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

Optimal. Leaf size=257 \[ -\frac {2 a^3 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]

[Out]

2*a^3*(3*c-d*(11+4*n))*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)/d^2/f/(3+2*n)/(5+2*n)/(a+a*sin(f*x+e))^(1/2)-2*a^3*(3
*c^2-2*c*d*(7+4*n)+d^2*(16*n^2+56*n+43))*cos(f*x+e)*hypergeom([1/2, -n],[3/2],d*(1-sin(f*x+e))/(c+d))*(c+d*sin
(f*x+e))^n/d^2/f/(3+2*n)/(5+2*n)/(((c+d*sin(f*x+e))/(c+d))^n)/(a+a*sin(f*x+e))^(1/2)-2*a^2*cos(f*x+e)*(c+d*sin
(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/f/(5+2*n)

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Rubi [A]  time = 0.48, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2763, 2981, 2776, 70, 69} \[ -\frac {2 a^3 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}+\frac {2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(3*c - d*(11 + 4*n))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d^2*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*S
in[e + f*x]]) - (2*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(5 + 2*n)) - (
2*a^3*(3*c^2 - 2*c*d*(7 + 4*n) + d^2*(43 + 56*n + 16*n^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, (d*(1
- Sin[e + f*x]))/(c + d)]*(c + d*Sin[e + f*x])^n)/(d^2*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*Sin[e + f*x]]*((c + d*
Sin[e + f*x])/(c + d))^n)

Rule 69

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

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]

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx &=-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {2 \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^n \left (\frac {1}{2} a^2 (c+d (7+4 n))-\frac {1}{2} a^2 (3 c-11 d-4 d n) \sin (e+f x)\right ) \, dx}{d (5+2 n)}\\ &=\frac {2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {\left (a^2 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right )\right ) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^n \, dx}{d^2 (3+2 n) (5+2 n)}\\ &=\frac {2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {\left (a^4 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {\left (a^4 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac {a (c+d \sin (e+f x))}{-a c-a d}\right )^{-n}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {c}{c+d}+\frac {d x}{c+d}\right )^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac {2 a^3 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 32.78, size = 190, normalized size = 0.74 \[ \frac {a^2 (\sin (e+f x)-1) \sec (e+f x) \sqrt {a (\sin (e+f x)+1)} (c+d \sin (e+f x))^n \left (\left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};-\frac {d (\sin (e+f x)-1)}{c+d}\right )-(3 c-d (4 n+11)) (c+d \sin (e+f x))+d (2 n+3) (\sin (e+f x)+1) (c+d \sin (e+f x))\right )}{d^2 f \left (n+\frac {5}{2}\right ) (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[e + f*x]*(-1 + Sin[e + f*x])*Sqrt[a*(1 + Sin[e + f*x])]*(c + d*Sin[e + f*x])^n*(-((3*c - d*(11 + 4*n)
)*(c + d*Sin[e + f*x])) + d*(3 + 2*n)*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x]) + ((3*c^2 - 2*c*d*(7 + 4*n) + d^
2*(43 + 56*n + 16*n^2))*Hypergeometric2F1[1/2, -n, 3/2, -((d*(-1 + Sin[e + f*x]))/(c + d))])/((c + d*Sin[e + f
*x])/(c + d))^n))/(d^2*f*(5/2 + n)*(3 + 2*n))

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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