\(\int \sec ^6(a+b x) \tan ^2(a+b x) \, dx\) [56]

Optimal result

Integrand size = 17, antiderivative size = 46 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\frac {\tan ^3(a+b x)}{3 b}+\frac {2 \tan ^5(a+b x)}{5 b}+\frac {\tan ^7(a+b x)}{7 b} \]

[Out]

1/3*tan(b*x+a)^3/b+2/5*tan(b*x+a)^5/b+1/7*tan(b*x+a)^7/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 = {2687, 276} \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\frac {\tan ^7(a+b x)}{7 b}+\frac {2 \tan ^5(a+b x)}{5 b}+\frac {\tan ^3(a+b x)}{3 b} \]

[In]

Int[Sec[a + b*x]^6*Tan[a + b*x]^2,x]

[Out]

Tan[a + b*x]^3/(3*b) + (2*Tan[a + b*x]^5)/(5*b) + Tan[a + b*x]^7/(7*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 )^2 \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (x^2+2 x^4+x^6\right ) \, dx,x,\tan (a+b x)\right )}{b} \\ & = \frac {\tan ^3(a+b x)}{3 b}+\frac {2 \tan ^5(a+b x)}{5 b}+\frac {\tan ^7(a+b x)}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.67 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=-\frac {8 \tan (a+b x)}{105 b}-\frac {4 \sec ^2(a+b x) \tan (a+b x)}{105 b}-\frac {\sec ^4(a+b x) \tan (a+b x)}{35 b}+\frac {\sec ^6(a+b x) \tan (a+b x)}{7 b} \]

[In]

Integrate[Sec[a + b*x]^6*Tan[a + b*x]^2,x]

[Out]

(-8*Tan[a + b*x])/(105*b) - (4*Sec[a + b*x]^2*Tan[a + b*x])/(105*b) - (Sec[a + b*x]^4*Tan[a + b*x])/(35*b) + (
Sec[a + b*x]^6*Tan[a + b*x])/(7*b)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {4 \left (\sin ^{3}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}+\frac {8 \left (\sin ^{3}\left (b x +a \right )\right )}{105 \cos \left (b x +a \right )^{3}}}{b}\)	\(60\)
default	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {4 \left (\sin ^{3}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}+\frac {8 \left (\sin ^{3}\left (b x +a \right )\right )}{105 \cos \left (b x +a \right )^{3}}}{b}\)	\(60\)
risch	\(-\frac {16 i \left (70 \,{\mathrm e}^{8 i \left (b x +a \right )}-35 \,{\mathrm e}^{6 i \left (b x +a \right )}+21 \,{\mathrm e}^{4 i \left (b x +a \right )}+7 \,{\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{105 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{7}}\)	\(66\)
parallelrisch	\(-\frac {8 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (35 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+28 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+114 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+28 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+35\right )}{105 b \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{7}}\)	\(86\)
norman	\(\frac {-\frac {8 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {32 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{15 b}-\frac {304 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{35 b}-\frac {32 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{15 b}-\frac {8 \left (\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{7}}\)	\(98\)

[In]

int(sec(b*x+a)^8*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/7*sin(b*x+a)^3/cos(b*x+a)^7+4/35*sin(b*x+a)^3/cos(b*x+a)^5+8/105*sin(b*x+a)^3/cos(b*x+a)^3)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.11 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=-\frac {{\left (8 \, \cos \left (b x + a\right )^{6} + 4 \, \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} - 15\right )} \sin \left (b x + a\right )}{105 \, b \cos \left (b x + a\right )^{7}} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/105*(8*cos(b*x + a)^6 + 4*cos(b*x + a)^4 + 3*cos(b*x + a)^2 - 15)*sin(b*x + a)/(b*cos(b*x + a)^7)

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(sec(b*x+a)**8*sin(b*x+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\frac {15 \, \tan \left (b x + a\right )^{7} + 42 \, \tan \left (b x + a\right )^{5} + 35 \, \tan \left (b x + a\right )^{3}}{105 \, b} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/105*(15*tan(b*x + a)^7 + 42*tan(b*x + a)^5 + 35*tan(b*x + a)^3)/b

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\frac {15 \, \tan \left (b x + a\right )^{7} + 42 \, \tan \left (b x + a\right )^{5} + 35 \, \tan \left (b x + a\right )^{3}}{105 \, b} \]

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/105*(15*tan(b*x + a)^7 + 42*tan(b*x + a)^5 + 35*tan(b*x + a)^3)/b

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \sec ^6(a+b x) \tan ^2(a+b x) \, dx=\frac {{\mathrm {tan}\left (a+b\,x\right )}^3\,\left (15\,{\mathrm {tan}\left (a+b\,x\right )}^4+42\,{\mathrm {tan}\left (a+b\,x\right )}^2+35\right )}{105\,b} \]

[In]

int(sin(a + b*x)^2/cos(a + b*x)^8,x)

[Out]

(tan(a + b*x)^3*(42*tan(a + b*x)^2 + 15*tan(a + b*x)^4 + 35))/(105*b)