3.2.43 \(\int x^m \sin ^2(a+b x^n) \, dx\) [143]

Optimal. Leaf size=139 \[ \frac {x^{1+m}}{2 (1+m)}+\frac {2^{-\frac {1+m+2 n}{n}} e^{2 i a} x^{1+m} \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 i b x^n\right )}{n}+\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 i a} x^{1+m} \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 i b x^n\right )}{n} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3506, 3505, 2250} \begin {gather*} \frac {e^{2 i a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (-i b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},-2 i b x^n\right )}{n}+\frac {e^{-2 i a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (i b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},2 i b x^n\right )}{n}+\frac {x^{m+1}}{2 (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*Sin[a + b*x^n]^2,x]

[Out]

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

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3505

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

Rule 3506

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.38, size = 129, normalized size = 0.93 \begin {gather*} \frac {x^{1+m} \left (2 n+2^{-\frac {1+m}{n}} e^{2 i a} (1+m) \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 i b x^n\right )+2^{-\frac {1+m}{n}} e^{-2 i a} (1+m) \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 i b x^n\right )\right )}{4 (1+m) n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*Sin[a + b*x^n]^2,x]

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int x^{m} \left (\sin ^{2}\left (a +b \,x^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

integral(-x^m*cos(b*x^n + a)^2 + x^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^m*sin(b*x^n + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sin(a + b*x^n)^2,x)

[Out]

int(x^m*sin(a + b*x^n)^2, x)

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