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

Optimal. Leaf size=157 \[ \frac {64 a c^4 (3 A-B) \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 (3 A-B) \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 (3 A-B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[Out]

64/315*a*(3*A-B)*c^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-2/9*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/f+16/
105*a*(3*A-B)*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)+2/21*a*(3*A-B)*c^2*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)
/f

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Rubi [A]  time = 0.41, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2967, 2856, 2674, 2673} \[ \frac {64 a c^4 (3 A-B) \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 (3 A-B) \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 (3 A-B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a*(3*A - B)*c^4*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (16*a*(3*A - B)*c^3*Cos[e + f*x]^3)/(
105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*(3*A - B)*c^2*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) - (2*a*B*
c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*f)

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\\ &=-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{3} (a (3 A-B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{21} \left (8 a (3 A-B) c^2\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac {1}{105} \left (32 a (3 A-B) c^3\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {64 a (3 A-B) c^4 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 123, normalized size = 0.78 \[ -\frac {a c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 ((648 A-741 B) \sin (e+f x)+30 (3 A-8 B) \cos (2 (e+f x))-942 A+35 B \sin (3 (e+f x))+664 B)}{630 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/630*(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-942*A + 664*B + 30*(3*A - 8*B
)*Cos[2*(e + f*x)] + (648*A - 741*B)*Sin[e + f*x] + 35*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*
x)/2]))

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fricas [A]  time = 0.43, size = 243, normalized size = 1.55 \[ \frac {2 \, {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{5} + 5 \, {\left (9 \, A - 10 \, B\right )} a c^{2} \cos \left (f x + e\right )^{4} + {\left (117 \, A - 109 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} - 8 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2} + {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{4} - 5 \, {\left (9 \, A - 17 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} + 24 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*a*c^2*cos(f*x + e)^5 + 5*(9*A - 10*B)*a*c^2*cos(f*x + e)^4 + (117*A - 109*B)*a*c^2*cos(f*x + e)^3
- 8*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*x + e) + 64*(3*A - B)*a*c^2 + (35*B*a*c^2*cos(f*
x + e)^4 - 5*(9*A - 17*B)*a*c^2*cos(f*x + e)^3 + 24*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*
x + e) + 64*(3*A - B)*a*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)

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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))*(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)sqrt(2*c)*(24*f*(2*A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a*c^2*sign(sin(1/2
*(f*x+exp(1))-1/4*pi)))*sin(1/4*(6*f*x+6*exp(1)-pi))/(24*f)^2+56*f*(2*A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi
))-2*B*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(14*f*x+14*exp(1)-pi))/(56*f)^2-8*f*(8*A*a*c^2*sign(s
in(1/2*(f*x+exp(1))-1/4*pi))-2*B*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/(8*f
)^2+8*f*(-2*A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(
2*f*x+2*exp(1)+pi))/(8*f)^2+40*f*(-2*A*a*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a*c^2*sign(sin(1/2*(f*x+ex
p(1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(1)+pi))/(40*f)^2+24*A*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/
4*(6*f*x+6*exp(1)+pi))/(12*f)^2-40*A*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(10*f*x+10*exp(1)-pi))
/(20*f)^2+224*B*a*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(14*f*x+14*exp(1)+pi))/(112*f)^2-288*B*a*c^
2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(18*f*x+18*exp(1)-pi))/(144*f)^2)

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maple [A]  time = 1.23, size = 103, normalized size = 0.66 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (-35 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-162 A +194 B \right ) \sin \left (f x +e \right )+\left (-45 A +120 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+258 A -226 B \right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(sin(f*x+e)-1)*c^3*(1+sin(f*x+e))^2*a*(-35*B*cos(f*x+e)^2*sin(f*x+e)+(-162*A+194*B)*sin(f*x+e)+(-45*A+1
20*B)*cos(f*x+e)^2+258*A-226*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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