3.56 \(\int \sec ^6(a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{\tan ^7(a+b x)}{7 b}+\frac{2 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

[Out]

Tan[a + b*x]^3/(3*b) + (2*Tan[a + b*x]^5)/(5*b) + Tan[a + b*x]^7/(7*b)

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Rubi [A]  time = 0.038507, 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} \[ \frac{\tan ^7(a+b x)}{7 b}+\frac{2 \tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^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 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 ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^2 \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{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]  time = 0.0426543, size = 77, normalized size = 1.67 \[ -\frac{8 \tan (a+b x)}{105 b}+\frac{\tan (a+b x) \sec ^6(a+b x)}{7 b}-\frac{\tan (a+b x) \sec ^4(a+b x)}{35 b}-\frac{4 \tan (a+b x) \sec ^2(a+b x)}{105 b} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.022, size = 60, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{7\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{35\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{105\, \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)^8*sin(b*x+a)^2,x)

[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)

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Maxima [A]  time = 0.989448, size = 49, normalized size = 1.07 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.84528, size = 138, normalized size = 3. \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [A]  time = 1.17382, size = 49, normalized size = 1.07 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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