Integrand size = 23, antiderivative size = 580 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=-\frac {3 b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (1-m)}-\frac {b^3 e \operatorname {AppellF1}\left (3-m,\frac {1-m}{2},\frac {1-m}{2},4-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (3-m) (b+a \cos (c+d x))^2}+\frac {3 b^2 e \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^3 d e (1+m) \sqrt {\cos ^2(c+d x)}} \]
-3*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^4/d/( 1-m)-b^3*e*AppellF1(3-m,-1/2*m+1/2,-1/2*m+1/2,4-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^4 /d/(3-m)/(b+a*cos(d*x+c))^2+3*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*c os(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^4/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^3/d/e/(1+m )/(cos(d*x+c)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2700\) vs. \(2(580)=1160\).
Time = 16.99 (sec) , antiderivative size = 2700, normalized size of antiderivative = 4.66 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\text {Result too large to show} \]
(-6*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])^2*Sec[c + d*x]^3*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^3*d*(a + b*Sec[c + d*x])^3*(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)* Tan[(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)))) + (6*b^2*((a + b)*App ellF1[(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)])*(b + a*Cos[c + d*x])*Sec[c + d*x]^3*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^3*d*(a + b*Sec[c + d*x]) ^3*(((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 + ...
Time = 0.96 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4360, 25, 25, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^m}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(-a \cos (c+d x)-b)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(b+a \cos (c+d x))^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(a \cos (c+d x)+b)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (-\frac {b^3 (e \sin (c+d x))^m}{a^3 (a \cos (c+d x)+b)^3}+\frac {3 b^2 (e \sin (c+d x))^m}{a^3 (a \cos (c+d x)+b)^2}-\frac {3 b (e \sin (c+d x))^m}{a^3 (a \cos (c+d x)+b)}+\frac {(e \sin (c+d x))^m}{a^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^3 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (3-m,\frac {1-m}{2},\frac {1-m}{2},4-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (3-m) (a \cos (c+d x)+b)^2}+\frac {3 b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (2-m) (a \cos (c+d x)+b)}-\frac {3 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{a^3 d e (m+1) \sqrt {\cos ^2(c+d x)}}\) |
(-3*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^4*d*(1 - m)) - (b^3*e*AppellF1[ 3 - m, (1 - m)/2, (1 - m)/2, 4 - 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^4*d*(3 - m)*(b + a*Cos[c + d*x])^2) + (3*b^2*e*A ppellF1[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*Sin[c + d*x])^(-1 + m))/(a^4*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^3*d*e*(1 + m)*Sqrt[Cos[c + d*x]^2])
3.3.62.3.1 Defintions of rubi rules used
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])
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]
\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +b \sec \left (d x +c \right )\right )^{3}}d x\]
\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
integral((e*sin(d*x + c))^m/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]