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

Optimal. Leaf size=97 \[ -\frac{(a \sec (c+d x)+a)^{n+3} \text{Hypergeometric2F1}(1,n+3,n+4,\sec (c+d x)+1)}{a^3 d (n+3)}-\frac{3 (a \sec (c+d x)+a)^{n+3}}{a^3 d (n+3)}+\frac{(a \sec (c+d x)+a)^{n+4}}{a^4 d (n+4)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0817055, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3880, 88, 65} \[ -\frac{(a \sec (c+d x)+a)^{n+3} \, _2F_1(1,n+3;n+4;\sec (c+d x)+1)}{a^3 d (n+3)}-\frac{3 (a \sec (c+d x)+a)^{n+3}}{a^3 d (n+3)}+\frac{(a \sec (c+d x)+a)^{n+4}}{a^4 d (n+4)} \]

Antiderivative was successfully verified.

[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 3880

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]

Rule 88

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 65

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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{\, _2F_1(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]  time = 0.144581, size = 72, normalized size = 0.74 \[ \frac{(\sec (c+d x)+1)^3 (a (\sec (c+d x)+1))^n (-(n+4) \text{Hypergeometric2F1}(1,n+3,n+4,\sec (c+d x)+1)+(n+3) \sec (c+d x)-2 n-9)}{d (n+3) (n+4)} \]

Antiderivative was successfully verified.

[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]  time = 0.309, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \tan \left ( dx+c \right ) \right ) ^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)