3.365 \(\int \sec ^2(a+b x) (d \tan (a+b x))^n \, dx\)

Optimal. Leaf size=24 \[ \frac {(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

[Out]

(d*tan(b*x+a))^(1+n)/b/d/(1+n)

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Rubi [A]  time = 0.04, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2607, 32} \[ \frac {(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^2*(d*Tan[a + b*x])^n,x]

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.04 \[ \frac {\tan (a+b x) (d \tan (a+b x))^n}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^2*(d*Tan[a + b*x])^n,x]

[Out]

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

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fricas [A]  time = 0.54, size = 40, normalized size = 1.67 \[ \frac {\left (\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}\right )^{n} \sin \left (b x + a\right )}{{\left (b n + b\right )} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*(d*tan(b*x+a))^n,x, algorithm="fricas")

[Out]

(d*sin(b*x + a)/cos(b*x + a))^n*sin(b*x + a)/((b*n + b)*cos(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{n} \sec \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*(d*tan(b*x+a))^n,x, algorithm="giac")

[Out]

integrate((d*tan(b*x + a))^n*sec(b*x + a)^2, x)

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maple [A]  time = 0.11, size = 25, normalized size = 1.04 \[ \frac {\left (d \tan \left (b x +a \right )\right )^{1+n}}{b d \left (1+n \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2*(d*tan(b*x+a))^n,x)

[Out]

(d*tan(b*x+a))^(1+n)/b/d/(1+n)

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maxima [A]  time = 0.73, size = 24, normalized size = 1.00 \[ \frac {\left (d \tan \left (b x + a\right )\right )^{n + 1}}{b d {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2*(d*tan(b*x+a))^n,x, algorithm="maxima")

[Out]

(d*tan(b*x + a))^(n + 1)/(b*d*(n + 1))

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mupad [B]  time = 2.64, size = 49, normalized size = 2.04 \[ \frac {\sin \left (2\,a+2\,b\,x\right )\,{\left (\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{2\,{\cos \left (a+b\,x\right )}^2}\right )}^n}{2\,b\,{\cos \left (a+b\,x\right )}^2\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^n/cos(a + b*x)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan {\left (a + b x \right )}\right )^{n} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2*(d*tan(b*x+a))**n,x)

[Out]

Integral((d*tan(a + b*x))**n*sec(a + b*x)**2, x)

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