3.363 \(\int \sec ^6(a+b x) (d \tan (a+b x))^n \, dx\)

Optimal. Leaf size=74 \[ \frac {(d \tan (a+b x))^{n+5}}{b d^5 (n+5)}+\frac {2 (d \tan (a+b x))^{n+3}}{b d^3 (n+3)}+\frac {(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

[Out]

(d*tan(b*x+a))^(1+n)/b/d/(1+n)+2*(d*tan(b*x+a))^(3+n)/b/d^3/(3+n)+(d*tan(b*x+a))^(5+n)/b/d^5/(5+n)

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Rubi [A]  time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, 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, 270} \[ \frac {2 (d \tan (a+b x))^{n+3}}{b d^3 (n+3)}+\frac {(d \tan (a+b x))^{n+5}}{b d^5 (n+5)}+\frac {(d \tan (a+b x))^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^6*(d*Tan[a + b*x])^n,x]

[Out]

(d*Tan[a + b*x])^(1 + n)/(b*d*(1 + n)) + (2*(d*Tan[a + b*x])^(3 + n))/(b*d^3*(3 + n)) + (d*Tan[a + b*x])^(5 +
n)/(b*d^5*(5 + n))

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]

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 ^6(a+b x) (d \tan (a+b x))^n \, dx &=\frac {\operatorname {Subst}\left (\int (d x)^n \left (1+x^2\right )^2 \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left ((d x)^n+\frac {2 (d x)^{2+n}}{d^2}+\frac {(d x)^{4+n}}{d^4}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {(d \tan (a+b x))^{1+n}}{b d (1+n)}+\frac {2 (d \tan (a+b x))^{3+n}}{b d^3 (3+n)}+\frac {(d \tan (a+b x))^{5+n}}{b d^5 (5+n)}\\ \end {align*}

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Mathematica [A]  time = 2.10, size = 101, normalized size = 1.36 \[ \frac {d (d \tan (a+b x))^{n-1} \left (\tan ^2(a+b x) \sec ^4(a+b x) \left (2 (n+3) \cos (2 (a+b x))+\cos (4 (a+b x))+n^2+6 n+8\right )+8 \left (-\tan ^2(a+b x)\right )^{\frac {1-n}{2}}\right )}{b (n+1) (n+3) (n+5)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^6*(d*Tan[a + b*x])^n,x]

[Out]

(d*(d*Tan[a + b*x])^(-1 + n)*((8 + 6*n + n^2 + 2*(3 + n)*Cos[2*(a + b*x)] + Cos[4*(a + b*x)])*Sec[a + b*x]^4*T
an[a + b*x]^2 + 8*(-Tan[a + b*x]^2)^((1 - n)/2)))/(b*(1 + n)*(3 + n)*(5 + n))

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fricas [A]  time = 0.56, size = 85, normalized size = 1.15 \[ \frac {{\left (8 \, \cos \left (b x + a\right )^{4} + 4 \, {\left (n + 1\right )} \cos \left (b x + a\right )^{2} + n^{2} + 4 \, n + 3\right )} \left (\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}\right )^{n} \sin \left (b x + a\right )}{{\left (b n^{3} + 9 \, b n^{2} + 23 \, b n + 15 \, b\right )} \cos \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*(d*tan(b*x+a))^n,x, algorithm="fricas")

[Out]

(8*cos(b*x + a)^4 + 4*(n + 1)*cos(b*x + a)^2 + n^2 + 4*n + 3)*(d*sin(b*x + a)/cos(b*x + a))^n*sin(b*x + a)/((b
*n^3 + 9*b*n^2 + 23*b*n + 15*b)*cos(b*x + a)^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (b x + a\right )\right )^{n} \sec \left (b x + a\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*(d*tan(b*x+a))^n,x, algorithm="giac")

[Out]

integrate((d*tan(b*x + a))^n*sec(b*x + a)^6, x)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{6}\left (b x +a \right )\right ) \left (d \tan \left (b x +a \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^6*(d*tan(b*x+a))^n,x)

[Out]

int(sec(b*x+a)^6*(d*tan(b*x+a))^n,x)

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maxima [A]  time = 0.95, size = 77, normalized size = 1.04 \[ \frac {\frac {d^{n} \tan \left (b x + a\right )^{n} \tan \left (b x + a\right )^{5}}{n + 5} + \frac {2 \, d^{n} \tan \left (b x + a\right )^{n} \tan \left (b x + a\right )^{3}}{n + 3} + \frac {\left (d \tan \left (b x + a\right )\right )^{n + 1}}{d {\left (n + 1\right )}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^6*(d*tan(b*x+a))^n,x, algorithm="maxima")

[Out]

(d^n*tan(b*x + a)^n*tan(b*x + a)^5/(n + 5) + 2*d^n*tan(b*x + a)^n*tan(b*x + a)^3/(n + 3) + (d*tan(b*x + a))^(n
+ 1)/(d*(n + 1)))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^n}{{\cos \left (a+b\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^n/cos(a + b*x)^6,x)

[Out]

int((d*tan(a + b*x))^n/cos(a + b*x)^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan {\left (a + b x \right )}\right )^{n} \sec ^{6}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**6*(d*tan(b*x+a))**n,x)

[Out]

Integral((d*tan(a + b*x))**n*sec(a + b*x)**6, x)

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