\(\int \sec (a+b x) \tan ^3(a+b x) \, dx\) [74]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b} \]

[Out]

-sec(b*x+a)/b+1/3*sec(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2686} \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=\frac {\sec ^3(a+b x)}{3 b}-\frac {\sec (a+b x)}{b} \]

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^3,x]

[Out]

-(Sec[a + b*x]/b) + Sec[a + b*x]^3/(3*b)

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])

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b} \]

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x]^3,x]

[Out]

-(Sec[a + b*x]/b) + Sec[a + b*x]^3/(3*b)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\left (\sec ^{3}\left (b x +a \right )\right )}{3}-\sec \left (b x +a \right )}{b}\)	\(24\)
default	\(\frac {\frac {\left (\sec ^{3}\left (b x +a \right )\right )}{3}-\sec \left (b x +a \right )}{b}\)	\(24\)
norman	\(\frac {-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {4}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3}}\)	\(39\)
parallelrisch	\(\frac {\frac {4}{3}-4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3}}\)	\(47\)
risch	\(-\frac {2 \left (3 \,{\mathrm e}^{5 i \left (b x +a \right )}+2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3}}\)	\(53\)

[In]

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

[Out]

1/b*(1/3*sec(b*x+a)^3-sec(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \]

[In]

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

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \]

[In]

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

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{2} - 1}{3 \, b \cos \left (b x + a\right )^{3}} \]

[In]

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

[Out]

-1/3*(3*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^3)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \sec (a+b x) \tan ^3(a+b x) \, dx=-\frac {{\cos \left (a+b\,x\right )}^2-\frac {1}{3}}{b\,{\cos \left (a+b\,x\right )}^3} \]

[In]

int(sin(a + b*x)^3/cos(a + b*x)^4,x)

[Out]

-(cos(a + b*x)^2 - 1/3)/(b*cos(a + b*x)^3)