\(\int x^m \cos ^2(a+b x) \, dx\) [114]

Optimal result

Integrand size = 12, antiderivative size = 103 \[ \int x^m \cos ^2(a+b x) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {i 2^{-3-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (1+m,-2 i b x)}{b}+\frac {i 2^{-3-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (1+m,2 i b x)}{b} \]

[Out]

1/2*x^(1+m)/(1+m)-I*2^(-3-m)*exp(2*I*a)*x^m*GAMMA(1+m,-2*I*b*x)/b/((-I*b*x)^m)+I*2^(-3-m)*x^m*GAMMA(1+m,2*I*b*
x)/b/exp(2*I*a)/((I*b*x)^m)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3393, 3388, 2212} \[ \int x^m \cos ^2(a+b x) \, dx=-\frac {i e^{2 i a} 2^{-m-3} x^m (-i b x)^{-m} \Gamma (m+1,-2 i b x)}{b}+\frac {i e^{-2 i a} 2^{-m-3} x^m (i b x)^{-m} \Gamma (m+1,2 i b x)}{b}+\frac {x^{m+1}}{2 (m+1)} \]

[In]

Int[x^m*Cos[a + b*x]^2,x]

[Out]

x^(1 + m)/(2*(1 + m)) - (I*2^(-3 - m)*E^((2*I)*a)*x^m*Gamma[1 + m, (-2*I)*b*x])/(b*((-I)*b*x)^m) + (I*2^(-3 -
m)*x^m*Gamma[1 + m, (2*I)*b*x])/(b*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*} \text {integral}& = \int \left (\frac {x^m}{2}+\frac {1}{2} x^m \cos (2 a+2 b x)\right ) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{2} \int x^m \cos (2 a+2 b x) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{4} \int e^{-i (2 a+2 b x)} x^m \, dx+\frac {1}{4} \int e^{i (2 a+2 b x)} x^m \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}-\frac {i 2^{-3-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (1+m,-2 i b x)}{b}+\frac {i 2^{-3-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (1+m,2 i b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int x^m \cos ^2(a+b x) \, dx=\frac {1}{8} x^m \left (\frac {4 x}{1+m}-2^{-m} e^{2 i a} x (-i b x)^{-1-m} \Gamma (1+m,-2 i b x)-2^{-m} e^{-2 i a} x (i b x)^{-1-m} \Gamma (1+m,2 i b x)\right ) \]

[In]

Integrate[x^m*Cos[a + b*x]^2,x]

[Out]

(x^m*((4*x)/(1 + m) - (E^((2*I)*a)*x*((-I)*b*x)^(-1 - m)*Gamma[1 + m, (-2*I)*b*x])/2^m - (x*(I*b*x)^(-1 - m)*G
amma[1 + m, (2*I)*b*x])/(2^m*E^((2*I)*a))))/8

Maple [F]

\[\int x^{m} \left (\cos ^{2}\left (b x +a \right )\right )d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.67 \[ \int x^m \cos ^2(a+b x) \, dx=\frac {4 \, b x x^{m} + {\left (i \, m + i\right )} e^{\left (-m \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 1, 2 i \, b x\right ) + {\left (-i \, m - i\right )} e^{\left (-m \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 1, -2 i \, b x\right )}{8 \, {\left (b m + b\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^m \cos ^2(a+b x) \, dx=\int x^{m} \cos ^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral(x**m*cos(a + b*x)**2, x)

Maxima [F]

\[ \int x^m \cos ^2(a+b x) \, dx=\int { x^{m} \cos \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^m \cos ^2(a+b x) \, dx=\int { x^{m} \cos \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

integrate(x^m*cos(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int x^m \cos ^2(a+b x) \, dx=\int x^m\,{\cos \left (a+b\,x\right )}^2 \,d x \]

[In]

int(x^m*cos(a + b*x)^2,x)

[Out]

int(x^m*cos(a + b*x)^2, x)