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

Optimal. Leaf size=59 \[ \frac{\tan ^5(e+f x)}{5 a^3 c^3 f}+\frac{2 \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac{\tan (e+f x)}{a^3 c^3 f} \]

[Out]

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

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Rubi [A]  time = 0.0725032, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2736, 3767} \[ \frac{\tan ^5(e+f x)}{5 a^3 c^3 f}+\frac{2 \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac{\tan (e+f x)}{a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^3 (c-c \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) \, dx}{a^3 c^3}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{a^3 c^3 f}\\ &=\frac{\tan (e+f x)}{a^3 c^3 f}+\frac{2 \tan ^3(e+f x)}{3 a^3 c^3 f}+\frac{\tan ^5(e+f x)}{5 a^3 c^3 f}\\ \end{align*}

Mathematica [A]  time = 0.12801, size = 41, normalized size = 0.69 \[ \frac{\frac{1}{5} \tan ^5(e+f x)+\frac{2}{3} \tan ^3(e+f x)+\tan (e+f x)}{a^3 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{3} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.13831, size = 54, normalized size = 0.92 \begin{align*} \frac{3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )}{15 \, a^{3} c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))/(a^3*c^3*f)

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Fricas [A]  time = 1.56539, size = 119, normalized size = 2.02 \begin{align*} \frac{{\left (8 \, \cos \left (f x + e\right )^{4} + 4 \, \cos \left (f x + e\right )^{2} + 3\right )} \sin \left (f x + e\right )}{15 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 61.7444, size = 687, normalized size = 11.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-30*tan(e/2 + f*x/2)**9/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 +
150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**
2 - 15*a**3*c**3*f) + 40*tan(e/2 + f*x/2)**7/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*tan(e/2 + f
*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3*f*tan(e/2
+ f*x/2)**2 - 15*a**3*c**3*f) - 116*tan(e/2 + f*x/2)**5/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a**3*c**3*f*
tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75*a**3*c**3
*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) + 40*tan(e/2 + f*x/2)**3/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10 - 75*a*
*3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)**4 + 75
*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f) - 30*tan(e/2 + f*x/2)/(15*a**3*c**3*f*tan(e/2 + f*x/2)**10
- 75*a**3*c**3*f*tan(e/2 + f*x/2)**8 + 150*a**3*c**3*f*tan(e/2 + f*x/2)**6 - 150*a**3*c**3*f*tan(e/2 + f*x/2)*
*4 + 75*a**3*c**3*f*tan(e/2 + f*x/2)**2 - 15*a**3*c**3*f), Ne(f, 0)), (x/((a*sin(e) + a)**3*(-c*sin(e) + c)**3
), True))

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Giac [A]  time = 2.01524, size = 58, normalized size = 0.98 \begin{align*} \frac{3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )}{15 \, a^{3} c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))/(a^3*c^3*f)