3.276 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=76 \[ \frac {4 \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac {4 \tan (e+f x)}{5 a^2 c^3 f}+\frac {\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

1/5*sec(f*x+e)^3/a^2/f/(c^3-c^3*sin(f*x+e))+4/5*tan(f*x+e)/a^2/c^3/f+4/15*tan(f*x+e)^3/a^2/c^3/f

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Rubi [A]  time = 0.11, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac {4 \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac {4 \tan (e+f x)}{5 a^2 c^3 f}+\frac {\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x]

[Out]

Sec[e + f*x]^3/(5*a^2*f*(c^3 - c^3*Sin[e + f*x])) + (4*Tan[e + f*x])/(5*a^2*c^3*f) + (4*Tan[e + f*x]^3)/(15*a^
2*c^3*f)

Rule 2672

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] + 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 2736

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 3767

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*} \int \frac {1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx &=\frac {\int \frac {\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=\frac {\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {4 \int \sec ^4(e+f x) \, dx}{5 a^2 c^3}\\ &=\frac {\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {4 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^2 c^3 f}\\ &=\frac {\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {4 \tan (e+f x)}{5 a^2 c^3 f}+\frac {4 \tan ^3(e+f x)}{15 a^2 c^3 f}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 131, normalized size = 1.72 \[ -\frac {18 \sin (e+f x)+512 \sin (2 (e+f x))+27 \sin (3 (e+f x))+128 \sin (4 (e+f x))+9 \sin (5 (e+f x))+128 \cos (e+f x)-72 \cos (2 (e+f x))+192 \cos (3 (e+f x))-18 \cos (4 (e+f x))+64 \cos (5 (e+f x))-54}{1920 a^2 c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x]

[Out]

-1/1920*(-54 + 128*Cos[e + f*x] - 72*Cos[2*(e + f*x)] + 192*Cos[3*(e + f*x)] - 18*Cos[4*(e + f*x)] + 64*Cos[5*
(e + f*x)] + 18*Sin[e + f*x] + 512*Sin[2*(e + f*x)] + 27*Sin[3*(e + f*x)] + 128*Sin[4*(e + f*x)] + 9*Sin[5*(e
+ f*x)])/(a^2*c^3*f*(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x])^2)

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fricas [A]  time = 0.44, size = 86, normalized size = 1.13 \[ -\frac {8 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right ) - 1}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(8*cos(f*x + e)^4 - 4*cos(f*x + e)^2 + 4*(2*cos(f*x + e)^2 + 1)*sin(f*x + e) - 1)/(a^2*c^3*f*cos(f*x + e
)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)

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giac [A]  time = 0.23, size = 133, normalized size = 1.75 \[ -\frac {\frac {5 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 24 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13\right )}}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {165 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 400 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/120*(5*(15*tan(1/2*f*x + 1/2*e)^2 + 24*tan(1/2*f*x + 1/2*e) + 13)/(a^2*c^3*(tan(1/2*f*x + 1/2*e) + 1)^3) +
(165*tan(1/2*f*x + 1/2*e)^4 - 480*tan(1/2*f*x + 1/2*e)^3 + 650*tan(1/2*f*x + 1/2*e)^2 - 400*tan(1/2*f*x + 1/2*
e) + 113)/(a^2*c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f

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maple [A]  time = 0.26, size = 133, normalized size = 1.75 \[ \frac {-\frac {2}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {5}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {11}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {1}{6 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {5}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{f \,c^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x)

[Out]

2/f/c^3/a^2*(-1/5/(tan(1/2*f*x+1/2*e)-1)^5-1/2/(tan(1/2*f*x+1/2*e)-1)^4-5/6/(tan(1/2*f*x+1/2*e)-1)^3-3/4/(tan(
1/2*f*x+1/2*e)-1)^2-11/16/(tan(1/2*f*x+1/2*e)-1)-1/12/(tan(1/2*f*x+1/2*e)+1)^3+1/8/(tan(1/2*f*x+1/2*e)+1)^2-5/
16/(tan(1/2*f*x+1/2*e)+1))

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maxima [B]  time = 0.97, size = 335, normalized size = 4.41 \[ \frac {2 \, {\left (\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + 3\right )}}{15 \, {\left (a^{2} c^{3} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{2} c^{3} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 25*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 15*sin(f*x + e)
^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 3)/((a^2*c^3 - 2*a^2*c^3*sin(f*x + e)/(cos(
f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 6*a^2*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
- 6*a^2*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 2*a^2*c^3*si
n(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2*c^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*f)

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mupad [B]  time = 7.83, size = 128, normalized size = 1.68 \[ -\frac {2\,\left (15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7-15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+25\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+13\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-21\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+9\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+3\right )}{15\,a^2\,c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^3),x)

[Out]

-(2*(9*tan(e/2 + (f*x)/2) - 21*tan(e/2 + (f*x)/2)^2 + 13*tan(e/2 + (f*x)/2)^3 + 25*tan(e/2 + (f*x)/2)^4 - 5*ta
n(e/2 + (f*x)/2)^5 - 15*tan(e/2 + (f*x)/2)^6 + 15*tan(e/2 + (f*x)/2)^7 + 3))/(15*a^2*c^3*f*(tan(e/2 + (f*x)/2)
- 1)^5*(tan(e/2 + (f*x)/2) + 1)^3)

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sympy [A]  time = 15.88, size = 1418, normalized size = 18.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*tan(e/2 + f*x/2)**7/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 -
30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 +
30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 30*tan(e/2 + f*x/2)*
*6/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)*
*6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)
**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 10*tan(e/2 + f*x/2)**5/(15*a**2*c**3*f*tan(e/2 + f*x
/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*
x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f
*x/2) - 15*a**2*c**3*f) - 50*tan(e/2 + f*x/2)**4/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2
+ f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2
+ f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 26*tan
(e/2 + f*x/2)**3/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan
(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*ta
n(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 42*tan(e/2 + f*x/2)**2/(15*a**2*c**3*f
*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*
f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3
*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 18*tan(e/2 + f*x/2)/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3
*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**
3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*
f) - 6/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x
/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*
x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f), Ne(f, 0)), (x/((a*sin(e) + a)**2*(-c*sin(e) + c)*
*3), True))

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