\(\int x^{-1+m} \sin (a+b x) \, dx\) [81]

Optimal result

Integrand size = 12, antiderivative size = 69 \[ \int x^{-1+m} \sin (a+b x) \, dx=\frac {1}{2} i e^{i a} x^m (-i b x)^{-m} \Gamma (m,-i b x)-\frac {1}{2} i e^{-i a} x^m (i b x)^{-m} \Gamma (m,i b x) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int x^{-1+m} \sin (a+b x) \, dx=\frac {1}{2} i e^{-i a} x^m \left (e^{2 i a} (-i b x)^{-m} \Gamma (m,-i b x)-(i b x)^{-m} \Gamma (m,i b x)\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 69, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int x^{m-1} \sin (a+b x)dx\)

\(\Big \downarrow \) 3789

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

\(\Big \downarrow \) 2612

\(\displaystyle \frac {1}{2} i e^{i a} x^m (-i b x)^{-m} \Gamma (m,-i b x)-\frac {1}{2} i e^{-i a} x^m (i b x)^{-m} \Gamma (m,i b x)\)

Input:

Int[x^(-1 + m)*Sin[a + b*x],x]
 

Output:

((I/2)*E^(I*a)*x^m*Gamma[m, (-I)*b*x])/((-I)*b*x)^m - ((I/2)*x^m*Gamma[m, 
I*b*x])/(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 = 426, normalized size of antiderivative = 6.17

method	result	size
meijerg	\(2^{-1+m} \left (b^{2}\right )^{-\frac {m}{2}} \sqrt {\pi }\, \left (\frac {3 x^{-1+m} 2^{-m} \left (b^{2}\right )^{\frac {m}{2}} \left (2 x^{2} b^{2}+2 m +4\right ) \sin \left (b x \right )}{\sqrt {\pi }\, m \left (6+3 m \right ) b}+\frac {2^{1-m} x^{-1+m} \left (b^{2}\right )^{\frac {m}{2}} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, m b}-\frac {3 x^{2+m} 2^{1-m} \left (b^{2}\right )^{\frac {m}{2}} b^{2} \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, m \left (6+3 m \right )}-\frac {x^{2+m} 2^{1-m} \left (b^{2}\right )^{\frac {m}{2}} b^{2} \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }\, m}\right ) \sin \left (a \right )+2^{-1+m} b^{-m} \sqrt {\pi }\, \left (\frac {2^{1-m} x^{m} b^{m} \sin \left (b x \right )}{\sqrt {\pi }\, \left (1+m \right )}-\frac {2^{1-m} x^{m} b^{m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, \left (1+m \right ) m}-\frac {x^{2+m} b^{2+m} 2^{1-m} \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (1+m \right )}+\frac {x^{2+m} b^{2+m} 2^{1-m} \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }\, \left (1+m \right ) m}\right ) \cos \left (a \right )\)	\(426\)

Input:

int(x^(-1+m)*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int x^{-1+m} \sin (a+b x) \, dx=-\frac {e^{\left (-{\left (m - 1\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m, i \, b x\right ) + e^{\left (-{\left (m - 1\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m, -i \, b x\right )}{2 \, b} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^{-1+m} \sin (a+b x) \, dx=\frac {x^{m} \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+x^{m} \sin \left (b x +a \right )-2 x^{m} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \left (\int \frac {x^{m} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x +x}d x \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} m +2 \left (\int \frac {x^{m} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x +x}d x \right ) m}{m \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right )} \] Input:

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

Output:

(x**m*sin(a + b*x)*tan((a + b*x)/2)**2 + x**m*sin(a + b*x) - 2*x**m*tan((a 
 + b*x)/2) + 2*int((x**m*tan((a + b*x)/2))/(tan((a + b*x)/2)**2*x + x),x)* 
tan((a + b*x)/2)**2*m + 2*int((x**m*tan((a + b*x)/2))/(tan((a + b*x)/2)**2 
*x + x),x)*m)/(m*(tan((a + b*x)/2)**2 + 1))