3.216 \(\int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx\)
Optimal. Leaf size=195 \[ \frac {b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \]
[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))
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Rubi [A] time = 0.25, antiderivative size = 195, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used =
{2823, 3189, 429} \[ \frac {b \cos (e+f x) \sin ^2(e+f x)^{-m/2} (d \sin (e+f x))^m F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a d \cos (e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{m-1} F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
[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, -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])^m)/((a^2 - b^2)*f*(Sin[e + f*x]^2)^
(m/2))
Rule 429
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 2823
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 3189
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*} \int \frac {(d \sin (e+f x))^m}{a+b \sin (e+f x)} \, dx &=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 ) \operatorname {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 ) \operatorname {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 F_1\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 F_1\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*}
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Mathematica [B] time = 17.69, size = 1590, normalized size = 8.15 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[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))))
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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maple [F] time = 0.75, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x +e \right )\right )^{m}}{a +b \sin \left (f x +e \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sin \left (f x + e\right )\right )^{m}}{b \sin \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^m}{a+b\,\sin \left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sin(f*x+e))**m/(a+b*sin(f*x+e)),x)
[Out]
Timed out
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