\(\int \sec (a+b x) \tan ^5(a+b x) \, dx\) [110]

Optimal result

Integrand size = 15, antiderivative size = 41 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {\sec (a+b x)}{b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \]

[Out]

sec(b*x+a)/b-2/3*sec(b*x+a)^3/b+1/5*sec(b*x+a)^5/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2686, 200} \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec (a+b x)}{b} \]

[In]

Int[Sec[a + b*x]*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]/b - (2*Sec[a + b*x]^3)/(3*b) + Sec[a + b*x]^5/(5*b)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 )^2 \, dx,x,\sec (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\sec (a+b x)\right )}{b} \\ & = \frac {\sec (a+b x)}{b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {\sec (a+b x)}{b}-\frac {2 \sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \]

[In]

Integrate[Sec[a + b*x]*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]/b - (2*Sec[a + b*x]^3)/(3*b) + Sec[a + b*x]^5/(5*b)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}-\frac {2 \left (\sec ^{3}\left (b x +a \right )\right )}{3}+\sec \left (b x +a \right )}{b}\)	\(32\)
default	\(\frac {\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}-\frac {2 \left (\sec ^{3}\left (b x +a \right )\right )}{3}+\sec \left (b x +a \right )}{b}\)	\(32\)
norman	\(\frac {-\frac {16}{15 b}+\frac {16 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {32 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5}}\)	\(55\)
parallelrisch	\(\frac {-\frac {16}{15}-\frac {32 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3}+\frac {16 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3}}{b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{5}}\)	\(60\)
risch	\(\frac {2 \,{\mathrm e}^{9 i \left (b x +a \right )}+\frac {8 \,{\mathrm e}^{7 i \left (b x +a \right )}}{3}+\frac {116 \,{\mathrm e}^{5 i \left (b x +a \right )}}{15}+\frac {8 \,{\mathrm e}^{3 i \left (b x +a \right )}}{3}+2 \,{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{5}}\)	\(75\)

[In]

int(sec(b*x+a)^6*sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/5*sec(b*x+a)^5-2/3*sec(b*x+a)^3+sec(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} + 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/15*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 + 3)/(b*cos(b*x + a)^5)

Sympy [F(-1)]

Timed out. \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**6*sin(b*x+a)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} + 3}{15 \, b \cos \left (b x + a\right )^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/15*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 + 3)/(b*cos(b*x + a)^5)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.76 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {16 \, {\left (\frac {5 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {10 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}}{15 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{5}} \]

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^5,x, algorithm="giac")

[Out]

16/15*(5*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 10*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)/(b*((cos(b*
x + a) - 1)/(cos(b*x + a) + 1) + 1)^5)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \sec (a+b x) \tan ^5(a+b x) \, dx=\frac {15\,{\cos \left (a+b\,x\right )}^4-10\,{\cos \left (a+b\,x\right )}^2+3}{15\,b\,{\cos \left (a+b\,x\right )}^5} \]

[In]

int(sin(a + b*x)^5/cos(a + b*x)^6,x)

[Out]

(15*cos(a + b*x)^4 - 10*cos(a + b*x)^2 + 3)/(15*b*cos(a + b*x)^5)