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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.195201, 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} \[ \frac{a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.454203, size = 153, normalized size = 1.58 \[ \frac{2 a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin (e+f x) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-1\right )}{c f (\sin (e+f x)-1) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.14, size = 377, normalized size = 3.9 \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 ) -2\,\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 ) \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\, \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 ) \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 ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +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\,\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 ) +2 \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( -c \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((a+a*sin(f*x+e))^(3/2)/(c-c*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)-2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f
*x+e)-cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+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)
*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))-2*cos(f*
x+e)^2-cos(f*x+e)*ln(2/(cos(f*x+e)+1))+2*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*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))+2)*(a*(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)/(-c*(-1+sin(f*x+e)))^(3/2)

________________________________________________________________________________________

Maxima [A]  time = 1.72081, size = 185, normalized size = 1.91 \begin{align*} -\frac{\frac{2 \, a^{\frac{3}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac{3}{2}}} - \frac{a^{\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 )}{c^{\frac{3}{2}}} + \frac{4 \, a^{\frac{3}{2}} \sqrt{c} \sin \left (f x + e\right )}{{\left (c^{2} - \frac{2 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c^{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((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

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