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

Optimal. Leaf size=145 \[ \frac {4 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}+\frac {\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}} \]

[Out]

1/2*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f/(a+a*sin(f*x+e))^(1/2)+4*c^2*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*si
n(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+2*c*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {4 c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}+\frac {\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.12, size = 134, normalized size = 0.92 \[ \frac {c (\sin (e+f x)-1) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (12 \sin (e+f x)+\cos (2 (e+f x))-32 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{4 f (a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-1 + Sin[e + f*x])*(Cos[2*(e + f*x)] - 32*Log[Cos[(e + f*x)/2] + S
in[(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])^3*(a*
(1 + Sin[e + f*x]))^(3/2))

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - c \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.35, size = 175, normalized size = 1.21 \[ \frac {\left (-\left (\cos ^{2}\left (f x +e \right )\right )+16 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-6 \sin \left (f x +e \right )-8 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+1\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right )}{2 f \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right )-\cos \left (f x +e \right )+2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.06, size = 844, normalized size = 5.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(16*c^(3/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(3/2) - 8*c^(3/2)*log(sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 1)/a^(3/2) - (10*c^(3/2) + 13*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 25*c^(3/2)*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 + 20*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e)
+ 1)^4 + 9*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(3/2) + 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) +
3*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^(3/2)*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 + 2*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(3/2)*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6) + (10*c^(3/2) + 11*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 15*c^(3/2)*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 20*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)
^4 + 7*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(3/2) + 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a
^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^(3/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 2*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(3/2)*sin(f*x + e)^6/(cos(f*x
+ e) + 1)^6) - 2*(5*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 +
8*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5*c^(3/2)*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(3/2) + 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^(3/2)*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 4*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^(3/2)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 2*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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