3.10 \(\int \frac{(d \sin (e+f x))^n (A+B \sin (e+f x))}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=152 \[ -\frac{(A-B) \cos (e+f x) \sin ^{-n}(e+f x) F_1\left (\frac{1}{2};-n,1;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{f \sqrt{a \sin (e+f x)+a}}-\frac{2 B \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right )}{f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(((A - B)*AppellF1[1/2, -n, 1, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(d*Sin[e + f*x])^n)/
(f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]])) - (2*B*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f
*x]]*(d*Sin[e + f*x])^n)/(f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.396167, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {2987, 2787, 2786, 2785, 130, 429, 2776, 67, 65} \[ -\frac{(A-B) \cos (e+f x) \sin ^{-n}(e+f x) F_1\left (\frac{1}{2};-n,1;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{f \sqrt{a \sin (e+f x)+a}}-\frac{2 B \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right )}{f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((d*Sin[e + f*x])^n*(A + B*Sin[e + f*x]))/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

-(((A - B)*AppellF1[1/2, -n, 1, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(d*Sin[e + f*x])^n)/
(f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]])) - (2*B*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f
*x]]*(d*Sin[e + f*x])^n)/(f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]])

Rule 2987

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

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 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])

Rule 2776

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

Rule 67

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

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \frac{(d \sin (e+f x))^n (A+B \sin (e+f x))}{\sqrt{a+a \sin (e+f x)}} \, dx &=(A-B) \int \frac{(d \sin (e+f x))^n}{\sqrt{a+a \sin (e+f x)}} \, dx+\frac{B \int (d \sin (e+f x))^n \sqrt{a+a \sin (e+f x)} \, dx}{a}\\ &=\frac{\left ((A-B) \sqrt{1+\sin (e+f x)}\right ) \int \frac{(d \sin (e+f x))^n}{\sqrt{1+\sin (e+f x)}} \, dx}{\sqrt{a+a \sin (e+f x)}}+\frac{(a B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(d x)^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left ((A-B) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \sqrt{1+\sin (e+f x)}\right ) \int \frac{\sin ^n(e+f x)}{\sqrt{1+\sin (e+f x)}} \, dx}{\sqrt{a+a \sin (e+f x)}}+\frac{\left (a B \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \operatorname{Subst}\left (\int \frac{x^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 B \cos (e+f x) \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f \sqrt{a+a \sin (e+f x)}}-\frac{\left ((A-B) \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) \sqrt{x}} \, dx,x,1-\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 B \cos (e+f x) \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f \sqrt{a+a \sin (e+f x)}}-\frac{\left (2 (A-B) \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}{2-x^2} \, dx,x,\sqrt{1-\sin (e+f x)}\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{(A-B) F_1\left (\frac{1}{2};-n,1;\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}{f \sqrt{a+a \sin (e+f x)}}-\frac{2 B \cos (e+f x) \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.70877, size = 250, normalized size = 1.64 \[ \frac{\cos (e+f x) \sqrt{a (\sin (e+f x)+1)} \sin ^n(e+f x) \left (-\sin ^2(e+f x)\right )^{-n} \left (1-\frac{1}{\sin (e+f x)+1}\right )^{-n} (d \sin (e+f x))^n \left (4 (A-B) \sqrt{\frac{\sin (e+f x)-1}{\sin (e+f x)+1}} (-\sin (e+f x))^n F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{1}{\sin (e+f x)+1}\right )-(2 n+1) (A+B) \sqrt{2-2 \sin (e+f x)} \left (1-\frac{1}{\sin (e+f x)+1}\right )^n F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\sin (e+f x)+1\right )\right )}{4 a f (2 n+1) (\sin (e+f x)-1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d*Sin[e + f*x])^n*(A + B*Sin[e + f*x]))/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^n*(d*Sin[e + f*x])^n*Sqrt[a*(1 + Sin[e + f*x])]*(4*(A - B)*AppellF1[-1/2 - n, -1/2,
-n, 1/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)]*(-Sin[e + f*x])^n*Sqrt[(-1 + Sin[e + f*x])/(1 + S
in[e + f*x])] - (A + B)*(1 + 2*n)*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sqrt[2 - 2*S
in[e + f*x]]*(1 - (1 + Sin[e + f*x])^(-1))^n))/(4*a*f*(1 + 2*n)*(-1 + Sin[e + f*x])*(-Sin[e + f*x]^2)^n*(1 - (
1 + Sin[e + f*x])^(-1))^n)

________________________________________________________________________________________

Maple [F]  time = 0.383, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( A+B\sin \left ( fx+e \right ) \right ){\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e))^n/sqrt(a*sin(f*x + e) + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*sin(f*x + e) + A)*(d*sin(f*x + e))^n/sqrt(a*sin(f*x + e) + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**n*(A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral((d*sin(e + f*x))**n*(A + B*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e))^n/sqrt(a*sin(f*x + e) + a), x)