\(\int (a+b \sec (c+d x))^n \tan (c+d x) \, dx\) [355]

Optimal result

Integrand size = 19, antiderivative size = 48 \[ \int (a+b \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3970, 67} \[ \int (a+b \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

[In]

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

[Out]

-((Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a*d*(1 + 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 3970

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int (a+b \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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