Integrand size = 26, antiderivative size = 91 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {f x}{4 a d}+\frac {f \cos (c+d x)}{a d^2}+\frac {(e+f x) \sin (c+d x)}{a d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 a d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d} \] Output:
1/4*f*x/a/d+f*cos(d*x+c)/a/d^2+(f*x+e)*sin(d*x+c)/a/d-1/4*f*cos(d*x+c)*sin (d*x+c)/a/d^2-1/2*(f*x+e)*sin(d*x+c)^2/a/d
Time = 7.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-f \cos (c+d x) (-4+\sin (c+d x))+d (e+f x) (\cos (2 (c+d x))+4 \sin (c+d x))}{4 a d^2} \] Input:
Integrate[((e + f*x)*Cos[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(-(f*Cos[c + d*x]*(-4 + Sin[c + d*x])) + d*(e + f*x)*(Cos[2*(c + d*x)] + 4 *Sin[c + d*x]))/(4*a*d^2)
Time = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5034, 3042, 3777, 25, 3042, 3118, 4904, 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.
\(\displaystyle \int \frac {(e+f x) \cos ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 5034 |
\(\displaystyle \frac {\int (e+f x) \cos (c+d x)dx}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x) \sin \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {\frac {f \int -\sin (c+d x)dx}{d}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x)dx}{a}\) |
\(\Big \downarrow \) 4904 |
\(\displaystyle \frac {\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\frac {(e+f x) \sin ^2(c+d x)}{2 d}-\frac {f \int \sin ^2(c+d x)dx}{2 d}}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\frac {(e+f x) \sin ^2(c+d x)}{2 d}-\frac {f \int \sin (c+d x)^2dx}{2 d}}{a}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\frac {(e+f x) \sin ^2(c+d x)}{2 d}-\frac {f \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d}}{a}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}}{a}-\frac {\frac {(e+f x) \sin ^2(c+d x)}{2 d}-\frac {f \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d}}{a}\) |
Input:
Int[((e + f*x)*Cos[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
((f*Cos[c + d*x])/d^2 + ((e + f*x)*Sin[c + d*x])/d)/a - (((e + f*x)*Sin[c + d*x]^2)/(2*d) - (f*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/(2*d))/a
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[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
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]
Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.) *Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Simp[1/b Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*Sin[ c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^ 2 - b^2, 0]
Time = 1.77 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76
method | result | size |
parallelrisch | \(\frac {2 d \left (f x +e \right ) \cos \left (2 d x +2 c \right )-\sin \left (2 d x +2 c \right ) f +8 d \left (f x +e \right ) \sin \left (d x +c \right )-2 d e +8 \cos \left (d x +c \right ) f -8 f}{8 d^{2} a}\) | \(69\) |
risch | \(\frac {f \cos \left (d x +c \right )}{a \,d^{2}}+\frac {\left (f x +e \right ) \sin \left (d x +c \right )}{a d}+\frac {\left (f x +e \right ) \cos \left (2 d x +2 c \right )}{4 a d}-\frac {f \sin \left (2 d x +2 c \right )}{8 d^{2} a}\) | \(74\) |
default | \(\frac {\frac {4 e \left (-\frac {\sin \left (d x +c \right )^{2}}{2}+\sin \left (d x +c \right )\right )}{d}+\frac {4 f \left (\frac {\left (d x +c \right ) \cos \left (d x +c \right )^{2}}{2}-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}-\frac {d x}{4}-\frac {c}{4}-\frac {c \cos \left (d x +c \right )^{2}}{2}+\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )-c \sin \left (d x +c \right )\right )}{d^{2}}}{4 a}\) | \(111\) |
norman | \(\frac {\frac {2 f}{d^{2} a}+\frac {\left (2 d e +2 f \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d^{2} a}+\frac {\left (2 d e +4 f \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d^{2} a}+\frac {f x}{4 a d}+\frac {5 f \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d^{2} a}+\frac {7 f \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d^{2} a}+\frac {\left (4 d e +f \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d^{2} a}+\frac {\left (4 d e +3 f \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d^{2} a}+\frac {9 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {3 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a d}+\frac {11 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a d}+\frac {11 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a d}+\frac {3 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a d}+\frac {9 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4 a d}+\frac {f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(337\) |
Input:
int((f*x+e)*cos(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/8*(2*d*(f*x+e)*cos(2*d*x+2*c)-sin(2*d*x+2*c)*f+8*d*(f*x+e)*sin(d*x+c)-2* d*e+8*cos(d*x+c)*f-8*f)/d^2/a
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {d f x - 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right )^{2} - 4 \, f \cos \left (d x + c\right ) - {\left (4 \, d f x + 4 \, d e - f \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, a d^{2}} \] Input:
integrate((f*x+e)*cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/4*(d*f*x - 2*(d*f*x + d*e)*cos(d*x + c)^2 - 4*f*cos(d*x + c) - (4*d*f*x + 4*d*e - f*cos(d*x + c))*sin(d*x + c))/(a*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 724 vs. \(2 (78) = 156\).
Time = 2.65 (sec) , antiderivative size = 724, normalized size of antiderivative = 7.96 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((f*x+e)*cos(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Piecewise((8*d*e*tan(c/2 + d*x/2)**3/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d **2*tan(c/2 + d*x/2)**2 + 4*a*d**2) - 8*d*e*tan(c/2 + d*x/2)**2/(4*a*d**2* tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*e*tan (c/2 + d*x/2)/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + d*f*x*tan(c/2 + d*x/2)**4/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8 *a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*f*x*tan(c/2 + d*x/2)**3/(4*a *d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) - 6*d *f*x*tan(c/2 + d*x/2)**2/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*f*x*tan(c/2 + d*x/2)/(4*a*d**2*tan(c/2 + d*x /2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + d*f*x/(4*a*d**2*tan(c/ 2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 2*f*tan(c/2 + d *x/2)**3/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4* a*d**2) + 8*f*tan(c/2 + d*x/2)**2/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2 *tan(c/2 + d*x/2)**2 + 4*a*d**2) - 2*f*tan(c/2 + d*x/2)/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*f/(4*a*d**2*tan (c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2), Ne(d, 0)), (( e*x + f*x**2/2)*cos(c)**3/(a*sin(c) + a), True))
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.25 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} e}{a} - \frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c f}{a d} - \frac {{\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} f}{a d}}{8 \, d} \] Input:
integrate((f*x+e)*cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/8*(4*(sin(d*x + c)^2 - 2*sin(d*x + c))*e/a - 4*(sin(d*x + c)^2 - 2*sin( d*x + c))*c*f/(a*d) - (2*(d*x + c)*cos(2*d*x + 2*c) + 8*(d*x + c)*sin(d*x + c) + 8*cos(d*x + c) - sin(2*d*x + 2*c))*f/(a*d))/d
Leaf count of result is larger than twice the leaf count of optimal. 947 vs. \(2 (85) = 170\).
Time = 0.17 (sec) , antiderivative size = 947, normalized size of antiderivative = 10.41 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((f*x+e)*cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/4*(d*f*x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*d*f*x*tan(1/2*d*x)^4*tan(1/2*c) ^3 - 8*d*f*x*tan(1/2*d*x)^3*tan(1/2*c)^4 + d*e*tan(1/2*d*x)^4*tan(1/2*c)^4 - 6*d*f*x*tan(1/2*d*x)^4*tan(1/2*c)^2 - 16*d*f*x*tan(1/2*d*x)^3*tan(1/2*c )^3 - 8*d*e*tan(1/2*d*x)^4*tan(1/2*c)^3 - 6*d*f*x*tan(1/2*d*x)^2*tan(1/2*c )^4 - 8*d*e*tan(1/2*d*x)^3*tan(1/2*c)^4 + 4*f*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*d*f*x*tan(1/2*d*x)^4*tan(1/2*c) - 6*d*e*tan(1/2*d*x)^4*tan(1/2*c)^2 - 16*d*e*tan(1/2*d*x)^3*tan(1/2*c)^3 + 2*f*tan(1/2*d*x)^4*tan(1/2*c)^3 - 8*d *f*x*tan(1/2*d*x)*tan(1/2*c)^4 - 6*d*e*tan(1/2*d*x)^2*tan(1/2*c)^4 + 2*f*t an(1/2*d*x)^3*tan(1/2*c)^4 + d*f*x*tan(1/2*d*x)^4 + 16*d*f*x*tan(1/2*d*x)^ 3*tan(1/2*c) - 8*d*e*tan(1/2*d*x)^4*tan(1/2*c) + 36*d*f*x*tan(1/2*d*x)^2*t an(1/2*c)^2 + 16*d*f*x*tan(1/2*d*x)*tan(1/2*c)^3 - 16*f*tan(1/2*d*x)^3*tan (1/2*c)^3 + d*f*x*tan(1/2*c)^4 - 8*d*e*tan(1/2*d*x)*tan(1/2*c)^4 + 8*d*f*x *tan(1/2*d*x)^3 + d*e*tan(1/2*d*x)^4 + 16*d*e*tan(1/2*d*x)^3*tan(1/2*c) - 2*f*tan(1/2*d*x)^4*tan(1/2*c) + 36*d*e*tan(1/2*d*x)^2*tan(1/2*c)^2 - 12*f* tan(1/2*d*x)^3*tan(1/2*c)^2 + 8*d*f*x*tan(1/2*c)^3 + 16*d*e*tan(1/2*d*x)*t an(1/2*c)^3 - 12*f*tan(1/2*d*x)^2*tan(1/2*c)^3 + d*e*tan(1/2*c)^4 - 2*f*ta n(1/2*d*x)*tan(1/2*c)^4 - 6*d*f*x*tan(1/2*d*x)^2 + 8*d*e*tan(1/2*d*x)^3 - 4*f*tan(1/2*d*x)^4 - 16*d*f*x*tan(1/2*d*x)*tan(1/2*c) - 16*f*tan(1/2*d*x)^ 3*tan(1/2*c) - 6*d*f*x*tan(1/2*c)^2 + 8*d*e*tan(1/2*c)^3 - 16*f*tan(1/2*d* x)*tan(1/2*c)^3 - 4*f*tan(1/2*c)^4 + 8*d*f*x*tan(1/2*d*x) - 6*d*e*tan(1...
Time = 38.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {f\,\sin \left (2\,c+2\,d\,x\right )}{2}+8\,f\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,d\,e\,\sin \left (c+d\,x\right )+2\,d\,e\,{\sin \left (c+d\,x\right )}^2-4\,d\,f\,x\,\sin \left (c+d\,x\right )+d\,f\,x\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )}{4\,a\,d^2} \] Input:
int((cos(c + d*x)^3*(e + f*x))/(a + a*sin(c + d*x)),x)
Output:
-((f*sin(2*c + 2*d*x))/2 + 8*f*sin(c/2 + (d*x)/2)^2 - 4*d*e*sin(c + d*x) + 2*d*e*sin(c + d*x)^2 - 4*d*f*x*sin(c + d*x) + d*f*x*(2*sin(c + d*x)^2 - 1 ))/(4*a*d^2)
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.99 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right ) f +4 \cos \left (d x +c \right ) f -2 \sin \left (d x +c \right )^{2} d e -2 \sin \left (d x +c \right )^{2} d f x +4 \sin \left (d x +c \right ) d e +4 \sin \left (d x +c \right ) d f x +c f +4 d e +d f x}{4 a \,d^{2}} \] Input:
int((f*x+e)*cos(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
( - cos(c + d*x)*sin(c + d*x)*f + 4*cos(c + d*x)*f - 2*sin(c + d*x)**2*d*e - 2*sin(c + d*x)**2*d*f*x + 4*sin(c + d*x)*d*e + 4*sin(c + d*x)*d*f*x + c *f + 4*d*e + d*f*x)/(4*a*d**2)