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

Optimal. Leaf size=97 \[ -\frac{c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.197259, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2739, 2737, 2667, 31} \[ -\frac{c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^2*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]])) - (c*Cos[e
+ f*x]*Sqrt[c - c*Sin[e + f*x]])/(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 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))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a+a \sin (e+f x))^{3/2}}-\frac{c \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a+a \sin (e+f x))^{3/2}}-\frac{\left (c^2 \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{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a+a \sin (e+f x))^{3/2}}-\frac{\left (c^2 \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{c^2 \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{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.453933, size = 134, normalized size = 1.38 \[ -\frac{2 c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+1\right )}{f (a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(1 + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]] + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x]))/(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 = 0.141, size = 388, normalized size = 4. \begin{align*}{\frac{1}{f \left ( \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +2 \right ) } \left ( \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +4\,\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,\sin \left ( fx+e \right ) +4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2 \right ) \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

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))^(3/2),x)

[Out]

1/f*(ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)+cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-2*sin(f*x+e)*cos(f*x+e)*ln(-
(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*sin(f*x+e)*
ln(2/(cos(f*x+e)+1))+cos(f*x+e)*ln(2/(cos(f*x+e)+1))+2*sin(f*x+e)*cos(f*x+e)+4*sin(f*x+e)*ln(-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))+2*cos(f*x+e)^2-2*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*ln(2/(cos(f*x+
e)+1))-2*sin(f*x+e)+4*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2)*(-c*(-1+sin(f*x+e)))^(3/2)/(sin(f*x+e)*cos
(f*x+e)-cos(f*x+e)^2-2*sin(f*x+e)-cos(f*x+e)+2)/(a*(1+sin(f*x+e)))^(3/2)

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Maxima [A]  time = 1.77917, size = 184, normalized size = 1.9 \begin{align*} \frac{\frac{2 \, c^{\frac{3}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{\frac{3}{2}}} - \frac{c^{\frac{3}{2}} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{a^{\frac{3}{2}}} - \frac{4 \, \sqrt{a} c^{\frac{3}{2}} \sin \left (f x + e\right )}{{\left (a^{2} + \frac{2 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{a \sin \left (f x + e\right ) + a}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{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))^(3/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- c \left (\sin{\left (e + f x \right )} - 1\right )\right )^{\frac{3}{2}}}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

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

[Out]

Integral((-c*(sin(e + f*x) - 1))**(3/2)/(a*(sin(e + f*x) + 1))**(3/2), x)

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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{3}{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))^(3/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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