3.214 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \, dx\)

Optimal. Leaf size=163 \[ \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{f (2 m+1)}-\frac {B 2^{\frac {1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+1);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2972, 2745, 2689, 70, 69} \[ \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{f (2 m+1)}-\frac {B 2^{\frac {1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+1);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]

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])^(-1 - m),x]

[Out]

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

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 2972

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

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Mathematica [C]  time = 11.62, size = 675, normalized size = 4.14 \[ -\frac {2^{-m} (2 m-3) \cos ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \cot \left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \sin ^{-2 m}\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-m-1} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-m-1)} \left (8 B (2 m+1) \tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )-(A+B) \left ((2 m+1) \tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) \, _2F_1\left (\frac {1}{2}-m,-2 m;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )+(2 m-1) \, _2F_1\left (-m-\frac {1}{2},-2 m;\frac {1}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )\right )}{f \left (4 m^2-1\right ) \left ((2 m-3) \left ((A+B) \left (\cos \left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right )+1\right ) \left (1-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )^{2 m}-4 B \sin ^2\left (\frac {1}{2} \left (-e-f x+\frac {\pi }{2}\right )\right ) F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )-64 B m \sin ^4\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) F_1\left (\frac {3}{2}-m;1-2 m,1;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )-32 B \sin ^4\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ) F_1\left (\frac {3}{2}-m;-2 m,2;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e-f x+\frac {\pi }{2}\right )\right )\right )\right )} \]

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])^(-1 - m),x]

[Out]

-(((-3 + 2*m)*Cos[(-e + Pi/2 - f*x)/4]^2*Cot[(-e + Pi/2 - f*x)/4]*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*
(c - c*Sin[e + f*x])^(-1 - m)*(8*B*(1 + 2*m)*AppellF1[1/2 - m, -2*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -
Tan[(-e + Pi/2 - f*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2 - (A + B)*((-1 + 2*m)*Hypergeometric2F1[-1/2 - m, -2*m,
1/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2] + (1 + 2*m)*Hypergeometric2F1[1/2 - m, -2*m, 3/2 - m, Tan[(-e + Pi/2 - f
*x)/4]^2]*Tan[(-e + Pi/2 - f*x)/4]^2)))/(2^m*f*(-1 + 4*m^2)*Sin[(-e + Pi/2 - f*x)/2]^(2*m)*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^(2*(-1 - m))*(-64*B*m*AppellF1[3/2 - m, 1 - 2*m, 1, 5/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -T
an[(-e + Pi/2 - f*x)/4]^2]*Sin[(-e + Pi/2 - f*x)/4]^4 - 32*B*AppellF1[3/2 - m, -2*m, 2, 5/2 - m, Tan[(-e + Pi/
2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sin[(-e + Pi/2 - f*x)/4]^4 + (-3 + 2*m)*(-4*B*AppellF1[1/2 - m, -2
*m, 1, 3/2 - m, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + (A + B)*
(1 + Cos[(-e + Pi/2 - f*x)/2])*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)))))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1}, x\right ) \]

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))^(-1-m),x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1}\,{d x} \]

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))^(-1-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 - 1), x)

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maple [F]  time = 2.87, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-1-m}\, dx \]

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))^(-1-m),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-m),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 1}\,{d x} \]

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))^(-1-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 - 1), x)

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

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 + 1),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(m + 1), 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*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(-1-m),x)

[Out]

Timed out

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