3.1.30 \(\int (a (b \sin (c+d x))^p)^n \, dx\) [30]

Optimal. Leaf size=79 \[ \frac {\cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(c+d x)\right ) \sin (c+d x) \left (a (b \sin (c+d x))^p\right )^n}{d (1+n p) \sqrt {\cos ^2(c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3287, 2722} \begin {gather*} \frac {\sin (c+d x) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);\sin ^2(c+d x)\right ) \left (a (b \sin (c+d x))^p\right )^n}{d (n p+1) \sqrt {\cos ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*(b*Sin[c + d*x])^p)^n,x]

[Out]

(Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[c + d*x]^2]*Sin[c + d*x]*(a*(b*Sin[c + d*x]
)^p)^n)/(d*(1 + n*p)*Sqrt[Cos[c + d*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 3287

Int[(u_.)*((b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sin[e + f*x
])^n)^FracPart[p]/(c*Sin[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Sin[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 73, normalized size = 0.92 \begin {gather*} \frac {\sqrt {\cos ^2(c+d x)} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);\sin ^2(c+d x)\right ) \left (a (b \sin (c+d x))^p\right )^n \tan (c+d x)}{d (1+n p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*(b*Sin[c + d*x])^p)^n,x]

[Out]

(Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/2, (1 + n*p)/2, (3 + n*p)/2, Sin[c + d*x]^2]*(a*(b*Sin[c + d*x])^p)^
n*Tan[c + d*x])/(d*(1 + n*p))

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (a \left (b \sin \left (d x +c \right )\right )^{p}\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*(b*sin(d*x+c))^p)^n,x)

[Out]

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

[Out]

integrate(((b*sin(d*x + c))^p*a)^n, x)

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Fricas [F]
time = 0.41, size = 16, normalized size = 0.20 \begin {gather*} {\rm integral}\left (\left (\left (b \sin \left (d x + c\right )\right )^{p} a\right )^{n}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((b*sin(d*x + c))^p*a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*(b*sin(d*x+c))**p)**n,x)

[Out]

Integral((a*(b*sin(c + d*x))**p)**n, 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((a*(b*sin(d*x+c))^p)^n,x, algorithm="giac")

[Out]

integrate(((b*sin(d*x + c))^p*a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*(b*sin(c + d*x))^p)^n,x)

[Out]

int((a*(b*sin(c + d*x))^p)^n, x)

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