\(\int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx\) [216]

Optimal result

Integrand size = 23, antiderivative size = 195 \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=-\frac {a d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^{-1+m} \sin ^2(e+f x)^{\frac {1-m}{2}}}{\left (a^2-b^2\right ) f}+\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \sin (e+f x))^m \sin ^2(e+f x)^{-m/2}}{\left (a^2-b^2\right ) f} \]

[Out]

-a*d*AppellF1(1/2,1/2-1/2*m,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)^(1/2-1/2*m)/(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))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2902, 3268, 440} \[ \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx=\frac {b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-\frac {m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \]

[In]

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

[Out]

-((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*S
in[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))

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 2902

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a, Int[(d*
Sin[e + f*x])^n/(a^2 - b^2*Sin[e + f*x]^2), x], x] - Dist[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]

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}& = 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))^{1+m}}{a^2-b^2 \sin ^2(e+f x)} \, dx}{d} \\ & = -\frac {\left (a 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)}}{a^2-b^2+b^2 x^2} \, dx,x,\cos (e+f x)\right )}{f}+\frac {\left (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}}{a^2-b^2+b^2 x^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\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} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

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=\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 b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )+(1+m) \left (\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),1,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )\right ) \tan (e+f x)\right )}{a^2 b f (1+m) (2+m) (a+b \sin (e+f x)) \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 b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )+(1+m) \left (\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),1,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )\right ) \tan (e+f x)\right )}{a^2 b (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 b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )+(1+m) \left (\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),1,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )\right ) \tan (e+f x)\right )}{a^2 b (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 (a b (2+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {m}{2},1,\frac {3+m}{2},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right )+(1+m) \left (\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),1,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )\right ) \tan (e+f x)\right ) \left (\sqrt {\sec ^2(e+f x)}-\frac {\tan ^2(e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )}{a^2 b (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 ((1+m) \left (\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {1}{2} (-1+m),1,\frac {4+m}{2},-\tan ^2(e+f x),\left (-1+\frac {b^2}{a^2}\right ) \tan ^2(e+f x)\right )-a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )\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},1,1+\frac {3+m}{2},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \sec ^2(e+f x) \tan (e+f x)}{3+m}+\frac {2 \left (-a^2+b^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),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \sec ^2(e+f x) \tan (e+f x)}{a^2 (3+m)}\right )+(1+m) \tan (e+f x) \left (\left (a^2-b^2\right ) \left (-\frac {(-1+m) (2+m) \operatorname {AppellF1}\left (1+\frac {2+m}{2},1+\frac {1}{2} (-1+m),1,1+\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 {2 \left (-1+\frac {b^2}{a^2}\right ) (2+m) \operatorname {AppellF1}\left (1+\frac {2+m}{2},\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}\right )-a^2 (2+m) \csc (e+f x) \sec (e+f x) \left (-\operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right )+\left (1+\tan ^2(e+f x)\right )^{\frac {1}{2} (-1-m)}\right )\right )\right )}{a^2 b (1+m) (2+m)}\right )} \]

[In]

Integrate[(d*Sin[e + f*x])^m/(a + b*Sin[e + f*x]),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*b*(2 + m)*App
ellF1[(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*Hypergeom
etric2F1[(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, -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]*(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 - 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])*(Sqrt[Sec[e + f*x]^2] - Tan[e + f*x]^2/Sqrt[Sec[e + f*x]^2]))/(a^2*b*(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*((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])*Sec[e + f*x]^2 + a*b*(2 + m)*(-((m*(1 + m)*AppellF1[1 + (1 + m)/2, 1 +
m/2, 1, 1 + (3 + m)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3 + m
)) + (2*(-a^2 + b^2)*(1 + m)*AppellF1[1 + (1 + m)/2, m/2, 2, 1 + (3 + m)/2, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan
[e + f*x]^2)/a^2]*Sec[e + f*x]^2*Tan[e + f*x])/(a^2*(3 + m))) + (1 + m)*Tan[e + f*x]*((a^2 - b^2)*(-(((-1 + m)
*(2 + m)*AppellF1[1 + (2 + m)/2, 1 + (-1 + m)/2, 1, 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)) + (2*(-1 + b^2/a^2)*(2 + m)*AppellF1[1 + (2 + m)/2, (-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)) - a^2*(2
+ m)*Csc[e + f*x]*Sec[e + f*x]*(-Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (4 + m)/2, -Tan[e + f*x]^2] + (1 + Ta
n[e + f*x]^2)^((-1 - m)/2)))))/(a^2*b*(1 + m)*(2 + m))))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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