Integrand size = 14, antiderivative size = 55 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {(c+d x)^2}{4 d}-\frac {(c+d x) \cos (a+b x) \sin (a+b x)}{2 b}+\frac {d \sin ^2(a+b x)}{4 b^2} \] Output:
1/4*(d*x+c)^2/d-1/2*(d*x+c)*cos(b*x+a)*sin(b*x+a)/b+1/4*d*sin(b*x+a)^2/b^2
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {-d \cos (2 (a+b x))+2 b (2 a c+b x (2 c+d x)-(c+d x) \sin (2 (a+b x)))}{8 b^2} \] Input:
Integrate[(c + d*x)*Sin[a + b*x]^2,x]
Output:
(-(d*Cos[2*(a + b*x)]) + 2*b*(2*a*c + b*x*(2*c + d*x) - (c + d*x)*Sin[2*(a + b*x)]))/(8*b^2)
Time = 0.21 (sec) , antiderivative size = 55, 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, 3791, 17}
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.
\(\displaystyle \int (c+d x) \sin ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x) \sin (a+b x)^2dx\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {1}{2} \int (c+d x)dx+\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^2}{4 d}\) |
Input:
Int[(c + d*x)*Sin[a + b*x]^2,x]
Output:
(c + d*x)^2/(4*d) - ((c + d*x)*Cos[a + b*x]*Sin[a + b*x])/(2*b) + (d*Sin[a + b*x]^2)/(4*b^2)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Time = 0.76 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {d \,x^{2}}{4}+\frac {c x}{2}-\frac {d \cos \left (2 b x +2 a \right )}{8 b^{2}}-\frac {\left (d x +c \right ) \sin \left (2 b x +2 a \right )}{4 b}\) | \(46\) |
derivativedivides | \(\frac {-\frac {d a \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )+\frac {d \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}+\frac {\sin \left (b x +a \right )^{2}}{4}\right )}{b}}{b}\) | \(112\) |
default | \(\frac {-\frac {d a \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )+\frac {d \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}+\frac {\sin \left (b x +a \right )^{2}}{4}\right )}{b}}{b}\) | \(112\) |
norman | \(\frac {\frac {c \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{b}+c x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\frac {d \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{b^{2}}+\frac {d x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{b}+\frac {c x}{2}+\frac {d \,x^{2}}{4}-\frac {c \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}+\frac {c x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{2}+\frac {d \,x^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{2}+\frac {d \,x^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{4}-\frac {d x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}}{\left (1+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )^{2}}\) | \(171\) |
orering | \(\frac {\left (2 b^{2} d^{3} x^{4}+8 b^{2} c \,d^{2} x^{3}+10 b^{2} c^{2} d \,x^{2}+4 b^{2} c^{3} x +3 d^{3} x^{2}+6 c \,d^{2} x +2 d \,c^{2}\right ) \sin \left (b x +a \right )^{2}}{4 b^{2} \left (d x +c \right )^{2}}-\frac {\left (2 x^{2} d^{2}+4 c d x +c^{2}\right ) \left (d \sin \left (b x +a \right )^{2}+2 \left (d x +c \right ) \sin \left (b x +a \right ) b \cos \left (b x +a \right )\right )}{4 \left (d x +c \right )^{2} b^{2}}+\frac {x \left (d x +2 c \right ) \left (4 d \sin \left (b x +a \right ) b \cos \left (b x +a \right )+2 \left (d x +c \right ) b^{2} \cos \left (b x +a \right )^{2}-2 \left (d x +c \right ) \sin \left (b x +a \right )^{2} b^{2}\right )}{8 b^{2} \left (d x +c \right )}\) | \(221\) |
Input:
int((d*x+c)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/4*d*x^2+1/2*c*x-1/8*d/b^2*cos(2*b*x+2*a)-1/4/b*(d*x+c)*sin(2*b*x+2*a)
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {b^{2} d x^{2} + 2 \, b^{2} c x - d \cos \left (b x + a\right )^{2} - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{2}} \] Input:
integrate((d*x+c)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
1/4*(b^2*d*x^2 + 2*b^2*c*x - d*cos(b*x + a)^2 - 2*(b*d*x + b*c)*cos(b*x + a)*sin(b*x + a))/b^2
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (48) = 96\).
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.29 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\begin {cases} \frac {c x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c x \cos ^{2}{\left (a + b x \right )}}{2} + \frac {d x^{2} \sin ^{2}{\left (a + b x \right )}}{4} + \frac {d x^{2} \cos ^{2}{\left (a + b x \right )}}{4} - \frac {c \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {d x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} + \frac {d \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*sin(b*x+a)**2,x)
Output:
Piecewise((c*x*sin(a + b*x)**2/2 + c*x*cos(a + b*x)**2/2 + d*x**2*sin(a + b*x)**2/4 + d*x**2*cos(a + b*x)**2/4 - c*sin(a + b*x)*cos(a + b*x)/(2*b) - d*x*sin(a + b*x)*cos(a + b*x)/(2*b) + d*sin(a + b*x)**2/(4*b**2), Ne(b, 0 )), ((c*x + d*x**2/2)*sin(a)**2, True))
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.75 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {2 \, {\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} c - \frac {2 \, {\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a d}{b} + \frac {{\left (2 \, {\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{8 \, b} \] Input:
integrate((d*x+c)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
1/8*(2*(2*b*x + 2*a - sin(2*b*x + 2*a))*c - 2*(2*b*x + 2*a - sin(2*b*x + 2 *a))*a*d/b + (2*(b*x + a)^2 - 2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2 *a))*d/b)/b
Time = 0.40 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {1}{4} \, d x^{2} + \frac {1}{2} \, c x - \frac {d \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} \] Input:
integrate((d*x+c)*sin(b*x+a)^2,x, algorithm="giac")
Output:
1/4*d*x^2 + 1/2*c*x - 1/8*d*cos(2*b*x + 2*a)/b^2 - 1/4*(b*d*x + b*c)*sin(2 *b*x + 2*a)/b^2
Time = 34.89 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {c\,x}{2}+\frac {d\,x^2}{4}-\frac {d\,\cos \left (2\,a+2\,b\,x\right )}{8\,b^2}-\frac {c\,\sin \left (2\,a+2\,b\,x\right )}{4\,b}-\frac {d\,x\,\sin \left (2\,a+2\,b\,x\right )}{4\,b} \] Input:
int(sin(a + b*x)^2*(c + d*x),x)
Output:
(c*x)/2 + (d*x^2)/4 - (d*cos(2*a + 2*b*x))/(8*b^2) - (c*sin(2*a + 2*b*x))/ (4*b) - (d*x*sin(2*a + 2*b*x))/(4*b)
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.31 \[ \int (c+d x) \sin ^2(a+b x) \, dx=\frac {-2 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b c -2 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b d x +\sin \left (b x +a \right )^{2} d +2 a b c +2 b^{2} c x +b^{2} d \,x^{2}-2 d}{4 b^{2}} \] Input:
int((d*x+c)*sin(b*x+a)^2,x)
Output:
( - 2*cos(a + b*x)*sin(a + b*x)*b*c - 2*cos(a + b*x)*sin(a + b*x)*b*d*x + sin(a + b*x)**2*d + 2*a*b*c + 2*b**2*c*x + b**2*d*x**2 - 2*d)/(4*b**2)