3.2.33 \(\int (1-\sin (e+f x))^m (-\sin (e+f x))^n \, dx\) [133]

Optimal. Leaf size=68 \[ \frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \cos (e+f x)}{f \sqrt {1-\sin (e+f x)}} \]

[Out]

2^(1/2+m)*AppellF1(1/2,-n,1/2-m,3/2,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*cos(f*x+e)/f/(1-sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2864, 138} \begin {gather*} \frac {2^{m+\frac {1}{2}} \cos (e+f x) F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f \sqrt {1-\sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + m)*AppellF1[1/2, -n, 1/2 - m, 3/2, 1 + Sin[e + f*x], (1 + Sin[e + f*x])/2]*Cos[e + f*x])/(f*Sqrt[1 -
Sin[e + f*x]])

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

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]

Rubi steps

\begin {align*} \int (1-\sin (e+f x))^m (-\sin (e+f x))^n \, dx &=\frac {\cos (e+f x) \text {Subst}\left (\int \frac {(1-x)^n (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1+\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=\frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \cos (e+f x)}{f \sqrt {1-\sin (e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(68)=136\).
time = 1.53, size = 300, normalized size = 4.41 \begin {gather*} -\frac {(3+2 m) F_1\left (\frac {1}{2}+m;-n,1+m+n;\frac {3}{2}+m;\cot ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right ) \cos (e+f x) (1-\sin (e+f x))^m (-\sin (e+f x))^n}{f (1+2 m) \left ((3+2 m) F_1\left (\frac {1}{2}+m;-n,1+m+n;\frac {3}{2}+m;\cot ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )-2 \left (n F_1\left (\frac {3}{2}+m;1-n,1+m+n;\frac {5}{2}+m;\cot ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )+(1+m+n) F_1\left (\frac {3}{2}+m;-n,2+m+n;\frac {5}{2}+m;\cot ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )\right ) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((3 + 2*m)*AppellF1[1/2 + m, -n, 1 + m + n, 3/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]
^2]*Cos[e + f*x]*(1 - Sin[e + f*x])^m*(-Sin[e + f*x])^n)/(f*(1 + 2*m)*((3 + 2*m)*AppellF1[1/2 + m, -n, 1 + m +
n, 3/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] - 2*(n*AppellF1[3/2 + m, 1 - n, 1 + m
+ n, 5/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] + (1 + m + n)*AppellF1[3/2 + m, -n, 2
+ m + n, 5/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2])*Tan[(2*e - Pi + 2*f*x)/4]^2)))

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (1-\sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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