\(\int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx\) [24]

Optimal result

Integrand size = 19, antiderivative size = 48 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=3 a^3 x+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d} \]

[Out]

3*a^3*x+3*a^3*arctanh(sin(d*x+c))/d+a^3*sin(d*x+c)/d+a^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3876, 2717, 3855, 3852, 8} \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+3 a^3 x \]

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^3,x]

[Out]

3*a^3*x + (3*a^3*ArcTanh[Sin[c + d*x]])/d + (a^3*Sin[c + d*x])/d + (a^3*Tan[c + d*x])/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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (3 a^3+a^3 \cos (c+d x)+3 a^3 \sec (c+d x)+a^3 \sec ^2(c+d x)\right ) \, dx \\ & = 3 a^3 x+a^3 \int \cos (c+d x) \, dx+a^3 \int \sec ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \, dx \\ & = 3 a^3 x+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d}-\frac {a^3 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = 3 a^3 x+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 \sin (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.69 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 (3 d x+3 \text {arctanh}(\sin (c+d x))+\sin (c+d x)+\tan (c+d x))}{d} \]

[In]

Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^3,x]

[Out]

(a^3*(3*d*x + 3*ArcTanh[Sin[c + d*x]] + Sin[c + d*x] + Tan[c + d*x]))/d

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {a^{3} \tan \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \left (d x +c \right )+a^{3} \sin \left (d x +c \right )}{d}\)	\(55\)
default	\(\frac {a^{3} \tan \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \left (d x +c \right )+a^{3} \sin \left (d x +c \right )}{d}\)	\(55\)
parallelrisch	\(\frac {a^{3} \left (6 d x \cos \left (d x +c \right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2 \sin \left (d x +c \right )+\sin \left (2 d x +2 c \right )\right )}{2 d \cos \left (d x +c \right )}\)	\(85\)
risch	\(3 a^{3} x -\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(108\)
norman	\(\frac {3 a^{3} x +\frac {4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-3 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(167\)

[In]

int(cos(d*x+c)*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*tan(d*x+c)+3*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*(d*x+c)+a^3*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.90 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {6 \, a^{3} d x \cos \left (d x + c\right ) + 3 \, a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^3*d*x*cos(d*x + c) + 3*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*a^3*cos(d*x + c)*log(-sin(d*x + c)
+ 1) + 2*(a^3*cos(d*x + c) + a^3)*sin(d*x + c))/(d*cos(d*x + c))

Sympy [F]

\[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos {\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**3,x)

[Out]

a**3*(Integral(3*cos(c + d*x)*sec(c + d*x), x) + Integral(3*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(cos(c
+ d*x)*sec(c + d*x)**3, x) + Integral(cos(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.33 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {6 \, {\left (d x + c\right )} a^{3} + 3 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3} \tan \left (d x + c\right )}{2 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(6*(d*x + c)*a^3 + 3*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*a^3*sin(d*x + c) + 2*a^3*tan(
d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.67 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {3 \, {\left (d x + c\right )} a^{3} + 3 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \, dx=3\,a^3\,x+\frac {6\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {4\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )} \]

[In]

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

[Out]

3*a^3*x + (6*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (4*a^3*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^4 - 1))