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

Optimal. Leaf size=139 \[ \frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}}+\frac{4 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.28084, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2740, 2737, 2667, 31} \[ \frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a \sin (e+f x)+a}}+\frac{4 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c^3*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]]) + (2*c^2*Cos[
e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]) + (c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/
(2*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))^{5/2}}{\sqrt{a+a \sin (e+f x)}} \, dx &=\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a+a \sin (e+f x)}}+(2 c) \int \frac{(c-c \sin (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a+a \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a+a \sin (e+f x)}}+\left (4 c^2\right ) \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=\frac{2 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{f \sqrt{a+a \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a+a \sin (e+f x)}}+\frac{\left (4 a 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)}}{f \sqrt{a+a \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a+a \sin (e+f x)}}+\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 )}{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))}{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)}}{f \sqrt{a+a \sin (e+f x)}}+\frac{c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.561192, size = 136, normalized size = 0.98 \[ -\frac{c^2 (\sin (e+f x)-1)^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (12 \sin (e+f x)+\cos (2 (e+f x))-32 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{4 f \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^2*(Cos[2*(e + f*x)] - 32*Log[Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]] + 12*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*
Sqrt[a*(1 + Sin[e + f*x])])

________________________________________________________________________________________

Maple [B]  time = 0.198, size = 320, normalized size = 2.3 \begin{align*} -{\frac{1}{2\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -2\,\cos \left ( fx+e \right ) -4\,\sin \left ( fx+e \right ) +4 \right ) } \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +6\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -16\,\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 ) -8\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +16\,\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 ) -\cos \left ( fx+e \right ) -8\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -5\,\sin \left ( fx+e \right ) +16\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -5 \right ) \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{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))^(5/2)/(a+a*sin(f*x+e))^(1/2),x)

[Out]

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

________________________________________________________________________________________

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}}}{\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))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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{-c \sin \left (f x + e\right ) + c}}{\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))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

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}}}{\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))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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