\(\int x^{2+m} \sin (a+b x) \, dx\) [78]

Optimal result

Integrand size = 12, antiderivative size = 75 \[ \int x^{2+m} \sin (a+b x) \, dx=\frac {e^{i a} x^m (-i b x)^{-m} \Gamma (3+m,-i b x)}{2 b^3}+\frac {e^{-i a} x^m (i b x)^{-m} \Gamma (3+m,i b x)}{2 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int x^{2+m} \sin (a+b x) \, dx=\frac {e^{i a} x^m (-i b x)^{-m} \Gamma (3+m,-i b x)}{2 b^3}+\frac {e^{-i a} x^m (i b x)^{-m} \Gamma (3+m,i b x)}{2 b^3} \] Input:

Integrate[x^(2 + m)*Sin[a + b*x],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3789, 2612}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^{m+2} \sin (a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int x^{m+2} \sin (a+b x)dx\)

\(\Big \downarrow \) 3789

\(\displaystyle \frac {1}{2} i \int e^{-i (a+b x)} x^{m+2}dx-\frac {1}{2} i \int e^{i (a+b x)} x^{m+2}dx\)

\(\Big \downarrow \) 2612

\(\displaystyle \frac {e^{i a} x^m (-i b x)^{-m} \Gamma (m+3,-i b x)}{2 b^3}+\frac {e^{-i a} x^m (i b x)^{-m} \Gamma (m+3,i b x)}{2 b^3}\)

Input:

Int[x^(2 + m)*Sin[a + b*x],x]
 

Output:

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

Defintions of rubi rules used

rule 2612

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3789

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
/2   Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2   Int[(c + d*x)^m*E 
^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.51 (sec) , antiderivative size = 353, normalized size of antiderivative = 4.71

method	result	size
meijerg	\(\frac {2^{2+m} \left (b^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \sqrt {\pi }\, \left (\frac {3 \,2^{-3-m} x^{2+m} \left (b^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (2+\frac {2 m}{3}\right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (3+m \right ) b}-\frac {2^{-2-m} x^{2+m} \left (b^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (2+m \right ) m \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, b}+\frac {2^{-2-m} x^{2+m} \left (b^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (2+m \right ) \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }\, b}\right ) \sin \left (a \right )}{b^{2}}+2^{2+m} b^{-3-m} \sqrt {\pi }\, \left (-\frac {2^{-2-m} x^{1+m} b^{1+m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }}+\frac {2^{-2-m} x^{2+m} b^{2+m} \left (m^{2}+5 m +4\right ) \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (4+m \right )}+\frac {2^{-2-m} x^{2+m} b^{2+m} \left (2+m \right ) \left (1+m \right ) \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }}\right ) \cos \left (a \right )\)	\(353\)

Input:

int(x^(2+m)*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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))*sin(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)*si 
n(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))*cos(a)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.69 \[ \int x^{2+m} \sin (a+b x) \, dx=-\frac {e^{\left (-{\left (m + 2\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m + 3, i \, b x\right ) + e^{\left (-{\left (m + 2\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m + 3, -i \, b x\right )}{2 \, b} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**(m + 2)*sin(a + b*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{2+m} \sin (a+b x) \, dx=\frac {-x^{m} \cos \left (b x +a \right ) b^{2} x^{2}+x^{m} \cos \left (b x +a \right ) m^{2}+3 x^{m} \cos \left (b x +a \right ) m +2 x^{m} \cos \left (b x +a \right )+x^{m} \sin \left (b x +a \right ) b m x +2 x^{m} \sin \left (b x +a \right ) b x +x^{m} m^{2}+3 x^{m} m +2 x^{m}-2 \left (\int \frac {x^{m}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x +x}d x \right ) m^{3}-6 \left (\int \frac {x^{m}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x +x}d x \right ) m^{2}-4 \left (\int \frac {x^{m}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x +x}d x \right ) m}{b^{3}} \] Input:

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

Output:

( - x**m*cos(a + b*x)*b**2*x**2 + x**m*cos(a + b*x)*m**2 + 3*x**m*cos(a + 
b*x)*m + 2*x**m*cos(a + b*x) + x**m*sin(a + b*x)*b*m*x + 2*x**m*sin(a + b* 
x)*b*x + x**m*m**2 + 3*x**m*m + 2*x**m - 2*int(x**m/(tan((a + b*x)/2)**2*x 
 + x),x)*m**3 - 6*int(x**m/(tan((a + b*x)/2)**2*x + x),x)*m**2 - 4*int(x** 
m/(tan((a + b*x)/2)**2*x + x),x)*m)/b**3