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

[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]])

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Rubi [A]  time = 0.646381, antiderivative size = 192, normalized size of antiderivative = 1., 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 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]

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 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 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 31

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

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*}

Mathematica [A]  time = 2.33449, 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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Maple [A]  time = 0.215, size = 272, normalized size = 1.4 \begin{align*}{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +2}{2\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{3}+2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\sin \left ( fx+e \right ) +2\,\cos \left ( fx+e \right ) -4 \right ) } \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +24\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -48\,\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +9\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-24\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +25\,\sin \left ( fx+e \right ) +48\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -9 \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)+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))+9*cos(f*x+e)^2-24*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)*(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*sin
(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*(-1+sin(f*x+e)))^
(5/2)

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Maxima [B]  time = 1.90576, size = 1512, normalized size = 7.88 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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]

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