3.2.98 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx\) [198]

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

[Out]

1/3*2^(1/2+m)*a^2*c*(B*(1-m)-A*(2+m))*cos(f*x+e)^3*hypergeom([3/2, 1/2-m],[5/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x
+e))^(1/2-m)*(a+a*sin(f*x+e))^(-2+m)/f/(2+m)-a*B*c*cos(f*x+e)^3*(a+a*sin(f*x+e))^(-1+m)/f/(2+m)

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Rubi [A]
time = 0.20, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 = {3046, 2939, 2768, 72, 71} \begin {gather*} \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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2768

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 2939

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] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 3046

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]))

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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.66, size = 462, normalized size = 3.32 \begin {gather*} \frac {i 4^{-1-m} c 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 (i+e^{i (e+f x)}\right )\right )^{2 m} \left (-\frac {i B e^{-i (2 e+f (2+m) x)} \, _2F_1\left (-2-m,-2 m;-1-m;-i e^{-i (e+f x)}\right )}{2+m}+\frac {2 (-i A+B) e^{-i (e+f (1+m) x)} \, _2F_1\left (-1-m,-2 m;-m;-i e^{-i (e+f x)}\right )}{1+m}+\frac {2 i A e^{i (e-f (-1+m) x)} \, _2F_1\left (1-m,-2 m;2-m;-i e^{-i (e+f x)}\right )}{-1+m}+\frac {2 B e^{i (e-f (-1+m) x)} \, _2F_1\left (1-m,-2 m;2-m;-i e^{-i (e+f x)}\right )}{-1+m}+\frac {i B e^{2 i e-i f (-2+m) x} \, _2F_1\left (2-m,-2 m;3-m;-i e^{-i (e+f x)}\right )}{-2+m}+\frac {4 A e^{-i f m x} \, _2F_1\left (-2 m,-m;1-m;-i e^{-i (e+f x)}\right )}{m}\right ) (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^m \sin ^{-2 m}\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \end {gather*}

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 = 1.31, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - c \left (\int \left (- A \left (a \sin {\left (e + f x \right )} + a\right )^{m}\right )\, dx + \int A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c-c\,\sin \left (e+f\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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