\(\int x^{-1+m} \cos ^2(a+b x) \, dx\) [115]

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\frac {x^m}{2 m}-2^{-2-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (m,-2 i b x)-2^{-2-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (m,2 i b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3393, 3388, 2212} \[ \int x^{-1+m} \cos ^2(a+b x) \, dx=e^{2 i a} \left (-2^{-m-2}\right ) x^m (-i b x)^{-m} \Gamma (m,-2 i b x)-e^{-2 i a} 2^{-m-2} x^m (i b x)^{-m} \Gamma (m,2 i b x)+\frac {x^m}{2 m} \]

[In]

Int[x^(-1 + m)*Cos[a + b*x]^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\frac {1}{4} x^m \left (\frac {2}{m}-2^{-m} e^{2 i a} (-i b x)^{-m} \Gamma (m,-2 i b x)-2^{-m} e^{-2 i a} (i b x)^{-m} \Gamma (m,2 i b x)\right ) \]

[In]

Integrate[x^(-1 + m)*Cos[a + b*x]^2,x]

[Out]

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

Maple [F]

\[\int x^{-1+m} \left (\cos ^{2}\left (b x +a \right )\right )d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\frac {4 \, b x x^{m - 1} + i \, m e^{\left (-{\left (m - 1\right )} \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m, 2 i \, b x\right ) - i \, m e^{\left (-{\left (m - 1\right )} \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m, -2 i \, b x\right )}{8 \, b m} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\int x^{m - 1} \cos ^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\int { x^{m - 1} \cos \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\int { x^{m - 1} \cos \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^{-1+m} \cos ^2(a+b x) \, dx=\int x^{m-1}\,{\cos \left (a+b\,x\right )}^2 \,d x \]

[In]

int(x^(m - 1)*cos(a + b*x)^2,x)

[Out]

int(x^(m - 1)*cos(a + b*x)^2, x)