3.113 \(\int x^{1+m} \cos ^2(a+b x) \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.137478, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {3312, 3307, 2181} \[ \frac{e^{2 i a} 2^{-m-4} x^m (-i b x)^{-m} \text{Gamma}(m+2,-2 i b x)}{b^2}+\frac{e^{-2 i a} 2^{-m-4} x^m (i b x)^{-m} \text{Gamma}(m+2,2 i b x)}{b^2}+\frac{x^{m+2}}{2 (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[x^(1 + m)*Cos[a + b*x]^2,x]

[Out]

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

Rule 3312

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])
)

Rule 3307

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 2181

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

Rubi steps

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

Mathematica [A]  time = 0.0975459, size = 90, normalized size = 0.93 \[ \frac{1}{16} x^m \left (\frac{e^{2 i a} 2^{-m} (-i b x)^{-m} \text{Gamma}(m+2,-2 i b x)}{b^2}+\frac{e^{-2 i a} 2^{-m} (i b x)^{-m} \text{Gamma}(m+2,2 i b x)}{b^2}+\frac{8 x^2}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(1 + m)*Cos[a + b*x]^2,x]

[Out]

(x^m*((8*x^2)/(2 + m) + (E^((2*I)*a)*Gamma[2 + m, (-2*I)*b*x])/(2^m*b^2*((-I)*b*x)^m) + Gamma[2 + m, (2*I)*b*x
]/(2^m*b^2*E^((2*I)*a)*(I*b*x)^m)))/16

________________________________________________________________________________________

Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{x}^{1+m} \left ( \cos \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1+m)*cos(b*x+a)^2,x)

[Out]

int(x^(1+m)*cos(b*x+a)^2,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (m + 2\right )} \int x x^{m} \cos \left (2 \, b x + 2 \, a\right )\,{d x} + e^{\left (m \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{2 \,{\left (m + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*cos(b*x+a)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.1073, size = 235, normalized size = 2.42 \begin{align*} \frac{4 \, b x x^{m + 1} +{\left (i \, m + 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 2, 2 i \, b x\right ) +{\left (-i \, m - 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 2, -2 i \, b x\right )}{8 \,{\left (b m + 2 \, b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*cos(b*x+a)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \cos ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1+m)*cos(b*x+a)**2,x)

[Out]

Integral(x**(m + 1)*cos(a + b*x)**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \cos \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+m)*cos(b*x+a)^2,x, algorithm="giac")

[Out]

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