\(\int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) [251]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\cos (c+d x)}{a d}+\frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {b^3 \log (b+a \cos (c+d x))}{a^2 \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4482, 2916, 12, 1643} \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {b^3 \log (a \cos (c+d x)+b)}{a^2 d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}-\frac {\log (\cos (c+d x)+1)}{2 d (a-b)}+\frac {\cos (c+d x)}{a d} \]

[In]

Int[Cos[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

Cos[c + d*x]/(a*d) + Log[1 - Cos[c + d*x]]/(2*(a + b)*d) - Log[1 + Cos[c + d*x]]/(2*(a - b)*d) + (b^3*Log[b +
a*Cos[c + d*x]])/(a^2*(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 1643

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 2916

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/b)*x)^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 4482

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^2(c+d x) \cot (c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^3}{a^3 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^3}{(b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{a^2 d} \\ & = -\frac {\text {Subst}\left (\int \left (-1+\frac {a^2}{2 (a+b) (a-x)}+\frac {a^2}{2 (a-b) (a+x)}+\frac {b^3}{(-a+b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {\cos (c+d x)}{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^3 \log (b+a \cos (c+d x))}{a^2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {\cos (c+d x)}{a}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{-a+b}+\frac {b^3 \log (b+a \cos (c+d x))}{a^4-a^2 b^2}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}}{d} \]

[In]

Integrate[Cos[c + d*x]^2/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

(Cos[c + d*x]/a + Log[Cos[(c + d*x)/2]]/(-a + b) + (b^3*Log[b + a*Cos[c + d*x]])/(a^4 - a^2*b^2) + Log[Sin[(c
+ d*x)/2]]/(a + b))/d

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\cos \left (d x +c \right )}{a}+\frac {b^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{2} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(85\)
default	\(\frac {\frac {\cos \left (d x +c \right )}{a}+\frac {b^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{2} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\)	\(85\)
risch	\(-\frac {i x b}{a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i b^{3} x}{a^{2} \left (a^{2}-b^{2}\right )}-\frac {2 i b^{3} c}{a^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d \left (a^{2}-b^{2}\right )}\)	\(228\)

[In]

int(cos(d*x+c)^2/(sin(d*x+c)*a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2 \, b^{3} \log \left (a \cos \left (d x + c\right ) + b\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{3} + a^{2} b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - a^{2} b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d} \]

[In]

integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*b^3*log(a*cos(d*x + c) + b) + 2*(a^3 - a*b^2)*cos(d*x + c) - (a^3 + a^2*b)*log(1/2*cos(d*x + c) + 1/2)
+ (a^3 - a^2*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^4 - a^2*b^2)*d)

Sympy [F]

\[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]

[In]

integrate(cos(d*x+c)**2/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2/(a*sin(c + d*x) + b*tan(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {b^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{4} - a^{2} b^{2}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {b \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{2}} + \frac {2}{a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]

[In]

integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

(b^3*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^4 - a^2*b^2) + log(sin(d*x + c)/(cos(d*x + c)
+ 1))/(a + b) + b*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/a^2 + 2/(a + a*sin(d*x + c)^2/(cos(d*x + c) +
1)^2))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (88) = 176\).

Time = 0.37 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {2 \, b^{3} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - a^{2} b^{2}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} + \frac {2 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (2 \, a - b + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*b^3*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))
)/(a^4 - a^2*b^2) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) + 2*b*log(abs(-(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) + 1))/a^2 - 2*(2*a - b + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/(a^2*((cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - 1)))/d

Mupad [B] (verification not implemented)

Time = 23.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2}{a\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}+\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {b^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^2\,d\,\left (a^2-b^2\right )} \]

[In]

int(cos(c + d*x)^2/(a*sin(c + d*x) + b*tan(c + d*x)),x)

[Out]

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