3.88 \(\int \sec ^6(a+b x) \tan ^4(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{\tan ^9(a+b x)}{9 b}+\frac{2 \tan ^7(a+b x)}{7 b}+\frac{\tan ^5(a+b x)}{5 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0357062, antiderivative size = 46, 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} \[ \frac{\tan ^9(a+b x)}{9 b}+\frac{2 \tan ^7(a+b x)}{7 b}+\frac{\tan ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^6*Tan[a + b*x]^4,x]

[Out]

Tan[a + b*x]^5/(5*b) + (2*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 ^6(a+b x) \tan ^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\tan ^5(a+b x)}{5 b}+\frac{2 \tan ^7(a+b x)}{7 b}+\frac{\tan ^9(a+b x)}{9 b}\\ \end{align*}

Mathematica [B]  time = 0.0388211, size = 98, normalized size = 2.13 \[ \frac{8 \tan (a+b x)}{315 b}+\frac{\tan (a+b x) \sec ^8(a+b x)}{9 b}-\frac{10 \tan (a+b x) \sec ^6(a+b x)}{63 b}+\frac{\tan (a+b x) \sec ^4(a+b x)}{105 b}+\frac{4 \tan (a+b x) \sec ^2(a+b x)}{315 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6*Tan[a + b*x]^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.023, size = 60, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{9\, \left ( \cos \left ( bx+a \right ) \right ) ^{9}}}+{\frac{4\, \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{63\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( bx+a \right ) \right ) ^{5}}{315\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^10*sin(b*x+a)^4,x)

[Out]

1/b*(1/9*sin(b*x+a)^5/cos(b*x+a)^9+4/63*sin(b*x+a)^5/cos(b*x+a)^7+8/315*sin(b*x+a)^5/cos(b*x+a)^5)

________________________________________________________________________________________

Maxima [A]  time = 0.98759, size = 49, normalized size = 1.07 \begin{align*} \frac{35 \, \tan \left (b x + a\right )^{9} + 90 \, \tan \left (b x + a\right )^{7} + 63 \, \tan \left (b x + a\right )^{5}}{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)^4,x, algorithm="maxima")

[Out]

1/315*(35*tan(b*x + a)^9 + 90*tan(b*x + a)^7 + 63*tan(b*x + a)^5)/b

________________________________________________________________________________________

Fricas [A]  time = 1.66066, size = 163, normalized size = 3.54 \begin{align*} \frac{{\left (8 \, \cos \left (b x + a\right )^{8} + 4 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} - 50 \, \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)^4,x, algorithm="fricas")

[Out]

1/315*(8*cos(b*x + a)^8 + 4*cos(b*x + a)^6 + 3*cos(b*x + a)^4 - 50*cos(b*x + a)^2 + 35)*sin(b*x + a)/(b*cos(b*
x + a)^9)

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.16883, size = 49, normalized size = 1.07 \begin{align*} \frac{35 \, \tan \left (b x + a\right )^{9} + 90 \, \tan \left (b x + a\right )^{7} + 63 \, \tan \left (b x + a\right )^{5}}{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)^4,x, algorithm="giac")

[Out]

1/315*(35*tan(b*x + a)^9 + 90*tan(b*x + a)^7 + 63*tan(b*x + a)^5)/b