\(\int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx\) [217]

Optimal result

Integrand size = 23, antiderivative size = 306 \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=-\frac {b^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1-m),2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{1+m} \sin ^2(e+f x)^{\frac {1}{2} (-1-m)}}{\left (a^2-b^2\right )^2 d f}-\frac {a^2 d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{\left (a^2-b^2\right )^2 f}+\frac {2 a b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right )^2 f} \]

[Out]

-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(1/2,1/2-1/2*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/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))

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2903, 3268, 440, 16} \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=-\frac {b^2 \cos (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (-m-1)} (d \sin (e+f x))^{m+1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-m-1),2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{d f \left (a^2-b^2\right )^2}-\frac {a^2 d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac {2 a b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},2,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

[In]

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

[Out]

-((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*Sin[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))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 2903

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(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]

Rule 3268

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]}, Dist[(-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]

Rubi steps \begin{align*} \text {integral}& = \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 (-a^2+b^2 \sin ^2(e+f x)\right )^2}\right ) \, dx \\ & = a^2 \int \frac {(d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx-(2 a b) \int \frac {\sin (e+f x) (d \sin (e+f x))^m}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx+b^2 \int \frac {\sin ^2(e+f x) (d \sin (e+f x))^m}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx \\ & = \frac {b^2 \int \frac {(d \sin (e+f x))^{2+m}}{\left (-a^2+b^2 \sin ^2(e+f x)\right )^2} \, dx}{d^2}-\frac {(2 a b) \int \frac {(d \sin (e+f x))^{1+m}}{\left (a^2-b^2 \sin ^2(e+f x)\right )^2} \, dx}{d}-\frac {\left (a^2 d (d \sin (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \sin ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+m)}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{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 {\left (b^2 (d \sin (e+f x))^{2 \left (\frac {1}{2}+\frac {m}{2}\right )} \sin ^2(e+f x)^{-\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1+m}{2}}}{\left (-a^2+b^2-b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{d f}+\frac {\left (2 a b (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{m/2}}{\left (a^2-b^2+b^2 x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\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} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

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=-\frac {\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^m \left (-a \left (a^2+b^2\right ) (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+2 b \left (a b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},2,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\left (a^2-b^2\right ) (1+m) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),2,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^3 \left (a^2-b^2\right ) f (1+m) (2+m) (a+b \sin (e+f x))^2 \left (-\frac {\sec ^2(e+f x)^{1+\frac {m}{2}} \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^m \left (-a \left (a^2+b^2\right ) (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+2 b \left (a b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},2,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\left (a^2-b^2\right ) (1+m) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),2,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^3 \left (a^2-b^2\right ) (1+m) (2+m)}-\frac {m \sec ^2(e+f x)^{m/2} \tan ^2(e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^m \left (-a \left (a^2+b^2\right ) (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+2 b \left (a b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},2,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\left (a^2-b^2\right ) (1+m) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),2,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^3 \left (a^2-b^2\right ) (1+m) (2+m)}-\frac {m \sec ^2(e+f x)^{m/2} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{-1+m} \left (\sqrt {\sec ^2(e+f x)}-\frac {\tan ^2(e+f x)}{\sqrt {\sec ^2(e+f x)}}\right ) \left (-a \left (a^2+b^2\right ) (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+2 b \left (a b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},2,\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )+\left (a^2-b^2\right ) (1+m) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),2,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{a^3 \left (a^2-b^2\right ) (1+m) (2+m)}-\frac {\sec ^2(e+f x)^{m/2} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^m \left (-a \left (a^2+b^2\right ) (2+m) \left (-\frac {m (1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},1+\frac {m}{2},1,1+\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+m}+\frac {2 \left (-1+\frac {b^2}{a^2}\right ) (1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},\frac {m}{2},2,1+\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+m}\right )+2 b \left (\left (a^2-b^2\right ) (1+m) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),2,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+a b (2+m) \left (-\frac {m (1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},1+\frac {m}{2},2,1+\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+m}+\frac {4 \left (-1+\frac {b^2}{a^2}\right ) (1+m) \operatorname {AppellF1}\left (1+\frac {1+m}{2},\frac {m}{2},3,1+\frac {3+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+m}\right )+\left (a^2-b^2\right ) (1+m) \tan (e+f x) \left (-\frac {(-1+m) (2+m) \operatorname {AppellF1}\left (1+\frac {2+m}{2},1+\frac {1}{2} (-1+m),2,1+\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{4+m}+\frac {4 \left (-1+\frac {b^2}{a^2}\right ) (2+m) \operatorname {AppellF1}\left (1+\frac {2+m}{2},\frac {1}{2} (-1+m),3,1+\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{4+m}\right )\right )\right )}{a^3 \left (a^2-b^2\right ) (1+m) (2+m)}\right )} \]

[In]

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

[Out]

-(((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*(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)*Ta
n[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*x]/Sqrt[Sec[e + f*x]^2])^(-1 + m)*(Sqrt[Sec[e + f*x]^2] - Tan[
e + f*x]^2/Sqrt[Sec[e + f*x]^2])*(-(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)) - ((Sec[e + f*x]^2)^(m
/2)*Tan[e + f*x]*(Tan[e + f*x]/Sqrt[Sec[e + f*x]^2])^m*(-(a*(a^2 + b^2)*(2 + m)*(-((m*(1 + m)*AppellF1[1 + (1
+ m)/2, 1 + m/2, 1, 1 + (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x]
)/(3 + m)) + (2*(-1 + b^2/a^2)*(1 + m)*AppellF1[1 + (1 + m)/2, m/2, 2, 1 + (3 + m)/2, -Tan[e + f*x]^2, (-1 + b
^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + m))) + 2*b*((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]*Sec[e + f*x]^2 + a*b*(2 + m)*(-((m*(
1 + m)*AppellF1[1 + (1 + m)/2, 1 + m/2, 2, 1 + (3 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[
e + f*x]^2*Tan[e + f*x])/(3 + m)) + (4*(-1 + b^2/a^2)*(1 + m)*AppellF1[1 + (1 + m)/2, m/2, 3, 1 + (3 + m)/2, -
Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + m)) + (a^2 - b^2)*(1 + m)*Tan
[e + f*x]*(-(((-1 + m)*(2 + m)*AppellF1[1 + (2 + m)/2, 1 + (-1 + m)/2, 2, 1 + (4 + m)/2, -Tan[e + f*x]^2, (-1
+ b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(4 + m)) + (4*(-1 + b^2/a^2)*(2 + m)*AppellF1[1 + (2 +
m)/2, (-1 + m)/2, 3, 1 + (4 + m)/2, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*
x])/(4 + m)))))/(a^3*(a^2 - b^2)*(1 + m)*(2 + m)))))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(d*sin(f*x + e))^m/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sin (e+f x))^m}{(a+b \sin (e+f x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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