Integrand size = 26, antiderivative size = 83 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {\sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {\sec (e+f x)}{5 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{5 a^3 c f} \] Output:
-1/5*sec(f*x+e)/a/c/f/(a+a*sin(f*x+e))^2-1/5*sec(f*x+e)/c/f/(a^3+a^3*sin(f *x+e))+2/5*tan(f*x+e)/a^3/c/f
Time = 1.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (425 \cos (e+f x)+128 \cos (2 (e+f x))-85 \cos (3 (e+f x))-160 \sin (e+f x)+340 \sin (2 (e+f x))+32 \sin (3 (e+f x)))}{320 a^3 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^3} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])),x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(425*Cos[e + f*x] + 128*Cos[2*(e + f*x)] - 85*Cos[3*(e + f*x)] - 160*Si n[e + f*x] + 340*Sin[2*(e + f*x)] + 32*Sin[3*(e + f*x)]))/(320*a^3*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3)
Time = 0.49 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3215, 3042, 3151, 3042, 3151, 3042, 4254, 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.
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{(\sin (e+f x) a+a)^2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\cos (e+f x)^2 (\sin (e+f x) a+a)^2}dx}{a c}\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {\frac {3 \int \frac {\sec ^2(e+f x)}{\sin (e+f x) a+a}dx}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\cos (e+f x)^2 (\sin (e+f x) a+a)}dx}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {\frac {3 \left (\frac {2 \int \sec ^2(e+f x)dx}{3 a}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (\frac {2 \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx}{3 a}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\frac {3 \left (-\frac {2 \int 1d(-\tan (e+f x))}{3 a f}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {3 \left (\frac {2 \tan (e+f x)}{3 a f}-\frac {\sec (e+f x)}{3 f (a \sin (e+f x)+a)}\right )}{5 a}-\frac {\sec (e+f x)}{5 f (a \sin (e+f x)+a)^2}}{a c}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])),x]
Output:
(-1/5*Sec[e + f*x]/(f*(a + a*Sin[e + f*x])^2) + (3*(-1/3*Sec[e + f*x]/(f*( a + a*Sin[e + f*x])) + (2*Tan[e + f*x])/(3*a*f)))/(5*a))/(a*c)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.74 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61
method | result | size |
parallelrisch | \(\frac {\left (2 \sec \left (f x +e \right )^{4}+\sec \left (f x +e \right )^{2}+2\right ) \tan \left (f x +e \right )-2 \sec \left (f x +e \right )^{5}-2}{5 a^{3} c f}\) | \(51\) |
risch | \(-\frac {4 i \left (4 i {\mathrm e}^{i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{5 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a^{3} c f}\) | \(66\) |
derivativedivides | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {4}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {5}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c f}\) | \(101\) |
default | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {4}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {5}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} c f}\) | \(101\) |
norman | \(\frac {-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a f c}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a c f}+\frac {4}{5 a c f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f c}+\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 a c f}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\) | \(129\) |
Input:
int(1/(a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e)),x,method=_RETURNVERBOSE)
Output:
1/5*((2*sec(f*x+e)^4+sec(f*x+e)^2+2)*tan(f*x+e)-2*sec(f*x+e)^5-2)/a^3/c/f
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {4 \, \cos \left (f x + e\right )^{2} + {\left (2 \, \cos \left (f x + e\right )^{2} - 3\right )} \sin \left (f x + e\right ) - 2}{5 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="fricas")
Output:
1/5*(4*cos(f*x + e)^2 + (2*cos(f*x + e)^2 - 3)*sin(f*x + e) - 2)/(a^3*c*f* cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 614 vs. \(2 (66) = 132\).
Time = 2.44 (sec) , antiderivative size = 614, normalized size of antiderivative = 7.40 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e)),x)
Output:
Piecewise((-10*tan(e/2 + f*x/2)**5/(5*a**3*c*f*tan(e/2 + f*x/2)**6 + 20*a* *3*c*f*tan(e/2 + f*x/2)**5 + 25*a**3*c*f*tan(e/2 + f*x/2)**4 - 25*a**3*c*f *tan(e/2 + f*x/2)**2 - 20*a**3*c*f*tan(e/2 + f*x/2) - 5*a**3*c*f) - 20*tan (e/2 + f*x/2)**4/(5*a**3*c*f*tan(e/2 + f*x/2)**6 + 20*a**3*c*f*tan(e/2 + f *x/2)**5 + 25*a**3*c*f*tan(e/2 + f*x/2)**4 - 25*a**3*c*f*tan(e/2 + f*x/2)* *2 - 20*a**3*c*f*tan(e/2 + f*x/2) - 5*a**3*c*f) - 20*tan(e/2 + f*x/2)**3/( 5*a**3*c*f*tan(e/2 + f*x/2)**6 + 20*a**3*c*f*tan(e/2 + f*x/2)**5 + 25*a**3 *c*f*tan(e/2 + f*x/2)**4 - 25*a**3*c*f*tan(e/2 + f*x/2)**2 - 20*a**3*c*f*t an(e/2 + f*x/2) - 5*a**3*c*f) + 6*tan(e/2 + f*x/2)/(5*a**3*c*f*tan(e/2 + f *x/2)**6 + 20*a**3*c*f*tan(e/2 + f*x/2)**5 + 25*a**3*c*f*tan(e/2 + f*x/2)* *4 - 25*a**3*c*f*tan(e/2 + f*x/2)**2 - 20*a**3*c*f*tan(e/2 + f*x/2) - 5*a* *3*c*f) + 4/(5*a**3*c*f*tan(e/2 + f*x/2)**6 + 20*a**3*c*f*tan(e/2 + f*x/2) **5 + 25*a**3*c*f*tan(e/2 + f*x/2)**4 - 25*a**3*c*f*tan(e/2 + f*x/2)**2 - 20*a**3*c*f*tan(e/2 + f*x/2) - 5*a**3*c*f), Ne(f, 0)), (x/((a*sin(e) + a)* *3*(-c*sin(e) + c)), True))
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (77) = 154\).
Time = 0.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.54 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {2 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 2\right )}}{5 \, {\left (a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="maxima")
Output:
-2/5*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 10*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f* x + e) + 1)^5 + 2)/((a^3*c + 4*a^3*c*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a ^3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*a^3*c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - a^3*c*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*f)
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {\frac {5}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 90 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 21}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{20 \, f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x, algorithm="giac")
Output:
-1/20*(5/(a^3*c*(tan(1/2*f*x + 1/2*e) - 1)) + (35*tan(1/2*f*x + 1/2*e)^4 + 90*tan(1/2*f*x + 1/2*e)^3 + 120*tan(1/2*f*x + 1/2*e)^2 + 70*tan(1/2*f*x + 1/2*e) + 21)/(a^3*c*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
Time = 17.64 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=-\frac {2\,\left (5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-2\right )}{5\,a^3\,c\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5} \] Input:
int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))),x)
Output:
-(2*(10*tan(e/2 + (f*x)/2)^3 - 3*tan(e/2 + (f*x)/2) + 10*tan(e/2 + (f*x)/2 )^4 + 5*tan(e/2 + (f*x)/2)^5 - 2))/(5*a^3*c*f*(tan(e/2 + (f*x)/2) - 1)*(ta n(e/2 + (f*x)/2) + 1)^5)
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx=\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )+4 \sin \left (f x +e \right )^{3}+8 \sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )-4}{10 \cos \left (f x +e \right ) a^{3} c f \left (\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1\right )} \] Input:
int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x)
Output:
(cos(e + f*x)*sin(e + f*x)**2 + 2*cos(e + f*x)*sin(e + f*x) + cos(e + f*x) + 4*sin(e + f*x)**3 + 8*sin(e + f*x)**2 + 2*sin(e + f*x) - 4)/(10*cos(e + f*x)*a**3*c*f*(sin(e + f*x)**2 + 2*sin(e + f*x) + 1))