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

Optimal. Leaf size=139 \[ \frac{a^2 c 2^{m+\frac{1}{2}} (B (1-m)-A (m+2)) \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{3}{2},\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (m+2)}-\frac{a B c \cos ^3(e+f x) (a \sin (e+f x)+a)^{m-1}}{f (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.289108, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2967, 2860, 2689, 70, 69} \[ \frac{a^2 c 2^{m+\frac{1}{2}} (B (1-m)-A (m+2)) \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m-2} \, _2F_1\left (\frac{3}{2},\frac{1}{2}-m;\frac{5}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (m+2)}-\frac{a B c \cos ^3(e+f x) (a \sin (e+f x)+a)^{m-1}}{f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 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 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx &=(a c) \int \cos ^2(e+f x) (a+a \sin (e+f x))^{-1+m} (A+B \sin (e+f x)) \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\left (a c \left (A-\frac{B (1-m)}{2+m}\right )\right ) \int \cos ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\frac{\left (a^3 c \left (A-\frac{B (1-m)}{2+m}\right ) \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{a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}+\frac{\left (2^{-\frac{1}{2}+m} a^3 c \left (A-\frac{B (1-m)}{2+m}\right ) \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 \left (A-\frac{B (1-m)}{2+m}\right ) \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}-\frac{a B c \cos ^3(e+f x) (a+a \sin (e+f x))^{-1+m}}{f (2+m)}\\ \end{align*}

Mathematica [C]  time = 4.20376, size = 462, normalized size = 3.32 \[ \frac{i c 4^{-m-1} e^{i f m x} \left (1+i e^{-i (e+f x)}\right )^{-2 m} \left (-(-1)^{3/4} e^{-\frac{1}{2} i (e+f x)} \left (e^{i (e+f x)}+i\right )\right )^{2 m} (\sin (e+f x)-1) \sin ^{-2 m}\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (\frac{2 (B-i A) e^{-i (e+f (m+1) x)} \, _2F_1\left (-m-1,-2 m;-m;-i e^{-i (e+f x)}\right )}{m+1}+\frac{2 i A e^{i (e-f (m-1) x)} \, _2F_1\left (1-m,-2 m;2-m;-i e^{-i (e+f x)}\right )}{m-1}+\frac{4 A e^{-i f m x} \, _2F_1\left (-2 m,-m;1-m;-i e^{-i (e+f x)}\right )}{m}-\frac{i B e^{-i (2 e+f (m+2) x)} \, _2F_1\left (-m-2,-2 m;-m-1;-i e^{-i (e+f x)}\right )}{m+2}+\frac{2 B e^{i (e-f (m-1) x)} \, _2F_1\left (1-m,-2 m;2-m;-i e^{-i (e+f x)}\right )}{m-1}+\frac{i B e^{2 i e-i f (m-2) x} \, _2F_1\left (2-m,-2 m;3-m;-i e^{-i (e+f x)}\right )}{m-2}\right )}{f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*4^(-1 - m)*c*E^(I*f*m*x)*(-(((-1)^(3/4)*(I + E^(I*(e + f*x))))/E^((I/2)*(e + f*x))))^(2*m)*(((-I)*B*Hyperge
ometric2F1[-2 - m, -2*m, -1 - m, (-I)/E^(I*(e + f*x))])/(E^(I*(2*e + f*(2 + m)*x))*(2 + m)) + (2*((-I)*A + B)*
Hypergeometric2F1[-1 - m, -2*m, -m, (-I)/E^(I*(e + f*x))])/(E^(I*(e + f*(1 + m)*x))*(1 + m)) + ((2*I)*A*E^(I*(
e - f*(-1 + m)*x))*Hypergeometric2F1[1 - m, -2*m, 2 - m, (-I)/E^(I*(e + f*x))])/(-1 + m) + (2*B*E^(I*(e - f*(-
1 + m)*x))*Hypergeometric2F1[1 - m, -2*m, 2 - m, (-I)/E^(I*(e + f*x))])/(-1 + m) + (I*B*E^((2*I)*e - I*f*(-2 +
m)*x)*Hypergeometric2F1[2 - m, -2*m, 3 - m, (-I)/E^(I*(e + f*x))])/(-2 + m) + (4*A*Hypergeometric2F1[-2*m, -m
, 1 - m, (-I)/E^(I*(e + f*x))])/(E^(I*f*m*x)*m))*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m)/((1 + I/E^(I*(e
+ f*x)))^(2*m)*f*(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 = 1.549, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \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*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int{\left (B \sin \left (f x + e\right ) + A\right )}{\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*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

-integrate((B*sin(f*x + e) + A)*(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 (B c \cos \left (f x + e\right )^{2} -{\left (A - B\right )} c \sin \left (f x + e\right ) +{\left (A - B\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*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*c*cos(f*x + e)^2 - (A - B)*c*sin(f*x + e) + (A - B)*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 - A \left (a \sin{\left (e + f x \right )} + a\right )^{m}\, dx + \int A \left (a \sin{\left (e + f x \right )} + a\right )^{m} \sin{\left (e + f x \right )}\, dx + \int - B \left (a \sin{\left (e + f x \right )} + a\right )^{m} \sin{\left (e + f x \right )}\, dx + \int B \left (a \sin{\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(Integral(-A*(a*sin(e + f*x) + a)**m, x) + Integral(A*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(-
B*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(B*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2, x))

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

[Out]

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