3.26 \(\int (c \sin ^2(a+b x))^p \, dx\)

Optimal. Leaf size=77 \[ \frac {\sin (a+b x) \cos (a+b x) \left (c \sin ^2(a+b x)\right )^p \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2 p+1);\frac {1}{2} (2 p+3);\sin ^2(a+b x)\right )}{b (2 p+1) \sqrt {\cos ^2(a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 77, 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 = {3207, 2643} \[ \frac {\sin (a+b x) \cos (a+b x) \left (c \sin ^2(a+b x)\right )^p \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2 p+1);\frac {1}{2} (2 p+3);\sin ^2(a+b x)\right )}{b (2 p+1) \sqrt {\cos ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^2)^p,x]

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3207

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

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Mathematica [A]  time = 0.07, size = 61, normalized size = 0.79 \[ \frac {\sqrt {\cos ^2(a+b x)} \tan (a+b x) \left (c \sin ^2(a+b x)\right )^p \, _2F_1\left (\frac {1}{2},p+\frac {1}{2};p+\frac {3}{2};\sin ^2(a+b x)\right )}{2 b p+b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x]^2)^p,x]

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-c \cos \left (b x + a\right )^{2} + c\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^2)^p,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )^{2}\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^2)^p,x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a)^2)^p, x)

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maple [F]  time = 1.84, size = 0, normalized size = 0.00 \[ \int \left (c \left (\sin ^{2}\left (b x +a \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a)^2)^p,x)

[Out]

int((c*sin(b*x+a)^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )^{2}\right )^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)^2)^p,x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a)^2)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,{\sin \left (a+b\,x\right )}^2\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^2)^p,x)

[Out]

int((c*sin(a + b*x)^2)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin ^{2}{\left (a + b x \right )}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a)**2)**p,x)

[Out]

Integral((c*sin(a + b*x)**2)**p, x)

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