\(\int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx\) [422]

Optimal result

Integrand size = 19, antiderivative size = 33 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=2 a b x+\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, 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 = {2825, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+2 a b x+\frac {b^2 \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2825

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2
)*(Cos[e + f*x]/(d*f)), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=2 a b x+\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \cos (d x) \sin (c)}{d}+\frac {b^2 \cos (c) \sin (d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \left (d x +c \right )+\sin \left (d x +c \right ) b^{2}}{d}\)	\(43\)
default	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \left (d x +c \right )+\sin \left (d x +c \right ) b^{2}}{d}\)	\(43\)
parts	\(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \sin \left (d x +c \right )}{d}+\frac {2 a b \left (d x +c \right )}{d}\)	\(48\)
parallelrisch	\(\frac {2 a b x d -a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right ) b^{2}}{d}\)	\(55\)
risch	\(2 a b x -\frac {i b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(84\)
norman	\(\frac {2 a b x +\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+4 a b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(131\)

[In]

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

[Out]

1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b*(d*x+c)+sin(d*x+c)*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=\frac {4 \, a b d x + a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b^{2} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} a b + a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + b^{2} \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

(2*(d*x + c)*a*b + a^2*log(sec(d*x + c) + tan(d*x + c)) + b^2*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (33) = 66\).

Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.36 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} a b + a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.21 \[ \int (a+b \cos (c+d x))^2 \sec (c+d x) \, dx=\frac {b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

[In]

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

[Out]

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