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

Optimal. Leaf size=57 \[ \frac {3 a^2 \cos (e+f x)}{c f}+\frac {2 a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^2}-\frac {3 a^2 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2736, 2680, 2682, 8} \[ \frac {3 a^2 \cos (e+f x)}{c f}+\frac {2 a^2 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^2}-\frac {3 a^2 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2680

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 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

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

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Mathematica [B]  time = 0.38, size = 130, normalized size = 2.28 \[ \frac {a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right ) (3 (e+f x)-\cos (e+f x))+\sin \left (\frac {1}{2} (e+f x)\right ) (\cos (e+f x)-3 e-3 f x-8)\right )}{c f (\sin (e+f x)-1) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2]*(3*(e + f*x) - Cos[e + f*x]) + (-8 - 3*e - 3*f*x
+ Cos[e + f*x])*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-1 + Sin
[e + f*x]))

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fricas [A]  time = 0.44, size = 105, normalized size = 1.84 \[ -\frac {3 \, a^{2} f x - a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2} + {\left (3 \, a^{2} f x - 5 \, a^{2}\right )} \cos \left (f x + e\right ) - {\left (3 \, a^{2} f x - a^{2} \cos \left (f x + e\right ) + 4 \, a^{2}\right )} \sin \left (f x + e\right )}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 103, normalized size = 1.81 \[ -\frac {\frac {3 \, {\left (f x + e\right )} a^{2}}{c} + \frac {2 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} c}}{f} \]

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 73, normalized size = 1.28 \[ -\frac {8 a^{2}}{f c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {2 a^{2}}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}-\frac {6 a^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f c} \]

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)),x)

[Out]

-8/f*a^2/c/(tan(1/2*f*x+1/2*e)-1)+2/f*a^2/c/(1+tan(1/2*f*x+1/2*e)^2)-6/f*a^2/c*arctan(tan(1/2*f*x+1/2*e))

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maxima [B]  time = 1.92, size = 218, normalized size = 3.82 \[ -\frac {2 \, {\left (a^{2} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + 2 \, a^{2} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {a^{2}}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.91, size = 118, normalized size = 2.07 \[ \frac {3\,\sqrt {2}\,a^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (e+f\,x\right )-\frac {\sqrt {2}\,a^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (6\,e+6\,f\,x-16\right )}{2}}{c\,f\,\left (\sqrt {2}\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sqrt {2}\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}-\frac {3\,a^2\,x}{c}+\frac {2\,a^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{c\,f} \]

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)),x)

[Out]

(3*2^(1/2)*a^2*sin(e/2 + (f*x)/2)*(e + f*x) - (2^(1/2)*a^2*sin(e/2 + (f*x)/2)*(6*e + 6*f*x - 16))/2)/(c*f*(2^(
1/2)*cos(e/2 + (f*x)/2) - 2^(1/2)*sin(e/2 + (f*x)/2))) - (3*a^2*x)/c + (2*a^2*cos(e/2 + (f*x)/2)^2)/(c*f)

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sympy [A]  time = 4.20, size = 454, normalized size = 7.96 \[ \begin {cases} - \frac {3 a^{2} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} + \frac {3 a^{2} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} - \frac {3 a^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} + \frac {3 a^{2} f x}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} - \frac {8 a^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} + \frac {2 a^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} - \frac {10 a^{2}}{c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - c f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\relax (e )} + a\right )^{2}}{- c \sin {\relax (e )} + c} & \text {otherwise} \end {cases} \]

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)),x)

[Out]

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

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