3.4.19 \(\int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx\) [319]

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

[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 = 0.14, 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 = {2815, 2753, 2752} \begin {gather*} \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} \end {gather*}

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 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2753

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 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))^{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 = 0.18, size = 88, normalized size = 1.47 \begin {gather*} -\frac {2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (3+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \end {gather*}

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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Maple [A]
time = 1.76, size = 49, normalized size = 0.82

method result size
default \(\frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (3+\sin \left (f x +e \right )\right )}{a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (60) = 120\).
time = 0.51, size = 158, normalized size = 2.63 \begin {gather*} \frac {2 \, {\left (3 \, c^{\frac {3}{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \end {gather*}

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="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 44, normalized size = 0.73 \begin {gather*} -\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 )} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\right )\, dx}{a} \end {gather*}

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)

[Out]

(Integral(c*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x) + Integral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f
*x)/(sin(e + f*x) + 1), x))/a

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Giac [A]
time = 0.50, size = 67, normalized size = 1.12 \begin {gather*} -\frac {8 \, \sqrt {2} c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}} \end {gather*}

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]

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

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Mupad [B]
time = 7.29, size = 90, normalized size = 1.50 \begin {gather*} -\frac {2\,c\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (22\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+4\,\sin \left (2\,e+2\,f\,x\right )-12\right )}{a\,f\,\left (4\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*c*(-c*(sin(e + f*x) - 1))^(1/2)*(4*sin(2*e + 2*f*x) + 22*sin(e/2 + (f*x)/2)^2 + 2*sin((3*e)/2 + (3*f*x)/2)
^2 - 12))/(a*f*(sin(e + f*x) + sin(3*e + 3*f*x) + 4*sin(e + f*x)^2 - 4))

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