3.140 \(\int \sin ^2(a+b x^n) \, dx\)

Optimal. Leaf size=100 \[ \frac {e^{2 i a} 2^{-\frac {1}{n}-2} x \left (-i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i b x^n\right )}{n}+\frac {e^{-2 i a} 2^{-\frac {1}{n}-2} x \left (i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i b x^n\right )}{n}+\frac {x}{2} \]

[Out]

1/2*x+2^(-2-1/n)*exp(2*I*a)*x*GAMMA(1/n,-2*I*b*x^n)/n/((-I*b*x^n)^(1/n))+2^(-2-1/n)*x*GAMMA(1/n,2*I*b*x^n)/exp
(2*I*a)/n/((I*b*x^n)^(1/n))

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3367, 3366, 2208} \[ \frac {e^{2 i a} 2^{-\frac {1}{n}-2} x \left (-i b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},-2 i b x^n\right )}{n}+\frac {e^{-2 i a} 2^{-\frac {1}{n}-2} x \left (i b x^n\right )^{-1/n} \text {Gamma}\left (\frac {1}{n},2 i b x^n\right )}{n}+\frac {x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x^n]^2,x]

[Out]

x/2 + (2^(-2 - n^(-1))*E^((2*I)*a)*x*Gamma[n^(-1), (-2*I)*b*x^n])/(n*((-I)*b*x^n)^n^(-1)) + (2^(-2 - n^(-1))*x
*Gamma[n^(-1), (2*I)*b*x^n])/(E^((2*I)*a)*n*(I*b*x^n)^n^(-1))

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 3366

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f, n}, x]

Rule 3367

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[p, 1]

Rubi steps

\begin {align*} \int \sin ^2\left (a+b x^n\right ) \, dx &=\int \left (\frac {1}{2}-\frac {1}{2} \cos \left (2 a+2 b x^n\right )\right ) \, dx\\ &=\frac {x}{2}-\frac {1}{2} \int \cos \left (2 a+2 b x^n\right ) \, dx\\ &=\frac {x}{2}-\frac {1}{4} \int e^{-2 i a-2 i b x^n} \, dx-\frac {1}{4} \int e^{2 i a+2 i b x^n} \, dx\\ &=\frac {x}{2}+\frac {2^{-2-\frac {1}{n}} e^{2 i a} x \left (-i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i b x^n\right )}{n}+\frac {2^{-2-\frac {1}{n}} e^{-2 i a} x \left (i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i b x^n\right )}{n}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 94, normalized size = 0.94 \[ \frac {x \left (e^{2 i a} 2^{-1/n} \left (-i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i b x^n\right )+e^{-2 i a} 2^{-1/n} \left (i b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i b x^n\right )+2 n\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x^n]^2,x]

[Out]

(x*(2*n + (E^((2*I)*a)*Gamma[n^(-1), (-2*I)*b*x^n])/(2^n^(-1)*((-I)*b*x^n)^n^(-1)) + Gamma[n^(-1), (2*I)*b*x^n
]/(2^n^(-1)*E^((2*I)*a)*(I*b*x^n)^n^(-1))))/(4*n)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\cos \left (b x^{n} + a\right )^{2} + 1, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-cos(b*x^n + a)^2 + 1, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x^n + a)^2, x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \sin ^{2}\left (a +b \,x^{n}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x - \frac {1}{2} \, \int \cos \left (2 \, b x^{n} + 2 \, a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x - 1/2*integrate(cos(2*b*x^n + 2*a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^n)^2,x)

[Out]

int(sin(a + b*x^n)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b*x**n)**2,x)

[Out]

Integral(sin(a + b*x**n)**2, x)

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