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

Optimal. Leaf size=192 \[ \frac {12 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.65, antiderivative size = 192, 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} \[ \frac {6 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {12 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(c*f*(c - c*Sin[e + f*x])^(3/2)) + (12*a^3*Cos[e + f*x]*Log[1 - Sin[
e + f*x]])/(c^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (6*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e +
f*x]])/(c^2*f*Sqrt[c - c*Sin[e + f*x]]) + (3*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*c^2*f*Sqrt[c - c*Si
n[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 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) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {3 \int \frac {(a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(6 a) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {6 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (12 a^2\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {6 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (12 a^3 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {6 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (12 a^3 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {12 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.38, size = 181, normalized size = 0.94 \[ \frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin (3 (e+f x))+18 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sin (e+f x) \left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+44\right )}{8 c^2 f (\sin (e+f x)-1)^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 2 \, a^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.37, size = 273, normalized size = 1.42 \[ -\frac {\left (-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+48 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-24 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-9 \left (\cos ^{2}\left (f x +e \right )\right )-48 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-25 \sin \left (f x +e \right )+9\right ) \left (\sin \left (f x +e \right ) \cos \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right )-\cos \left (f x +e \right )+2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{2 f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+2 \cos \left (f x +e \right )-4\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(-cos(f*x+e)^2*sin(f*x+e)+48*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-24*sin(f*x+e)*ln(2/(
cos(f*x+e)+1))-9*cos(f*x+e)^2-48*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)+9)*(sin(f*x+e)*cos(f*x+e)-cos(f*x+e)^2-2*sin(f*x+e)-cos(f*x+e)+2)*(a*(1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^2*s
in(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)/(-c*(sin(f*x+e)-1))
^(5/2)

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maxima [B]  time = 0.51, size = 1120, normalized size = 5.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(144*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) - 72*a^(5/2)*log(sin(f*x + e)^2/(cos(f*x +
e) + 1)^2 + 1)/c^(5/2) + (46*a^(5/2) - 121*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 149*a^(5/2)*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 - 179*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 148*a^(5/2)*sin(f*x + e)^4/(cos(f*x
+ e) + 1)^4 - 43*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 33*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6
+ 15*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(5/2) - 4*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*c^
(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 12*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*c^(5/2)*sin(f*
x + e)^4/(cos(f*x + e) + 1)^4 - 12*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 8*c^(5/2)*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6 - 4*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8
) - (46*a^(5/2) - 199*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 335*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 - 509*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 496*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 373*a^
(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 219*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 63*a^(5/2)*sin(f
*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(5/2) - 4*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*c^(5/2)*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 - 12*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*c^(5/2)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 - 12*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 8*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 -
4*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) + 6*(13*a^(5/2)*s
in(f*x + e)/(cos(f*x + e) + 1) - 39*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 55*a^(5/2)*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 74*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 55*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e)
+ 1)^5 - 39*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 13*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(
5/2) - 4*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 12*c^(5/2)*
sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 12*c^(5/2)*sin(f*x + e)
^5/(cos(f*x + e) + 1)^5 + 8*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 4*c^(5/2)*sin(f*x + e)^7/(cos(f*x +
e) + 1)^7 + c^(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 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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