3.81 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=198 \[ \frac {256 a c^5 (11 A-5 B) \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 (11 A-5 B) \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 (11 A-5 B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a c^2 (11 A-5 B) \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

[Out]

256/3465*a*(11*A-5*B)*c^5*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2/99*a*(11*A-5*B)*c^2*cos(f*x+e)^3*(c-c*sin(f*
x+e))^(3/2)/f-2/11*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/f+64/1155*a*(11*A-5*B)*c^4*cos(f*x+e)^3/f/(c-c*si
n(f*x+e))^(1/2)+8/231*a*(11*A-5*B)*c^3*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.49, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2967, 2856, 2674, 2673} \[ \frac {256 a c^5 (11 A-5 B) \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a c^4 (11 A-5 B) \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a c^3 (11 A-5 B) \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a c^2 (11 A-5 B) \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(256*a*(11*A - 5*B)*c^5*Cos[e + f*x]^3)/(3465*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*(11*A - 5*B)*c^4*Cos[e + f
*x]^3)/(1155*f*Sqrt[c - c*Sin[e + f*x]]) + (8*a*(11*A - 5*B)*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(231
*f) + (2*a*(11*A - 5*B)*c^2*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(99*f) - (2*a*B*c*Cos[e + f*x]^3*(c - c
*Sin[e + f*x])^(5/2))/(11*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))^{7/2} \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{11} (a (11 A-5 B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx\\ &=\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{33} \left (4 a (11 A-5 B) c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {1}{231} \left (32 a (11 A-5 B) c^3\right ) \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac {\left (128 a (11 A-5 B) c^4\right ) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{1155}\\ &=\frac {256 a (11 A-5 B) c^5 \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt {c-c \sin (e+f x)}}+\frac {8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{231 f}+\frac {2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}\\ \end {align*}

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Mathematica [A]  time = 2.99, size = 149, normalized size = 0.75 \[ -\frac {a c^3 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 (60 (121 A-202 B) \cos (2 (e+f x))+30558 A \sin (e+f x)-770 A \sin (3 (e+f x))-35332 A-31530 B \sin (e+f x)+2870 B \sin (3 (e+f x))+315 B \cos (4 (e+f x))+27085 B)}{13860 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-1/13860*(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-35332*A + 27085*B + 60*(121
*A - 202*B)*Cos[2*(e + f*x)] + 315*B*Cos[4*(e + f*x)] + 30558*A*Sin[e + f*x] - 31530*B*Sin[e + f*x] - 770*A*Si
n[3*(e + f*x)] + 2870*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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fricas [A]  time = 0.44, size = 287, normalized size = 1.45 \[ \frac {2 \, {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{6} - 35 \, {\left (11 \, A - 32 \, B\right )} a c^{3} \cos \left (f x + e\right )^{5} + 5 \, {\left (209 \, A - 221 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 2 \, {\left (1243 \, A - 1195 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 32 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} + 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) + 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3} - {\left (315 \, B a c^{3} \cos \left (f x + e\right )^{5} + 35 \, {\left (11 \, A - 23 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 10 \, {\left (143 \, A - 191 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 96 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 128 \, {\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) - 256 \, {\left (11 \, A - 5 \, B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3465 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

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))^(7/2),x, algorithm="fricas")

[Out]

2/3465*(315*B*a*c^3*cos(f*x + e)^6 - 35*(11*A - 32*B)*a*c^3*cos(f*x + e)^5 + 5*(209*A - 221*B)*a*c^3*cos(f*x +
e)^4 + 2*(1243*A - 1195*B)*a*c^3*cos(f*x + e)^3 - 32*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 + 128*(11*A - 5*B)*a*c
^3*cos(f*x + e) + 256*(11*A - 5*B)*a*c^3 - (315*B*a*c^3*cos(f*x + e)^5 + 35*(11*A - 23*B)*a*c^3*cos(f*x + e)^4
+ 10*(143*A - 191*B)*a*c^3*cos(f*x + e)^3 - 96*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 - 128*(11*A - 5*B)*a*c^3*cos
(f*x + e) - 256*(11*A - 5*B)*a*c^3)*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))^(7/2),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)sqrt(2*c)*(-112*f*
(2*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-4*B*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(14*f*x+14
*exp(1)+pi))/(112*f)^2+48*f*(8*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-6*B*a*c^3*sign(sin(1/2*(f*x+exp(1))-
1/4*pi)))*sin(1/4*(6*f*x+6*exp(1)-pi))/(48*f)^2-8*f*(10*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-4*B*a*c^3*s
ign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/(8*f)^2+224*f*(16*A*a*c^3*sign(sin(1/2*(f*x+
exp(1))-1/4*pi))-10*B*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(14*f*x+14*exp(1)-pi))/(224*f)^2-144*f
*(-2*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+4*B*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(18*f*x+
18*exp(1)-pi))/(144*f)^2+16*f*(-8*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+6*B*a*c^3*sign(sin(1/2*(f*x+exp(1
))-1/4*pi)))*sin(1/4*(2*f*x+2*exp(1)+pi))/(16*f)^2+160*f*(-16*A*a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+10*B*
a*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(1)+pi))/(160*f)^2+24*A*a*c^3*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^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos
(1/4*(10*f*x+10*exp(1)-pi))/(20*f)^2-576*B*a*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(18*f*x+18*exp(1
)+pi))/(288*f)^2+704*B*a*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(22*f*x+22*exp(1)-pi))/(352*f)^2)

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maple [A]  time = 1.30, size = 119, normalized size = 0.60 \[ \frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (\left (-385 A +1435 B \right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (3916 A -4300 B \right ) \sin \left (f x +e \right )+315 B \left (\cos ^{4}\left (f x +e \right )\right )+\left (1815 A -3345 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )-5324 A +4940 B \right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^(7/2),x)

[Out]

2/3465*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^2*a*((-385*A+1435*B)*sin(f*x+e)*cos(f*x+e)^2+(3916*A-4300*B)*sin(f*x+
e)+315*B*cos(f*x+e)^4+(1815*A-3345*B)*cos(f*x+e)^2-5324*A+4940*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 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}\,{d x} \]

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))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2} \,d x \]

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))^(7/2),x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2), 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))**(7/2),x)

[Out]

Timed out

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