3.57 \(\int \sec ^8(a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=61 \[ \frac{\tan ^9(a+b x)}{9 b}+\frac{3 \tan ^7(a+b x)}{7 b}+\frac{3 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

[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)

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Rubi [A]  time = 0.039758, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2607, 270} \[ \frac{\tan ^9(a+b x)}{9 b}+\frac{3 \tan ^7(a+b x)}{7 b}+\frac{3 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[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 2607

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])

Rule 270

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]

Rubi steps

\begin{align*} \int \sec ^8(a+b x) \tan ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{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]  time = 0.0346972, size = 98, normalized size = 1.61 \[ -\frac{16 \tan (a+b x)}{315 b}+\frac{\tan (a+b x) \sec ^8(a+b x)}{9 b}-\frac{\tan (a+b x) \sec ^6(a+b x)}{63 b}-\frac{2 \tan (a+b x) \sec ^4(a+b x)}{105 b}-\frac{8 \tan (a+b x) \sec ^2(a+b x)}{315 b} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.023, size = 78, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{9\, \left ( \cos \left ( bx+a \right ) \right ) ^{9}}}+{\frac{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{21\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{105\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{315\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/9*sin(b*x+a)^3/cos(b*x+a)^9+2/21*sin(b*x+a)^3/cos(b*x+a)^7+8/105*sin(b*x+a)^3/cos(b*x+a)^5+16/315*sin(b
*x+a)^3/cos(b*x+a)^3)

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Maxima [A]  time = 0.979227, size = 62, normalized size = 1.02 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.59504, size = 165, normalized size = 2.7 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.20449, size = 62, normalized size = 1.02 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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