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

Optimal. Leaf size=96 \[ -\frac{2 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.191691, antiderivative size = 96, 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 = {2740, 2737, 2667, 31} \[ -\frac{2 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.317503, size = 113, normalized size = 1.18 \[ -\frac{(a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin (e+f x)+4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.17, size = 252, normalized size = 2.6 \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 ( \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -2\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \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 ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +4\,\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 ) -\sin \left ( fx+e \right ) +2\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -4\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +1 \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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