3.408 \(\int \frac {(a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx\)
Optimal. Leaf size=76 \[ \frac {2^{m+\frac {1}{2}} \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{c f} \]
[Out]
2^(1/2+m)*hypergeom([-1/2, 1/2-m],[1/2],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(1+sin(f*x+e))^(1/2-m)*(a+a*sin(f*x+e))
^m/c/f
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Rubi [A] time = 0.13, antiderivative size = 76, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used =
{2736, 2689, 70, 69} \[ \frac {2^{m+\frac {1}{2}} \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{c f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x]),x]
[Out]
(2^(1/2 + m)*Hypergeometric2F1[-1/2, 1/2 - m, 1/2, (1 - Sin[e + f*x])/2]*Sec[e + f*x]*(1 + Sin[e + f*x])^(1/2
- m)*(a + a*Sin[e + f*x])^m)/(c*f)
Rule 69
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))
Rule 70
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] &&
(RationalQ[m] || !SimplerQ[n + 1, m + 1])
Rule 2689
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Rule 2736
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx &=\frac {\int \sec ^2(e+f x) (a+a \sin (e+f x))^{1+m} \, dx}{a c}\\ &=\frac {\left (a \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c f}\\ &=\frac {\left (2^{-\frac {1}{2}+m} a \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^m \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c f}\\ &=\frac {2^{\frac {1}{2}+m} \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-m;\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{c f}\\ \end {align*}
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Mathematica [C] time = 6.39, size = 3844, normalized size = 50.58 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x]),x]
[Out]
-1/2*((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Cot[(-e + Pi/2 - f*x)/4]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*(a +
a*Sin[e + f*x])^m*(-(AppellF1[-1/2, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*
(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*
m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(
-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/
4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2)))/(f*(c - c*Sin[e + f*x])*(Cos[Pi/4 + (e - Pi/
2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^2*(-1/2*(m*(Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(-(AppellF1[-1/2,
-2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)) +
(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*
x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/
2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*T
an[(-e + Pi/2 - f*x)/4]^2))) - ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Csc[(-e + Pi/2 - f*x)/4]^2*(-(AppellF1[-1/2
, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m))
+ (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 -
f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2,
-Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + P
i/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])
*Tan[(-e + Pi/2 - f*x)/4]^2)))/8 + ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Cot[(-e + Pi/2 - f*x)/4]*(-(m*AppellF1[
-1/2, -2*m, 2*m, 1/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2
*m)*Tan[(-e + Pi/2 - f*x)/4]) - (Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(m*AppellF1[1/2, 1 - 2*m, 2*m, 3/2, Tan[(-e
+ Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4] + m*Appe
llF1[1/2, -2*m, 1 + 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4
]^2*Tan[(-e + Pi/2 - f*x)/4]) + (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(2*(3*
AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1
- 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2)) + (3*Tan[(-e + Pi/2 - f
*x)/4]^2*(-1/3*(m*AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Se
c[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) - (m*AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x
)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/3)*(1 - Tan[(-e + Pi
/2 - f*x)/4]^2)^(2*m))/(3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^
2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1
[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2
) - (3*m*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2
- f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]^3*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m))/(3*AppellF1[1/2, -2*m, 2*
m, 3/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(
-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/
4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2) - (3*AppellF1[1/2, -2*m, 2*m, 3/2, Tan[(-e + P
i/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m
)*(-2*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[
3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Sec[(-e + Pi/2 - f*x)/4]^2*
Tan[(-e + Pi/2 - f*x)/4] + 3*(-1/3*(m*AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) - (m*AppellF1[3/2, -2*m, 1 + 2*m, 5/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/
3) - 4*m*Tan[(-e + Pi/2 - f*x)/4]^2*((-6*m*AppellF1[5/2, 1 - 2*m, 1 + 2*m, 7/2, Tan[(-e + Pi/2 - f*x)/4]^2, -T
an[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/5 + (3*(1 - 2*m)*AppellF1[5/2,
2 - 2*m, 2*m, 7/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-
e + Pi/2 - f*x)/4])/10 - (3*(1 + 2*m)*AppellF1[5/2, -2*m, 2 + 2*m, 7/2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/10)))/(3*AppellF1[1/2, -2*m, 2*m, 3/2,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*m*(AppellF1[3/2, 1 - 2*m, 2*m, 5/2, Tan[(-e + Pi
/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[3/2, -2*m, 1 + 2*m, 5/2, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]^2)^2))/2))
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="fricas")
[Out]
integral(-(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="giac")
[Out]
integrate(-(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)
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maple [F] time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{c -c \sin \left (f x +e \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x)
[Out]
int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e)),x, algorithm="maxima")
[Out]
-integrate((a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{c-c\,\sin \left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x)),x)
[Out]
int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\sin {\left (e + f x \right )} - 1}\, dx}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**m/(c-c*sin(f*x+e)),x)
[Out]
-Integral((a*sin(e + f*x) + a)**m/(sin(e + f*x) - 1), x)/c
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