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]

-1/5*a*A*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-1/6*B*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c
-c*sin(f*x+e))^(5/2)/f-2/15*a^3*A*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/5*a^2*A*cos(f*x
+e)*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.47, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 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)]

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 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]

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*}

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Mathematica [A]  time = 0.80, 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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fricas [A]  time = 0.46, size = 117, normalized size = 0.65 \[ -\frac {{\left (5 \, B a^{2} c^{2} \cos \left (f x + e\right )^{6} - 5 \, B a^{2} c^{2} - 2 \, {\left (3 \, A a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, A a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, A a^{2} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)sqrt(2*a)*sqrt(2*c)*(-80*A*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))
*sin(f*x+exp(1))/(16*f)^2-480*A*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)
)*sin(3*f*x+3*exp(1))/(96*f)^2-160*A*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/
4*pi))*sin(5*f*x+5*exp(1))/(160*f)^2+64*B*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1
))-1/4*pi))*cos(2*f*x+2*exp(1))/(64*f)^2+768*B*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+
exp(1))-1/4*pi))*cos(4*f*x+4*exp(1))/(256*f)^2+384*B*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2
*(f*x+exp(1))-1/4*pi))*cos(6*f*x+6*exp(1))/(384*f)^2+384*B*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(c
os(1/2*(f*x+exp(1))-1/4*pi))*cos(-2*f*x-2*exp(1))/(-128*f)^2+256*B*a^2*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi)
)*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(-4*f*x-4*exp(1))/(-256*f)^2)

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maple [A]  time = 0.77, size = 114, normalized size = 0.63 \[ \frac {\left (5 B \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+6 A \left (\cos ^{4}\left (f x +e \right )\right )+5 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 A \left (\cos ^{2}\left (f x +e \right )\right )+5 B \sin \left (f x +e \right )+16 A \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{30 f \cos \left (f x +e \right )^{5}} \]

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*(sin(f*x+e)-1))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

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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mupad [B]  time = 16.00, size = 131, normalized size = 0.73 \[ -\frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,B\,\cos \left (e+f\,x\right )+105\,B\,\cos \left (3\,e+3\,f\,x\right )+35\,B\,\cos \left (5\,e+5\,f\,x\right )+5\,B\,\cos \left (7\,e+7\,f\,x\right )-700\,A\,\sin \left (2\,e+2\,f\,x\right )-112\,A\,\sin \left (4\,e+4\,f\,x\right )-12\,A\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(75*B*cos(e + f*x) + 105*B*cos(3*e + 3*f*
x) + 35*B*cos(5*e + 5*f*x) + 5*B*cos(7*e + 7*f*x) - 700*A*sin(2*e + 2*f*x) - 112*A*sin(4*e + 4*f*x) - 12*A*sin
(6*e + 6*f*x)))/(960*f*(cos(2*e + 2*f*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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