3.9.21 \(\int (c (d \sin (e+f x))^p)^n (a+a \sin (e+f x))^2 \, dx\) [821]

Optimal. Leaf size=222 \[ -\frac {a^2 \cos (e+f x) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p)}+\frac {a^2 (3+2 n p) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (1+n p) (2+n p) \sqrt {\cos ^2(e+f x)}}+\frac {2 a^2 \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2+n p);\frac {1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin ^2(e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (2+n p) \sqrt {\cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2905, 2842, 2827, 2722} \begin {gather*} \frac {2 a^2 \sin ^2(e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+2);\frac {1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2) \sqrt {\cos ^2(e+f x)}}+\frac {a^2 (2 n p+3) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(e+f x)\right ) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+1) (n p+2) \sqrt {\cos ^2(e+f x)}}-\frac {a^2 \sin (e+f x) \cos (e+f x) \left (c (d \sin (e+f x))^p\right )^n}{f (n p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Cos[e + f*x]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p))) + (a^2*(3 + 2*n*p)*Cos[e + f*x]*Hype
rgeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n)/(f*(1 + n*
p)*(2 + n*p)*Sqrt[Cos[e + f*x]^2]) + (2*a^2*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n*p)/2, (4 + n*p)/2, Sin[
e + f*x]^2]*Sin[e + f*x]^2*(c*(d*Sin[e + f*x])^p)^n)/(f*(2 + n*p)*Sqrt[Cos[e + f*x]^2])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 2905

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

Rubi steps

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

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Mathematica [A]
time = 0.44, size = 222, normalized size = 1.00 \begin {gather*} -\frac {a^2 \cos (e+f x) \sqrt {\cos ^2(e+f x)} \sin (e+f x) \left (c (d \sin (e+f x))^p\right )^n \left (\left (6+5 n p+n^2 p^2\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(e+f x)\right )+(1+n p) \sin (e+f x) \left (2 (3+n p) \, _2F_1\left (\frac {1}{2},1+\frac {n p}{2};2+\frac {n p}{2};\sin ^2(e+f x)\right )+(2+n p) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (3+n p);\frac {1}{2} (5+n p);\sin ^2(e+f x)\right ) \sin (e+f x)\right )\right )}{f (1+n p) (2+n p) (3+n p) (-1+\sin (e+f x)) (1+\sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Cos[e + f*x]*Sqrt[Cos[e + f*x]^2]*Sin[e + f*x]*(c*(d*Sin[e + f*x])^p)^n*((6 + 5*n*p + n^2*p^2)*Hypergeo
metric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2] + (1 + n*p)*Sin[e + f*x]*(2*(3 + n*p)*Hypergeometric2
F1[1/2, 1 + (n*p)/2, 2 + (n*p)/2, Sin[e + f*x]^2] + (2 + n*p)*Hypergeometric2F1[1/2, (3 + n*p)/2, (5 + n*p)/2,
Sin[e + f*x]^2]*Sin[e + f*x])))/(f*(1 + n*p)*(2 + n*p)*(3 + n*p)*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])))

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Maple [F]
time = 0.55, size = 0, normalized size = 0.00 \[\int \left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int 2 \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n} \sin {\left (e + f x \right )}\, dx + \int \left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral((c*(d*sin(e + f*x))**p)**n, x) + Integral(2*(c*(d*sin(e + f*x))**p)**n*sin(e + f*x), x) + Integ
ral((c*(d*sin(e + f*x))**p)**n*sin(e + f*x)**2, x))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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