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

Optimal. Leaf size=51 \[ \frac{\cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

[Out]

(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]])

________________________________________________________________________________________

Rubi [A]  time = 0.333223, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2841, 2737, 2667, 31} \[ \frac{\cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]])

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

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

Mathematica [A]  time = 0.432335, size = 102, normalized size = 2. \[ \frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f (a (\sin (e+f x)+1))^{3/2} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.208, size = 136, normalized size = 2.7 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) \right ) \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 ) }{2\,f \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ( -\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +2\,\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \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(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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