3.75 \(\int \sec ^2(a+b x) \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=15 \[ \frac{\tan ^4(a+b x)}{4 b} \]

[Out]

Tan[a + b*x]^4/(4*b)

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Rubi [A]  time = 0.0278485, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2607, 30} \[ \frac{\tan ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

Tan[a + b*x]^4/(4*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0053825, size = 15, normalized size = 1. \[ \frac{\tan ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^2*Tan[a + b*x]^3,x]

[Out]

Tan[a + b*x]^4/(4*b)

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Maple [A]  time = 0.019, size = 22, normalized size = 1.5 \begin{align*}{\frac{ \left ( \sin \left ( bx+a \right ) \right ) ^{4}}{4\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5*sin(b*x+a)^3,x)

[Out]

1/4/b*sin(b*x+a)^4/cos(b*x+a)^4

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Maxima [B]  time = 0.978591, size = 53, normalized size = 3.53 \begin{align*} \frac{2 \, \sin \left (b x + a\right )^{2} - 1}{4 \,{\left (\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/4*(2*sin(b*x + a)^2 - 1)/((sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1)*b)

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Fricas [A]  time = 1.59917, size = 65, normalized size = 4.33 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/4*(2*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^4)

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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)**5*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.19368, size = 34, normalized size = 2.27 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/4*(2*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^4)