3.390 \(\int \frac{(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a \sin (e+f x)+a}}-\frac{4 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.28775, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2739, 2740, 2737, 2667, 31} \[ -\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a \sin (e+f x)+a}}-\frac{4 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

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

Rule 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac{(2 c) \int \frac{(c-c \sin (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 c^2\right ) \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 c^3 \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac{\left (4 c^3 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{4 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}-\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f \sqrt{a+a \sin (e+f x)}}-\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.780627, size = 153, normalized size = 1.07 \[ -\frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos (2 (e+f x))+16 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+2 \sin (e+f x) \left (8 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-1\right )+7\right )}{2 f (a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(7 + Cos[2*(e + f*x)] + 16*Log[Cos[(e + f
*x)/2] + Sin[(e + f*x)/2]] + 2*(-1 + 8*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x]))/(2*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B]  time = 0.157, size = 446, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

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))^(3/2),x)

[Out]

1/f*(cos(f*x+e)^2*sin(f*x+e)-8*sin(f*x+e)*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+4*ln(2/(cos(f*
x+e)+1))*sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^3-8*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+4*cos(f*
x+e)^2*ln(2/(cos(f*x+e)+1))+5*sin(f*x+e)*cos(f*x+e)+16*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-8
*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+6*cos(f*x+e)^2-8*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+4*cos(
f*x+e)*ln(2/(cos(f*x+e)+1))-6*sin(f*x+e)+cos(f*x+e)+16*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-8*ln(2/(cos(
f*x+e)+1))-6)*(-c*(-1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^3+cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2+2*sin(f*x+e)*cos
(f*x+e)-2*cos(f*x+e)-4*sin(f*x+e)+4)/(a*(1+sin(f*x+e)))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

integral((c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/
(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

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