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

Optimal. Leaf size=106 \[ -\frac {2 a^2 (4 n+5) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right )}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \cos (e+f x) \sin ^{n+1}(e+f x)}{f (2 n+3) \sqrt {a \sin (e+f x)+a}} \]

[Out]

-2*a^2*(5+4*n)*cos(f*x+e)*hypergeom([1/2, -n],[3/2],1-sin(f*x+e))/f/(3+2*n)/(a+a*sin(f*x+e))^(1/2)-2*a^2*cos(f
*x+e)*sin(f*x+e)^(1+n)/f/(3+2*n)/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 = {2763, 21, 2776, 65} \[ -\frac {2 a^2 (4 n+5) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right )}{f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \cos (e+f x) \sin ^{n+1}(e+f x)}{f (2 n+3) \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(5 + 4*n)*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f*x]])/(f*(3 + 2*n)*Sqrt[a + a*Sin[
e + f*x]]) - (2*a^2*Cos[e + f*x]*Sin[e + f*x]^(1 + n))/(f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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])

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], 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] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[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]

Rubi steps

\begin {align*} \int \sin ^n(e+f x) (a+a \sin (e+f x))^{3/2} \, dx &=-\frac {2 a^2 \cos (e+f x) \sin ^{1+n}(e+f x)}{f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 \int \frac {\sin ^n(e+f x) \left (\frac {1}{2} a^2 (5+4 n)+\frac {1}{2} a^2 (5+4 n) \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{3+2 n}\\ &=-\frac {2 a^2 \cos (e+f x) \sin ^{1+n}(e+f x)}{f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {(a (5+4 n)) \int \sin ^n(e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{3+2 n}\\ &=-\frac {2 a^2 \cos (e+f x) \sin ^{1+n}(e+f x)}{f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^3 (5+4 n) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=-\frac {2 a^2 (5+4 n) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};1-\sin (e+f x)\right )}{f (3+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sin ^{1+n}(e+f x)}{f (3+2 n) \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 6.33, size = 5111, normalized size = 48.22 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{n}, x\right ) \]

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

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

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

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{n}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{n}{\left (e + f x \right )}\, dx \]

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

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