\(\int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx\) [521]

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {1}{2} \left (a^2+b^2\right ) x-\frac {\cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3587, 737, 209} \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {1}{2} x \left (a^2+b^2\right )-\frac {\cos ^2(c+d x) (b-a \tan (c+d x)) (a+b \tan (c+d x))}{2 d} \]

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 3587

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \left (a^2+b^2\right ) (c+d x)-2 a b \cos (2 (c+d x))+\left (a^2-b^2\right ) \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.31

method	result	size
risch	\(\frac {a^{2} x}{2}+\frac {b^{2} x}{2}-\frac {a b \cos \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}-\frac {\sin \left (2 d x +2 c \right ) b^{2}}{4 d}\)	\(64\)
derivativedivides	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a b \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(70\)
default	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a b \left (\cos ^{2}\left (d x +c \right )\right )+a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {2 \, a b \cos \left (d x + c\right )^{2} - {\left (a^{2} + b^{2}\right )} d x - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\left (a^{2} + b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, a b - {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

1/2*((a^2 + b^2)*(d*x + c) - (2*a*b - (a^2 - b^2)*tan(d*x + c))/(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. 245 vs. \(2 (46) = 92\).

Time = 0.50 (sec) , antiderivative size = 245, normalized size of antiderivative = 5.00 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^{2} d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + b^{2} d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + a^{2} d x \tan \left (d x\right )^{2} + b^{2} d x \tan \left (d x\right )^{2} + a^{2} d x \tan \left (c\right )^{2} + b^{2} d x \tan \left (c\right )^{2} - a b \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - a^{2} \tan \left (d x\right )^{2} \tan \left (c\right ) + b^{2} \tan \left (d x\right )^{2} \tan \left (c\right ) - a^{2} \tan \left (d x\right ) \tan \left (c\right )^{2} + b^{2} \tan \left (d x\right ) \tan \left (c\right )^{2} + a^{2} d x + b^{2} d x + a b \tan \left (d x\right )^{2} + 4 \, a b \tan \left (d x\right ) \tan \left (c\right ) + a b \tan \left (c\right )^{2} + a^{2} \tan \left (d x\right ) - b^{2} \tan \left (d x\right ) + a^{2} \tan \left (c\right ) - b^{2} \tan \left (c\right ) - a b}{2 \, {\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + d \tan \left (d x\right )^{2} + d \tan \left (c\right )^{2} + d\right )}} \]

[In]

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

[Out]

1/2*(a^2*d*x*tan(d*x)^2*tan(c)^2 + b^2*d*x*tan(d*x)^2*tan(c)^2 + a^2*d*x*tan(d*x)^2 + b^2*d*x*tan(d*x)^2 + a^2
*d*x*tan(c)^2 + b^2*d*x*tan(c)^2 - a*b*tan(d*x)^2*tan(c)^2 - a^2*tan(d*x)^2*tan(c) + b^2*tan(d*x)^2*tan(c) - a
^2*tan(d*x)*tan(c)^2 + b^2*tan(d*x)*tan(c)^2 + a^2*d*x + b^2*d*x + a*b*tan(d*x)^2 + 4*a*b*tan(d*x)*tan(c) + a*
b*tan(c)^2 + a^2*tan(d*x) - b^2*tan(d*x) + a^2*tan(c) - b^2*tan(c) - a*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 = 3.99 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=x\,\left (\frac {a^2}{2}+\frac {b^2}{2}\right )-\frac {{\cos \left (c+d\,x\right )}^2\,\left (a\,b-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )\right )}{d} \]

[In]

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

[Out]

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