3.407 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=84 \[ -\frac{a^2 c 2^{m+\frac{1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{3}{2},\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f} \]

[Out]

-(2^(1/2 + m)*a^2*c*Cos[e + f*x]^3*Hypergeometric2F1[3/2, 1/2 - m, 5/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x
])^(1/2 - m)*(a + a*Sin[e + f*x])^(-2 + m))/(3*f)

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Rubi [A]  time = 0.112976, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2736, 2689, 70, 69} \[ -\frac{a^2 c 2^{m+\frac{1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{3}{2},\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f} \]

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)*a^2*c*Cos[e + f*x]^3*Hypergeometric2F1[3/2, 1/2 - m, 5/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x
])^(1/2 - m)*(a + a*Sin[e + f*x])^(-2 + m))/(3*f)

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]))

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)) \, dx &=(a c) \int \cos ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \, dx\\ &=\frac{\left (a^3 c \cos ^3(e+f x)\right ) \operatorname{Subst}\left (\int \sqrt{a-a x} (a+a x)^{-\frac{1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}\\ &=\frac{\left (2^{-\frac{1}{2}+m} a^3 c \cos ^3(e+f x) (a+a \sin (e+f x))^{-2+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{1}{2}+m} \sqrt{a-a x} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{3/2}}\\ &=-\frac{2^{\frac{1}{2}+m} a^2 c \cos ^3(e+f x) \, _2F_1\left (\frac{3}{2},\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac{1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{3 f}\\ \end{align*}

Mathematica [C]  time = 1.65244, size = 261, normalized size = 3.11 \[ -\frac{(-1)^{3/4} c 2^{-2 m-1} e^{-\frac{3}{2} i (e+f x)} \left (-(-1)^{3/4} e^{-\frac{1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m+1} (\sin (e+f x)-1) \left ((m-1) m e^{2 i (e+f x)} \, _2F_1\left (1,m;-m;-i e^{-i (e+f x)}\right )-(m+1) \left (2 (m-1) e^{i (e+f x)} \, _2F_1\left (1,m+1;1-m;-i e^{-i (e+f x)}\right )-m \, _2F_1\left (1,m+2;2-m;-i e^{-i (e+f x)}\right )\right )\right ) \sin ^{-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m}{f (m-1) m (m+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]),x]

[Out]

-(((-1)^(3/4)*2^(-1 - 2*m)*c*(-(((-1)^(3/4)*(I + E^(I*(e + f*x))))/E^((I/2)*(e + f*x))))^(1 + 2*m)*(E^((2*I)*(
e + f*x))*(-1 + m)*m*Hypergeometric2F1[1, m, -m, (-I)/E^(I*(e + f*x))] - (1 + m)*(2*E^(I*(e + f*x))*(-1 + m)*H
ypergeometric2F1[1, 1 + m, 1 - m, (-I)/E^(I*(e + f*x))] - m*Hypergeometric2F1[1, 2 + m, 2 - m, (-I)/E^(I*(e +
f*x))]))*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m)/(E^(((3*I)/2)*(e + f*x))*f*(-1 + m)*m*(1 + m)*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(2*e + Pi + 2*f*x)/4]^(2*m)))

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Maple [F]  time = 0.977, 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 ) \, 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)),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., size = 0, normalized size = 0. \begin{align*} -\int{\left (c \sin \left (f x + e\right ) - c\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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)),x, algorithm="maxima")

[Out]

-integrate((c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (c \sin \left (f x + e\right ) - c\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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)),x, algorithm="fricas")

[Out]

integral(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - c \left (\int \left (a \sin{\left (e + f x \right )} + a\right )^{m} \sin{\left (e + f x \right )}\, dx + \int - \left (a \sin{\left (e + f x \right )} + a\right )^{m}\, dx\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)),x)

[Out]

-c*(Integral((a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(-(a*sin(e + f*x) + a)**m, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (c \sin \left (f x + e\right ) - c\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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)),x, algorithm="giac")

[Out]

integrate(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)