Integrand size = 23, antiderivative size = 195 \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=-\frac {a d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{\left (a^2-b^2\right ) f}+\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right ) f} \]
-a*d*AppellF1(1/2,-1/2*m+1/2,1,3/2,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)/ f+b*AppellF1(1/2,-1/2*m,1,3/2,cos(f*x+e)^2,-b^2*cos(f*x+e)^2/(a^2-b^2))*co s(f*x+e)*(d*sin(f*x+e))^m/(a^2-b^2)/f/((sin(f*x+e)^2)^(1/2*m))
Leaf count is larger than twice the leaf count of optimal. \(1590\) vs. \(2(195)=390\).
Time = 16.44 (sec) , antiderivative size = 1590, normalized size of antiderivative = 8.15 \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \]
((Sec[e + f*x]^2)^(m/2)*(d*Sin[e + f*x])^m*Tan[e + f*x]*(Tan[e + f*x]/Sqrt [Sec[e + f*x]^2])^m*(a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -T an[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (1 + m)*((a^2 - b^2)*A ppellF1[(2 + m)/2, (-1 + m)/2, 1, (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^ 2)*Tan[e + f*x]^2] - a^2*Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (4 + m)/2 , -Tan[e + f*x]^2])*Tan[e + f*x]))/(a^2*b*f*(1 + m)*(2 + m)*(a + b*Sin[e + f*x])*(((Sec[e + f*x]^2)^(1 + m/2)*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^m* (a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -Tan[e + f*x]^2, ((-a^ 2 + b^2)*Tan[e + f*x]^2)/a^2] + (1 + m)*((a^2 - b^2)*AppellF1[(2 + m)/2, ( -1 + m)/2, 1, (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - a^2*Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (4 + m)/2, -Tan[e + f*x]^2])* Tan[e + f*x]))/(a^2*b*(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*b*(2 + m)*AppellF1[(1 + m )/2, m/2, 1, (3 + m)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2 ] + (1 + m)*((a^2 - b^2)*AppellF1[(2 + m)/2, (-1 + m)/2, 1, (4 + m)/2, -Ta n[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] - a^2*Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (4 + m)/2, -Tan[e + f*x]^2])*Tan[e + f*x]))/(a^2*b*(1 + m )*(2 + m)) + (m*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec [e + f*x]^2])^(-1 + m)*(a*b*(2 + m)*AppellF1[(1 + m)/2, m/2, 1, (3 + m)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2] + (1 + m)*((a^2 - ...
Time = 0.52 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3302, 3042, 3668, 333}
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.
\(\displaystyle \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)}dx\) |
\(\Big \downarrow \) 3302 |
\(\displaystyle a \int \frac {(d \sin (e+f x))^m}{a^2-b^2 \sin ^2(e+f x)}dx-\frac {b \int \frac {(d \sin (e+f x))^{m+1}}{a^2-b^2 \sin ^2(e+f x)}dx}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {(d \sin (e+f x))^m}{a^2-b^2 \sin (e+f x)^2}dx-\frac {b \int \frac {(d \sin (e+f x))^{m+1}}{a^2-b^2 \sin (e+f x)^2}dx}{d}\) |
\(\Big \downarrow \) 3668 |
\(\displaystyle \frac {b \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \int \frac {\left (1-\cos ^2(e+f x)\right )^{m/2}}{a^2-b^2+b^2 \cos ^2(e+f x)}d\cos (e+f x)}{f}-\frac {a d \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \int \frac {\left (1-\cos ^2(e+f x)\right )^{\frac {m-1}{2}}}{a^2-b^2+b^2 \cos ^2(e+f x)}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}\) |
-((a*d*AppellF1[1/2, (1 - m)/2, 1, 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)*f)) + (b*AppellF1[1/2, -1/2*m, 1, 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)*f*(Sin[e + f*x]^2)^(m/2))
3.3.16.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[a Int[(d*Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x] ^2), x], x] - Simp[b/d Int[(d*Sin[e + f*x])^(n + 1)/(a^2 - b^2*Sin[e + f* x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( -ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) /(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2])) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && !IntegerQ[m]
\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{m}}{a +b \sin \left (f x +e \right )}d x\]
\[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^m}{a+b\,\sin \left (e+f\,x\right )} \,d x \]