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

Optimal. Leaf size=71 \[ \frac {(2 A-B+C) \tan (c+d x)}{a d}-\frac {(A-B+C) \tan (c+d x)}{d (a \cos (c+d x)+a)}-\frac {(A-B) \tanh ^{-1}(\sin (c+d x))}{a d} \]

[Out]

-(A-B)*arctanh(sin(d*x+c))/a/d+(2*A-B+C)*tan(d*x+c)/a/d-(A-B+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3041, 2748, 3767, 8, 3770} \[ \frac {(2 A-B+C) \tan (c+d x)}{a d}-\frac {(A-B+C) \tan (c+d x)}{d (a \cos (c+d x)+a)}-\frac {(A-B) \tanh ^{-1}(\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A - B)*ArcTanh[Sin[c + d*x]])/(a*d)) + ((2*A - B + C)*Tan[c + d*x])/(a*d) - ((A - B + C)*Tan[c + d*x])/(d*
(a + a*Cos[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int (a (2 A-B+C)-a (A-B) \cos (c+d x)) \sec ^2(c+d x) \, dx}{a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac {(A-B) \int \sec (c+d x) \, dx}{a}+\frac {(2 A-B+C) \int \sec ^2(c+d x) \, dx}{a}\\ &=-\frac {(A-B) \tanh ^{-1}(\sin (c+d x))}{a d}-\frac {(A-B+C) \tan (c+d x)}{d (a+a \cos (c+d x))}-\frac {(2 A-B+C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=-\frac {(A-B) \tanh ^{-1}(\sin (c+d x))}{a d}+\frac {(2 A-B+C) \tan (c+d x)}{a d}-\frac {(A-B+C) \tan (c+d x)}{d (a+a \cos (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.39, size = 256, normalized size = 3.61 \[ \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (\sec \left (\frac {c}{2}\right ) (A-B+C) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left ((A-B) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\frac {A \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{a d (\cos (c+d x)+1) (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*Cos[c + d*x]^2*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*((A - B + C)*Sec[c/2]*Sin[(d*x)/2]
+ Cos[(c + d*x)/2]*((A - B)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]]) + (A*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])))))/(a*d*(1 + Cos[c + d*x])*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.43, size = 128, normalized size = 1.80 \[ -\frac {{\left ({\left (A - B\right )} \cos \left (d x + c\right )^{2} + {\left (A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{2} + {\left (A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (2 \, A - B + C\right )} \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(((A - B)*cos(d*x + c)^2 + (A - B)*cos(d*x + c))*log(sin(d*x + c) + 1) - ((A - B)*cos(d*x + c)^2 + (A - B
)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*((2*A - B + C)*cos(d*x + c) + A)*sin(d*x + c))/(a*d*cos(d*x + c)^2
+ a*d*cos(d*x + c))

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giac [A]  time = 0.42, size = 121, normalized size = 1.70 \[ -\frac {\frac {{\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

-((A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - (A*tan(1/2*d*x
+ 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*
c)^2 - 1)*a))/d

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maple [B]  time = 0.22, size = 180, normalized size = 2.54 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{a d}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*B*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-1/a/d*A/(tan(1/2*d*x+1/2*c)-1
)+1/a/d*A*ln(tan(1/2*d*x+1/2*c)-1)-1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*B-1/a/d*A/(tan(1/2*d*x+1/2*c)+1)-1/a/d*A*ln(
tan(1/2*d*x+1/2*c)+1)+1/a/d*ln(tan(1/2*d*x+1/2*c)+1)*B

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maxima [B]  time = 0.36, size = 218, normalized size = 3.07 \[ -\frac {A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - \frac {C \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

-(A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/
((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - B*(l
og(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(cos(
d*x + c) + 1))) - C*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 1.19, size = 79, normalized size = 1.11 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(A - B + C))/(a*d) + (2*A*tan(c/2 + (d*x)/2))/(d*(a - a*tan(c/2 + (d*x)/2)^2)) - (2*atanh(
tan(c/2 + (d*x)/2))*(A - B))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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