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

Optimal. Leaf size=142 \[ \frac{c^2 (5 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{30 f \sqrt{c-c \sin (e+f x)}}+\frac{c (5 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}}{20 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.362424, antiderivative size = 142, 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} \[ \frac{c^2 (5 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{30 f \sqrt{c-c \sin (e+f x)}}+\frac{c (5 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}}{20 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.8909, size = 165, normalized size = 1.16 \[ -\frac{c (\sin (e+f x)-1) (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)} (4 (100 A+11 B) \sin (e+f x)+4 \cos (2 (e+f x)) (4 (5 A-2 B) \sin (e+f x)-15 (A+B))-3 \cos (4 (e+f x)) (5 (A+B)+4 B \sin (e+f x)))}{480 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.302, 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{3}{2}}} \left ( a \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((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),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)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(
5/2)/(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{5}{2}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.091, size = 304, normalized size = 2.14 \begin{align*} -\frac{{\left (15 \,{\left (A + B\right )} a^{2} c \cos \left (f x + e\right )^{4} - 15 \,{\left (A + B\right )} a^{2} c + 4 \,{\left (3 \, B a^{2} c \cos \left (f x + e\right )^{4} -{\left (5 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{2} - 2 \,{\left (5 \, A + B\right )} a^{2} c\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))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2