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

Optimal. Leaf size=79 \[ \frac{i e^{i a} x^m (-i b x)^{-m} \text{Gamma}(m+3,-i b x)}{2 b^3}-\frac{i e^{-i a} x^m (i b x)^{-m} \text{Gamma}(m+3,i b x)}{2 b^3} \]

[Out]

((I/2)*E^(I*a)*x^m*Gamma[3 + m, (-I)*b*x])/(b^3*((-I)*b*x)^m) - ((I/2)*x^m*Gamma[3 + m, I*b*x])/(b^3*E^(I*a)*(
I*b*x)^m)

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Rubi [A]  time = 0.0770637, antiderivative size = 79, normalized size of antiderivative = 1., 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} \[ \frac{i e^{i a} x^m (-i b x)^{-m} \text{Gamma}(m+3,-i b x)}{2 b^3}-\frac{i e^{-i a} x^m (i b x)^{-m} \text{Gamma}(m+3,i b x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^(2 + m)*Cos[a + b*x],x]

[Out]

((I/2)*E^(I*a)*x^m*Gamma[3 + m, (-I)*b*x])/(b^3*((-I)*b*x)^m) - ((I/2)*x^m*Gamma[3 + m, I*b*x])/(b^3*E^(I*a)*(
I*b*x)^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]

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

Mathematica [A]  time = 0.0158999, size = 79, normalized size = 1. \[ \frac{i e^{i a} x^m (-i b x)^{-m} \text{Gamma}(m+3,-i b x)}{2 b^3}-\frac{i e^{-i a} x^m (i b x)^{-m} \text{Gamma}(m+3,i b x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(2 + m)*Cos[a + b*x],x]

[Out]

((I/2)*E^(I*a)*x^m*Gamma[3 + m, (-I)*b*x])/(b^3*((-I)*b*x)^m) - ((I/2)*x^m*Gamma[3 + m, I*b*x])/(b^3*E^(I*a)*(
I*b*x)^m)

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Maple [C]  time = 0.078, size = 354, normalized size = 4.5 \begin{align*}{\frac{{2}^{2+m}\sqrt{\pi }\cos \left ( a \right ) }{{b}^{2}} \left ({b}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 3\,{\frac{{2}^{-3-m}{x}^{2+m} \left ({b}^{2} \right ) ^{3/2+m/2} \left ( 2+2/3\,m \right ) \sin \left ( bx \right ) }{\sqrt{\pi } \left ( 3+m \right ) b}}-{\frac{{2}^{-2-m}{x}^{2+m} \left ( 2+m \right ) m\sin \left ( bx \right ) }{\sqrt{\pi }b} \left ({b}^{2} \right ) ^{{\frac{3}{2}}+{\frac{m}{2}}} \left ( bx \right ) ^{-{\frac{3}{2}}-m}{\it LommelS1} \left ( m+{\frac{1}{2}},{\frac{3}{2}},bx \right ) }+{\frac{{2}^{-2-m}{x}^{2+m} \left ( 2+m \right ) \left ( \cos \left ( bx \right ) xb-\sin \left ( bx \right ) \right ) }{\sqrt{\pi }b} \left ({b}^{2} \right ) ^{{\frac{3}{2}}+{\frac{m}{2}}} \left ( bx \right ) ^{-{\frac{5}{2}}-m}{\it LommelS1} \left ( m+{\frac{3}{2}},{\frac{1}{2}},bx \right ) } \right ) }-{2}^{2+m}{b}^{-3-m}\sqrt{\pi } \left ( -{\frac{{2}^{-2-m}{x}^{1+m}{b}^{1+m} \left ( \cos \left ( bx \right ) xb-\sin \left ( bx \right ) \right ) }{\sqrt{\pi }}}+{\frac{{2}^{-2-m}{x}^{2+m}{b}^{2+m} \left ({m}^{2}+5\,m+4 \right ) \sin \left ( bx \right ) }{\sqrt{\pi } \left ( 4+m \right ) } \left ( bx \right ) ^{-{\frac{3}{2}}-m}{\it LommelS1} \left ( m+{\frac{3}{2}},{\frac{3}{2}},bx \right ) }+{\frac{{2}^{-2-m}{x}^{2+m}{b}^{2+m} \left ( 2+m \right ) \left ( 1+m \right ) \left ( \cos \left ( bx \right ) xb-\sin \left ( bx \right ) \right ) }{\sqrt{\pi }} \left ( bx \right ) ^{-{\frac{5}{2}}-m}{\it LommelS1} \left ( m+{\frac{1}{2}},{\frac{1}{2}},bx \right ) } \right ) \sin \left ( a \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m + 2)*cos(b*x + a), x)

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Fricas [A]  time = 1.68011, size = 153, normalized size = 1.94 \begin{align*} \frac{i \, e^{\left (-{\left (m + 2\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m + 3, i \, b x\right ) - i \, e^{\left (-{\left (m + 2\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m + 3, -i \, b x\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(m + 2)*cos(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m + 2)*cos(b*x + a), x)