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

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

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 97, 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, 2739, 2737, 2667, 31} \[ -\frac {c \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(c \cos (e+f x)) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(c \cos (e+f x)) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {c \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.98, size = 100, normalized size = 1.03 \[ \frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)} \left (-2 \log \left (e^{i (e+f x)}+i\right )+\left (i f x-2 \log \left (e^{i (e+f x)}+i\right )\right ) \sin (e+f x)+i f x-2\right )}{a^2 f \sqrt {a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \cos \left (f x + e\right )^{2}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e)^2/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a
^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), 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*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.36, size = 192, normalized size = 1.98 \[ \frac {\left (2 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \sin \left (f x +e \right )-\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right )}{f \left (-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x)

[Out]

1/f*(2*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-sin(f*x+e)*ln(2/(cos(f*x+e)+1))+2*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)))*(-c*(sin(f*x+e)-1))^(1/2)*(cos(f*x+e)^2+sin(f*x
+e)*cos(f*x+e)+cos(f*x+e)-2*sin(f*x+e)-2)/(-1+cos(f*x+e)+sin(f*x+e))/(a*(1+sin(f*x+e)))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-c \sin \left (f x + e\right ) + c} \cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^2*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((cos(e + f*x)^2*(c - c*sin(e + f*x))^(1/2))/(a + a*sin(e + f*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \cos ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Integral(sqrt(-c*(sin(e + f*x) - 1))*cos(e + f*x)**2/(a*(sin(e + f*x) + 1))**(5/2), x)

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