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

Optimal. Leaf size=161 \[ \frac{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{n+1}}{b^5 d (n+1)}-\frac{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^{n+2}}{b^5 d (n+2)}+\frac{2 \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^{n+3}}{b^5 d (n+3)}-\frac{4 a (a+b \tan (c+d x))^{n+4}}{b^5 d (n+4)}+\frac{(a+b \tan (c+d x))^{n+5}}{b^5 d (n+5)} \]

[Out]

((a^2 + b^2)^2*(a + b*Tan[c + d*x])^(1 + n))/(b^5*d*(1 + n)) - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^(2 + n))/
(b^5*d*(2 + n)) + (2*(3*a^2 + b^2)*(a + b*Tan[c + d*x])^(3 + n))/(b^5*d*(3 + n)) - (4*a*(a + b*Tan[c + d*x])^(
4 + n))/(b^5*d*(4 + n)) + (a + b*Tan[c + d*x])^(5 + n)/(b^5*d*(5 + n))

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Rubi [A]  time = 0.123486, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 697} \[ \frac{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{n+1}}{b^5 d (n+1)}-\frac{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^{n+2}}{b^5 d (n+2)}+\frac{2 \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^{n+3}}{b^5 d (n+3)}-\frac{4 a (a+b \tan (c+d x))^{n+4}}{b^5 d (n+4)}+\frac{(a+b \tan (c+d x))^{n+5}}{b^5 d (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + b^2)^2*(a + b*Tan[c + d*x])^(1 + n))/(b^5*d*(1 + n)) - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^(2 + n))/
(b^5*d*(2 + n)) + (2*(3*a^2 + b^2)*(a + b*Tan[c + d*x])^(3 + n))/(b^5*d*(3 + n)) - (4*a*(a + b*Tan[c + d*x])^(
4 + n))/(b^5*d*(4 + n)) + (a + b*Tan[c + d*x])^(5 + n)/(b^5*d*(5 + n))

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 2.96177, size = 161, normalized size = 1. \[ \frac{(a+b \tan (c+d x))^{n+1} \left (4 \left (a^2+b^2\right ) \left (\frac{a^2+b^2}{n+1}+\frac{(a+b \tan (c+d x))^2}{n+3}-\frac{2 a (a+b \tan (c+d x))}{n+2}\right )-4 a (a+b \tan (c+d x)) \left (\frac{a^2+b^2}{n+2}+\frac{(a+b \tan (c+d x))^2}{n+4}-\frac{2 a (a+b \tan (c+d x))}{n+3}\right )+b^4 \sec ^4(c+d x)\right )}{b^5 d (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Tan[c + d*x])^(1 + n)*(b^4*Sec[c + d*x]^4 + 4*(a^2 + b^2)*((a^2 + b^2)/(1 + n) - (2*a*(a + b*Tan[c + d
*x]))/(2 + n) + (a + b*Tan[c + d*x])^2/(3 + n)) - 4*a*(a + b*Tan[c + d*x])*((a^2 + b^2)/(2 + n) - (2*a*(a + b*
Tan[c + d*x]))/(3 + n) + (a + b*Tan[c + d*x])^2/(4 + n))))/(b^5*d*(5 + n))

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Maple [F]  time = 0.245, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{6} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.44236, size = 899, normalized size = 5.58 \begin{align*} \frac{{\left (8 \,{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} -{\left (a^{3} b^{2} - 3 \, a b^{4}\right )} n^{2} + 3 \,{\left (a^{3} b^{2} + 5 \, a b^{4}\right )} n\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (2 \, a b^{4} n^{3} + 3 \,{\left (a^{3} b^{2} + 3 \, a b^{4}\right )} n^{2} +{\left (3 \, a^{3} b^{2} + 7 \, a b^{4}\right )} n\right )} \cos \left (d x + c\right )^{3} +{\left (a b^{4} n^{4} + 6 \, a b^{4} n^{3} + 11 \, a b^{4} n^{2} + 6 \, a b^{4} n\right )} \cos \left (d x + c\right ) +{\left (b^{5} n^{4} + 10 \, b^{5} n^{3} + 35 \, b^{5} n^{2} + 50 \, b^{5} n + 24 \, b^{5} + 8 \,{\left (8 \, b^{5} -{\left (3 \, a^{2} b^{3} - b^{5}\right )} n^{2} - 3 \,{\left (a^{4} b + 3 \, a^{2} b^{3} - 2 \, b^{5}\right )} n\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (8 \, b^{5} -{\left (a^{2} b^{3} - b^{5}\right )} n^{3} -{\left (3 \, a^{2} b^{3} - 7 \, b^{5}\right )} n^{2} - 2 \,{\left (a^{2} b^{3} - 7 \, b^{5}\right )} n\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \left (\frac{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{n}}{{\left (b^{5} d n^{5} + 15 \, b^{5} d n^{4} + 85 \, b^{5} d n^{3} + 225 \, b^{5} d n^{2} + 274 \, b^{5} d n + 120 \, b^{5} d\right )} \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(3*a^5 + 10*a^3*b^2 + 15*a*b^4 - (a^3*b^2 - 3*a*b^4)*n^2 + 3*(a^3*b^2 + 5*a*b^4)*n)*cos(d*x + c)^5 + 4*(2*a
*b^4*n^3 + 3*(a^3*b^2 + 3*a*b^4)*n^2 + (3*a^3*b^2 + 7*a*b^4)*n)*cos(d*x + c)^3 + (a*b^4*n^4 + 6*a*b^4*n^3 + 11
*a*b^4*n^2 + 6*a*b^4*n)*cos(d*x + c) + (b^5*n^4 + 10*b^5*n^3 + 35*b^5*n^2 + 50*b^5*n + 24*b^5 + 8*(8*b^5 - (3*
a^2*b^3 - b^5)*n^2 - 3*(a^4*b + 3*a^2*b^3 - 2*b^5)*n)*cos(d*x + c)^4 + 4*(8*b^5 - (a^2*b^3 - b^5)*n^3 - (3*a^2
*b^3 - 7*b^5)*n^2 - 2*(a^2*b^3 - 7*b^5)*n)*cos(d*x + c)^2)*sin(d*x + c))*((a*cos(d*x + c) + b*sin(d*x + c))/co
s(d*x + c))^n/((b^5*d*n^5 + 15*b^5*d*n^4 + 85*b^5*d*n^3 + 225*b^5*d*n^2 + 274*b^5*d*n + 120*b^5*d)*cos(d*x + c
)^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(d*x+c)**6*(a+b*tan(d*x+c))**n,x)

[Out]

Timed out

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

[Out]

Exception raised: NotImplementedError