3.318 \(\int \frac {(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=98 \[ -\frac {64 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac {16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f} \]

[Out]

16/3*c*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f+2/3*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f-64/3*c^2*sec(f*x+e)*(c-
c*sin(f*x+e))^(1/2)/a/f

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Rubi [A]  time = 0.27, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ -\frac {64 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac {16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x]),x]

[Out]

(-64*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*a*f) + (16*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(3*a*
f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*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))^{5/2}}{a+a \sin (e+f x)} \, dx &=\frac {\int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a c}\\ &=\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac {8 \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{3 a}\\ &=\frac {16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac {(32 c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{3 a}\\ &=-\frac {64 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}+\frac {16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 102, normalized size = 1.04 \[ -\frac {c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (20 \sin (e+f x)+\cos (2 (e+f x))+45)}{3 a f (\sin (e+f x)+1) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x]),x]

[Out]

-1/3*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(45 + Cos[2*(e + f*x)] + 20*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 = 58, normalized size = 0.59 \[ -\frac {2 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + 10 \, c^{2} \sin \left (f x + e\right ) + 22 \, c^{2}\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-2/3*(c^2*cos(f*x + e)^2 + 10*c^2*sin(f*x + e) + 22*c^2)*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.70, size = 59, normalized size = 0.60 \[ -\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (\sin ^{2}\left (f x +e \right )-10 \sin \left (f x +e \right )-23\right )}{3 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x)

[Out]

-2/3*c^3/a*(sin(f*x+e)-1)*(sin(f*x+e)^2-10*sin(f*x+e)-23)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [B]  time = 0.66, size = 192, normalized size = 1.96 \[ \frac {2 \, {\left (23 \, c^{\frac {5}{2}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {40 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {23 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \, {\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

2/3*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 40*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(5/2)
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/((a + a*sin(f*x + e)/(c
os(f*x + e) + 1))*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{a+a\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x)),x)

[Out]

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e)),x)

[Out]

Timed out

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