3.13 \(\int (d \sin (e+f x))^n (a-a \sin (e+f x)) (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=114 \[ \frac{\sec (e+f x) (a-a \sin (e+f x)) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{n+1} F_1\left (n+1;-\frac{1}{2},\frac{1}{2}-m;n+2;\sin (e+f x),-\sin (e+f x)\right )}{d f (n+1) \sqrt{1-\sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.155902, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3008, 135, 133} \[ \frac{\sec (e+f x) (a-a \sin (e+f x)) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{n+1} F_1\left (n+1;-\frac{1}{2},\frac{1}{2}-m;n+2;\sin (e+f x),-\sin (e+f x)\right )}{d f (n+1) \sqrt{1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3008

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

Rule 135

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

Rule 133

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

Rubi steps

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

Mathematica [F]  time = 10.8855, size = 0, normalized size = 0. \[ \int (d \sin (e+f x))^n (a-a \sin (e+f x)) (a+a \sin (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 4.162, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a-a\sin \left ( fx+e \right ) \right ) \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int{\left (a \sin \left (f x + e\right ) - a\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a \sin \left (f x + e\right ) - a\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (a \sin \left (f x + e\right ) - a\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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