3.3.61 \(\int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx\) [261]

3.3.61.1 Optimal result

Integrand size = 23, antiderivative size = 405 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}+\frac {b^2 e \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^2 d e (1+m) \sqrt {\cos ^2(c+d x)}} \]

output

-2*b*e*AppellF1(1-m,-1/2*m+1/2,-1/2*m+1/2,2-m,(-a+b)/(b+a*cos(d*x+c)),(a+b 
)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c))/(b+a*cos(d*x+c)))^(-1/2*m+1/2)*(a*( 
1+cos(d*x+c))/(b+a*cos(d*x+c)))^(-1/2*m+1/2)*(e*sin(d*x+c))^(-1+m)/a^3/d/( 
1-m)+b^2*e*AppellF1(2-m,-1/2*m+1/2,-1/2*m+1/2,3-m,(-a+b)/(b+a*cos(d*x+c)), 
(a+b)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c))/(b+a*cos(d*x+c)))^(-1/2*m+1/2)* 
(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(-1/2*m+1/2)*(e*sin(d*x+c))^(-1+m)/a^3 
/d/(2-m)/(b+a*cos(d*x+c))+cos(d*x+c)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m 
],sin(d*x+c)^2)*(e*sin(d*x+c))^(1+m)/a^2/d/e/(1+m)/(cos(d*x+c)^2)^(1/2)
 

3.3.61.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1433\) vs. \(2(405)=810\).

Time = 11.66 (sec) , antiderivative size = 1433, normalized size of antiderivative = 3.54 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \]

input

Integrate[(e*Sin[c + d*x])^m/(a + b*Sec[c + d*x])^2,x]
 

output

(-4*b*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*T 
an[(c + d*x)/2]^2)/(a + b)]*(b + a*Cos[c + d*x])*Sec[c + d*x]^2*(e*Sin[c + 
 d*x])^m*Tan[(c + d*x)/2])/(a^2*d*(a + b*Sec[c + d*x])^2*(AppellF1[(1 + m) 
/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a 
+ b)]*Sec[(c + d*x)/2]^2 + 2*m*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[( 
c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Cot[c + d*x]*Tan[(c + 
 d*x)/2] + 2*m*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ( 
(a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Tan[(c + d*x)/2]^2 - (2*(1 + m)*((-a 
+ b)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Ta 
n[(c + d*x)/2]^2)/(a + b)] + (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 1, (5 + 
m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + 
 d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a + b)*(3 + m)))) + (2*b^2*((a + b)*Appel 
lF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x 
)/2]^2)/(a + b)] - 2*a*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x) 
/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[c + d*x]^2*(e*Sin[c + d* 
x])^m*Tan[(c + d*x)/2])/(a^2*d*(a + b*Sec[c + d*x])^2*(((a + b)*AppellF1[( 
1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^ 
2)/(a + b)] - 2*a*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2 
, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2 + 2*m*((a + b) 
*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan...
 

3.3.61.3 Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4360, 3042, 3391, 2009}

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.

\(\displaystyle \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3391

\(\displaystyle \int \left (\frac {b^2 (e \sin (c+d x))^m}{a^2 (a \cos (c+d x)+b)^2}-\frac {2 b (e \sin (c+d x))^m}{a^2 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^m}{a^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (2-m) (a \cos (c+d x)+b)}-\frac {2 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{a^2 d e (m+1) \sqrt {\cos ^2(c+d x)}}\)

input

Int[(e*Sin[c + d*x])^m/(a + b*Sec[c + d*x])^2,x]
 

output

(-2*b*e*AppellF1[1 - m, (1 - m)/2, (1 - m)/2, 2 - m, -((a - b)/(b + a*Cos[ 
c + d*x])), (a + b)/(b + a*Cos[c + d*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a 
*Cos[c + d*x])))^((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x])) 
^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^3*d*(1 - m)) + (b^2*e*AppellF1[ 
2 - m, (1 - m)/2, (1 - m)/2, 3 - m, -((a - b)/(b + a*Cos[c + d*x])), (a + 
b)/(b + a*Cos[c + d*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x])))^ 
((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x]))^((1 - m)/2)*(e*S 
in[c + d*x])^(-1 + m))/(a^3*d*(2 - m)*(b + a*Cos[c + d*x])) + (Cos[c + d*x 
]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + 
d*x])^(1 + m))/(a^2*d*e*(1 + m)*Sqrt[Cos[c + d*x]^2])
 

3.3.61.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3391

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (G 
tQ[m, 0] || IntegerQ[n])
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.3.61.4 Maple [F]

\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +b \sec \left (d x +c \right )\right )^{2}}d x\]

input

int((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,x)
 

output

int((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,x)
 

3.3.61.5 Fricas [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

input

integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
 

output

integral((e*sin(d*x + c))^m/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2 
), x)
 

3.3.61.6 Sympy [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

input

integrate((e*sin(d*x+c))**m/(a+b*sec(d*x+c))**2,x)
 

output

Integral((e*sin(c + d*x))**m/(a + b*sec(c + d*x))**2, x)
 

3.3.61.7 Maxima [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

input

integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
 

output

integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^2, x)
 

3.3.61.8 Giac [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

input

integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,x, algorithm="giac")
 

output

integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^2, x)
 

3.3.61.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

input

int((e*sin(c + d*x))^m/(a + b/cos(c + d*x))^2,x)
 

output

int((cos(c + d*x)^2*(e*sin(c + d*x))^m)/(b + a*cos(c + d*x))^2, x)