\(\int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\) [462]

Optimal result

Integrand size = 21, antiderivative size = 50 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {1}{2} \left (a^2+2 b^2\right ) x+\frac {2 a b \sin (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3873, 2717, 4130, 8} \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {1}{2} x \left (a^2+2 b^2\right )+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {2 a b \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 \left (a^2+2 b^2\right ) (c+d x)+8 a b \sin (c+d x)+a^2 \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(2*(a^2 + 2*b^2)*(c + d*x) + 8*a*b*Sin[c + d*x] + a^2*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {a^{2} x}{2}+x \,b^{2}+\frac {2 a b \sin \left (d x +c \right )}{d}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\)	\(43\)
parallelrisch	\(\frac {a^{2} \sin \left (2 d x +2 c \right )+8 a b \sin \left (d x +c \right )+2 \left (a^{2}+2 b^{2}\right ) x d}{4 d}\)	\(43\)
derivativedivides	\(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \sin \left (d x +c \right )+b^{2} \left (d x +c \right )}{d}\)	\(51\)
default	\(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \sin \left (d x +c \right )+b^{2} \left (d x +c \right )}{d}\)	\(51\)
norman	\(\frac {\left (-\frac {a^{2}}{2}-b^{2}\right ) x +\left (-\frac {a^{2}}{2}-b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a^{2}}{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {a^{2}}{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a \left (a -4 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a \left (4 b +a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(175\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (a^{2} + 2 \, b^{2}\right )} d x + {\left (a^{2} \cos \left (d x + c\right ) + 4 \, a b\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 4 \, {\left (d x + c\right )} b^{2} + 8 \, a b \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 + 4*(d*x + c)*b^2 + 8*a*b*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (46) = 92\).

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.92 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (a^{2} + 2 \, b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.48 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2\,x}{2}+b^2\,x+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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