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

Optimal. Leaf size=191 \[ -\frac {12 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f/(a+a*sin(f*x+e))^(3/2)-3/2*c*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f/(a
+a*sin(f*x+e))^(1/2)-12*c^3*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)-6*
c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.64, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2841, 2739, 2740, 2737, 2667, 31} \[ -\frac {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {12 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*c^3*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]]) - (6*c^
2*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (3*c*Cos[e + f*x]*(c - c*Sin[e + f
*x])^(3/2))/(2*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(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 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))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {3 \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(6 c) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^2\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^3 \cos (e+f x)\right ) \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 {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^3 \cos (e+f x)\right ) \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 {12 c^3 \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 {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 2.49, size = 164, normalized size = 0.86 \[ \frac {c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin (3 (e+f x))-18 \cos (2 (e+f x))-192 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+\sin (e+f x) \left (39-192 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-44\right )}{8 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-44 - 18*Cos[2*(e + f*x)] - 192*Log[Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]] + (39 - 192*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3*(
e + f*x)]))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2))

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fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} \cos \left (f x + e\right )^{4} + 2 \, c^{2} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 2 \, c^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.41, size = 270, normalized size = 1.41 \[ \frac {\left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-9 \left (\cos ^{2}\left (f x +e \right )\right )+24 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-48 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+25 \sin \left (f x +e \right )-48 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+9\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right )}{2 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\cos ^{3}\left (f x +e \right )+2 \sin \left (f x +e \right ) \cos \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )-2 \cos \left (f x +e \right )+4\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(cos(f*x+e)^2*sin(f*x+e)-9*cos(f*x+e)^2+24*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-48*sin(f*x+e)*ln(-(-1+cos(f*x
+e)-sin(f*x+e))/sin(f*x+e))+24*ln(2/(cos(f*x+e)+1))+25*sin(f*x+e)-48*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e)
)+9)*(-c*(sin(f*x+e)-1))^(5/2)*(cos(f*x+e)^2+sin(f*x+e)*cos(f*x+e)+cos(f*x+e)-2*sin(f*x+e)-2)/(cos(f*x+e)^2*si
n(f*x+e)+cos(f*x+e)^3+2*sin(f*x+e)*cos(f*x+e)-3*cos(f*x+e)^2-4*sin(f*x+e)-2*cos(f*x+e)+4)/(a*(1+sin(f*x+e)))^(
5/2)

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maxima [B]  time = 1.61, size = 1120, normalized size = 5.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(144*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) - 72*c^(5/2)*log(sin(f*x + e)^2/(cos(f*x + e
) + 1)^2 + 1)/a^(5/2) - (46*c^(5/2) + 199*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 335*c^(5/2)*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 + 509*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 496*c^(5/2)*sin(f*x + e)^4/(cos(f*x
+ e) + 1)^4 + 373*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 219*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^
6 + 63*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(a^(5/2) + 4*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*a
^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 12*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a^(5/2)*sin(f
*x + e)^4/(cos(f*x + e) + 1)^4 + 12*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 8*a^(5/2)*sin(f*x + e)^6/(co
s(f*x + e) + 1)^6 + 4*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^
8) + (46*c^(5/2) + 121*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 149*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 179*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 148*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 43*c^
(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 33*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 15*c^(5/2)*sin(f*
x + e)^7/(cos(f*x + e) + 1)^7)/(a^(5/2) + 4*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*a^(5/2)*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 + 12*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a^(5/2)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 12*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 8*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4
*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) - 6*(13*c^(5/2)*si
n(f*x + e)/(cos(f*x + e) + 1) + 39*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 55*c^(5/2)*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + 74*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 55*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) +
1)^5 + 39*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 13*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(a^(5
/2) + 4*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 12*a^(5/2)*s
in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 12*a^(5/2)*sin(f*x + e)^
5/(cos(f*x + e) + 1)^5 + 8*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e
) + 1)^7 + a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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))**(5/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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