3.7.53 \(\int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\) [653]

Optimal. Leaf size=116 \[ -\frac {\cos (e+f x) \, _2F_1\left (\frac {1}{2}-m,-m;1-m;-\frac {2 \sin (e+f x)}{1-\sin (e+f x)}\right ) (d \sin (e+f x))^{-m} \left (\frac {1+\sin (e+f x)}{1-\sin (e+f x)}\right )^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d f m (1+\sin (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2866, 2865, 2864, 134} \begin {gather*} -\frac {\cos (e+f x) \left (\frac {\sin (e+f x)+1}{1-\sin (e+f x)}\right )^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m} \, _2F_1\left (\frac {1}{2}-m,-m;1-m;-\frac {2 \sin (e+f x)}{1-\sin (e+f x)}\right )}{d f m (\sin (e+f x)+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

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

Rule 134

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c
*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f*x))))^n, x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 2864

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(-b)*(
d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])), Subst[Int[(a - x)^n*((2*a - x)^(m
- 1/2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !
IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]

Rule 2865

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(d/b)
^IntPart[n]*((d*Sin[e + f*x])^FracPart[n]/(b*Sin[e + f*x])^FracPart[n]), Int[(a + b*Sin[e + f*x])^m*(b*Sin[e +
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] &&  !Gt
Q[d/b, 0]

Rule 2866

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Sin[e + f*x])^FracPart[m]/(1 + (b/a)*Sin[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Sin[e + f*x])^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps

\begin {align*} \int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx &=\left ((1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (d \sin (e+f x))^{-1-m} (1+\sin (e+f x))^m \, dx\\ &=\frac {\left (\sin ^m(e+f x) (d \sin (e+f x))^{-m} (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int \sin ^{-1-m}(e+f x) (1+\sin (e+f x))^m \, dx}{d}\\ &=-\frac {\left (\cos (e+f x) \sin ^m(e+f x) (d \sin (e+f x))^{-m} (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-m} (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1-\sin (e+f x)\right )}{d f \sqrt {1-\sin (e+f x)}}\\ &=-\frac {\cos (e+f x) \, _2F_1\left (\frac {1}{2}-m,-m;1-m;-\frac {2 \sin (e+f x)}{1-\sin (e+f x)}\right ) (d \sin (e+f x))^{-m} \left (\frac {1+\sin (e+f x)}{1-\sin (e+f x)}\right )^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d f m (1+\sin (e+f x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.61, size = 194, normalized size = 1.67 \begin {gather*} \frac {(1-i) 2^m \, _2F_1\left (1+m,1+2 m;2 (1+m);\sqrt {2} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right ) \csc \left (\frac {1}{2} (e+f x)\right )\right ) ((1-i) (1+\cos (e+f x)-i \sin (e+f x)))^m ((1+i) (1-\cos (e+f x)+i \sin (e+f x)))^{-m} (d \sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m (\cos (e+f x)-i (1+\sin (e+f x))) (\cosh (m \log (2))-\sinh (m \log (2)))}{d f (1+2 m) (-1+\cos (e+f x)-i \sin (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

((1 - I)*2^m*Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), Sqrt[2]*Cos[(2*e - Pi + 2*f*x)/4]*Csc[(e + f*x)/2]]*
((1 - I)*(1 + Cos[e + f*x] - I*Sin[e + f*x]))^m*(a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] - I*(1 + Sin[e + f*x]))
*(Cosh[m*Log[2]] - Sinh[m*Log[2]]))/(d*f*(1 + 2*m)*(-1 + Cos[e + f*x] - I*Sin[e + f*x])*((1 + I)*(1 - Cos[e +
f*x] + I*Sin[e + f*x]))^m*(d*Sin[e + f*x])^m)

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (d \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

[Out]

int((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^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((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), 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((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (d \sin {\left (e + f x \right )}\right )^{- m - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**(-1-m)*(a+a*sin(f*x+e))**m,x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(d*sin(e + f*x))**(-m - 1), 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((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^m/(d*sin(e + f*x))^(m + 1),x)

[Out]

int((a + a*sin(e + f*x))^m/(d*sin(e + f*x))^(m + 1), x)

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