3.140 \(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=146 \[ -\frac{a^2 (5 A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{30 f \sqrt{a \sin (e+f x)+a}}-\frac{a (5 A-B) \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{20 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \]

[Out]

-(a^2*(5*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(30*f*Sqrt[a + a*Sin[e + f*x]]) - (a*(5*A - B)*Cos[e
+ f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(20*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2
)*(c - c*Sin[e + f*x])^(5/2))/(5*f)

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Rubi [A]  time = 0.361314, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ -\frac{a^2 (5 A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{30 f \sqrt{a \sin (e+f x)+a}}-\frac{a (5 A-B) \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{20 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(5*A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(30*f*Sqrt[a + a*Sin[e + f*x]]) - (a*(5*A - B)*Cos[e
+ f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(20*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2
)*(c - c*Sin[e + f*x])^(5/2))/(5*f)

Rule 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(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)] && NeQ[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 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac{1}{5} (5 A-B) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{a (5 A-B) \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{20 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac{1}{10} (a (5 A-B)) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{a^2 (5 A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{30 f \sqrt{a+a \sin (e+f x)}}-\frac{a (5 A-B) \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{20 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\\ \end{align*}

Mathematica [A]  time = 1.69345, size = 172, normalized size = 1.18 \[ \frac{c^2 (\sin (e+f x)-1)^2 (a (\sin (e+f x)+1))^{3/2} \sqrt{c-c \sin (e+f x)} (4 (100 A-11 B) \sin (e+f x)+3 \cos (4 (e+f x)) (5 A+4 B \sin (e+f x)-5 B)+4 \cos (2 (e+f x)) (4 (5 A+2 B) \sin (e+f x)+15 (A-B)))}{480 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(-1 + Sin[e + f*x])^2*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(4*(100*A - 11*B)*Sin[e + f*x
] + 3*Cos[4*(e + f*x)]*(5*A - 5*B + 4*B*Sin[e + f*x]) + 4*Cos[2*(e + f*x)]*(15*(A - B) + 4*(5*A + 2*B)*Sin[e +
f*x])))/(480*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [A]  time = 0.303, size = 147, normalized size = 1. \begin{align*}{\frac{ \left ( -12\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+15\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -20\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+15\,A\sin \left ( fx+e \right ) -15\,B\sin \left ( fx+e \right ) -40\,A+8\,B \right ) \sin \left ( fx+e \right ) }{60\,f \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x)

[Out]

1/60/f*(-12*B*cos(f*x+e)^4+15*A*cos(f*x+e)^2*sin(f*x+e)-15*B*cos(f*x+e)^2*sin(f*x+e)-20*A*cos(f*x+e)^2+4*B*cos
(f*x+e)^2+15*A*sin(f*x+e)-15*B*sin(f*x+e)-40*A+8*B)*(-c*(-1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(
3/2)/(-1+sin(f*x+e))/cos(f*x+e)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{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((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2), x)

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Fricas [A]  time = 1.81677, size = 302, normalized size = 2.07 \begin{align*} \frac{{\left (15 \,{\left (A - B\right )} a c^{2} \cos \left (f x + e\right )^{4} - 15 \,{\left (A - B\right )} a c^{2} + 4 \,{\left (3 \, B a c^{2} \cos \left (f x + e\right )^{4} +{\left (5 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (5 \, A - B\right )} a 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}}{60 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/60*(15*(A - B)*a*c^2*cos(f*x + e)^4 - 15*(A - B)*a*c^2 + 4*(3*B*a*c^2*cos(f*x + e)^4 + (5*A - B)*a*c^2*cos(f
*x + e)^2 + 2*(5*A - B)*a*c^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e
))

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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+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(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*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

sage0*x