3.70 \(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=145 \[ -\frac {8 (20 A-C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(55 A-8 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

A*arctanh(sin(d*x+c))/a^4/d-1/105*(55*A-8*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-8/105*(20*A-C)*sin(d*x+c)/a^4/d
/(1+cos(d*x+c))-1/7*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-2/35*(5*A-2*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3

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Rubi [A]  time = 0.47, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 2978, 12, 3770} \[ -\frac {8 (20 A-C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)}-\frac {(55 A-8 C) \sin (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A+C) \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^4,x]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((55*A - 8*C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - (8*(20*A -
C)*Sin[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])) - ((A + C)*Sin[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*(5*
A - 2*C)*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*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 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 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

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 {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(7 a A-a (3 A-4 C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (35 a^2 A-4 a^2 (5 A-2 C) \cos (c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(55 A-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \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-8 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(55 A-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {8 (20 A-C) \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-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {8 (20 A-C) \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-8 C) \sin (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A+C) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 (5 A-2 C) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {8 (20 A-C) \sin (c+d x)}{105 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.76, size = 245, normalized size = 1.69 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (70 (31 A-2 C) \sin \left (c+\frac {d x}{2}\right )-2625 A \sin \left (c+\frac {3 d x}{2}\right )+735 A \sin \left (2 c+\frac {3 d x}{2}\right )-1015 A \sin \left (2 c+\frac {5 d x}{2}\right )+105 A \sin \left (3 c+\frac {5 d x}{2}\right )-160 A \sin \left (3 c+\frac {7 d x}{2}\right )-70 (49 A-2 C) \sin \left (\frac {d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )+56 C \sin \left (2 c+\frac {5 d x}{2}\right )+8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )-6720 A \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*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 - 2*C)*Sin[(d*x)/2] + 70*(31*A - 2*C)*Sin[c + (d*x)/2] - 2625*A*Sin
[c + (3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] + 735*A*Sin[2*c + (3*d*x)/2] - 1015*A*Sin[2*c + (5*d*x)/2] + 56*C*S
in[2*c + (5*d*x)/2] + 105*A*Sin[3*c + (5*d*x)/2] - 160*A*Sin[3*c + (7*d*x)/2] + 8*C*Sin[3*c + (7*d*x)/2]))/(42
0*a^4*d*(1 + Cos[c + d*x])^4)

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fricas [A]  time = 1.25, size = 237, normalized size = 1.63 \[ \frac {105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 6 \, A \cos \left (d x + c\right )^{2} + 4 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, {\left (20 \, A - C\right )} \cos \left (d x + c\right )^{3} + {\left (535 \, A - 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (155 \, A - 13 \, C\right )} \cos \left (d x + c\right ) + 260 \, A - 13 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*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*(8*(20*A - C)*cos(d*x + c)^3 + (535*A - 32*C)*cos(d*x + c)^2 + 4*(155*A - 13*C)*cos(d*x + c) + 2
60*A - 13*C)*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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giac [A]  time = 0.61, size = 182, normalized size = 1.26 \[ \frac {\frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*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*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 21*C*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 - 35*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(
1/2*d*x + 1/2*c) - 105*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.18, size = 199, normalized size = 1.37 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{24 d \,a^{4}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^4,x)

[Out]

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

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maxima [A]  time = 0.34, size = 228, normalized size = 1.57 \[ -\frac {5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*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) - C*(105*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 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7)/a^4)/d

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mupad [B]  time = 0.90, size = 156, normalized size = 1.08 \[ \frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{8\,a^4}+\frac {A}{a^4}+\frac {6\,A-2\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{24\,a^4}+\frac {A}{6\,a^4}+\frac {6\,A-2\,C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A+C}{40\,a^4}+\frac {A}{10\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + a*cos(c + d*x))^4),x)

[Out]

(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (tan(c/2 + (d*x)/2)*((A + C)/(8*a^4) + A/a^4 + (6*A - 2*C)/(8*a^4)))
/d - (tan(c/2 + (d*x)/2)^3*((A + C)/(24*a^4) + A/(6*a^4) + (6*A - 2*C)/(24*a^4)))/d - (tan(c/2 + (d*x)/2)^5*((
A + C)/(40*a^4) + A/(10*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)/(a+a*cos(d*x+c))**4,x)

[Out]

Timed out

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