3.17 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=269 \[ \frac {c 2^{n+\frac {1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1) (m+n+2)}-\frac {(B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1) (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)} \]

[Out]

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

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Rubi [A]  time = 0.75, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3039, 2973, 2745, 2689, 70, 69} \[ \frac {c 2^{n+\frac {1}{2}} \cos (e+f x) ((m+n+1) (A (m+n+2)+C (-m+n+1))+(m-n) (B (m+n+2)+2 C m+C)) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1) (m+n+2)}-\frac {(B (m+n+2)+2 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1) (m+n+2)}+\frac {C \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+1}}{c f (m+n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

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

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 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 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 3039

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*
sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (b*B*d*(m + n + 2) - b*c*C*(2*m
+ 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}-\frac {\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n (-a c (C (1-m+n)+A (2+m+n))-a c (C+2 C m+B (2+m+n)) \sin (e+f x)) \, dx}{a c (2+m+n)}\\ &=-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac {((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx}{(1+m+n) (2+m+n)}\\ &=-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac {\left (((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \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))^{-m+n} \, dx}{(1+m+n) (2+m+n)}\\ &=-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac {\left (c^2 ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \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)^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f (1+m+n) (2+m+n)}\\ &=-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}+\frac {\left (2^{-\frac {1}{2}+n} c^2 ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \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)+m+n} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}-n} (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 )^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f (1+m+n) (2+m+n)}\\ &=\frac {2^{\frac {1}{2}+n} c ((1+m+n) (C (1-m+n)+A (2+m+n))+(m-n) (C+2 C m+B (2+m+n))) \cos (e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n);\frac {1}{2} (3+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m) (1+m+n) (2+m+n)}-\frac {(C+2 C m+B (2+m+n)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n) (2+m+n)}+\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n}}{c f (2+m+n)}\\ \end {align*}

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Mathematica [C]  time = 16.50, size = 6226, normalized size = 23.14 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]

[Out]

Result too large to show

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]

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))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

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))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, x)

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maple [F]  time = 13.73, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

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))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

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))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right ) \,d x \]

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))^n*(A + B*sin(e + f*x) + C*sin(e + f*x)^2),x)

[Out]

int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n*(A + B*sin(e + f*x) + C*sin(e + f*x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**n*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)

[Out]

Timed out

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