Integrand size = 27, antiderivative size = 76 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^3 \log (a+b \sin (c+d x))}{b^4 d}+\frac {a^2 \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\sin ^3(c+d x)}{3 b d} \] Output:
-a^3*ln(a+b*sin(d*x+c))/b^4/d+a^2*sin(d*x+c)/b^3/d-1/2*a*sin(d*x+c)^2/b^2/ d+1/3*sin(d*x+c)^3/b/d
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-6 a^3 \log (a+b \sin (c+d x))+6 a^2 b \sin (c+d x)-3 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)}{6 b^4 d} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(-6*a^3*Log[a + b*Sin[c + d*x]] + 6*a^2*b*Sin[c + d*x] - 3*a*b^2*Sin[c + d *x]^2 + 2*b^3*Sin[c + d*x]^3)/(6*b^4*d)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x)}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-\frac {a^3}{a+b \sin (c+d x)}+a^2-b \sin (c+d x) a+b^2 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (-\log (a+b \sin (c+d x)))+a^2 b \sin (c+d x)-\frac {1}{2} a b^2 \sin ^2(c+d x)+\frac {1}{3} b^3 \sin ^3(c+d x)}{b^4 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(-(a^3*Log[a + b*Sin[c + d*x]]) + a^2*b*Sin[c + d*x] - (a*b^2*Sin[c + d*x] ^2)/2 + (b^3*Sin[c + d*x]^3)/3)/(b^4*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.53 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {b a \sin \left (d x +c \right )^{2}}{2}+a^{2} \sin \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(65\) |
default | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {b a \sin \left (d x +c \right )^{2}}{2}+a^{2} \sin \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\) | \(65\) |
parallelrisch | \(\frac {-12 a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+12 a^{3} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+3 a \,b^{2} \cos \left (2 d x +2 c \right )-b^{3} \sin \left (3 d x +3 c \right )+3 \left (4 a^{2} b +b^{3}\right ) \sin \left (d x +c \right )-3 a \,b^{2}}{12 d \,b^{4}}\) | \(112\) |
risch | \(\frac {i a^{3} x}{b^{4}}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {2 i a^{3} c}{b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {\sin \left (3 d x +3 c \right )}{12 b d}\) | \(195\) |
norman | \(\frac {\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{3} d}+\frac {2 \left (9 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{3} d}+\frac {2 \left (9 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 b^{3} d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,b^{2}}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2} d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{2} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {a^{3} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,b^{4}}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{4}}\) | \(244\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^3*(1/3*sin(d*x+c)^3*b^2-1/2*b*a*sin(d*x+c)^2+a^2*sin(d*x+c))-a^3/ b^4*ln(a+b*sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/6*(3*a*b^2*cos(d*x + c)^2 - 6*a^3*log(b*sin(d*x + c) + a) - 2*(b^3*cos(d *x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c))/(b^4*d)
Time = 0.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.38 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\begin {cases} \frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} & \text {for}\: b = 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {a^{3} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b^{4} d} + \frac {a^{2} \sin {\left (c + d x \right )}}{b^{3} d} + \frac {a \cos ^{2}{\left (c + d x \right )}}{2 b^{2} d} + \frac {\sin ^{3}{\left (c + d x \right )}}{3 b d} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Piecewise((x*sin(c)**3*cos(c)/a, Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)**4/(4 *a*d), Eq(b, 0)), (x*sin(c)**3*cos(c)/(a + b*sin(c)), Eq(d, 0)), (-a**3*lo g(a/b + sin(c + d*x))/(b**4*d) + a**2*sin(c + d*x)/(b**3*d) + a*cos(c + d* x)**2/(2*b**2*d) + sin(c + d*x)**3/(3*b*d), True))
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}} - \frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{b^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/6*(6*a^3*log(b*sin(d*x + c) + a)/b^4 - (2*b^2*sin(d*x + c)^3 - 3*a*b*si n(d*x + c)^2 + 6*a^2*sin(d*x + c))/b^3)/d
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4} d} + \frac {2 \, b^{2} d^{2} \sin \left (d x + c\right )^{3} - 3 \, a b d^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} d^{2} \sin \left (d x + c\right )}{6 \, b^{3} d^{3}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-a^3*log(abs(b*sin(d*x + c) + a))/(b^4*d) + 1/6*(2*b^2*d^2*sin(d*x + c)^3 - 3*a*b*d^2*sin(d*x + c)^2 + 6*a^2*d^2*sin(d*x + c))/(b^3*d^3)
Time = 19.68 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3}{3\,b}-\frac {a^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^4}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2\,b^2}+\frac {a^2\,\sin \left (c+d\,x\right )}{b^3}}{d} \] Input:
int((cos(c + d*x)*sin(c + d*x)^3)/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)^3/(3*b) - (a^3*log(a + b*sin(c + d*x)))/b^4 - (a*sin(c + d*x )^2)/(2*b^2) + (a^2*sin(c + d*x))/b^3)/d
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \cos \left (d x +c \right )^{2} a \,b^{2}-6 \,\mathrm {log}\left (\sin \left (d x +c \right ) b +a \right ) a^{3}+2 \sin \left (d x +c \right )^{3} b^{3}+6 \sin \left (d x +c \right ) a^{2} b}{6 b^{4} d} \] Input:
int(cos(d*x+c)*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
(3*cos(c + d*x)**2*a*b**2 - 6*log(sin(c + d*x)*b + a)*a**3 + 2*sin(c + d*x )**3*b**3 + 6*sin(c + d*x)*a**2*b)/(6*b**4*d)