Integrand size = 14, antiderivative size = 97 \[ \int x^{3+m} \sin ^2(a+b x) \, dx=\frac {x^{4+m}}{2 (4+m)}+\frac {2^{-6-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (4+m,-2 i b x)}{b^4}+\frac {2^{-6-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (4+m,2 i b x)}{b^4} \] Output:
x^(4+m)/(8+2*m)+2^(-6-m)*exp(2*I*a)*x^m*GAMMA(4+m,-2*I*b*x)/b^4/((-I*b*x)^ m)+2^(-6-m)*x^m*GAMMA(4+m,2*I*b*x)/b^4/exp(2*I*a)/((I*b*x)^m)
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int x^{3+m} \sin ^2(a+b x) \, dx=\frac {2^{-6-m} x^m \left (b^2 x^2\right )^{-m} \left (2^{5+m} b^4 x^4 \left (b^2 x^2\right )^m+(4+m) (-i b x)^m \Gamma (4+m,2 i b x) (\cos (a)-i \sin (a))^2+(4+m) (i b x)^m \Gamma (4+m,-2 i b x) (\cos (a)+i \sin (a))^2\right )}{b^4 (4+m)} \] Input:
Integrate[x^(3 + m)*Sin[a + b*x]^2,x]
Output:
(2^(-6 - m)*x^m*(2^(5 + m)*b^4*x^4*(b^2*x^2)^m + (4 + m)*((-I)*b*x)^m*Gamm a[4 + m, (2*I)*b*x]*(Cos[a] - I*Sin[a])^2 + (4 + m)*(I*b*x)^m*Gamma[4 + m, (-2*I)*b*x]*(Cos[a] + I*Sin[a])^2))/(b^4*(4 + m)*(b^2*x^2)^m)
Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3793, 2009}
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.
\(\displaystyle \int x^{m+3} \sin ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int x^{m+3} \sin (a+b x)^2dx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \int \left (\frac {x^{m+3}}{2}-\frac {1}{2} x^{m+3} \cos (2 a+2 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 i a} 2^{-m-6} x^m (-i b x)^{-m} \Gamma (m+4,-2 i b x)}{b^4}+\frac {e^{-2 i a} 2^{-m-6} x^m (i b x)^{-m} \Gamma (m+4,2 i b x)}{b^4}+\frac {x^{m+4}}{2 (m+4)}\) |
Input:
Int[x^(3 + m)*Sin[a + b*x]^2,x]
Output:
x^(4 + m)/(2*(4 + m)) + (2^(-6 - m)*E^((2*I)*a)*x^m*Gamma[4 + m, (-2*I)*b* x])/(b^4*((-I)*b*x)^m) + (2^(-6 - m)*x^m*Gamma[4 + m, (2*I)*b*x])/(b^4*E^( (2*I)*a)*(I*b*x)^m)
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
\[\int x^{3+m} \sin \left (b x +a \right )^{2}d x\]
Input:
int(x^(3+m)*sin(b*x+a)^2,x)
Output:
int(x^(3+m)*sin(b*x+a)^2,x)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int x^{3+m} \sin ^2(a+b x) \, dx=\frac {4 \, b x x^{m + 3} + {\left (-i \, m - 4 i\right )} e^{\left (-{\left (m + 3\right )} \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 4, 2 i \, b x\right ) + {\left (i \, m + 4 i\right )} e^{\left (-{\left (m + 3\right )} \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 4, -2 i \, b x\right )}{8 \, {\left (b m + 4 \, b\right )}} \] Input:
integrate(x^(3+m)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/8*(4*b*x*x^(m + 3) + (-I*m - 4*I)*e^(-(m + 3)*log(2*I*b) - 2*I*a)*gamma( m + 4, 2*I*b*x) + (I*m + 4*I)*e^(-(m + 3)*log(-2*I*b) + 2*I*a)*gamma(m + 4 , -2*I*b*x))/(b*m + 4*b)
\[ \int x^{3+m} \sin ^2(a+b x) \, dx=\int x^{m + 3} \sin ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(x**(3+m)*sin(b*x+a)**2,x)
Output:
Integral(x**(m + 3)*sin(a + b*x)**2, x)
\[ \int x^{3+m} \sin ^2(a+b x) \, dx=\int { x^{m + 3} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^(3+m)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
-1/2*((m + 4)*integrate(x^3*x^m*cos(2*b*x + 2*a), x) - e^(m*log(x) + 4*log (x)))/(m + 4)
\[ \int x^{3+m} \sin ^2(a+b x) \, dx=\int { x^{m + 3} \sin \left (b x + a\right )^{2} \,d x } \] Input:
integrate(x^(3+m)*sin(b*x+a)^2,x, algorithm="giac")
Output:
integrate(x^(m + 3)*sin(b*x + a)^2, x)
Timed out. \[ \int x^{3+m} \sin ^2(a+b x) \, dx=\int x^{m+3}\,{\sin \left (a+b\,x\right )}^2 \,d x \] Input:
int(x^(m + 3)*sin(a + b*x)^2,x)
Output:
int(x^(m + 3)*sin(a + b*x)^2, x)
\[ \int x^{3+m} \sin ^2(a+b x) \, dx =\text {Too large to display} \] Input:
int(x^(3+m)*sin(b*x+a)^2,x)
Output:
( - 6*x**m*cos(a + b*x)*sin(a + b*x)*b**3*m*x**3 - 24*x**m*cos(a + b*x)*si n(a + b*x)*b**3*x**3 + 2*x**m*cos(a + b*x)*sin(a + b*x)*b*m**3*x + 18*x**m *cos(a + b*x)*sin(a + b*x)*b*m**2*x + 52*x**m*cos(a + b*x)*sin(a + b*x)*b* m*x + 48*x**m*cos(a + b*x)*sin(a + b*x)*b*x + 2*x**m*cos(a + b*x)*m**4 + 2 0*x**m*cos(a + b*x)*m**3 + 70*x**m*cos(a + b*x)*m**2 + 100*x**m*cos(a + b* x)*m + 48*x**m*cos(a + b*x) + 3*x**m*sin(a + b*x)**2*b**2*m**2*x**2 + 21*x **m*sin(a + b*x)**2*b**2*m*x**2 + 36*x**m*sin(a + b*x)**2*b**2*x**2 - x**m *sin(a + b*x)**2*m**4 - 10*x**m*sin(a + b*x)**2*m**3 - 35*x**m*sin(a + b*x )**2*m**2 - 50*x**m*sin(a + b*x)**2*m - 24*x**m*sin(a + b*x)**2 + 2*x**m*s in(a + b*x)*b*m**3*x + 18*x**m*sin(a + b*x)*b*m**2*x + 52*x**m*sin(a + b*x )*b*m*x + 48*x**m*sin(a + b*x)*b*x + 6*x**m*b**4*x**4 + 2*x**m*m**4 + 20*x **m*m**3 + 70*x**m*m**2 + 100*x**m*m + 48*x**m - 4*int(x**m/(tan((a + b*x) /2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*m**5 - 40*int(x**m/(tan((a + b* x)/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*m**4 - 140*int(x**m/(tan((a + b*x)/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*m**3 - 200*int(x**m/(tan(( a + b*x)/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*m**2 - 96*int(x**m/(tan ((a + b*x)/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*m - 4*int((x**m*x)/(t an((a + b*x)/2)**4 + 2*tan((a + b*x)/2)**2 + 1),x)*b**2*m**3 - 36*int((x** m*x)/(tan((a + b*x)/2)**4 + 2*tan((a + b*x)/2)**2 + 1),x)*b**2*m**2 - 104* int((x**m*x)/(tan((a + b*x)/2)**4 + 2*tan((a + b*x)/2)**2 + 1),x)*b**2*...