\(\int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx\) [29]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=a^3 x+\frac {7 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

a^3*x+7/2*a^3*arctanh(sin(d*x+c))/d+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2836, 3855, 3852, 8, 3853} \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=\frac {7 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}+a^3 x \]

[In]

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

[Out]

a^3*x + (7*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 2836

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

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 3853

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

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}& = \int \left (a^3+3 a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+a^3 \sec ^3(c+d x)\right ) \, dx \\ & = a^3 x+a^3 \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx \\ & = a^3 x+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^3 \int \sec (c+d x) \, dx-\frac {\left (3 a^3\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = a^3 x+\frac {7 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=a^3 \left (x+\frac {7 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 \tan (c+d x)}{d}+\frac {\sec (c+d x) \tan (c+d x)}{2 d}\right ) \]

[In]

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

[Out]

a^3*(x + (7*ArcTanh[Sin[c + d*x]])/(2*d) + (3*Tan[c + d*x])/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {a^{3} \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(80\)
default	\(\frac {a^{3} \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(80\)
parts	\(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{3} \left (d x +c \right )}{d}+\frac {3 a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(88\)
risch	\(a^{3} x -\frac {i a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}\)	\(104\)
parallelrisch	\(\frac {a^{3} \left (2 d x \cos \left (2 d x +2 c \right )+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )+2 d x +2 \sin \left (d x +c \right )+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \sin \left (2 d x +2 c \right )\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(133\)
norman	\(\frac {a^{3} x +a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{3} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {16 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-2 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {7 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {7 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(254\)

[In]

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

[Out]

1/d*(a^3*(d*x+c)+3*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*tan(d*x+c)+a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*
x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.66 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=\frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} + 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*(4*a^3*d*x*cos(d*x + c)^2 + 7*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 7*a^3*cos(d*x + c)^2*log(-sin(d*x
+ c) + 1) + 2*(6*a^3*cos(d*x + c) + a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} a^{3} - a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*a^3 - a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) + 6*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*a^3*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.69 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} a^{3} + 7 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 7 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*a^3 + 7*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 7*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2
*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*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 = 14.63 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.49 \[ \int (a+a \cos (c+d x))^3 \sec ^3(c+d x) \, dx=a^3\,x+\frac {7\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

a^3*x + (7*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (5*a^3*tan(c/2 + (d*x)/2)^3 - 7*a^3*tan(c/2 + (d*x)/2))/(d*(tan(
c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))