\(\int x^{3+m} \sin (a+b x) \, dx\) [77]

Optimal result

Integrand size = 12, antiderivative size = 79 \[ \int x^{3+m} \sin (a+b x) \, dx=\frac {i e^{i a} x^m (-i b x)^{-m} \Gamma (4+m,-i b x)}{2 b^4}-\frac {i e^{-i a} x^m (i b x)^{-m} \Gamma (4+m,i b x)}{2 b^4} \] Output:

1/2*I*exp(I*a)*x^m*GAMMA(4+m,-I*b*x)/b^4/((-I*b*x)^m)-1/2*I*x^m*GAMMA(4+m, 
I*b*x)/b^4/exp(I*a)/((I*b*x)^m)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int x^{3+m} \sin (a+b x) \, dx=\frac {i e^{i a} x^m (-i b x)^{-m} \Gamma (4+m,-i b x)}{2 b^4}-\frac {i e^{-i a} x^m (i b x)^{-m} \Gamma (4+m,i b x)}{2 b^4} \] Input:

Integrate[x^(3 + m)*Sin[a + b*x],x]
 

Output:

((I/2)*E^(I*a)*x^m*Gamma[4 + m, (-I)*b*x])/(b^4*((-I)*b*x)^m) - ((I/2)*x^m 
*Gamma[4 + m, I*b*x])/(b^4*E^(I*a)*(I*b*x)^m)
 

Rubi [A] (verified)

Time = 0.28 (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

\(\displaystyle \frac {i e^{i a} x^m (-i b x)^{-m} \Gamma (m+4,-i b x)}{2 b^4}-\frac {i e^{-i a} x^m (i b x)^{-m} \Gamma (m+4,i b x)}{2 b^4}\)

Input:

Int[x^(3 + m)*Sin[a + b*x],x]
 

Output:

((I/2)*E^(I*a)*x^m*Gamma[4 + m, (-I)*b*x])/(b^4*((-I)*b*x)^m) - ((I/2)*x^m 
*Gamma[4 + m, I*b*x])/(b^4*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 = 0.56 (sec) , antiderivative size = 454, normalized size of antiderivative = 5.75

method	result	size
meijerg	\(\frac {2^{3+m} \left (b^{2}\right )^{-\frac {m}{2}} \sqrt {\pi }\, \left (\frac {3 \,2^{-4-m} x^{3+m} b^{3} \left (b^{2}\right )^{\frac {m}{2}} \left (\frac {8}{3}+\frac {2 m}{3}\right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (4+m \right )}-\frac {2^{-3-m} x^{1+m} b \left (b^{2}\right )^{\frac {m}{2}} \left (-m^{2}-7 m -12\right ) \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }\, \left (4+m \right )}+\frac {2^{-3-m} x^{2+m} b^{2} \left (b^{2}\right )^{\frac {m}{2}} \left (-m^{3}-8 m^{2}-19 m -12\right ) \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (4+m \right )}-\frac {2^{-3-m} x^{2+m} b^{2} \left (b^{2}\right )^{\frac {m}{2}} \left (2+m \right ) \left (1+m \right ) \left (3+m \right ) \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }}\right ) \sin \left (a \right )}{b^{4}}+2^{3+m} b^{-4-m} \sqrt {\pi }\, \left (\frac {2^{-3-m} x^{2+m} b^{2+m} \left (m^{2}+7 m +10\right ) \sin \left (b x \right )}{\sqrt {\pi }\, \left (5+m \right )}-\frac {2^{-3-m} x^{2+m} b^{2+m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right )}{\sqrt {\pi }}-\frac {2^{-3-m} x^{2+m} b^{2+m} m \left (3+m \right ) \left (2+m \right ) \left (b x \right )^{-\frac {3}{2}-m} \operatorname {LommelS1}\left (m +\frac {1}{2}, \frac {3}{2}, b x \right ) \sin \left (b x \right )}{\sqrt {\pi }}+\frac {2^{-3-m} x^{2+m} b^{2+m} \left (3+m \right ) \left (2+m \right ) \left (b x \right )^{-\frac {5}{2}-m} \left (\cos \left (b x \right ) x b -\sin \left (b x \right )\right ) \operatorname {LommelS1}\left (m +\frac {3}{2}, \frac {1}{2}, b x \right )}{\sqrt {\pi }}\right ) \cos \left (a \right )\)	\(454\)

Input:

int(x^(3+m)*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

2^(3+m)/b^4*(b^2)^(-1/2*m)*Pi^(1/2)*(3*2^(-4-m)/Pi^(1/2)/(4+m)*x^(3+m)*b^3 
*(b^2)^(1/2*m)*(8/3+2/3*m)*sin(b*x)-2^(-3-m)/Pi^(1/2)/(4+m)*x^(1+m)*b*(b^2 
)^(1/2*m)*(-m^2-7*m-12)*(cos(b*x)*x*b-sin(b*x))+2^(-3-m)/Pi^(1/2)/(4+m)*x^ 
(2+m)*b^2*(b^2)^(1/2*m)*(-m^3-8*m^2-19*m-12)*(b*x)^(-3/2-m)*LommelS1(m+3/2 
,3/2,b*x)*sin(b*x)-2^(-3-m)/Pi^(1/2)*x^(2+m)*b^2*(b^2)^(1/2*m)*(2+m)*(1+m) 
*(3+m)*(b*x)^(-5/2-m)*(cos(b*x)*x*b-sin(b*x))*LommelS1(m+1/2,1/2,b*x))*sin 
(a)+2^(3+m)*b^(-4-m)*Pi^(1/2)*(2^(-3-m)/Pi^(1/2)/(5+m)*x^(2+m)*b^(2+m)*(m^ 
2+7*m+10)*sin(b*x)-2^(-3-m)/Pi^(1/2)*x^(2+m)*b^(2+m)*(cos(b*x)*x*b-sin(b*x 
))-2^(-3-m)/Pi^(1/2)*x^(2+m)*b^(2+m)*m*(3+m)*(2+m)*(b*x)^(-3/2-m)*LommelS1 
(m+1/2,3/2,b*x)*sin(b*x)+2^(-3-m)/Pi^(1/2)*x^(2+m)*b^(2+m)*(3+m)*(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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int x^{3+m} \sin (a+b x) \, dx=-\frac {e^{\left (-{\left (m + 3\right )} \log \left (i \, b\right ) - i \, a\right )} \Gamma \left (m + 4, i \, b x\right ) + e^{\left (-{\left (m + 3\right )} \log \left (-i \, b\right ) + i \, a\right )} \Gamma \left (m + 4, -i \, b x\right )}{2 \, b} \] 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 + 4, I*b*x) + e^(-(m + 3)*log(-I 
*b) + I*a)*gamma(m + 4, -I*b*x))/b
 

Sympy [F]

\[ \int x^{3+m} \sin (a+b x) \, dx=\int x^{m + 3} \sin {\left (a + b x \right )}\, dx \] Input:

integrate(x**(3+m)*sin(b*x+a),x)
 

Output:

Integral(x**(m + 3)*sin(a + b*x), x)
 

Maxima [F]

\[ \int x^{3+m} \sin (a+b x) \, dx=\int { x^{m + 3} \sin \left (b x + a\right ) \,d x } \] Input:

integrate(x^(3+m)*sin(b*x+a),x, algorithm="maxima")
 

Output:

integrate(x^(m + 3)*sin(b*x + a), x)
 

Giac [F]

\[ \int x^{3+m} \sin (a+b x) \, dx=\int { x^{m + 3} \sin \left (b x + a\right ) \,d x } \] Input:

integrate(x^(3+m)*sin(b*x+a),x, algorithm="giac")
 

Output:

integrate(x^(m + 3)*sin(b*x + a), x)
 

Mupad [F(-1)]

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)
 

Reduce [F]

\[ \int x^{3+m} \sin (a+b x) \, dx =\text {Too large to display} \] Input:

int(x^(3+m)*sin(b*x+a),x)
                                                                                    
                                                                                    
 

Output:

( - x**m*cos(a + b*x)*tan((a + b*x)/2)**2*b**3*x**3 + x**m*cos(a + b*x)*ta 
n((a + b*x)/2)**2*b*m**2*x + 5*x**m*cos(a + b*x)*tan((a + b*x)/2)**2*b*m*x 
 + 6*x**m*cos(a + b*x)*tan((a + b*x)/2)**2*b*x - x**m*cos(a + b*x)*b**3*x* 
*3 + x**m*cos(a + b*x)*b*m**2*x + 5*x**m*cos(a + b*x)*b*m*x + 6*x**m*cos(a 
 + b*x)*b*x + x**m*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*m*x**2 + 3*x**m*s 
in(a + b*x)*tan((a + b*x)/2)**2*b**2*x**2 + x**m*sin(a + b*x)*b**2*m*x**2 
+ 3*x**m*sin(a + b*x)*b**2*x**2 - 2*x**m*tan((a + b*x)/2)*m**3 - 12*x**m*t 
an((a + b*x)/2)*m**2 - 22*x**m*tan((a + b*x)/2)*m - 12*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**4 + 12*int((x**m*tan((a + b*x)/2))/(tan((a + b*x)/2)**2*x + 
 x),x)*tan((a + b*x)/2)**2*m**3 + 22*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 + 12*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*t 
an((a + b*x)/2))/(tan((a + b*x)/2)**2*x + x),x)*m**4 + 12*int((x**m*tan((a 
 + b*x)/2))/(tan((a + b*x)/2)**2*x + x),x)*m**3 + 22*int((x**m*tan((a + b* 
x)/2))/(tan((a + b*x)/2)**2*x + x),x)*m**2 + 12*int((x**m*tan((a + b*x)/2) 
)/(tan((a + b*x)/2)**2*x + x),x)*m)/(b**4*(tan((a + b*x)/2)**2 + 1))