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

Optimal. Leaf size=34 \[ \frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2736, 2671} \[ \frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2671

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*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^3} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}\\ \end {align*}

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Mathematica [B]  time = 0.40, size = 81, normalized size = 2.38 \[ \frac {a^2 \left (-10 \sin \left (\frac {1}{2} (e+f x)\right )-5 \sin \left (\frac {3}{2} (e+f x)\right )+\sin \left (\frac {5}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{10 c^3 f (\sin (e+f x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-10*Sin[(e + f*x)/2] - 5*Sin[(3*(e + f*x))/2] + Sin[(5*(e + f*x))/
2]))/(10*c^3*f*(-1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.46, size = 168, normalized size = 4.94 \[ \frac {a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} + {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 60, normalized size = 1.76 \[ -\frac {2 \, {\left (5 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 10 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{2}\right )}}{5 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(5*a^2*tan(1/2*f*x + 1/2*e)^4 + 10*a^2*tan(1/2*f*x + 1/2*e)^2 + a^2)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)

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maple [B]  time = 0.27, size = 88, normalized size = 2.59 \[ \frac {2 a^{2} \left (-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\right )}{f \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 2.03, size = 557, normalized size = 16.38 \[ -\frac {2 \, {\left (\frac {a^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 7\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {6 \, a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(a^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*a^2*(5*sin(f*x + e)/(cos(f*x + e) + 1)
- 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)
/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*a^2*(5*sin(f*x + e)
/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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mupad [B]  time = 6.99, size = 92, normalized size = 2.71 \[ \frac {2\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{5\,c^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*cos(e/2 + (f*x)/2)*(cos(e/2 + (f*x)/2)^4 + 5*sin(e/2 + (f*x)/2)^4 + 10*cos(e/2 + (f*x)/2)^2*sin(e/2 + (
f*x)/2)^2))/(5*c^3*f*(cos(e/2 + (f*x)/2) - sin(e/2 + (f*x)/2))^5)

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sympy [A]  time = 14.88, size = 354, normalized size = 10.41 \[ \begin {cases} - \frac {10 a^{2} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} - \frac {20 a^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} - \frac {2 a^{2}}{5 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 25 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 50 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 5 c^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\relax (e )} + a\right )^{2}}{\left (- c \sin {\relax (e )} + c\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-10*a**2*tan(e/2 + f*x/2)**4/(5*c**3*f*tan(e/2 + f*x/2)**5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**
3*f*tan(e/2 + f*x/2)**3 - 50*c**3*f*tan(e/2 + f*x/2)**2 + 25*c**3*f*tan(e/2 + f*x/2) - 5*c**3*f) - 20*a**2*tan
(e/2 + f*x/2)**2/(5*c**3*f*tan(e/2 + f*x/2)**5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**3*f*tan(e/2 + f*x/2)**3
- 50*c**3*f*tan(e/2 + f*x/2)**2 + 25*c**3*f*tan(e/2 + f*x/2) - 5*c**3*f) - 2*a**2/(5*c**3*f*tan(e/2 + f*x/2)*
*5 - 25*c**3*f*tan(e/2 + f*x/2)**4 + 50*c**3*f*tan(e/2 + f*x/2)**3 - 50*c**3*f*tan(e/2 + f*x/2)**2 + 25*c**3*f
*tan(e/2 + f*x/2) - 5*c**3*f), Ne(f, 0)), (x*(a*sin(e) + a)**2/(-c*sin(e) + c)**3, True))

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