\(\int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx\) [144]

Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx=-\frac {e \operatorname {AppellF1}\left (1-n,\frac {1-m}{2},\frac {1}{2} (1-m-2 n),2-n,\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{\frac {1}{2} (1-m-2 n)} (a+a \sec (c+d x))^n (e \sin (c+d x))^{-1+m}}{d (1-n)} \]

[Out]

-e*AppellF1(1-n,1/2-1/2*m-n,1/2-1/2*m,2-n,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x+c))^(1/2-1/2*m)*cos(d*x+c)*(1+cos
(d*x+c))^(1/2-1/2*m-n)*(a+a*sec(d*x+c))^n*(e*sin(d*x+c))^(-1+m)/d/(1-n)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3961, 2965, 140, 138} \[ \int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx=-\frac {e \cos (c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac {1}{2} (-m-2 n+1)} \operatorname {AppellF1}\left (1-n,\frac {1-m}{2},\frac {1}{2} (-m-2 n+1),2-n,\cos (c+d x),-\cos (c+d x)\right )}{d (1-n)} \]

[In]

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

[Out]

-((e*AppellF1[1 - n, (1 - m)/2, (1 - m - 2*n)/2, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 -
m)/2)*Cos[c + d*x]*(1 + Cos[c + d*x])^((1 - m - 2*n)/2)*(a + a*Sec[c + d*x])^n*(e*Sin[c + d*x])^(-1 + m))/(d*(
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 3961

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Sin[e
+ f*x]^FracPart[m]*((a + b*Csc[e + f*x])^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]), Int[(g*Cos[e + f*x])^
p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(276\) vs. \(2(130)=260\).

Time = 2.06 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.12 \[ \int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^n \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left ((3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))-4 \left ((1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},n,2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3+m}{2},1+n,1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(4*(3 + m)*AppellF1[(1 + m)/2, n, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^
3*(a*(1 + Sec[c + d*x]))^n*Sin[(c + d*x)/2]*(e*Sin[c + d*x])^m)/(d*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, n, 1 +
m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]) - 4*((1 + m)*AppellF1[(3 + m)/2, n,
2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[(3 + m)/2, 1 + n, 1 + m, (5 + m)/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Sin[(c + d*x)/2]^2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**n*(e*sin(c + d*x))**m, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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