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]

-1/5*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-1/30*a^2*(5*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))
^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/20*a*(5*A-B)*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 = 0.36, antiderivative size = 146, normalized size of antiderivative = 1.00, 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 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))^{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*}

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Mathematica [A]  time = 1.61, 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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fricas [A]  time = 0.45, size = 126, normalized size = 0.86 \[ \frac {{\left (15 \, {\left (A - B\right )} a c^{2} \cos \left (f x + e\right )^{4} - 15 \, {\left (A - B\right )} a c^{2} + 4 \, {\left (3 \, B a c^{2} \cos \left (f x + e\right )^{4} + {\left (5 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, A - B\right )} a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{60 \, f \cos \left (f x + e\right )} \]

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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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))^(3/2)*(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)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)sqrt(2*a)*sqrt(2*c)*(-32*f*(A*a*c^2*sign(sin(1/2*(f*x+exp(1))-
1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-B*a*c^2*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))/(32*f)^2-64*f*(A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+e
xp(1))-1/4*pi))-B*a*c^2*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))/(64*f)^2+32*f*(-A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+B*a*c^2*sig
n(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))/(-32*f)^2+96*f*(-4*A*
a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-B*a*c^2*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+16*f*(-6*A*a*c^2*sign(sin(1/2*(f*x+e
xp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+B*a*c^2*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)^2-160*B*a*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)

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maple [A]  time = 0.79, size = 147, normalized size = 1.01 \[ \frac {\left (-12 B \left (\cos ^{4}\left (f x +e \right )\right )+15 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-20 A \left (\cos ^{2}\left (f x +e \right )\right )+4 B \left (\cos ^{2}\left (f x +e \right )\right )+15 A \sin \left (f x +e \right )-15 B \sin \left (f x +e \right )-40 A +8 B \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{60 f \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right )^{3}} \]

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*(sin(f*x+e)-1))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3
/2)/(sin(f*x+e)-1)/cos(f*x+e)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(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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mupad [B]  time = 16.57, size = 174, normalized size = 1.19 \[ \frac {a\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (60\,A\,\cos \left (e+f\,x\right )-60\,B\,\cos \left (e+f\,x\right )+75\,A\,\cos \left (3\,e+3\,f\,x\right )+15\,A\,\cos \left (5\,e+5\,f\,x\right )-75\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )+400\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )-50\,B\,\sin \left (2\,e+2\,f\,x\right )+16\,B\,\sin \left (4\,e+4\,f\,x\right )+6\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{480\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(60*A*cos(e + f*x) - 60*B*cos(e + f*x) + 75*
A*cos(3*e + 3*f*x) + 15*A*cos(5*e + 5*f*x) - 75*B*cos(3*e + 3*f*x) - 15*B*cos(5*e + 5*f*x) + 400*A*sin(2*e + 2
*f*x) + 40*A*sin(4*e + 4*f*x) - 50*B*sin(2*e + 2*f*x) + 16*B*sin(4*e + 4*f*x) + 6*B*sin(6*e + 6*f*x)))/(480*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))**(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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