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

Optimal. Leaf size=211 \[ \frac{c^2 (A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a \sin (e+f x)+a}}+\frac{c^3 (A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c (A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

((A - 5*B)*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]]) +
((A - 5*B)*c^2*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(2*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + ((A - 5*B)*c*Cos[e
+ f*x]*(c - c*Sin[e + f*x])^(3/2))/(4*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e +
f*x])^(5/2))/(4*f*(a + a*Sin[e + f*x])^(5/2))

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Rubi [A]  time = 0.49444, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2972, 2739, 2740, 2737, 2667, 31} \[ \frac{c^2 (A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a \sin (e+f x)+a}}+\frac{c^3 (A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{c (A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 5*B)*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]]) +
((A - 5*B)*c^2*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(2*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + ((A - 5*B)*c*Cos[e
+ f*x]*(c - c*Sin[e + f*x])^(3/2))/(4*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e +
f*x])^(5/2))/(4*f*(a + a*Sin[e + f*x])^(5/2))

Rule 2972

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), 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, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 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 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{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(A-5 B) \int \frac{(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a}\\ &=\frac{(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{((A-5 B) c) \int \frac{(c-c \sin (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=\frac{(A-5 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\left ((A-5 B) c^2\right ) \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac{(A-5 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\left ((A-5 B) 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{(A-5 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\left ((A-5 B) 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{(A-5 B) 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{(A-5 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 f \sqrt{a+a \sin (e+f x)}}+\frac{(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.1438, size = 199, normalized size = 0.94 \[ \frac{(c-c \sin (e+f x))^{5/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (4 (A-2 B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+2 (A-5 B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-2 A+B \sin (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+2 B\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^(5/2)*(-2*A + 2*B + 4*(A - 2*B)*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^2 + 2*(A - 5*B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2])^4 + B*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(
a*(1 + Sin[e + f*x]))^(5/2))

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Maple [B]  time = 0.269, size = 1106, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2*A-14*B+2*A*sin(f*x+e)-8*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+40*B*ln((1-cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))-2*A*cos(f*x+e)^2-2*A*cos(f*x+e)^2*sin(f*x+e)-2*A*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+15*B*cos(f*x+e)^2*l
n(2/(cos(f*x+e)+1))+9*B*cos(f*x+e)^2*sin(f*x+e)-2*A*cos(f*x+e)-2*A*cos(f*x+e)*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-
B*cos(f*x+e)^3*sin(f*x+e)+2*A*cos(f*x+e)^3-8*B*cos(f*x+e)^3+8*B*cos(f*x+e)+10*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e
)*cos(f*x+e)+4*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+6*B*sin(f*x+e)*cos(f*x+e)+10*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1)
)-20*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-3*A*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+A*ln(2/(cos(f*x+e)+1))*cos(f*x+e)
^3+10*B*cos(f*x+e)^3*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*A*cos(f*x+e)^3*ln((1-cos(f*x+e)+sin(f*x+e))/si
n(f*x+e))-30*B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+6*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+
e))*cos(f*x+e)^2-A*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^2*sin(f*x+e)+2*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*c
os(f*x+e)^2*sin(f*x+e)-10*B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)*ln
((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-20*B*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-8*A*sin(f*x+e)
*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+40*B*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-20*B*cos(f*
x+e)*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+4*A*cos(f*x+e)*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e)
)/sin(f*x+e))-B*cos(f*x+e)^4+15*B*cos(f*x+e)^2+4*A*ln(2/(cos(f*x+e)+1))-20*B*ln(2/(cos(f*x+e)+1))+5*B*cos(f*x+
e)^2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-5*B*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-14*B*sin(f*x+e))*(-c*(-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)/(a*(1+sin(f*x+e)))^(5/2)

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Maxima [B]  time = 1.59875, size = 680, normalized size = 3.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*sqrt(a)*c^(5/2)*sin(f*x + e)^2/((a^3 + 4*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 6*a^3*sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + 4*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*x
+ e) + 1)^2) - 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) + c^(5/2)*log(sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + 1)/a^(5/2))*A + B*(10*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) - 5*c^(5/2)*lo
g(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/a^(5/2) - 2*(5*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 16*c^(5/2)
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 14*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 16*c^(5/2)*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + 5*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(5/2) + 4*a^(5/2)*sin(f*x + e)/(c
os(f*x + e) + 1) + 7*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 + 7*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(5/2)*s
in(f*x + e)^6/(cos(f*x + e) + 1)^6)))/f

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (A - 2 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A - B\right )} c^{2} +{\left (B c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (A - B\right )} c^{2}\right )} \sin \left (f x + e\right )\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((A - 2*B)*c^2*cos(f*x + e)^2 - 2*(A - B)*c^2 + (B*c^2*cos(f*x + e)^2 + 2*(A - B)*c^2)*sin(f*x + e))*
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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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((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*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 (B \sin \left (f x + e\right ) + A\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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