3.127 \(\int \frac{(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{\cos (e+f x) \sin ^{-n}(e+f x) F_1\left (\frac{1}{2};-n,2;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{2 f \sqrt{\sin (e+f x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.135221, antiderivative size = 80, 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 = {2786, 2785, 130, 429} \[ -\frac{\cos (e+f x) \sin ^{-n}(e+f x) F_1\left (\frac{1}{2};-n,2;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{2 f \sqrt{\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sin[e + f*x])^n/(1 + 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]*(d*Sin[e + f*x])^n)/(2*f*Sin[
e + f*x]^n*Sqrt[1 + Sin[e + f*x]])

Rule 2786

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

Mathematica [B]  time = 0.857067, size = 265, normalized size = 3.31 \[ \frac{\sec (e+f x) \left (\sqrt{2-2 \sin (e+f x)} (\sin (e+f x)+1)^2 (-\sin (e+f x))^{-n} F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )-\frac{4 (\sin (e+f x)+1) \sqrt{1-\frac{2}{\sin (e+f x)+1}} \left (1-\frac{1}{\sin (e+f x)+1}\right )^{-n} \left (2 (2 n+1) F_1\left (\frac{1}{2}-n;-\frac{1}{2},-n;\frac{3}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{1}{\sin (e+f x)+1}\right )+(2 n-1) (\sin (e+f x)+1) F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{1}{\sin (e+f x)+1}\right )\right )}{4 n^2-1}\right ) (d \sin (e+f x))^n}{8 f \sqrt{\sin (e+f x)+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sin[e + f*x])^n/(1 + Sin[e + f*x])^(3/2),x]

[Out]

(Sec[e + f*x]*(d*Sin[e + f*x])^n*((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*(1 + Sin[e + f*x])*Sqrt[1 - 2/(1 + Sin[e + f*x])]*(
2*(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)] + (-1 + 2*n)*A
ppellF1[-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)*(1 - (1 + Sin[e + f*x])^(-1))^n)))/(8*f*Sqrt[1 + Sin[e + f*x]])

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^n/(1+sin(f*x+e))^(3/2),x)

[Out]

int((d*sin(f*x+e))^n/(1+sin(f*x+e))^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n/(1+sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e))^n/(sin(f*x + e) + 1)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (d \sin \left (f x + e\right )\right )^{n} \sqrt{\sin \left (f x + e\right ) + 1}}{\cos \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n/(1+sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-(d*sin(f*x + e))^n*sqrt(sin(f*x + e) + 1)/(cos(f*x + e)^2 - 2*sin(f*x + e) - 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**n/(1+sin(f*x+e))**(3/2),x)

[Out]

Integral((d*sin(e + f*x))**n/(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{\left (d \sin \left (f x + e\right )\right )^{n}}{{\left (\sin \left (f x + e\right ) + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n/(1+sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e))^n/(sin(f*x + e) + 1)^(3/2), x)