∫ (c−c sin(e+fx))3∕2 ___________________________

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

Optimal. Leaf size=60 \[ \frac {2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}-\frac {8 c \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a f} \]

[Out]

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

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Rubi [A] time = 0.20, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac {2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}-\frac {8 c \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-8*c*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(a*f)

Rule 2673

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 - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq

Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

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 - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(

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

&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + 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 {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx &=\frac {\int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}+\frac {4 \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{a}\\ &=-\frac {8 c \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}\\ \end {align*}

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Mathematica [A] time = 0.31, size = 88, normalized size = 1.47 \[ -\frac {2 c (\sin (e+f x)+3) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{a f (\sin (e+f x)+1) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

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fricas [A] time = 0.44, size = 41, normalized size = 0.68 \[ -\frac {2 \, {\left (c \sin \left (f x + e\right ) + 3 \, c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\right )} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/