\(\int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx\) [798]

Optimal result

Integrand size = 25, antiderivative size = 60 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=a x-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2917, 2686, 3554, 8} \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \sec (c+d x)}{d}+a x \]

[In]

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

[Out]

a*x - (a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

Rule 8

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

Rule 2686

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

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = a \int \sec (c+d x) \tan ^3(c+d x) \, dx+a \int \tan ^4(c+d x) \, dx \\ & = \frac {a \tan ^3(c+d x)}{3 d}-a \int \tan ^2(c+d x) \, dx+\frac {a \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d}+a \int 1 \, dx \\ & = a x-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \arctan (\tan (c+d x))}{d}-\frac {a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

(a*ArcTan[Tan[c + d*x]])/d - (a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d
*x]^3)/(3*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07

method	result	size
risch	\(a x -\frac {2 a \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}-4 i+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(64\)
derivativedivides	\(\frac {a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+a \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(88\)
default	\(\frac {a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+a \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(88\)
parallelrisch	\(\frac {a \left (3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d -6 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x +3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x -12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 d x +4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(119\)
norman	\(\frac {a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a x +\frac {4 a}{3 d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {14 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {14 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+2 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(187\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

a*x-2/3*a*(3*exp(3*I*(d*x+c))-4*I+5*exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)/(exp(I*(d*x+c))-I)^3/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {3 \, a d x \cos \left (d x + c\right ) - 4 \, a \cos \left (d x + c\right )^{2} - {\left (3 \, a d x \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a}{3 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {6 \, {\left (d x + c\right )} a + \frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*a + 3*a/(tan(1/2*d*x + 1/2*c) + 1) + (9*a*tan(1/2*d*x + 1/2*c)^2 - 24*a*tan(1/2*d*x + 1/2*c)
+ 11*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 11.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.95 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=a\,x-\frac {\left (\frac {a\,\left (6\,d\,x-6\right )}{3}-2\,a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (2\,a\,d\,x-\frac {a\,\left (6\,d\,x-2\right )}{3}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (3\,d\,x-4\right )}{3}-a\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

[In]

int((sin(c + d*x)^3*(a + a*sin(c + d*x)))/cos(c + d*x)^4,x)

[Out]

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