3.3.12 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx\) [212]

Optimal. Leaf size=191 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (5+2 m) \left (3+8 m+4 m^2\right )} \]

[Out]

(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m)/f/(5+2*m)+(2*A-B*(3+2*m))*cos(f*x+e)*(a+a*sin(f*x+
e))^m*(c-c*sin(f*x+e))^(-2-m)/c/f/(4*m^2+16*m+15)+(2*A-B*(3+2*m))*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e
))^(-1-m)/c^2/f/(8*m^3+36*m^2+46*m+15)

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Rubi [A]
time = 0.22, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3051, 2822, 2821} \begin {gather*} \frac {(2 A-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{c^2 f (2 m+5) \left (4 m^2+8 m+3\right )}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-3}}{f (2 m+5)}+\frac {(2 A-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-2}}{c f (2 m+3) (2 m+5)} \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])^(-3 - m),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m))/(f*(5 + 2*m)) + ((2*A - B*(3 + 2*m
))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 - m))/(c*f*(3 + 2*m)*(5 + 2*m)) + ((2*A - B*(3
+ 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c^2*f*(5 + 2*m)*(3 + 8*m + 4*m^2)
)

Rule 2821

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rule 2822

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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/(a*f*(2*m + 1))), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(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)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 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))^{-3-m} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx}{c (5+2 m)}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m} \, dx}{c^2 (3+2 m) (5+2 m)}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (1+2 m) (3+2 m) (5+2 m)}\\ \end {align*}

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Mathematica [A]
time = 14.86, size = 269, normalized size = 1.41 \begin {gather*} \frac {2^{-13-m} \cos \left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right ) \csc ^{15}\left (\frac {1}{8} \left (-e+\frac {\pi }{2}-f x\right )\right ) \sec ^5\left (\frac {1}{8} \left (-e+\frac {\pi }{2}-f x\right )\right ) \sin ^{-2 m}\left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-3-m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m} \left (16 A-9 B+24 A m-6 B m+8 A m^2+(2 A-3 B-2 B m) \cos \left (2 \left (-e+\frac {\pi }{2}-f x\right )\right )+2 (3+2 m) (-2 A+B (3+2 m)) \sin (e+f x)\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (-1+\cot ^2\left (\frac {1}{8} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )^5} \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])^(-3 - m),x]

[Out]

(2^(-13 - m)*Cos[(-e + Pi/2 - f*x)/2]*Csc[(-e + Pi/2 - f*x)/8]^15*Sec[(-e + Pi/2 - f*x)/8]^5*(a + a*Sin[e + f*
x])^m*(c - c*Sin[e + f*x])^(-3 - m)*(16*A - 9*B + 24*A*m - 6*B*m + 8*A*m^2 + (2*A - 3*B - 2*B*m)*Cos[2*(-e + P
i/2 - f*x)] + 2*(3 + 2*m)*(-2*A + B*(3 + 2*m))*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(-1 + Cot[(-e +
Pi/2 - f*x)/8]^2)^5*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(2*(-3 - m)))

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Maple [F]
time = 1.33, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-3-m}\, dx\]

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))^(-3-m),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-m),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))^(-3-m),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3), x)

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Fricas [A]
time = 0.41, size = 143, normalized size = 0.75 \begin {gather*} \frac {{\left ({\left (2 \, B m - 2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} + {\left (4 \, B m^{2} - 4 \, {\left (A - 3 \, B\right )} m - 6 \, A + 9 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (4 \, A m^{2} + 4 \, {\left (3 \, A - B\right )} m + 9 \, A - 6 \, B\right )} \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f} \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))^(-3-m),x, algorithm="fricas")

[Out]

((2*B*m - 2*A + 3*B)*cos(f*x + e)^3 + (4*B*m^2 - 4*(A - 3*B)*m - 6*A + 9*B)*cos(f*x + e)*sin(f*x + e) + (4*A*m
^2 + 4*(3*A - B)*m + 9*A - 6*B)*cos(f*x + e))*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)/(8*f*m^3 +
36*f*m^2 + 46*f*m + 15*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(-3-m),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8011 deep

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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))^(-3-m),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3), x)

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Mupad [B]
time = 15.48, size = 239, normalized size = 1.25 \begin {gather*} -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (30\,A\,\cos \left (e+f\,x\right )-15\,B\,\cos \left (e+f\,x\right )-2\,A\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (3\,e+3\,f\,x\right )-12\,A\,\sin \left (2\,e+2\,f\,x\right )+18\,B\,\sin \left (2\,e+2\,f\,x\right )+8\,B\,m^2\,\sin \left (2\,e+2\,f\,x\right )+48\,A\,m\,\cos \left (e+f\,x\right )-10\,B\,m\,\cos \left (e+f\,x\right )+16\,A\,m^2\,\cos \left (e+f\,x\right )+2\,B\,m\,\cos \left (3\,e+3\,f\,x\right )-8\,A\,m\,\sin \left (2\,e+2\,f\,x\right )+24\,B\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^3\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (8\,m^3+36\,m^2+46\,m+15\right )\,\left (15\,\sin \left (e+f\,x\right )+6\,\cos \left (2\,e+2\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )-10\right )} \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))^(m + 3),x)

[Out]

-((a*(sin(e + f*x) + 1))^m*(30*A*cos(e + f*x) - 15*B*cos(e + f*x) - 2*A*cos(3*e + 3*f*x) + 3*B*cos(3*e + 3*f*x
) - 12*A*sin(2*e + 2*f*x) + 18*B*sin(2*e + 2*f*x) + 8*B*m^2*sin(2*e + 2*f*x) + 48*A*m*cos(e + f*x) - 10*B*m*co
s(e + f*x) + 16*A*m^2*cos(e + f*x) + 2*B*m*cos(3*e + 3*f*x) - 8*A*m*sin(2*e + 2*f*x) + 24*B*m*sin(2*e + 2*f*x)
))/(c^3*f*(-c*(sin(e + f*x) - 1))^m*(46*m + 36*m^2 + 8*m^3 + 15)*(15*sin(e + f*x) + 6*cos(2*e + 2*f*x) - sin(3
*e + 3*f*x) - 10))

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