\(\int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx\) [800]

Optimal result

Integrand size = 25, antiderivative size = 33 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2917, 2686, 30, 2687} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \sec ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2687

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(36\)
default	\(\frac {\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(36\)
parallelrisch	\(-\frac {2 a \left (3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(59\)
risch	\(-\frac {2 i \left (-2 i a \,{\mathrm e}^{i \left (d x +c \right )}-a +3 a \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d}\)	\(64\)
norman	\(\frac {-\frac {2 a}{3 d}-\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(124\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \cos \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) - 2 \, a}{3 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a*(Integral(sin(c + d*x)*sec(c + d*x)**4, x) + Integral(sin(c + d*x)**2*sec(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \tan \left (d x + c\right )^{3} + \frac {a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=\frac {\frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} - \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.61 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx=-\frac {2\,a\,\left (\cos \left (c+d\,x\right )+1\right )\,\left (\cos \left (c+d\,x\right )+\sin \left (c+d\,x\right )-2\right )}{3\,d\,\left (2\,\cos \left (c+d\,x\right )-\sin \left (2\,c+2\,d\,x\right )\right )} \]

[In]

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

[Out]

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