3.424 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx\)
Optimal. Leaf size=114 \[ \frac{c^3 2^{\frac{5}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m-3),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
[Out]
(2^(5/2 - m)*c^3*Cos[e + f*x]*Hypergeometric2F1[(-3 + 2*m)/2, (1 + 2*m)/2, (3 + 2*m)/2, (1 + Sin[e + f*x])/2]*
(1 - Sin[e + f*x])^(1/2 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(f*(1 + 2*m))
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Rubi [A] time = 0.182276, antiderivative size = 114, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{2745, 2689, 70, 69} \[ \frac{c^3 2^{\frac{5}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac{1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac{1}{2} (2 m-3),\frac{1}{2} (2 m+1);\frac{1}{2} (2 m+3);\frac{1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(2 - m),x]
[Out]
(2^(5/2 - m)*c^3*Cos[e + f*x]*Hypergeometric2F1[(-3 + 2*m)/2, (1 + 2*m)/2, (3 + 2*m)/2, (1 + Sin[e + f*x])/2]*
(1 - Sin[e + f*x])^(1/2 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(f*(1 + 2*m))
Rule 2745
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^FracPart[m])/Cos[e + f*x]^(2
*FracPart[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, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
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 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 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]))
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx &=\left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{2-2 m} \, dx\\ &=\frac{\left (c^2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac{1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int (c-c x)^{2-2 m+\frac{1}{2} (-1+2 m)} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (2^{\frac{3}{2}-m} c^4 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac{1}{2}+\frac{1}{2} (-1-2 m)} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}+m} (c+c \sin (e+f x))^{\frac{1}{2} (-1-2 m)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{2-2 m+\frac{1}{2} (-1+2 m)} (c+c x)^{\frac{1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{2^{\frac{5}{2}-m} c^3 \cos (e+f x) \, _2F_1\left (\frac{1}{2} (-3+2 m),\frac{1}{2} (1+2 m);\frac{1}{2} (3+2 m);\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}\\ \end{align*}
Mathematica [C] time = 12.7401, size = 1201, normalized size = 10.54 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(2 - m),x]
[Out]
(2^(4 - m)*(-3 + 2*m)*(AppellF1[1/2 - m, -2*m, 3, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/
4]^2] - 2*AppellF1[1/2 - m, -2*m, 4, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + Appel
lF1[1/2 - m, -2*m, 5, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Cos[(-e + Pi/2 - f*x)
/4]*Sin[(-e + Pi/2 - f*x)/4]*Sin[(-e + Pi/2 - f*x)/2]^(4 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(2
- m))/(f*(-1 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(2 - m))*((-3 + 2*m)*AppellF1[1/2 - m, -2*m, 3,
3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (6 - 4*m)*AppellF1[1/2 - m, -2*m, 4, 3/2 -
m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 3*AppellF1[1/2 - m, -2*m, 5, 3/2 - m, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 2*m*AppellF1[1/2 - m, -2*m, 5, 3/2 - m, Tan[(-e + Pi/2 - f*x
)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 8*AppellF1[3/2 - m, -2*m, 5, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[
(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2 + 8*AppellF1[3/2 - m, -2*m, 5, 5/2 - m, Tan[(-e + Pi/2 - f*
x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cos[(-e + Pi/2 - f*x)/2]*Sec[(-e + Pi/2 - f*x)/4]^2 + 4*m*AppellF1[3/2 -
m, 1 - 2*m, 3, 5/2 - m, 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*m*AppellF1[3/2 - m, 1 - 2*m, 4, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e +
Pi/2 - f*x)/4]^2 + 4*m*AppellF1[3/2 - m, 1 - 2*m, 5, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f
*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 + 6*AppellF1[3/2 - m, -2*m, 4, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[
(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 + 10*AppellF1[3/2 - m, -2*m, 6, 5/2 - m, Tan[(-e + Pi/2 - f
*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2))
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Maple [F] time = 0.289, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{2-m}\, dx \end{align*}
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))^(2-m),x)
[Out]
int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(2-m),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2}\,{d x} \end{align*}
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))^(2-m),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m + 2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2}, x\right ) \end{align*}
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))^(2-m),x, algorithm="fricas")
[Out]
integral((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m + 2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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))**(2-m),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2}\,{d x} \end{align*}
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))^(2-m),x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m + 2), x)