3.55 \(\int \sec ^4(a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0340102, antiderivative size = 31, 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, 14} \[ \frac{\tan ^5(a+b x)}{5 b}+\frac{\tan ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4*Tan[a + b*x]^2,x]

[Out]

Tan[a + b*x]^3/(3*b) + Tan[a + b*x]^5/(5*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sec ^4(a+b x) \tan ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\tan ^3(a+b x)}{3 b}+\frac{\tan ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0411832, size = 56, normalized size = 1.81 \[ -\frac{2 \tan (a+b x)}{15 b}+\frac{\tan (a+b x) \sec ^4(a+b x)}{5 b}-\frac{\tan (a+b x) \sec ^2(a+b x)}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4*Tan[a + b*x]^2,x]

[Out]

(-2*Tan[a + b*x])/(15*b) - (Sec[a + b*x]^2*Tan[a + b*x])/(15*b) + (Sec[a + b*x]^4*Tan[a + b*x])/(5*b)

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Maple [A]  time = 0.02, size = 42, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}{15\, \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)^6*sin(b*x+a)^2,x)

[Out]

1/b*(1/5*sin(b*x+a)^3/cos(b*x+a)^5+2/15*sin(b*x+a)^3/cos(b*x+a)^3)

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Maxima [A]  time = 0.991552, size = 35, normalized size = 1.13 \begin{align*} \frac{3 \, \tan \left (b x + a\right )^{5} + 5 \, \tan \left (b x + a\right )^{3}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/15*(3*tan(b*x + a)^5 + 5*tan(b*x + a)^3)/b

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Fricas [A]  time = 1.79198, size = 107, normalized size = 3.45 \begin{align*} -\frac{{\left (2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right )}{15 \, b \cos \left (b x + a\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/15*(2*cos(b*x + a)^4 + cos(b*x + a)^2 - 3)*sin(b*x + a)/(b*cos(b*x + a)^5)

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

[Out]

Timed out

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Giac [A]  time = 1.15766, size = 35, normalized size = 1.13 \begin{align*} \frac{3 \, \tan \left (b x + a\right )^{5} + 5 \, \tan \left (b x + a\right )^{3}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/15*(3*tan(b*x + a)^5 + 5*tan(b*x + a)^3)/b