\(\int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx\) [514]

Optimal result

Integrand size = 19, antiderivative size = 43 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a x}{2}-\frac {b \cos ^2(c+d x)}{2 d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3567, 2715, 8} \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a x}{2}-\frac {b \cos ^2(c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a (c+d x)}{2 d}-\frac {b \cos ^2(c+d x)}{2 d}+\frac {a \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {a x}{2}-\frac {b \cos \left (2 d x +2 c \right )}{4 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\)	\(36\)
derivativedivides	\(\frac {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(41\)
default	\(\frac {-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(41\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a d x - b \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {{\left (d x + c\right )} a + \frac {a \tan \left (d x + c\right ) - b}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (37) = 74\).

Time = 0.37 (sec) , antiderivative size = 146, normalized size of antiderivative = 3.40 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, a d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 2 \, a d x \tan \left (d x\right )^{2} + 2 \, a d x \tan \left (c\right )^{2} - b \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, a \tan \left (d x\right )^{2} \tan \left (c\right ) - 2 \, a \tan \left (d x\right ) \tan \left (c\right )^{2} + 2 \, a d x + b \tan \left (d x\right )^{2} + 4 \, b \tan \left (d x\right ) \tan \left (c\right ) + b \tan \left (c\right )^{2} + 2 \, a \tan \left (d x\right ) + 2 \, a \tan \left (c\right ) - b}{4 \, {\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + d \tan \left (d x\right )^{2} + d \tan \left (c\right )^{2} + d\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a\,x}{2}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {b}{2}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{2}\right )}{d} \]

[In]

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

[Out]

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