3.1.84 \(\int x^{3+m} \sin ^2(a+b x) \, dx\) [84]

Optimal. Leaf size=97 \[ \frac {x^{4+m}}{2 (4+m)}+\frac {2^{-6-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (4+m,-2 i b x)}{b^4}+\frac {2^{-6-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (4+m,2 i b x)}{b^4} \]

[Out]

1/2*x^(4+m)/(4+m)+2^(-6-m)*exp(2*I*a)*x^m*GAMMA(4+m,-2*I*b*x)/b^4/((-I*b*x)^m)+2^(-6-m)*x^m*GAMMA(4+m,2*I*b*x)
/b^4/exp(2*I*a)/((I*b*x)^m)

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Rubi [A]
time = 0.11, antiderivative size = 97, 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 = {3393, 3388, 2212} \begin {gather*} \frac {e^{2 i a} 2^{-m-6} x^m (-i b x)^{-m} \text {Gamma}(m+4,-2 i b x)}{b^4}+\frac {e^{-2 i a} 2^{-m-6} x^m (i b x)^{-m} \text {Gamma}(m+4,2 i b x)}{b^4}+\frac {x^{m+4}}{2 (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3 + m)*Sin[a + b*x]^2,x]

[Out]

x^(4 + m)/(2*(4 + m)) + (2^(-6 - m)*E^((2*I)*a)*x^m*Gamma[4 + m, (-2*I)*b*x])/(b^4*((-I)*b*x)^m) + (2^(-6 - m)
*x^m*Gamma[4 + m, (2*I)*b*x])/(b^4*E^((2*I)*a)*(I*b*x)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin {align*} \int x^{3+m} \sin ^2(a+b x) \, dx &=\int \left (\frac {x^{3+m}}{2}-\frac {1}{2} x^{3+m} \cos (2 a+2 b x)\right ) \, dx\\ &=\frac {x^{4+m}}{2 (4+m)}-\frac {1}{2} \int x^{3+m} \cos (2 a+2 b x) \, dx\\ &=\frac {x^{4+m}}{2 (4+m)}-\frac {1}{4} \int e^{-i (2 a+2 b x)} x^{3+m} \, dx-\frac {1}{4} \int e^{i (2 a+2 b x)} x^{3+m} \, dx\\ &=\frac {x^{4+m}}{2 (4+m)}+\frac {2^{-6-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (4+m,-2 i b x)}{b^4}+\frac {2^{-6-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (4+m,2 i b x)}{b^4}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 118, normalized size = 1.22 \begin {gather*} \frac {2^{-6-m} x^m \left (b^2 x^2\right )^{-m} \left (2^{5+m} b^4 x^4 \left (b^2 x^2\right )^m+(4+m) (-i b x)^m \Gamma (4+m,2 i b x) (\cos (a)-i \sin (a))^2+(4+m) (i b x)^m \Gamma (4+m,-2 i b x) (\cos (a)+i \sin (a))^2\right )}{b^4 (4+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3 + m)*Sin[a + b*x]^2,x]

[Out]

(2^(-6 - m)*x^m*(2^(5 + m)*b^4*x^4*(b^2*x^2)^m + (4 + m)*((-I)*b*x)^m*Gamma[4 + m, (2*I)*b*x]*(Cos[a] - I*Sin[
a])^2 + (4 + m)*(I*b*x)^m*Gamma[4 + m, (-2*I)*b*x]*(Cos[a] + I*Sin[a])^2))/(b^4*(4 + m)*(b^2*x^2)^m)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int x^{3+m} \left (\sin ^{2}\left (b x +a \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3+m)*sin(b*x+a)^2,x)

[Out]

int(x^(3+m)*sin(b*x+a)^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^(3+m)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/2*((m + 4)*integrate(x^3*x^m*cos(2*b*x + 2*a), x) - e^(m*log(x) + 4*log(x)))/(m + 4)

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Fricas [A]
time = 0.13, size = 77, normalized size = 0.79 \begin {gather*} \frac {4 \, b x x^{m + 3} + {\left (-i \, m - 4 i\right )} e^{\left (-{\left (m + 3\right )} \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 4, 2 i \, b x\right ) + {\left (i \, m + 4 i\right )} e^{\left (-{\left (m + 3\right )} \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 4, -2 i \, b x\right )}{8 \, {\left (b m + 4 \, b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3+m)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(4*b*x*x^(m + 3) + (-I*m - 4*I)*e^(-(m + 3)*log(2*I*b) - 2*I*a)*gamma(m + 4, 2*I*b*x) + (I*m + 4*I)*e^(-(m
+ 3)*log(-2*I*b) + 2*I*a)*gamma(m + 4, -2*I*b*x))/(b*m + 4*b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m + 3} \sin ^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3+m)*sin(b*x+a)**2,x)

[Out]

Integral(x**(m + 3)*sin(a + b*x)**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^(3+m)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^(m + 3)*sin(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{m+3}\,{\sin \left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m + 3)*sin(a + b*x)^2,x)

[Out]

int(x^(m + 3)*sin(a + b*x)^2, x)

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