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

Optimal. Leaf size=65 \[ -\frac {1}{2} e^{i a} x^m (-i b x)^{-m} \Gamma (m,-i b x)-\frac {1}{2} e^{-i a} x^m (i b x)^{-m} \Gamma (m,i b x) \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3307, 2181} \[ -\frac {1}{2} e^{i a} x^m (-i b x)^{-m} \text {Gamma}(m,-i b x)-\frac {1}{2} e^{-i a} x^m (i b x)^{-m} \text {Gamma}(m,i b x) \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + m)*Cos[a + b*x],x]

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 62, normalized size = 0.95 \[ \frac {1}{2} e^{-i a} x^m \left (-e^{2 i a} (-i b x)^{-m} \Gamma (m,-i b x)-(i b x)^{-m} \Gamma (m,i b x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + m)*Cos[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.68, size = 50, normalized size = 0.77 \[ \frac {i \, e^{\left (-{\left (m - 1\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m, i \, b x\right ) - i \, e^{\left (-{\left (m - 1\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m, -i \, b x\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m - 1} \cos \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m - 1)*cos(b*x + a), x)

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maple [C]  time = 0.11, size = 427, normalized size = 6.57 \[ 2^{-1+m} \left (b^{2}\right )^{-\frac {m}{2}} \sqrt {\pi }\, \left (\frac {3 x^{-1+m} 2^{-m} \left (b^{2}\right )^{\frac {m}{2}} \left (2 x^{2} b^{2}+2 m +4\right ) \sin \left (b x \right )}{\sqrt {\pi }\, m \left (6+3 m \right ) b}+\frac {2^{1-m} x^{-1+m} \left (b^{2}\right )^{\frac {m}{2}} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, m b}-\frac {3 x^{2+m} 2^{1-m} \left (b^{2}\right )^{\frac {m}{2}} b^{2} \left (b x \right )^{-\frac {3}{2}-m} \LommelS 1 \left (m +\frac {3}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, m \left (6+3 m \right )}-\frac {x^{2+m} 2^{1-m} \left (b^{2}\right )^{\frac {m}{2}} b^{2} \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \LommelS 1 \left (m +\frac {1}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }\, m}\right ) \cos \relax (a )-2^{-1+m} b^{-m} \sqrt {\pi }\, \left (\frac {2^{1-m} x^{m} b^{m} \sin \left (b x \right )}{\sqrt {\pi }\, \left (1+m \right )}-\frac {2^{1-m} x^{m} b^{m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, \left (1+m \right ) m}-\frac {x^{2+m} b^{2+m} 2^{1-m} \left (b x \right )^{-\frac {3}{2}-m} \LommelS 1 \left (m +\frac {1}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (1+m \right )}+\frac {x^{2+m} b^{2+m} 2^{1-m} \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \LommelS 1 \left (m +\frac {3}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }\, \left (1+m \right ) m}\right ) \sin \relax (a ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^(-1+m)*(b^2)^(-1/2*m)*Pi^(1/2)*(3/Pi^(1/2)/m*x^(-1+m)*2^(-m)*(b^2)^(1/2*m)*(2*b^2*x^2+2*m+4)/(6+3*m)/b*sin(b
*x)+2^(1-m)/Pi^(1/2)/m*x^(-1+m)*(b^2)^(1/2*m)/b*(cos(b*x)*x*b-sin(b*x))-3/Pi^(1/2)/m*x^(2+m)*2^(1-m)*(b^2)^(1/
2*m)*b^2/(6+3*m)*(b*x)^(-3/2-m)*LommelS1(m+3/2,3/2,b*x)*sin(b*x)-1/Pi^(1/2)/m*x^(2+m)*2^(1-m)*(b^2)^(1/2*m)*b^
2*(b*x)^(-5/2-m)*(cos(b*x)*x*b-sin(b*x))*LommelS1(m+1/2,1/2,b*x))*cos(a)-2^(-1+m)*b^(-m)*Pi^(1/2)*(2^(1-m)/Pi^
(1/2)/(1+m)*x^m*b^m*sin(b*x)-2^(1-m)/Pi^(1/2)/(1+m)*x^m*b^m/m*(cos(b*x)*x*b-sin(b*x))-1/Pi^(1/2)/(1+m)*x^(2+m)
*b^(2+m)*2^(1-m)*(b*x)^(-3/2-m)*LommelS1(m+1/2,3/2,b*x)*sin(b*x)+1/Pi^(1/2)/(1+m)*x^(2+m)*b^(2+m)*2^(1-m)/m*(b
*x)^(-5/2-m)*(cos(b*x)*x*b-sin(b*x))*LommelS1(m+3/2,1/2,b*x))*sin(a)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m - 1} \cos \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m - 1)*cos(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^{m-1}\,\cos \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m - 1)*cos(a + b*x),x)

[Out]

int(x^(m - 1)*cos(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m - 1} \cos {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(m - 1)*cos(a + b*x), x)

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