\(\int (a+b \sec (c+d x))^n \sin (c+d x) \, dx\) [270]

Optimal result

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

[Out]

b*hypergeom([2, 1+n],[2+n],1+b*sec(d*x+c)/a)*(a+b*sec(d*x+c))^(1+n)/a^2/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 = {3959, 67} \[ \int (a+b \sec (c+d x))^n \sin (c+d x) \, dx=\frac {b (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {b \sec (c+d x)}{a}+1\right )}{a^2 d (n+1)} \]

[In]

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

[Out]

(b*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^2*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 3959

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[-f^(-1), Subs
t[Int[(-1 + x)^((p - 1)/2)*(1 + x)^((p - 1)/2)*((a + b*x)^m/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[{a,
b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.50 \[ \int (a+b \sec (c+d x))^n \sin (c+d x) \, dx=\frac {b \cos (c+d x) \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right ) (a+b \sec (c+d x))^n}{d (-1+n) (b+a \cos (c+d x))} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52 \[ \int (a+b \sec (c+d x))^n \sin (c+d x) \, dx=\frac {\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{{}}_2{\mathrm {F}}_1\left (1-n,-n;\ 2-n;\ -\frac {a\,\cos \left (c+d\,x\right )}{b}\right )}{d\,{\left (\frac {a\,\cos \left (c+d\,x\right )}{b}+1\right )}^n\,\left (n-1\right )} \]

[In]

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

[Out]

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