Integrand size = 23, antiderivative size = 306 \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=-\frac {b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-m),2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{1+m} \sin ^2(e+f x)^{\frac {1}{2} (-1-m)}}{\left (a^2-b^2\right )^2 d f}-\frac {a^2 d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{\left (a^2-b^2\right )^2 f}+\frac {2 a b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right )^2 f} \]
-b^2*AppellF1(1/2,-1/2-1/2*m,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2 ))*cos(f*x+e)*(d*sin(f*x+e))^(1+m)*(sin(f*x+e)^2)^(-1/2-1/2*m)/(a^2-b^2)^2 /d/f-a^2*d*AppellF1(1/2,-1/2*m+1/2,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a ^2-b^2))*cos(f*x+e)*(d*sin(f*x+e))^(-1+m)*(sin(f*x+e)^2)^(-1/2*m+1/2)/(a^2 -b^2)^2/f+2*a*b*AppellF1(1/2,-1/2*m,2,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/( a^2-b^2))*cos(f*x+e)*(d*sin(f*x+e))^m/(a^2-b^2)^2/f/((sin(f*x+e)^2)^(1/2*m ))
Leaf count is larger than twice the leaf count of optimal. \(1790\) vs. \(2(306)=612\).
Time = 18.66 (sec) , antiderivative size = 1790, normalized size of antiderivative = 5.85 \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx =\text {Too large to display} \]
-(((Sec[e + f*x]^2)^(m/2)*(d*Sin[e + f*x])^m*Tan[e + f*x]*(Tan[e + f*x]/Sq rt[Sec[e + f*x]^2])^m*(-(a*(a^2 + b^2)*(2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]) + 2*b*(a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 2, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^ 2)*Tan[e + f*x]^2] + (a^2 - b^2)*(1 + m)*AppellF1[(2 + m)/2, (-1 + m)/2, 2 , (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]) ))/(a^3*(a^2 - b^2)*f*(1 + m)*(2 + m)*(a + b*Sin[e + f*x])^2*(-(((Sec[e + f*x]^2)^(1 + m/2)*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^m*(-(a*(a^2 + b^2)*( 2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a ^2)*Tan[e + f*x]^2]) + 2*b*(a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 2, (3 + m )/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] + (a^2 - b^2)*(1 + m) *AppellF1[(2 + m)/2, (-1 + m)/2, 2, (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/ a^2)*Tan[e + f*x]^2]*Tan[e + f*x])))/(a^3*(a^2 - b^2)*(1 + m)*(2 + m))) - (m*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]^2*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2 ])^m*(-(a*(a^2 + b^2)*(2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -Tan[ e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]) + 2*b*(a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 2, (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2 ] + (a^2 - b^2)*(1 + m)*AppellF1[(2 + m)/2, (-1 + m)/2, 2, (4 + m)/2, -Tan [e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x])))/(a^3*(a^2 - b^ 2)*(1 + m)*(2 + m)) - (m*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]*(Tan[e + f...
Time = 0.69 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3303, 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.
\(\displaystyle \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2}dx\) |
\(\Big \downarrow \) 3303 |
\(\displaystyle \int \left (\frac {a^2 (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2}-\frac {2 a b \sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2}+\frac {b^2 \sin ^2(e+f x) (d \sin (e+f x))^m}{\left (b^2 \sin ^2(e+f x)-a^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-m-1)} (d \sin (e+f x))^{m+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-m-1),2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^2}-\frac {a^2 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac {2 a b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}\) |
-((b^2*AppellF1[1/2, (-1 - m)/2, 2, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f* x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + m)*(Sin[e + f*x]^2) ^((-1 - m)/2))/((a^2 - b^2)^2*d*f)) - (a^2*d*AppellF1[1/2, (1 - m)/2, 2, 3 /2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/(a^2 - b^2))]*Cos[e + f*x]*(d*S in[e + f*x])^(-1 + m)*(Sin[e + f*x]^2)^((1 - m)/2))/((a^2 - b^2)^2*f) + (2 *a*b*AppellF1[1/2, -1/2*m, 2, 3/2, Cos[e + f*x]^2, -((b^2*Cos[e + f*x]^2)/ (a^2 - b^2))]*Cos[e + f*x]*(d*Sin[e + f*x])^m)/((a^2 - b^2)^2*f*(Sin[e + f *x]^2)^(m/2))
3.3.17.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(1/((a - b*sin[ e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m)), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]
\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{m}}{\left (a +b \sin \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]