin \left (3 c+\frac{7 d x}{2}\right )+126 B \sin \left (c+\frac{3 d x}{2}\right )+42 B \sin \left (2 c+\frac{5 d x}{2}\right )+6 B \sin \left (3 c+\frac{7 d x}{2}\right )\right )-6720 A \cos ^8\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6720*A*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]]) + Cos[(c + d*x)/2]*Sec[c/2]*(-70*(49*A - 3*B)*Sin[(d*x)/2] + 2170*A*Sin[c + (d*x)/2] - 2625*A*Sin[c + (3*d
*x)/2] + 126*B*Sin[c + (3*d*x)/2] + 735*A*Sin[2*c + (3*d*x)/2] - 1015*A*Sin[2*c + (5*d*x)/2] + 42*B*Sin[2*c +
(5*d*x)/2] + 105*A*Sin[3*c + (5*d*x)/2] - 160*A*Sin[3*c + (7*d*x)/2] + 6*B*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(
1 + Cos[c + d*x])^4)

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Maple [A]  time = 0.095, size = 199, normalized size = 1.4 \begin{align*}{\frac{B}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{11\,A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{A}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{15\,A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{3\,B}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03875, size = 308, normalized size = 2.1 \begin{align*} -\frac{5 \, A{\left (\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac{3 \, B{\left (\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) +
1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7)/a^4)/d

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Fricas [A]  time = 1.48182, size = 624, normalized size = 4.24 \begin{align*} \frac{105 \,{\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (80 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (535 \, A - 24 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (620 \, A - 39 \, B\right )} \cos \left (d x + c\right ) + 260 \, A - 36 \, B\right )} \sin \left (d x + c\right )}{210 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.25902, size = 246, normalized size = 1.67 \begin{align*} \frac{\frac{840 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{840 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 63 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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