3.123 \(\int \frac{\sin ^n(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{\cos (e+f x) F_1\left (\frac{1}{2};-n,2;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.136025, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2787, 2785, 130, 429} \[ -\frac{\cos (e+f x) F_1\left (\frac{1}{2};-n,2;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right )}{2 a f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^n/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

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

Rule 2787

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
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]

Rule 2785

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] &&  !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

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

Mathematica [B]  time = 2.2251, size = 274, normalized size = 4.22 \[ \frac{\sec (e+f x) \sin ^n(e+f x) \left (a^2 \sqrt{2-2 \sin (e+f x)} (\sin (e+f x)+1)^2 (-\sin (e+f x))^{-n} F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )-\frac{4 a (\sin (e+f x)-1) \left (1-\frac{1}{\sin (e+f x)+1}\right )^{-n} \left (2 a (2 n+1) F_1\left (\frac{1}{2}-n;-\frac{1}{2},-n;\frac{3}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{1}{\sin (e+f x)+1}\right )+a (2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{1}{\sin (e+f x)+1}\right )\right )}{\left (4 n^2-1\right ) \sqrt{1-\frac{2}{\sin (e+f x)+1}}}\right )}{8 a^3 f \sqrt{a (\sin (e+f x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^n/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(Sec[e + f*x]*Sin[e + f*x]^n*((a^2*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sqrt[2 - 2*
Sin[e + f*x]]*(1 + Sin[e + f*x])^2)/(-Sin[e + f*x])^n - (4*a*(-1 + Sin[e + f*x])*(2*a*(1 + 2*n)*AppellF1[1/2 -
n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2,
-n, 1/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*Sqrt[1 - 2/(1 +
Sin[e + f*x])]*(1 - (1 + Sin[e + f*x])^(-1))^n)))/(8*a^3*f*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [F]  time = 0.123, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^n/(a+a*sin(f*x+e))^(3/2),x)

[Out]

int(sin(f*x+e)^n/(a+a*sin(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^n/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sin \left (f x + e\right ) + a} \sin \left (f x + e\right )^{n}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*sin(f*x + e)^n/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{n}{\left (e + f x \right )}}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**n/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral(sin(e + f*x)**n/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^n/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^n/(a*sin(f*x + e) + a)^(3/2), x)