\(\int \sec ^5(a+b x) \tan ^5(a+b x) \, dx\) [114]

Optimal result

Integrand size = 17, antiderivative size = 46 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^7(a+b x)}{7 b}+\frac {\sec ^9(a+b x)}{9 b} \]

[Out]

1/5*sec(b*x+a)^5/b-2/7*sec(b*x+a)^7/b+1/9*sec(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2686, 276} \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {\sec ^9(a+b x)}{9 b}-\frac {2 \sec ^7(a+b x)}{7 b}+\frac {\sec ^5(a+b x)}{5 b} \]

[In]

Int[Sec[a + b*x]^5*Tan[a + b*x]^5,x]

[Out]

Sec[a + b*x]^5/(5*b) - (2*Sec[a + b*x]^7)/(7*b) + Sec[a + b*x]^9/(9*b)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && 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 x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (a+b x)\right )}{b} \\ & = \frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^7(a+b x)}{7 b}+\frac {\sec ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {\sec ^5(a+b x)}{5 b}-\frac {2 \sec ^7(a+b x)}{7 b}+\frac {\sec ^9(a+b x)}{9 b} \]

[In]

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

[Out]

Sec[a + b*x]^5/(5*b) - (2*Sec[a + b*x]^7)/(7*b) + Sec[a + b*x]^9/(9*b)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {\left (\sec ^{9}\left (b x +a \right )\right )}{9}-\frac {2 \left (\sec ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(36\)
default	\(\frac {\frac {\left (\sec ^{9}\left (b x +a \right )\right )}{9}-\frac {2 \left (\sec ^{7}\left (b x +a \right )\right )}{7}+\frac {\left (\sec ^{5}\left (b x +a \right )\right )}{5}}{b}\)	\(36\)
risch	\(\frac {\frac {32 \,{\mathrm e}^{13 i \left (b x +a \right )}}{5}-\frac {384 \,{\mathrm e}^{11 i \left (b x +a \right )}}{35}+\frac {6976 \,{\mathrm e}^{9 i \left (b x +a \right )}}{315}-\frac {384 \,{\mathrm e}^{7 i \left (b x +a \right )}}{35}+\frac {32 \,{\mathrm e}^{5 i \left (b x +a \right )}}{5}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{9}}\)	\(75\)
parallelrisch	\(\frac {-\frac {16}{315}-\frac {32 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3}-16 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\frac {112 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5}-\frac {32 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5}-\frac {64 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35}+\frac {16 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35}}{b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{9}}\)	\(101\)

[In]

int(sec(b*x+a)^10*sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/9*sec(b*x+a)^9-2/7*sec(b*x+a)^7+1/5*sec(b*x+a)^5)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {63 \, \cos \left (b x + a\right )^{4} - 90 \, \cos \left (b x + a\right )^{2} + 35}{315 \, b \cos \left (b x + a\right )^{9}} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/315*(63*cos(b*x + a)^4 - 90*cos(b*x + a)^2 + 35)/(b*cos(b*x + a)^9)

Sympy [F(-1)]

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

[In]

integrate(sec(b*x+a)**10*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 = 0.76 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {63 \, \cos \left (b x + a\right )^{4} - 90 \, \cos \left (b x + a\right )^{2} + 35}{315 \, b \cos \left (b x + a\right )^{9}} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/315*(63*cos(b*x + a)^4 - 90*cos(b*x + a)^2 + 35)/(b*cos(b*x + a)^9)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (40) = 80\).

Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.48 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {16 \, {\left (\frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {36 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - \frac {126 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {441 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - \frac {315 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac {210 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} + 1\right )}}{315 \, b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{9}} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^5,x, algorithm="giac")

[Out]

16/315*(9*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 36*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 126*(cos(b*x
+ a) - 1)^3/(cos(b*x + a) + 1)^3 + 441*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 315*(cos(b*x + a) - 1)^5/(c
os(b*x + a) + 1)^5 + 210*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 + 1)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) +
1) + 1)^9)

Mupad [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \sec ^5(a+b x) \tan ^5(a+b x) \, dx=\frac {63\,{\cos \left (a+b\,x\right )}^4-90\,{\cos \left (a+b\,x\right )}^2+35}{315\,b\,{\cos \left (a+b\,x\right )}^9} \]

[In]

int(sin(a + b*x)^5/cos(a + b*x)^10,x)

[Out]

(63*cos(a + b*x)^4 - 90*cos(a + b*x)^2 + 35)/(315*b*cos(a + b*x)^9)