\(\int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx\) [911]

Optimal result

Integrand size = 33, antiderivative size = 127 \[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {1+p}{2},\frac {1}{2} (1-2 m+p),2+n,\sin (e+f x),-\sin (e+f x)\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1+p}{2}} (d \sin (e+f x))^{1+n} (1+\sin (e+f x))^{\frac {1}{2} (1-2 m+p)} (a+a \sin (e+f x))^m}{d f (1+n)} \]

[Out]

AppellF1(1+n,1/2-m+1/2*p,1/2+1/2*p,2+n,-sin(f*x+e),sin(f*x+e))*sec(f*x+e)*(g*sec(f*x+e))^p*(1-sin(f*x+e))^(1/2
+1/2*p)*(d*sin(f*x+e))^(1+n)*(1+sin(f*x+e))^(1/2-m+1/2*p)*(a+a*sin(f*x+e))^m/d/f/(1+n)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3005, 2965, 140, 138} \[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\frac {\sec (e+f x) (1-\sin (e+f x))^{\frac {p+1}{2}} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{n+1} (g \sec (e+f x))^p (\sin (e+f x)+1)^{\frac {1}{2} (-2 m+p+1)} \operatorname {AppellF1}\left (n+1,\frac {p+1}{2},\frac {1}{2} (-2 m+p+1),n+2,\sin (e+f x),-\sin (e+f x)\right )}{d f (n+1)} \]

[In]

Int[(g*Sec[e + f*x])^p*(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^m,x]

[Out]

(AppellF1[1 + n, (1 + p)/2, (1 - 2*m + p)/2, 2 + n, Sin[e + f*x], -Sin[e + f*x]]*Sec[e + f*x]*(g*Sec[e + f*x])
^p*(1 - Sin[e + f*x])^((1 + p)/2)*(d*Sin[e + f*x])^(1 + n)*(1 + Sin[e + f*x])^((1 - 2*m + p)/2)*(a + a*Sin[e +
f*x])^m)/(d*f*(1 + n))

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 2965

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[g*((g*Cos[e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2))), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 3005

Int[((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(2*IntPart[p])*(g*Cos[e + f*x])^FracPart[p]*(g*Sec[e + f*x])^FracP
art[p], Int[(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(g*Cos[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e
, f, g, m, n, p}, x] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \left ((g \cos (e+f x))^p (g \sec (e+f x))^p\right ) \int (g \cos (e+f x))^{-p} (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx \\ & = \frac {\left (\sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{\frac {1+p}{2}} (a+a \sin (e+f x))^{\frac {1+p}{2}}\right ) \text {Subst}\left (\int (d x)^n (a-a x)^{\frac {1}{2} (-1-p)} (a+a x)^{m+\frac {1}{2} (-1-p)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1}{2}+\frac {p}{2}} (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+p}{2}} (a+a \sin (e+f x))^{\frac {1+p}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1-p)} (d x)^n (a+a x)^{m+\frac {1}{2} (-1-p)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1}{2}+\frac {p}{2}} (1+\sin (e+f x))^{\frac {1}{2}-m+\frac {p}{2}} (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+p}{2}} (a+a \sin (e+f x))^{-\frac {1}{2}+m-\frac {p}{2}+\frac {1+p}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1-p)} (d x)^n (1+x)^{m+\frac {1}{2} (-1-p)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\operatorname {AppellF1}\left (1+n,\frac {1+p}{2},\frac {1}{2} (1-2 m+p),2+n,\sin (e+f x),-\sin (e+f x)\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1+p}{2}} (d \sin (e+f x))^{1+n} (1+\sin (e+f x))^{\frac {1}{2} (1-2 m+p)} (a+a \sin (e+f x))^m}{d f (1+n)} \\ \end{align*}

Mathematica [F]

\[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx \]

[In]

Integrate[(g*Sec[e + f*x])^p*(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^m,x]

[Out]

Integrate[(g*Sec[e + f*x])^p*(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^m, x]

Maple [F]

\[\int \left (g \sec \left (f x +e \right )\right )^{p} \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

[In]

int((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x)

[Out]

int((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x)

Fricas [F]

\[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x)

Sympy [F(-1)]

Timed out. \[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\text {Timed out} \]

[In]

integrate((g*sec(f*x+e))**p*(d*sin(f*x+e))**n*(a+a*sin(f*x+e))**m,x)

[Out]

Timed out

Maxima [F]

\[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x)

Giac [F]

\[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\int { \left (g \sec \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x)

Mupad [F(-1)]

Timed out. \[ \int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

[In]

int((d*sin(e + f*x))^n*(g/cos(e + f*x))^p*(a + a*sin(e + f*x))^m,x)

[Out]

int((d*sin(e + f*x))^n*(g/cos(e + f*x))^p*(a + a*sin(e + f*x))^m, x)