3.3.63 \(\int \frac {(c-c \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx\) [263]

Optimal. Leaf size=56 \[ -\frac {3 c^2 x}{a}-\frac {3 c^2 \cos (e+f x)}{a f}-\frac {2 a c^2 \cos ^3(e+f x)}{f (a+a \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 56, 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 = {2815, 2759, 2761, 8} \begin {gather*} -\frac {3 c^2 \cos (e+f x)}{a f}-\frac {2 a c^2 \cos ^3(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac {3 c^2 x}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2761

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(129\) vs. \(2(56)=112\).
time = 0.25, size = 129, normalized size = 2.30 \begin {gather*} -\frac {c^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))+(-8+3 e+3 f x+\cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2}{a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (1+\sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.27, size = 57, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {2 c^{2} \left (-\frac {1}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}-3 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\)	\(57\)
default	\(\frac {2 c^{2} \left (-\frac {1}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}-3 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f a}\)	\(57\)
risch	\(-\frac {3 c^{2} x}{a}-\frac {c^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 a f}-\frac {c^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 a f}-\frac {8 c^{2}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}\)	\(76\)
norman	\(\frac {-\frac {10 c^{2}}{a f}-\frac {8 c^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {2 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {3 c^{2} x}{a}-\frac {3 c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}-\frac {6 c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {6 c^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {3 c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {3 c^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}-\frac {18 c^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {2 c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (59) = 118\).
time = 0.55, size = 228, normalized size = 4.07 \begin {gather*} -\frac {2 \, {\left (c^{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}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \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 )}{a}\right )} + 2 \, c^{2} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac {c^{2}}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.32, size = 107, normalized size = 1.91 \begin {gather*} -\frac {3 \, c^{2} f x + c^{2} \cos \left (f x + e\right )^{2} + 4 \, c^{2} + {\left (3 \, c^{2} f x + 5 \, c^{2}\right )} \cos \left (f x + e\right ) + {\left (3 \, c^{2} f x + c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (53) = 106\).
time = 1.17, size = 456, normalized size = 8.14 \begin {gather*} \begin {cases} - \frac {3 c^{2} f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {3 c^{2} f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {3 c^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {3 c^{2} f x}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {8 c^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {2 c^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} - \frac {10 c^{2}}{a f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\left (e \right )} + c\right )^{2}}{a \sin {\left (e \right )} + a} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 100, normalized size = 1.79 \begin {gather*} -\frac {\frac {3 \, {\left (f x + e\right )} c^{2}}{a} + \frac {2 \, {\left (4 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, c^{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 )} a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.99, size = 118, normalized size = 2.11 \begin {gather*} -\frac {3\,c^2\,x}{a}-\frac {3\,\sqrt {2}\,c^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (e+f\,x\right )-\frac {\sqrt {2}\,c^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (6\,e+6\,f\,x+16\right )}{2}}{a\,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 {2\,c^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{a\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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