\(\int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) [250]

Optimal result

Integrand size = 28, antiderivative size = 113 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b \cos (c+d x)}{a^2 d}+\frac {\cos ^2(c+d x)}{2 a d}+\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^4 \log (b+a \cos (c+d x))}{a^3 \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 113, 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} \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b \cos (c+d x)}{a^2 d}-\frac {b^4 \log (a \cos (c+d x)+b)}{a^3 d \left (a^2-b^2\right )}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)}+\frac {\log (\cos (c+d x)+1)}{2 d (a-b)}+\frac {\cos ^2(c+d x)}{2 a d} \]

[In]

Int[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

-((b*Cos[c + d*x])/(a^2*d)) + Cos[c + d*x]^2/(2*a*d) + Log[1 - Cos[c + d*x]]/(2*(a + b)*d) + Log[1 + Cos[c + d
*x]]/(2*(a - b)*d) - (b^4*Log[b + a*Cos[c + d*x]])/(a^3*(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 ^3(c+d x) \cot (c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^4}{a^4 (b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^4}{(b+x) \left (a^2-x^2\right )} \, dx,x,a \cos (c+d x)\right )}{a^3 d} \\ & = -\frac {\text {Subst}\left (\int \left (b+\frac {a^3}{2 (a+b) (a-x)}-x-\frac {a^3}{2 (a-b) (a+x)}+\frac {b^4}{(a-b) (a+b) (b+x)}\right ) \, dx,x,a \cos (c+d x)\right )}{a^3 d} \\ & = -\frac {b \cos (c+d x)}{a^2 d}+\frac {\cos ^2(c+d x)}{2 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^4 \log (b+a \cos (c+d x))}{a^3 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {-\frac {4 b \cos (c+d x)}{a^2}+\frac {\cos (2 (c+d x))}{a}+4 \left (\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {b^4 \log (b+a \cos (c+d x))}{a^3 \left (-a^2+b^2\right )}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}\right )}{4 d} \]

[In]

Integrate[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b}{a^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{4} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{3} \left (a +b \right ) \left (a -b \right )}}{d}\)	\(100\)
default	\(\frac {\frac {\frac {a \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b}{a^{2}}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}-\frac {b^{4} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{3} \left (a +b \right ) \left (a -b \right )}}{d}\)	\(100\)
risch	\(\frac {i x}{a}+\frac {i x \,b^{2}}{a^{3}}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{4} x}{a^{3} \left (a^{2}-b^{2}\right )}+\frac {2 i b^{4} c}{d \,a^{3} \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \,a^{3} \left (a^{2}-b^{2}\right )}\)	\(273\)

[In]

int(cos(d*x+c)^3/(sin(d*x+c)*a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, b^{4} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - {\left (a^{4} + a^{3} b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{4} - a^{3} b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d} \]

[In]

integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*b^4*log(a*cos(d*x + c) + b) - (a^4 - a^2*b^2)*cos(d*x + c)^2 + 2*(a^3*b - a*b^3)*cos(d*x + c) - (a^4 +
a^3*b)*log(1/2*cos(d*x + c) + 1/2) - (a^4 - a^3*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^5 - a^3*b^2)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {b^{4} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{5} - a^{3} b^{2}} + \frac {2 \, {\left (b + \frac {{\left (a + b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{3}}}{d} \]

[In]

integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-(b^4*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^5 - a^3*b^2) + 2*(b + (a + b)*sin(d*x + c)^2
/(cos(d*x + c) + 1)^2)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4) - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b) + (a^2 + b^2)*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)
/a^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (107) = 214\).

Time = 0.37 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.68 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, b^{4} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{5} - a^{3} b^{2}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac {3 \, a^{2} - 4 \, a b + 3 \, b^{2} - \frac {2 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {6 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(2*b^4*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^5 - a^3*b^2) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) + 2*(a^2 + b^2)*log(abs(-(cos(d
*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 - (3*a^2 - 4*a*b + 3*b^2 - 2*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 4*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^2*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a^3*((cos(d*x + c) - 1)
/(cos(d*x + c) + 1) - 1)^2))/d

Mupad [B] (verification not implemented)

Time = 24.40 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\frac {2\,b}{a^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a+b\right )}{a^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^5-a^3\,b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+b^2\right )}{a^3\,d} \]

[In]

int(cos(c + d*x)^3/(a*sin(c + d*x) + b*tan(c + d*x)),x)

[Out]

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