\(\int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx\) [267]

Optimal result

Integrand size = 26, antiderivative size = 85 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{5 a c^3 f} \]

[Out]

1/5*sec(f*x+e)/a/c/f/(c-c*sin(f*x+e))^2+1/5*sec(f*x+e)/a/f/(c^3-c^3*sin(f*x+e))+2/5*tan(f*x+e)/a/c^3/f

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2751, 3852, 8} \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {2 \tan (e+f x)}{5 a c^3 f}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]

[In]

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^3),x]

[Out]

Sec[e + f*x]/(5*a*c*f*(c - c*Sin[e + f*x])^2) + Sec[e + f*x]/(5*a*f*(c^3 - c^3*Sin[e + f*x])) + (2*Tan[e + f*x
])/(5*a*c^3*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[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[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[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] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{a c} \\ & = \frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {3 \int \frac {\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{5 a c^2} \\ & = \frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 \int \sec ^2(e+f x) \, dx}{5 a c^3} \\ & = \frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {2 \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{5 a c^3 f} \\ & = \frac {\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac {\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {2 \tan (e+f x)}{5 a c^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\frac {425 \cos (e+f x)+128 \cos (2 (e+f x))-85 \cos (3 (e+f x))+160 \sin (e+f x)-340 \sin (2 (e+f x))-32 \sin (3 (e+f x))}{320 a c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^3),x]

[Out]

(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*c^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {4 i \left (-4 i {\mathrm e}^{i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{5 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a \,c^{3} f}\)	\(66\)
parallelrisch	\(\frac {-\frac {4}{5}-2 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+4 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-4 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5}}{f a \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(90\)
derivativedivides	\(\frac {-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {5}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a \,c^{3} f}\)	\(103\)
default	\(\frac {-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {5}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{a \,c^{3} f}\)	\(103\)
norman	\(\frac {\frac {4 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {4 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}-\frac {4}{5 a c f}-\frac {2 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}+\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{5 a c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(129\)

[In]

int(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

-4/5*I*(-4*I*exp(I*(f*x+e))+5*exp(2*I*(f*x+e))-1)/(exp(I*(f*x+e))-I)^5/(exp(I*(f*x+e))+I)/a/c^3/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {4 \, \cos \left (f x + e\right )^{2} - {\left (2 \, \cos \left (f x + e\right )^{2} - 3\right )} \sin \left (f x + e\right ) - 2}{5 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/5*(4*cos(f*x + e)^2 - (2*cos(f*x + e)^2 - 3)*sin(f*x + e) - 2)/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x
+ e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 614 vs. \(2 (66) = 132\).

Time = 2.36 (sec) , antiderivative size = 614, normalized size of antiderivative = 7.22 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=\begin {cases} - \frac {10 \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} + \frac {20 \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} - \frac {20 \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} + \frac {6 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} - \frac {4}{5 a c^{3} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 20 a c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 a c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 20 a c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 a c^{3} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-10*tan(e/2 + f*x/2)**5/(5*a*c**3*f*tan(e/2 + f*x/2)**6 - 20*a*c**3*f*tan(e/2 + f*x/2)**5 + 25*a*c*
*3*f*tan(e/2 + f*x/2)**4 - 25*a*c**3*f*tan(e/2 + f*x/2)**2 + 20*a*c**3*f*tan(e/2 + f*x/2) - 5*a*c**3*f) + 20*t
an(e/2 + f*x/2)**4/(5*a*c**3*f*tan(e/2 + f*x/2)**6 - 20*a*c**3*f*tan(e/2 + f*x/2)**5 + 25*a*c**3*f*tan(e/2 + f
*x/2)**4 - 25*a*c**3*f*tan(e/2 + f*x/2)**2 + 20*a*c**3*f*tan(e/2 + f*x/2) - 5*a*c**3*f) - 20*tan(e/2 + f*x/2)*
*3/(5*a*c**3*f*tan(e/2 + f*x/2)**6 - 20*a*c**3*f*tan(e/2 + f*x/2)**5 + 25*a*c**3*f*tan(e/2 + f*x/2)**4 - 25*a*
c**3*f*tan(e/2 + f*x/2)**2 + 20*a*c**3*f*tan(e/2 + f*x/2) - 5*a*c**3*f) + 6*tan(e/2 + f*x/2)/(5*a*c**3*f*tan(e
/2 + f*x/2)**6 - 20*a*c**3*f*tan(e/2 + f*x/2)**5 + 25*a*c**3*f*tan(e/2 + f*x/2)**4 - 25*a*c**3*f*tan(e/2 + f*x
/2)**2 + 20*a*c**3*f*tan(e/2 + f*x/2) - 5*a*c**3*f) - 4/(5*a*c**3*f*tan(e/2 + f*x/2)**6 - 20*a*c**3*f*tan(e/2
+ f*x/2)**5 + 25*a*c**3*f*tan(e/2 + f*x/2)**4 - 25*a*c**3*f*tan(e/2 + f*x/2)**2 + 20*a*c**3*f*tan(e/2 + f*x/2)
- 5*a*c**3*f), Ne(f, 0)), (x/((a*sin(e) + a)*(-c*sin(e) + c)**3), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (81) = 162\).

Time = 0.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.48 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - 2\right )}}{5 \, {\left (a c^{3} - \frac {4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} f} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-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*c^3 - 4*a*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 5
*a*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a*c^3*sin(f*x + e
)^5/(cos(f*x + e) + 1)^5 - a*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*f)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {\frac {5}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 90 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 21}{a c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{20 \, f} \]

[In]

integrate(1/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/20*(5/(a*c^3*(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*c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f

Mupad [B] (verification not implemented)

Time = 6.61 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx=-\frac {2\,\left (5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\right )}{5\,a\,c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )} \]

[In]

int(1/((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^3),x)

[Out]

-(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*c^3*f*(tan(e/2 + (f*x)/2) - 1)^5*(tan(e/2 + (f*x)/2) + 1))