\(\int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx\) [220]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=-\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}-\frac {\operatorname {Hypergeometric2F1}(1,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)} \]

[Out]

-3*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)-hypergeom([1, 3+n],[4+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n
)+(a+a*sec(d*x+c))^(4+n)/a^4/d/(4+n)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3965, 90, 67} \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\frac {(a \sec (c+d x)+a)^{n+4}}{a^4 d (n+4)}-\frac {(a \sec (c+d x)+a)^{n+3} \operatorname {Hypergeometric2F1}(1,n+3,n+4,\sec (c+d x)+1)}{a^3 d (n+3)}-\frac {3 (a \sec (c+d x)+a)^{n+3}}{a^3 d (n+3)} \]

[In]

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]

[Out]

(-3*(a + a*Sec[c + d*x])^(3 + n))/(a^3*d*(3 + n)) - (Hypergeometric2F1[1, 3 + n, 4 + n, 1 + Sec[c + d*x]]*(a +
a*Sec[c + d*x])^(3 + n))/(a^3*d*(3 + n)) + (a + a*Sec[c + d*x])^(4 + n)/(a^4*d*(4 + n))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3965

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(d*b^(m - 1)
)^(-1), Subst[Int[(-a + b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(-a+a x)^2 (a+a x)^{2+n}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (-3 a^2 (a+a x)^{2+n}+\frac {a^2 (a+a x)^{2+n}}{x}+a (a+a x)^{3+n}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d} \\ & = -\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)}+\frac {\text {Subst}\left (\int \frac {(a+a x)^{2+n}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = -\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}-\frac {\operatorname {Hypergeometric2F1}(1,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\frac {(1+\sec (c+d x))^3 (a (1+\sec (c+d x)))^n (-9-2 n-(4+n) \operatorname {Hypergeometric2F1}(1,3+n,4+n,1+\sec (c+d x))+(3+n) \sec (c+d x))}{d (3+n) (4+n)} \]

[In]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]

[Out]

((1 + Sec[c + d*x])^3*(a*(1 + Sec[c + d*x]))^n*(-9 - 2*n - (4 + n)*Hypergeometric2F1[1, 3 + n, 4 + n, 1 + Sec[
c + d*x]] + (3 + n)*Sec[c + d*x]))/(d*(3 + n)*(4 + n))

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{5}d x\]

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**5,x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**5, x)

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n, x)