3.667 \(\int \frac{(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{\cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};-n,3;\frac{3}{2};\frac{d (1-\sin (e+f x))}{c+d},\frac{1}{2} (1-\sin (e+f x))\right )}{4 a^2 f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.177884, antiderivative size = 137, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2788, 137, 136} \[ -\frac{d^2 \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} F_1\left (n+1;\frac{1}{2},3;n+2;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right )}{f (n+1) (c-d)^3 \sqrt{a \sin (e+f x)+a} \left (a^2-a^2 \sin (e+f x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

Int[(c + d*Sin[e + f*x])^n/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[(a^2*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[a - b*x], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[m]

Rule 137

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

Rule 136

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

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{5/2}} \, dx &=\frac{\left (a^2 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{\sqrt{a-a x} (a+a x)^3} \, 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^2 \cos (e+f x) \sqrt{\frac{d (a-a \sin (e+f x))}{a c+a d}}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{(a+a x)^3 \sqrt{\frac{a d}{a c+a d}-\frac{a d x}{a c+a d}}} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{d^2 F_1\left (1+n;\frac{1}{2},3;2+n;\frac{c+d \sin (e+f x)}{c+d},\frac{c+d \sin (e+f x)}{c-d}\right ) \cos (e+f x) \sqrt{\frac{d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{1+n}}{(c-d)^3 f (1+n) \sqrt{a+a \sin (e+f x)} \left (a^2-a^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 9.46624, size = 414, normalized size = 3.98 \[ \frac{\sec (e+f x) (c+d \sin (e+f x))^n \left (a^3 \sqrt{2-2 \sin (e+f x)} (\sin (e+f x)+1)^3 \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (1;\frac{1}{2},-n;2;\frac{1}{2} (\sin (e+f x)+1),\frac{d (\sin (e+f x)+1)}{d-c}\right )-\frac{4 a^2 (\sin (e+f x)+1) \sqrt{1-\frac{2}{\sin (e+f x)+1}} \left (\frac{c-d}{d \sin (e+f x)+d}+1\right )^{-n} \left (a \left (4 n^2-8 n+3\right ) (\sin (e+f x)+1)^2 F_1\left (-n-\frac{1}{2};-\frac{1}{2},-n;\frac{1}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )+2 (2 n+1) \left (2 a (2 n-1) F_1\left (\frac{3}{2}-n;-\frac{1}{2},-n;\frac{5}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )+a (2 n-3) (\sin (e+f x)+1) F_1\left (\frac{1}{2}-n;-\frac{1}{2},-n;\frac{3}{2}-n;\frac{2}{\sin (e+f x)+1},\frac{d-c}{\sin (e+f x) d+d}\right )\right )\right )}{(2 n-3) (2 n-1) (2 n+1)}\right )}{16 a^4 f (a (\sin (e+f x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*Sin[e + f*x])^n/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(Sec[e + f*x]*(c + d*Sin[e + f*x])^n*((a^3*AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, (d*(1 + Sin[e + f*x])
)/(-c + d)]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x])^3)/((c + d*Sin[e + f*x])/(c - d))^n - (4*a^2*(1 + Sin[
e + f*x])*Sqrt[1 - 2/(1 + Sin[e + f*x])]*(a*(3 - 8*n + 4*n^2)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin
[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])^2 + 2*(1 + 2*n)*(2*a*(-1 + 2*n)*AppellF1[3/2 - n
, -1/2, -n, 5/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])] + a*(-3 + 2*n)*AppellF1[1/2 - n, -1/
2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x]))))/((-3 + 2*n)*(-1 + 2
*n)*(1 + 2*n)*(1 + (c - d)/(d + d*Sin[e + f*x]))^n)))/(16*a^4*f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [F]  time = 0.158, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(5/2),x)

[Out]

int((c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(5/2),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 ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^(5/2), 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}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2
- 4*a^3)*sin(f*x + e)), 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((c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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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 ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^(5/2), x)