3.252 \(\int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=80 \[ -\frac {b^2 \log (a \cos (c+d x)+b)}{a d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)} \]

[Out]

1/2*ln(1-cos(d*x+c))/(a+b)/d+1/2*ln(1+cos(d*x+c))/(a-b)/d-b^2*ln(b+a*cos(d*x+c))/a/(a^2-b^2)/d

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Rubi [A]  time = 0.24, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4397, 2837, 12, 1629} \[ -\frac {b^2 \log (a \cos (c+d x)+b)}{a d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + Log[1 + Cos[c + d*x]]/(2*(a - b)*d) - (b^2*Log[b + a*Cos[c + d*x]])/(a*(
a^2 - b^2)*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx &=\int \frac {\cos (c+d x) \cot (c+d x)}{b+a \cos (c+d x)} \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {x^2}{a^2 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{(b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{a d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {a}{2 (a+b) (a-x)}-\frac {a}{2 (a-b) (a+x)}+\frac {b^2}{(a-b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{a d}\\ &=\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^2 \log (b+a \cos (c+d x))}{a \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 70, normalized size = 0.88 \[ \frac {b^2 (-\log (a \cos (c+d x)+b))+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a (a+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

(a*(a + b)*Log[Cos[(c + d*x)/2]] - b^2*Log[b + a*Cos[c + d*x]] + a*(a - b)*Log[Sin[(c + d*x)/2]])/(a*(a - b)*(
a + b)*d)

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fricas [A]  time = 0.56, size = 75, normalized size = 0.94 \[ -\frac {2 \, b^{2} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*b^2*log(a*cos(d*x + c) + b) - (a^2 + a*b)*log(1/2*cos(d*x + c) + 1/2) - (a^2 - a*b)*log(-1/2*cos(d*x +
c) + 1/2))/((a^3 - a*b^2)*d)

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giac [B]  time = 0.39, size = 257, normalized size = 3.21 \[ -\frac {\frac {a \log \left ({\left | -a - b + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{2} - b^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | -2 \, b - \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{2} - b^{2}\right )} {\left | a \right |}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(a*log(abs(-a - b + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)
^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/(a^2 - b^2) - (a^2 - 2*b^2)*log(abs(-2*b - 2*a*(cos(d*x + c
) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*abs(a))/abs(-2*b - 2*a*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(a)))/((a^2 - b^2)*abs(a)) - log(a
bs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b))/d

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maple [A]  time = 0.15, size = 80, normalized size = 1.00 \[ -\frac {b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right ) \left (a -b \right ) a}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{d \left (2 a +2 b \right )}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{d \left (2 a -2 b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

-1/d*b^2/(a+b)/(a-b)/a*ln(b+a*cos(d*x+c))+1/d/(2*a+2*b)*ln(cos(d*x+c)-1)+1/d/(2*a-2*b)*ln(1+cos(d*x+c))

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maxima [A]  time = 0.42, size = 102, normalized size = 1.28 \[ -\frac {\frac {b^{2} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{3} - a b^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {\log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-(b^2*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^3 - a*b^2) - log(sin(d*x + c)/(cos(d*x + c)
+ 1))/(a + b) + log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/a)/d

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mupad [B]  time = 0.87, size = 93, normalized size = 1.16 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a\,b^2-a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x)),x)

[Out]

log(tan(c/2 + (d*x)/2))/(d*(a + b)) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) + (b^2*log(a + b - a*tan(c/2 + (d*x)
/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(d*(a*b^2 - a^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(cos(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x)), x)

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