3.7.63 \(\int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^n \, dx\) [663]

Optimal. Leaf size=160 \[ -\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 (c-d (5+4 n)) \cos (e+f x) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 160, 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 = {2842, 21, 2855, 72, 71} \begin {gather*} \frac {2 a^2 (c-d (4 n+5)) \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 f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 2855

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

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Mathematica [A]
time = 4.29, size = 133, normalized size = 0.83 \begin {gather*} -\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \left ((-c+d (5+4 n)) \, _2F_1\left (\frac {1}{2},-n;\frac {3}{2};-\frac {d (-1+\sin (e+f x))}{c+d}\right )+(c+d) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{1+n}\right )}{d f (3+2 n) \sqrt {a (1+\sin (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

Timed out

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Giac [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((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^n,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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