\(\int \sec ^8(a+b x) \tan ^2(a+b x) \, dx\) [57]

Optimal result

Integrand size = 17, antiderivative size = 61 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\frac {\tan ^3(a+b x)}{3 b}+\frac {3 \tan ^5(a+b x)}{5 b}+\frac {3 \tan ^7(a+b x)}{7 b}+\frac {\tan ^9(a+b x)}{9 b} \]

[Out]

1/3*tan(b*x+a)^3/b+3/5*tan(b*x+a)^5/b+3/7*tan(b*x+a)^7/b+1/9*tan(b*x+a)^9/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, 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 = {2687, 276} \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\frac {\tan ^9(a+b x)}{9 b}+\frac {3 \tan ^7(a+b x)}{7 b}+\frac {3 \tan ^5(a+b x)}{5 b}+\frac {\tan ^3(a+b x)}{3 b} \]

[In]

Int[Sec[a + b*x]^8*Tan[a + b*x]^2,x]

[Out]

Tan[a + b*x]^3/(3*b) + (3*Tan[a + b*x]^5)/(5*b) + (3*Tan[a + b*x]^7)/(7*b) + Tan[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 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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (x^2+3 x^4+3 x^6+x^8\right ) \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\tan ^3(a+b x)}{3 b}+\frac {3 \tan ^5(a+b x)}{5 b}+\frac {3 \tan ^7(a+b x)}{7 b}+\frac {\tan ^9(a+b x)}{9 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.61 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=-\frac {16 \tan (a+b x)}{315 b}-\frac {8 \sec ^2(a+b x) \tan (a+b x)}{315 b}-\frac {2 \sec ^4(a+b x) \tan (a+b x)}{105 b}-\frac {\sec ^6(a+b x) \tan (a+b x)}{63 b}+\frac {\sec ^8(a+b x) \tan (a+b x)}{9 b} \]

[In]

Integrate[Sec[a + b*x]^8*Tan[a + b*x]^2,x]

[Out]

(-16*Tan[a + b*x])/(315*b) - (8*Sec[a + b*x]^2*Tan[a + b*x])/(315*b) - (2*Sec[a + b*x]^4*Tan[a + b*x])/(105*b)
- (Sec[a + b*x]^6*Tan[a + b*x])/(63*b) + (Sec[a + b*x]^8*Tan[a + b*x])/(9*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {32 i \left (315 \,{\mathrm e}^{10 i \left (b x +a \right )}-189 \,{\mathrm e}^{8 i \left (b x +a \right )}+84 \,{\mathrm e}^{6 i \left (b x +a \right )}+36 \,{\mathrm e}^{4 i \left (b x +a \right )}+9 \,{\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{315 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{9}}\)	\(77\)
derivativedivides	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{9 \cos \left (b x +a \right )^{9}}+\frac {2 \left (\sin ^{3}\left (b x +a \right )\right )}{21 \cos \left (b x +a \right )^{7}}+\frac {8 \left (\sin ^{3}\left (b x +a \right )\right )}{105 \cos \left (b x +a \right )^{5}}+\frac {16 \left (\sin ^{3}\left (b x +a \right )\right )}{315 \cos \left (b x +a \right )^{3}}}{b}\)	\(78\)
default	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{9 \cos \left (b x +a \right )^{9}}+\frac {2 \left (\sin ^{3}\left (b x +a \right )\right )}{21 \cos \left (b x +a \right )^{7}}+\frac {8 \left (\sin ^{3}\left (b x +a \right )\right )}{105 \cos \left (b x +a \right )^{5}}+\frac {16 \left (\sin ^{3}\left (b x +a \right )\right )}{315 \cos \left (b x +a \right )^{3}}}{b}\)	\(78\)
parallelrisch	\(-\frac {8 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (105 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+126 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+711 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+356 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+711 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+126 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+105\right )}{315 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{9}}\)	\(112\)

[In]

int(sec(b*x+a)^10*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-32/315*I*(315*exp(10*I*(b*x+a))-189*exp(8*I*(b*x+a))+84*exp(6*I*(b*x+a))+36*exp(4*I*(b*x+a))+9*exp(2*I*(b*x+a
))+1)/b/(exp(2*I*(b*x+a))+1)^9

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=-\frac {{\left (16 \, \cos \left (b x + a\right )^{8} + 8 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} + 5 \, \cos \left (b x + a\right )^{2} - 35\right )} \sin \left (b x + a\right )}{315 \, b \cos \left (b x + a\right )^{9}} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/315*(16*cos(b*x + a)^8 + 8*cos(b*x + a)^6 + 6*cos(b*x + a)^4 + 5*cos(b*x + a)^2 - 35)*sin(b*x + a)/(b*cos(b
*x + a)^9)

Sympy [F(-1)]

Timed out. \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**10*sin(b*x+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\frac {35 \, \tan \left (b x + a\right )^{9} + 135 \, \tan \left (b x + a\right )^{7} + 189 \, \tan \left (b x + a\right )^{5} + 105 \, \tan \left (b x + a\right )^{3}}{315 \, b} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/315*(35*tan(b*x + a)^9 + 135*tan(b*x + a)^7 + 189*tan(b*x + a)^5 + 105*tan(b*x + a)^3)/b

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\frac {35 \, \tan \left (b x + a\right )^{9} + 135 \, \tan \left (b x + a\right )^{7} + 189 \, \tan \left (b x + a\right )^{5} + 105 \, \tan \left (b x + a\right )^{3}}{315 \, b} \]

[In]

integrate(sec(b*x+a)^10*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/315*(35*tan(b*x + a)^9 + 135*tan(b*x + a)^7 + 189*tan(b*x + a)^5 + 105*tan(b*x + a)^3)/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {{\mathrm {tan}\left (a+b\,x\right )}^9}{9}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^7}{7}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^5}{5}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^3}{3}}{b} \]

[In]

int(sin(a + b*x)^2/cos(a + b*x)^10,x)

[Out]

(tan(a + b*x)^3/3 + (3*tan(a + b*x)^5)/5 + (3*tan(a + b*x)^7)/7 + tan(a + b*x)^9/9)/b