Integrand size = 12, antiderivative size = 79 \[ \int x^{-3+m} \sin (a+b x) \, dx=-\frac {1}{2} i b^2 e^{i a} x^m (-i b x)^{-m} \Gamma (-2+m,-i b x)+\frac {1}{2} i b^2 e^{-i a} x^m (i b x)^{-m} \Gamma (-2+m,i b x) \] Output:
-1/2*I*b^2*exp(I*a)*x^m*GAMMA(-2+m,-I*b*x)/((-I*b*x)^m)+1/2*I*b^2*x^m*GAMM A(-2+m,I*b*x)/exp(I*a)/((I*b*x)^m)
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int x^{-3+m} \sin (a+b x) \, dx=-\frac {1}{2} i b^2 e^{i a} x^m (-i b x)^{-m} \Gamma (-2+m,-i b x)+\frac {1}{2} i b^2 e^{-i a} x^m (i b x)^{-m} \Gamma (-2+m,i b x) \] Input:
Integrate[x^(-3 + m)*Sin[a + b*x],x]
Output:
((-1/2*I)*b^2*E^(I*a)*x^m*Gamma[-2 + m, (-I)*b*x])/((-I)*b*x)^m + ((I/2)*b ^2*x^m*Gamma[-2 + m, I*b*x])/(E^(I*a)*(I*b*x)^m)
Time = 0.27 (sec) , antiderivative size = 79, 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.
\(\displaystyle \int x^{m-3} \sin (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^{m-3} \sin (a+b x)dx\) |
\(\Big \downarrow \) 3789 |
\(\displaystyle \frac {1}{2} i \int e^{-i (a+b x)} x^{m-3}dx-\frac {1}{2} i \int e^{i (a+b x)} x^{m-3}dx\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {1}{2} i e^{-i a} b^2 x^m (i b x)^{-m} \Gamma (m-2,i b x)-\frac {1}{2} i e^{i a} b^2 x^m (-i b x)^{-m} \Gamma (m-2,-i b x)\) |
Input:
Int[x^(-3 + m)*Sin[a + b*x],x]
Output:
((-1/2*I)*b^2*E^(I*a)*x^m*Gamma[-2 + m, (-I)*b*x])/((-I)*b*x)^m + ((I/2)*b ^2*x^m*Gamma[-2 + m, I*b*x])/(E^(I*a)*(I*b*x)^m)
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.52 (sec) , antiderivative size = 599, normalized size of antiderivative = 7.58
method | result | size |
meijerg | \(2^{-3+m} b^{2} \left (b^{2}\right )^{-\frac {m}{2}} \sqrt {\pi }\, \left (\frac {2^{2-m} x^{-3+m} \left (b^{2}\right )^{\frac {m}{2}} \left (-2 x^{4} b^{4}+2 x^{2} b^{2} m^{2}+2 x^{2} b^{2} m -4 x^{2} b^{2}+2 m^{3}+2 m^{2}-4 m \right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (-2+m \right ) b^{3} m \left (2+m \right ) \left (-1+m \right )}-\frac {2^{-m +3} x^{-3+m} \left (b^{2}\right )^{\frac {m}{2}} \left (x^{2} b^{2}-m^{2}+m \right ) \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, \left (-2+m \right ) b^{3} m \left (-1+m \right )}+\frac {2^{-m +3} x^{2+m} b^{2} \left (b^{2}\right )^{\frac {m}{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 }\, \left (-2+m \right ) m \left (2+m \right ) \left (-1+m \right )}+\frac {2^{-m +3} x^{2+m} b^{2} \left (b^{2}\right )^{\frac {m}{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 }\, \left (-2+m \right ) m \left (-1+m \right )}\right ) \sin \left (a \right )+2^{-3+m} b^{2-m} \sqrt {\pi }\, \left (\frac {2^{2-m} x^{-2+m} b^{-2+m} \left (-2 x^{2} b^{2}+2 m^{2}-2 m -4\right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (-1+m \right ) \left (1+m \right ) \left (-2+m \right )}+\frac {2^{-m +3} x^{-2+m} b^{-2+m} \left (x^{2} b^{2}-m^{2}-m \right ) \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, \left (-1+m \right ) \left (1+m \right ) \left (-2+m \right ) m}+\frac {2^{-m +3} x^{2+m} b^{2+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 ) \left (1+m \right ) \left (-2+m \right )}-\frac {2^{-m +3} x^{2+m} b^{2+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 ) \left (1+m \right ) \left (-2+m \right ) m}\right ) \cos \left (a \right )\) | \(599\) |
Input:
int(x^(-3+m)*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
2^(-3+m)*b^2*(b^2)^(-1/2*m)*Pi^(1/2)*(2^(2-m)/Pi^(1/2)/(-2+m)*x^(-3+m)/b^3 *(b^2)^(1/2*m)*(-2*b^4*x^4+2*b^2*m^2*x^2+2*b^2*m*x^2-4*b^2*x^2+2*m^3+2*m^2 -4*m)/m/(2+m)/(-1+m)*sin(b*x)-2^(-m+3)/Pi^(1/2)/(-2+m)*x^(-3+m)/b^3*(b^2)^ (1/2*m)*(b^2*x^2-m^2+m)/m/(-1+m)*(cos(b*x)*x*b-sin(b*x))+2^(-m+3)/Pi^(1/2) /(-2+m)*x^(2+m)*b^2*(b^2)^(1/2*m)/m/(2+m)/(-1+m)*(b*x)^(-3/2-m)*LommelS1(m +3/2,3/2,b*x)*sin(b*x)+2^(-m+3)/Pi^(1/2)/(-2+m)*x^(2+m)*b^2*(b^2)^(1/2*m)/ m/(-1+m)*(b*x)^(-5/2-m)*(cos(b*x)*x*b-sin(b*x))*LommelS1(m+1/2,1/2,b*x))*s in(a)+2^(-3+m)*b^(2-m)*Pi^(1/2)*(2^(2-m)/Pi^(1/2)/(-1+m)*x^(-2+m)*b^(-2+m) *(-2*b^2*x^2+2*m^2-2*m-4)/(1+m)/(-2+m)*sin(b*x)+2^(-m+3)/Pi^(1/2)/(-1+m)*x ^(-2+m)*b^(-2+m)*(b^2*x^2-m^2-m)/(1+m)/(-2+m)/m*(cos(b*x)*x*b-sin(b*x))+2^ (-m+3)/Pi^(1/2)/(-1+m)*x^(2+m)*b^(2+m)/(1+m)/(-2+m)*(b*x)^(-3/2-m)*LommelS 1(m+1/2,3/2,b*x)*sin(b*x)-2^(-m+3)/Pi^(1/2)/(-1+m)*x^(2+m)*b^(2+m)/(1+m)/( -2+m)/m*(b*x)^(-5/2-m)*(cos(b*x)*x*b-sin(b*x))*LommelS1(m+3/2,1/2,b*x))*co s(a)
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int x^{-3+m} \sin (a+b x) \, dx=-\frac {e^{\left (-{\left (m - 3\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m - 2, i \, b x\right ) + e^{\left (-{\left (m - 3\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m - 2, -i \, b x\right )}{2 \, b} \] Input:
integrate(x^(-3+m)*sin(b*x+a),x, algorithm="fricas")
Output:
-1/2*(e^(-(m - 3)*log(I*b) - I*a)*gamma(m - 2, I*b*x) + e^(-(m - 3)*log(-I *b) + I*a)*gamma(m - 2, -I*b*x))/b
\[ \int x^{-3+m} \sin (a+b x) \, dx=\int x^{m - 3} \sin {\left (a + b x \right )}\, dx \] Input:
integrate(x**(-3+m)*sin(b*x+a),x)
Output:
Integral(x**(m - 3)*sin(a + b*x), x)
\[ \int x^{-3+m} \sin (a+b x) \, dx=\int { x^{m - 3} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(x^(-3+m)*sin(b*x+a),x, algorithm="maxima")
Output:
integrate(x^(m - 3)*sin(b*x + a), x)
\[ \int x^{-3+m} \sin (a+b x) \, dx=\int { x^{m - 3} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(x^(-3+m)*sin(b*x+a),x, algorithm="giac")
Output:
integrate(x^(m - 3)*sin(b*x + a), x)
Timed out. \[ \int x^{-3+m} \sin (a+b x) \, dx=\int x^{m-3}\,\sin \left (a+b\,x\right ) \,d x \] Input:
int(x^(m - 3)*sin(a + b*x),x)
Output:
int(x^(m - 3)*sin(a + b*x), x)
\[ \int x^{-3+m} \sin (a+b x) \, dx=\int \frac {x^{m} \sin \left (b x +a \right )}{x^{3}}d x \] Input:
int(x^(-3+m)*sin(b*x+a),x)
Output:
int((x**m*sin(a + b*x))/x**3,x)