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

Optimal. Leaf size=180 \[ -\frac{2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{a^2 A \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{a A \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \]

[Out]

(-2*a^3*A*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*A*Cos[e + f*x]*Sqrt[
a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*f) - (a*A*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Si
n[e + f*x])^(5/2))/(5*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2))/(6*f)

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Rubi [A]  time = 0.465897, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ -\frac{2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{a^2 A \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{a A \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \]

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])^(5/2),x]

[Out]

(-2*a^3*A*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*A*Cos[e + f*x]*Sqrt[
a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*f) - (a*A*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Si
n[e + f*x])^(5/2))/(5*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2))/(6*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))^{5/2} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+A \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+\frac{1}{5} (4 a A) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{a^2 A \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}+\frac{1}{5} \left (2 a^2 A\right ) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 A \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.841482, size = 113, normalized size = 0.63 \[ \frac{a^2 c^2 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (600 A \sin (e+f x)+100 A \sin (3 (e+f x))+12 A \sin (5 (e+f x))-75 B \cos (2 (e+f x))-30 B \cos (4 (e+f x))-5 B \cos (6 (e+f x)))}{960 f} \]

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])^(5/2),x]

[Out]

(a^2*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-75*B*Cos[2*(e + f*x)] - 30*B*Cos[4
*(e + f*x)] - 5*B*Cos[6*(e + f*x)] + 600*A*Sin[e + f*x] + 100*A*Sin[3*(e + f*x)] + 12*A*Sin[5*(e + f*x)]))/(96
0*f)

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Maple [A]  time = 0.285, size = 114, normalized size = 0.6 \begin{align*}{\frac{ \left ( 5\,B\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+5\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +8\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\,B\sin \left ( fx+e \right ) +16\,A \right ) \sin \left ( fx+e \right ) }{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \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{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))^(5/2),x)

[Out]

1/30/f*(5*B*sin(f*x+e)*cos(f*x+e)^4+6*A*cos(f*x+e)^4+5*B*cos(f*x+e)^2*sin(f*x+e)+8*A*cos(f*x+e)^2+5*B*sin(f*x+
e)+16*A)*(-c*(-1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/cos(f*x+e)^5

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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{5}{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))^(5/2),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)^(5/2), x)

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Fricas [A]  time = 2.18402, size = 279, normalized size = 1.55 \begin{align*} -\frac{{\left (5 \, B a^{2} c^{2} \cos \left (f x + e\right )^{6} - 5 \, B a^{2} c^{2} - 2 \,{\left (3 \, A a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, A a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, A a^{2} 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}}{30 \, 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))^(5/2),x, algorithm="fricas")

[Out]

-1/30*(5*B*a^2*c^2*cos(f*x + e)^6 - 5*B*a^2*c^2 - 2*(3*A*a^2*c^2*cos(f*x + e)^4 + 4*A*a^2*c^2*cos(f*x + e)^2 +
8*A*a^2*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))**(5/2)*(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*} \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{5}{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))^(5/2),x, algorithm="giac")

[Out]

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