3.61 \(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=115 \[ -\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

A*arctanh(sin(d*x+c))/a^3/d-1/5*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(7*A-3*C)*sin(d*x+c)/a/d/(a+a*cos(d
*x+c))^2-1/15*(22*A-3*C)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.31, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 2978, 12, 3770} \[ -\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((A + C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((7*A - 3*C)*Sin[c +
d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((22*A - 3*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*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 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))^3} \, dx &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(5 a A-a (2 A-3 C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (15 a^2 A-a^2 (7 A-3 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int 15 a^3 A \sec (c+d x) \, dx}{15 a^6}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {A \int \sec (c+d x) \, dx}{a^3}\\ &=\frac {A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 203, normalized size = 1.77 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (15 (5 A-C) \sin \left (c+\frac {d x}{2}\right )-95 A \sin \left (c+\frac {3 d x}{2}\right )+15 A \sin \left (2 c+\frac {3 d x}{2}\right )-22 A \sin \left (2 c+\frac {5 d x}{2}\right )-5 (29 A-3 C) \sin \left (\frac {d x}{2}\right )+15 C \sin \left (c+\frac {3 d x}{2}\right )+3 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )-240 A \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{30 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

(-240*A*Cos[(c + d*x)/2]^6*(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]*(-5*(29*A - 3*C)*Sin[(d*x)/2] + 15*(5*A - C)*Sin[c + (d*x)/2] - 95*A*Sin[c + (3
*d*x)/2] + 15*C*Sin[c + (3*d*x)/2] + 15*A*Sin[2*c + (3*d*x)/2] - 22*A*Sin[2*c + (5*d*x)/2] + 3*C*Sin[2*c + (5*
d*x)/2]))/(30*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.63, size = 184, normalized size = 1.60 \[ \frac {15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (22 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 3 \, C\right )} \cos \left (d x + c\right ) + 32 \, A - 3 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/30*(15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - 15*(A*cos(d*x
+ c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*((22*A - 3*C)*cos(d*x + c)^2 +
3*(17*A - 3*C)*cos(d*x + c) + 32*A - 3*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3
*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 0.51, size = 131, normalized size = 1.14 \[ \frac {\frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/60*(60*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - (3*A*a^12*ta
n(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2
*d*x + 1/2*c) - 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 0.19, size = 139, normalized size = 1.21 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]

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))^3,x)

[Out]

-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-1/3/d/a^3*tan(1/2*d*x+1/2*c)^3*A-7/4/d/a^
3*A*tan(1/2*d*x+1/2*c)+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)-1/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^3*A*ln(tan(1/2*
d*x+1/2*c)+1)

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maxima [A]  time = 0.34, size = 167, normalized size = 1.45 \[ -\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^3) - 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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mupad [B]  time = 0.89, size = 114, normalized size = 0.99 \[ \frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{12\,a^3}+\frac {3\,A-C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{4\,a^3}+\frac {3\,A-C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

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))^3),x)

[Out]

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

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

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))**3,x)

[Out]

(Integral(A*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(C*cos(c + d
*x)**2*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3

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