\(\int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx\) [260]

Optimal result

Integrand size = 23, antiderivative size = 232 \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx=-\frac {b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a d e (1+m) \sqrt {\cos ^2(c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(687\) vs. \(2(232)=464\).

Time = 3.99 (sec) , antiderivative size = 687, normalized size of antiderivative = 2.96 \[ \int \frac {(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(-(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)]) + (a + b)*Hypergeometric2F1[(1 + m)/2, 1 + 
m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/( 
d*(a + b*Sec[c + d*x])*((-(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)]) + (a + b)*Hypergeometr 
ic2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Sec[(c + d*x)/2]^ 
2 + 2*m*Cot[c + d*x]*(-(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)]) + (a + b)*Hypergeometric2 
F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2] + 2 
*m*(-(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)]) + (a + b)*Hypergeometric2F1[(1 + m)/2, 1 + 
m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2 + ((1 + m)*Sec[(c + 
 d*x)/2]^2*(-((a + b)^2*(Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -T 
an[(c + d*x)/2]^2] - (Sec[(c + d*x)/2]^2)^(-1 - m))) + (2*b*((-a + b)*Appe 
llF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d* 
x)/2]^2)/(a + b)] + (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Ta 
n[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Tan[(c + d*x)/2]^ 
2)/(3 + m)))/(a + b)))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4360, 25, 25, 3042, 3346, 3042, 3122, 3182}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 3182

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

Input:

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

Output:

-((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^2*d*(1 - m))) + (Cos[c + d*x]*Hy 
pergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x] 
)^(1 + m))/(a*d*e*(1 + m)*Sqrt[Cos[c + d*x]^2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3182

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p)*((-b)*((1 - Sin[e + f*x])/(a + b*Sin[e + f*x])))^(( 
p - 1)/2)*(b*((1 + Sin[e + f*x])/(a + b*Sin[e + f*x])))^((p - 1)/2)))*Appel 
lF1[-p - m, (1 - p)/2, (1 - p)/2, 1 - p - m, (a + b)/(a + b*Sin[e + f*x]), 
(a - b)/(a + b*Sin[e + f*x])], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^ 
2 - b^2, 0] && ILtQ[m, 0] &&  !IGtQ[m + p + 1, 0]
 

rule 3346

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* 
(x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b   Int 
[(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b   Int[(g*Cos[e + f*x])^p/( 
a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - 
 b^2, 0]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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