\(\int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx\) [221]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(a+a \sec (c+d x))^{2+n}}{a^2 d (2+n)}+\frac {\operatorname {Hypergeometric2F1}(1,2+n,3+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{2+n}}{a^2 d (2+n)} \]

[Out]

(a+a*sec(d*x+c))^(2+n)/a^2/d/(2+n)+hypergeom([1, 2+n],[3+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(2+n)/a^2/d/(2+n)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3965, 81, 67} \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(a \sec (c+d x)+a)^{n+2} \operatorname {Hypergeometric2F1}(1,n+2,n+3,\sec (c+d x)+1)}{a^2 d (n+2)}+\frac {(a \sec (c+d x)+a)^{n+2}}{a^2 d (n+2)} \]

[In]

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^3,x]

[Out]

(a + a*Sec[c + d*x])^(2 + n)/(a^2*d*(2 + n)) + (Hypergeometric2F1[1, 2 + n, 3 + n, 1 + Sec[c + d*x]]*(a + a*Se
c[c + d*x])^(2 + n))/(a^2*d*(2 + 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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.71 \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(1+\operatorname {Hypergeometric2F1}(1,2+n,3+n,1+\sec (c+d x))) (1+\sec (c+d x))^2 (a (1+\sec (c+d x)))^n}{d (2+n)} \]

[In]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^3,x]

[Out]

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

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{3}d x\]

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^3,x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^3,x)

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**3,x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**3, x)

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

int(tan(c + d*x)^3*(a + a/cos(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^3*(a + a/cos(c + d*x))^n, x)