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

Optimal. Leaf size=99 \[ -\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f \sqrt {c-c \sin (e+f x)}}-\frac {2 a \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2841, 2740, 2737, 2667, 31} \[ -\frac {\cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f \sqrt {c-c \sin (e+f x)}}-\frac {2 a \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 1.09, size = 115, normalized size = 1.16 \[ -\frac {\sqrt {a (\sin (e+f x)+1)} \left (4 \log \left (i-e^{i (e+f x)}\right )+\sin (e+f x)-2 i f x\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (c-c \sin (e+f x))^{3/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \cos \left (f x + e\right )^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.38, size = 138, normalized size = 1.39 \[ \frac {\left (\sin \left (f x +e \right )+4 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )\right ) \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right )-\cos \left (f x +e \right )+2\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{f \left (1-\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (f x + e\right ) + a} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \cos ^{2}{\left (e + f x \right )}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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