3.7.66 \(\int \frac {(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx\) [666]

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

[Out]

-1/2*AppellF1(1/2,-n,2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n/a/f/(((c+d
*sin(f*x+e))/(c+d))^n)/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.25, 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 = {2867, 142, 141} \begin {gather*} \frac {d \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},2;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)^2 (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 142

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 2867

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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(104)=208\).
time = 4.59, size = 319, normalized size = 3.07 \begin {gather*} \frac {\sec (e+f x) (c+d \sin (e+f x))^n \left (a^2 F_1\left (1;\frac {1}{2},-n;2;\frac {1}{2} (1+\sin (e+f x)),\frac {d (1+\sin (e+f x))}{-c+d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x))^2 \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 a (1+\sin (e+f x)) \sqrt {1-\frac {2}{1+\sin (e+f x)}} \left (2 a (1+2 n) F_1\left (\frac {1}{2}-n;-\frac {1}{2},-n;\frac {3}{2}-n;\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+a (-1+2 n) F_1\left (-\frac {1}{2}-n;-\frac {1}{2},-n;\frac {1}{2}-n;\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right ) \left (1+\frac {c-d}{d+d \sin (e+f x)}\right )^{-n}}{-1+4 n^2}\right )}{8 a^3 f \sqrt {a (1+\sin (e+f x))}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

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))^(3/2),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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