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

3.2.44.1 Optimal result

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

output

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

3.2.44.2 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 \[ \int (a+a \sec (c+d x))^n (e \sin (c+d x))^m \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^n \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left ((3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))-4 \left ((1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},n,2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3+m}{2},1+n,1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

input

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

output

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

3.2.44.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.52, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4364, 3042, 3365, 152, 152, 150}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^n \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 4364

\(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int (-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^n (e \sin (c+d x))^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )^{-n} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^ndx\)

\(\Big \downarrow \) 3365

\(\displaystyle -\frac {e (-\cos (c+d x))^n (a \cos (c+d x)-a)^{\frac {1-m}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}-n} \int (-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{\frac {m-1}{2}+n} (a \cos (c+d x)-a)^{\frac {m-1}{2}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 152

\(\displaystyle -\frac {e (-\cos (c+d x))^n (a \cos (c+d x)-a)^{\frac {1-m}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac {1}{2} (-m-2 n+1)} (a (-\cos (c+d x))-a)^{\frac {1}{2} (m+2 n-1)+\frac {1-m}{2}-n} \int (-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{\frac {m-1}{2}+n} (a \cos (c+d x)-a)^{\frac {m-1}{2}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 152

\(\displaystyle -\frac {e (1-\cos (c+d x))^{\frac {1-m}{2}} (-\cos (c+d x))^n (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac {1}{2} (-m-2 n+1)} (a (-\cos (c+d x))-a)^{\frac {1}{2} (m+2 n-1)+\frac {1-m}{2}-n} \int (1-\cos (c+d x))^{\frac {m-1}{2}} (-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{\frac {m-1}{2}+n}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {e \cos (c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} (a \sec (c+d x)+a)^n (e \sin (c+d x))^{m-1} (\cos (c+d x)+1)^{\frac {1}{2} (-m-2 n+1)} (a (-\cos (c+d x))-a)^{\frac {1}{2} (m+2 n-1)+\frac {1-m}{2}-n} \operatorname {AppellF1}\left (1-n,\frac {1-m}{2},\frac {1}{2} (-m-2 n+1),2-n,\cos (c+d x),-\cos (c+d x)\right )}{d (1-n)}\)

input

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

output

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

3.2.44.3.1 Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> 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] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3365

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[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 4364

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[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])
 

3.2.44.4 Maple [F]

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

input

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

output

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

3.2.44.5 Fricas [F]

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

input

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

output

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

3.2.44.6 Sympy [F]

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

input

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

output

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

3.2.44.7 Maxima [F]

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

input

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

output

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

3.2.44.8 Giac [F]

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

input

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

output

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

3.2.44.9 Mupad [F(-1)]

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

input

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

output

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