3.1.27 \(\int (c \sin ^3(a+b x))^p \, dx\) [27]

Optimal. Leaf size=75 \[ \frac {\cos (a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+3 p);\frac {3 (1+p)}{2};\sin ^2(a+b x)\right ) \sin (a+b x) \left (c \sin ^3(a+b x)\right )^p}{b (1+3 p) \sqrt {\cos ^2(a+b x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3286, 2722} \begin {gather*} \frac {\sin (a+b x) \cos (a+b x) \left (c \sin ^3(a+b x)\right )^p \, _2F_1\left (\frac {1}{2},\frac {1}{2} (3 p+1);\frac {3 (p+1)}{2};\sin ^2(a+b x)\right )}{b (3 p+1) \sqrt {\cos ^2(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^p,x]

[Out]

(Cos[a + b*x]*Hypergeometric2F1[1/2, (1 + 3*p)/2, (3*(1 + p))/2, Sin[a + b*x]^2]*Sin[a + b*x]*(c*Sin[a + b*x]^
3)^p)/(b*(1 + 3*p)*Sqrt[Cos[a + b*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 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, 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 (c \sin ^3(a+b x)\right )^p \, dx &=\left (\sin ^{-3 p}(a+b x) \left (c \sin ^3(a+b x)\right )^p\right ) \int \sin ^{3 p}(a+b x) \, dx\\ &=\frac {\cos (a+b x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+3 p);\frac {3 (1+p)}{2};\sin ^2(a+b x)\right ) \sin (a+b x) \left (c \sin ^3(a+b x)\right )^p}{b (1+3 p) \sqrt {\cos ^2(a+b x)}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 67, normalized size = 0.89 \begin {gather*} \frac {\sqrt {\cos ^2(a+b x)} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1+3 p);\frac {3 (1+p)}{2};\sin ^2(a+b x)\right ) \left (c \sin ^3(a+b x)\right )^p \tan (a+b x)}{b+3 b p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x]^3)^p,x]

[Out]

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

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \left (c \left (\sin ^{3}\left (b x +a \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a)^3)^p,x)

[Out]

int((c*sin(b*x+a)^3)^p,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*sin(b*x+a)^3)^p,x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a)^3)^p, x)

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Fricas [F]
time = 0.39, size = 26, normalized size = 0.35 \begin {gather*} {\rm integral}\left (\left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{p}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^3)^p,x, algorithm="fricas")

[Out]

integral((-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^p, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \sin ^{3}{\left (a + b x \right )}\right )^{p}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)**3)**p,x)

[Out]

Integral((c*sin(a + b*x)**3)**p, 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*sin(b*x+a)^3)^p,x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^3)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^3)^p,x)

[Out]

int((c*sin(a + b*x)^3)^p, x)

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