Integrand size = 20, antiderivative size = 72 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {3 d x}{32 b}+\frac {3 d \cos (a+b x) \sin (a+b x)}{32 b^2}+\frac {d \cos (a+b x) \sin ^3(a+b x)}{16 b^2}+\frac {(c+d x) \sin ^4(a+b x)}{4 b} \] Output:
-3/32*d*x/b+3/32*d*cos(b*x+a)*sin(b*x+a)/b^2+1/16*d*cos(b*x+a)*sin(b*x+a)^ 3/b^2+1/4*(d*x+c)*sin(b*x+a)^4/b
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {c \sin ^4(a+b x)}{4 b}+\frac {d (-2 b x \cos (2 (a+b x))+\sin (2 (a+b x)))}{16 b^2}-\frac {d (-4 b x \cos (4 (a+b x))+\sin (4 (a+b x)))}{128 b^2} \] Input:
Integrate[(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^3,x]
Output:
(c*Sin[a + b*x]^4)/(4*b) + (d*(-2*b*x*Cos[2*(a + b*x)] + Sin[2*(a + b*x)]) )/(16*b^2) - (d*(-4*b*x*Cos[4*(a + b*x)] + Sin[4*(a + b*x)]))/(128*b^2)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4904, 3042, 3115, 3042, 3115, 24}
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.
\(\displaystyle \int (c+d x) \sin ^3(a+b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 4904 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \int \sin ^4(a+b x)dx}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \int \sin (a+b x)^4dx}{4 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \left (\frac {3}{4} \int \sin ^2(a+b x)dx-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \left (\frac {3}{4} \int \sin (a+b x)^2dx-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )}{4 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )}{4 b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {(c+d x) \sin ^4(a+b x)}{4 b}-\frac {d \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (a+b x) \cos (a+b x)}{2 b}\right )-\frac {\sin ^3(a+b x) \cos (a+b x)}{4 b}\right )}{4 b}\) |
Input:
Int[(c + d*x)*Cos[a + b*x]*Sin[a + b*x]^3,x]
Output:
((c + d*x)*Sin[a + b*x]^4)/(4*b) - (d*(-1/4*(Cos[a + b*x]*Sin[a + b*x]^3)/ b + (3*(x/2 - (Cos[a + b*x]*Sin[a + b*x])/(2*b)))/4))/(4*b)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x _)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) , x] - Simp[d*(m/(b*(n + 1))) Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Time = 1.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {-16 b \left (d x +c \right ) \cos \left (2 b x +2 a \right )+4 b \left (d x +c \right ) \cos \left (4 b x +4 a \right )+12 c b +8 d \sin \left (2 b x +2 a \right )-d \sin \left (4 b x +4 a \right )}{128 b^{2}}\) | \(69\) |
risch | \(\frac {\left (d x +c \right ) \cos \left (4 b x +4 a \right )}{32 b}-\frac {d \sin \left (4 b x +4 a \right )}{128 b^{2}}-\frac {\left (d x +c \right ) \cos \left (2 b x +2 a \right )}{8 b}+\frac {d \sin \left (2 b x +2 a \right )}{16 b^{2}}\) | \(70\) |
derivativedivides | \(\frac {-\frac {d a \sin \left (b x +a \right )^{4}}{4 b}+\frac {c \sin \left (b x +a \right )^{4}}{4}+\frac {d \left (\frac {\left (b x +a \right ) \sin \left (b x +a \right )^{4}}{4}+\frac {\left (\sin \left (b x +a \right )^{3}+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{16}-\frac {3 b x}{32}-\frac {3 a}{32}\right )}{b}}{b}\) | \(85\) |
default | \(\frac {-\frac {d a \sin \left (b x +a \right )^{4}}{4 b}+\frac {c \sin \left (b x +a \right )^{4}}{4}+\frac {d \left (\frac {\left (b x +a \right ) \sin \left (b x +a \right )^{4}}{4}+\frac {\left (\sin \left (b x +a \right )^{3}+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{16}-\frac {3 b x}{32}-\frac {3 a}{32}\right )}{b}}{b}\) | \(85\) |
norman | \(\frac {\frac {3 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{16 b^{2}}+\frac {11 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}}{16 b^{2}}-\frac {11 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{5}}{16 b^{2}}-\frac {3 d \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{7}}{16 b^{2}}-\frac {3 d x}{32 b}+\frac {4 c \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}}{b}-\frac {3 d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}}{8 b}+\frac {55 d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}}{16 b}-\frac {3 d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6}}{8 b}-\frac {3 d x \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{8}}{32 b}}{\left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )^{4}}\) | \(180\) |
orering | \(\frac {d \left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+d^{2}\right ) \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3}}{8 b^{4} \left (d x +c \right )^{2}}-\frac {\left (5 x^{2} d^{2} b^{2}+10 b^{2} c d x +5 b^{2} c^{2}+2 d^{2}\right ) \left (d \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3}-\left (d x +c \right ) b \sin \left (b x +a \right )^{4}+3 \left (d x +c \right ) \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2} b \right )}{16 b^{4} \left (d x +c \right )^{2}}+\frac {d \left (-2 d b \sin \left (b x +a \right )^{4}+6 d \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2} b -10 \left (d x +c \right ) b^{2} \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )+6 \left (d x +c \right ) \cos \left (b x +a \right )^{3} \sin \left (b x +a \right ) b^{2}\right )}{16 b^{4} \left (d x +c \right )}-\frac {-30 d \,b^{2} \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )+18 d \cos \left (b x +a \right )^{3} \sin \left (b x +a \right ) b^{2}-48 \left (d x +c \right ) b^{3} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{2}+10 \left (d x +c \right ) b^{3} \sin \left (b x +a \right )^{4}+6 \left (d x +c \right ) \cos \left (b x +a \right )^{4} b^{3}}{64 b^{4}}\) | \(364\) |
Input:
int((d*x+c)*cos(b*x+a)*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/128*(-16*b*(d*x+c)*cos(2*b*x+2*a)+4*b*(d*x+c)*cos(4*b*x+4*a)+12*c*b+8*d* sin(2*b*x+2*a)-d*sin(4*b*x+4*a))/b^2
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {8 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} + 5 \, b d x - 16 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} - {\left (2 \, d \cos \left (b x + a\right )^{3} - 5 \, d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{32 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="fricas")
Output:
1/32*(8*(b*d*x + b*c)*cos(b*x + a)^4 + 5*b*d*x - 16*(b*d*x + b*c)*cos(b*x + a)^2 - (2*d*cos(b*x + a)^3 - 5*d*cos(b*x + a))*sin(b*x + a))/b^2
Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.92 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\begin {cases} \frac {c \sin ^{4}{\left (a + b x \right )}}{4 b} + \frac {5 d x \sin ^{4}{\left (a + b x \right )}}{32 b} - \frac {3 d x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{16 b} - \frac {3 d x \cos ^{4}{\left (a + b x \right )}}{32 b} + \frac {5 d \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{32 b^{2}} + \frac {3 d \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{3}{\left (a \right )} \cos {\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)**3,x)
Output:
Piecewise((c*sin(a + b*x)**4/(4*b) + 5*d*x*sin(a + b*x)**4/(32*b) - 3*d*x* sin(a + b*x)**2*cos(a + b*x)**2/(16*b) - 3*d*x*cos(a + b*x)**4/(32*b) + 5* d*sin(a + b*x)**3*cos(a + b*x)/(32*b**2) + 3*d*sin(a + b*x)*cos(a + b*x)** 3/(32*b**2), Ne(b, 0)), ((c*x + d*x**2/2)*sin(a)**3*cos(a), True))
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {32 \, c \sin \left (b x + a\right )^{4} - \frac {32 \, a d \sin \left (b x + a\right )^{4}}{b} + \frac {{\left (4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) - 16 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{128 \, b} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="maxima")
Output:
1/128*(32*c*sin(b*x + a)^4 - 32*a*d*sin(b*x + a)^4/b + (4*(b*x + a)*cos(4* b*x + 4*a) - 16*(b*x + a)*cos(2*b*x + 2*a) - sin(4*b*x + 4*a) + 8*sin(2*b* x + 2*a))*d/b)/b
Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {{\left (b d x + b c\right )} \cos \left (4 \, b x + 4 \, a\right )}{32 \, b^{2}} - \frac {{\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} - \frac {d \sin \left (4 \, b x + 4 \, a\right )}{128 \, b^{2}} + \frac {d \sin \left (2 \, b x + 2 \, a\right )}{16 \, b^{2}} \] Input:
integrate((d*x+c)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="giac")
Output:
1/32*(b*d*x + b*c)*cos(4*b*x + 4*a)/b^2 - 1/8*(b*d*x + b*c)*cos(2*b*x + 2* a)/b^2 - 1/128*d*sin(4*b*x + 4*a)/b^2 + 1/16*d*sin(2*b*x + 2*a)/b^2
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.31 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {2\,d\,\sin \left (2\,a+2\,b\,x\right )-\frac {d\,\sin \left (4\,a+4\,b\,x\right )}{4}-2\,b\,c\,{\sin \left (2\,a+2\,b\,x\right )}^2+8\,b\,c\,{\sin \left (a+b\,x\right )}^2+4\,b\,d\,x\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )-b\,d\,x\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )}{32\,b^2} \] Input:
int(cos(a + b*x)*sin(a + b*x)^3*(c + d*x),x)
Output:
(2*d*sin(2*a + 2*b*x) - (d*sin(4*a + 4*b*x))/4 - 2*b*c*sin(2*a + 2*b*x)^2 + 8*b*c*sin(a + b*x)^2 + 4*b*d*x*(2*sin(a + b*x)^2 - 1) - b*d*x*(2*sin(2*a + 2*b*x)^2 - 1))/(32*b^2)
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int (c+d x) \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {6 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3} d +9 \cos \left (b x +a \right ) \sin \left (b x +a \right ) d +24 \sin \left (b x +a \right )^{4} b c +24 \sin \left (b x +a \right )^{4} b d x -9 a d -64 b c -9 b d x}{96 b^{2}} \] Input:
int((d*x+c)*cos(b*x+a)*sin(b*x+a)^3,x)
Output:
(6*cos(a + b*x)*sin(a + b*x)**3*d + 9*cos(a + b*x)*sin(a + b*x)*d + 24*sin (a + b*x)**4*b*c + 24*sin(a + b*x)**4*b*d*x - 9*a*d - 64*b*c - 9*b*d*x)/(9 6*b**2)