3.364 \(\int \sec ^4(a+b x) (d \tan (a+b x))^n \, dx\)

Optimal. Leaf size=49 \[ \frac{(d \tan (a+b x))^{n+3}}{b d^3 (n+3)}+\frac{(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

[Out]

(d*Tan[a + b*x])^(1 + n)/(b*d*(1 + n)) + (d*Tan[a + b*x])^(3 + n)/(b*d^3*(3 + n))

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Rubi [A]  time = 0.0511222, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2607, 14} \[ \frac{(d \tan (a+b x))^{n+3}}{b d^3 (n+3)}+\frac{(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4*(d*Tan[a + b*x])^n,x]

[Out]

(d*Tan[a + b*x])^(1 + n)/(b*d*(1 + n)) + (d*Tan[a + b*x])^(3 + n)/(b*d^3*(3 + n))

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

Mathematica [A]  time = 1.14592, size = 78, normalized size = 1.59 \[ \frac{d (d \tan (a+b x))^{n-1} \left (2 \left (-\tan ^2(a+b x)\right )^{\frac{1-n}{2}}+\tan ^2(a+b x) \sec ^2(a+b x) (\cos (2 (a+b x))+n+2)\right )}{b (n+1) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4*(d*Tan[a + b*x])^n,x]

[Out]

(d*(d*Tan[a + b*x])^(-1 + n)*((2 + n + Cos[2*(a + b*x)])*Sec[a + b*x]^2*Tan[a + b*x]^2 + 2*(-Tan[a + b*x]^2)^(
(1 - n)/2)))/(b*(1 + n)*(3 + n))

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Maple [F]  time = 0.139, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( bx+a \right ) \right ) ^{4} \left ( d\tan \left ( bx+a \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4*(d*tan(b*x+a))^n,x)

[Out]

int(sec(b*x+a)^4*(d*tan(b*x+a))^n,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^n,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71338, size = 151, normalized size = 3.08 \begin{align*} \frac{{\left (2 \, \cos \left (b x + a\right )^{2} + n + 1\right )} \left (\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}\right )^{n} \sin \left (b x + a\right )}{{\left (b n^{2} + 4 \, b n + 3 \, b\right )} \cos \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^n,x, algorithm="fricas")

[Out]

(2*cos(b*x + a)^2 + n + 1)*(d*sin(b*x + a)/cos(b*x + a))^n*sin(b*x + a)/((b*n^2 + 4*b*n + 3*b)*cos(b*x + a)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan{\left (a + b x \right )}\right )^{n} \sec ^{4}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4*(d*tan(b*x+a))**n,x)

[Out]

Integral((d*tan(a + b*x))**n*sec(a + b*x)**4, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^n,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError