3.3.42 \(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx\) [242]

Optimal. Leaf size=72 \[ \frac {a^2 x}{c^2}+\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 72, 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 = {2815, 2759, 8} \begin {gather*} -\frac {2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {a^2 x}{c^2}+\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

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]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^2} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-a^2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {a^2 \int 1 \, dx}{c^2}\\ &=\frac {a^2 x}{c^2}+\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {2 a^2 \cos (e+f x)}{f \left (c^2-c^2 \sin (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 121, normalized size = 1.68 \begin {gather*} -\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-3 (8+3 e+3 f x) \cos \left (\frac {1}{2} (e+f x)\right )+(16+3 e+3 f x) \cos \left (\frac {3}{2} (e+f x)\right )+6 (2 (2+e+f x)+(e+f x) \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{6 c^2 f (-1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-3*(8 + 3*e + 3*f*x)*Cos[(e + f*x)/2] + (16 + 3*e + 3*f*x)*Co
s[(3*(e + f*x))/2] + 6*(2*(2 + e + f*x) + (e + f*x)*Cos[e + f*x])*Sin[(e + f*x)/2]))/(c^2*f*(-1 + Sin[e + f*x]
)^2)

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Maple [A]
time = 0.33, size = 53, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(53\)
default	\(\frac {2 a^{2} \left (-\frac {8}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{2}}\)	\(53\)
risch	\(\frac {a^{2} x}{c^{2}}-\frac {8 \left (-3 i a^{2} {\mathrm e}^{i \left (f x +e \right )}+3 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-2 a^{2}\right )}{3 \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{3} f \,c^{2}}\)	\(67\)
norman	\(\frac {\frac {a^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {a^{2} x}{c}+\frac {8 a^{2}}{3 c f}+\frac {3 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}-\frac {5 a^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {7 a^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {7 a^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {3 a^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {16 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\)	\(299\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (76) = 152\).
time = 0.55, size = 394, normalized size = 5.47 \begin {gather*} \frac {2 \, {\left (a^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 4}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}}\right )} - \frac {a^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {2 \, a^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac {3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

2/3*(a^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x
+ e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 2*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*
sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (76) = 152\).
time = 0.32, size = 166, normalized size = 2.31 \begin {gather*} -\frac {6 \, a^{2} f x - {\left (3 \, a^{2} f x + 8 \, a^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, a^{2} + {\left (3 \, a^{2} f x - 4 \, a^{2}\right )} \cos \left (f x + e\right ) - {\left (6 \, a^{2} f x - 4 \, a^{2} + {\left (3 \, a^{2} f x - 8 \, a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

-1/3*(6*a^2*f*x - (3*a^2*f*x + 8*a^2)*cos(f*x + e)^2 + 4*a^2 + (3*a^2*f*x - 4*a^2)*cos(f*x + e) - (6*a^2*f*x -
4*a^2 + (3*a^2*f*x - 8*a^2)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f
+ (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (65) = 130\).
time = 2.56, size = 473, normalized size = 6.57 \begin {gather*} \begin {cases} \frac {3 a^{2} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {9 a^{2} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} + \frac {9 a^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {3 a^{2} f x}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} - \frac {24 a^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} + \frac {8 a^{2}}{3 c^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 c^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 c^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )^{2}}{\left (- c \sin {\left (e \right )} + c\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

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))**2,x)

[Out]

Piecewise((3*a**2*f*x*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**
2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 9*a**2*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan
(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 9*a**2*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x
/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 3*a**2*f*x/(3*c**2*f*tan(e/2 +
f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 24*a**2*tan(e/2 + f*x/2)/(
3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 8*a**2/(
3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f), Ne(f, 0))
, (x*(a*sin(e) + a)**2/(-c*sin(e) + c)**2, True))

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Giac [A]
time = 0.42, size = 60, normalized size = 0.83 \begin {gather*} \frac {\frac {3 \, {\left (f x + e\right )} a^{2}}{c^{2}} - \frac {8 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a^{2}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.87, size = 90, normalized size = 1.25 \begin {gather*} \frac {a^2\,x}{c^2}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,a^2\,\left (e+f\,x\right )-\frac {a^2\,\left (9\,e+9\,f\,x-24\right )}{3}\right )-a^2\,\left (e+f\,x\right )+\frac {a^2\,\left (3\,e+3\,f\,x-8\right )}{3}}{c^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^3} \end {gather*}

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))^2,x)

[Out]

(a^2*x)/c^2 - (tan(e/2 + (f*x)/2)*(3*a^2*(e + f*x) - (a^2*(9*e + 9*f*x - 24))/3) - a^2*(e + f*x) + (a^2*(3*e +
3*f*x - 8))/3)/(c^2*f*(tan(e/2 + (f*x)/2) - 1)^3)

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