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

Optimal. Leaf size=93 \[ \frac{2 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.187189, antiderivative size = 93, 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} \[ \frac{2 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.292943, size = 119, normalized size = 1.28 \[ \frac{c (\sin (e+f x)-1) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin (e+f x)-4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{f \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.174, size = 261, normalized size = 2.8 \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 ) +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\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}-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 ) +2\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -\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\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -1 \right ) \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}}} \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))^(1/2),x)

[Out]

-1/f*(sin(f*x+e)*cos(f*x+e)+4*sin(f*x+e)*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)^2-4*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+2*cos(f*x+e)*ln(2/(cos(f*x+e)+1))
-sin(f*x+e)+4*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*ln(2/(cos(f*x+e)+1))-1)*(-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)))^(1/2)

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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{3}{2}}}{\sqrt{a \sin \left (f x + e\right ) + a}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{\sqrt{a \sin \left (f x + e\right ) + a}}, 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))^(1/2),x, algorithm="fricas")

[Out]

integral((-c*sin(f*x + e) + c)^(3/2)/sqrt(a*sin(f*x + e) + a), 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}}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, 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))**(1/2),x)

[Out]

Integral((-c*(sin(e + f*x) - 1))**(3/2)/sqrt(a*(sin(e + f*x) + 1)), 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}}}{\sqrt{a \sin \left (f x + e\right ) + a}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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