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3.99 \(\int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=161 \[ \frac {64 a^3 c^6 (13 A+B) \cos ^7(e+f x)}{9009 f (c-c \sin (e+f x))^{7/2}}+\frac {16 a^3 c^5 (13 A+B) \cos ^7(e+f x)}{1287 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 c^4 (13 A+B) \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

64/9009*a^3*(13*A+B)*c^6*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+16/1287*a^3*(13*A+B)*c^5*cos(f*x+e)^7/f/(c-c*si
n(f*x+e))^(5/2)+2/143*a^3*(13*A+B)*c^4*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(3/2)-2/13*a^3*B*c^3*cos(f*x+e)^7/f/(c-
c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2967, 2856, 2674, 2673} \[ \frac {2 a^3 c^4 (13 A+B) \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a^3 c^5 (13 A+B) \cos ^7(e+f x)}{1287 f (c-c \sin (e+f x))^{5/2}}+\frac {64 a^3 c^6 (13 A+B) \cos ^7(e+f x)}{9009 f (c-c \sin (e+f x))^{7/2}}-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(64*a^3*(13*A + B)*c^6*Cos[e + f*x]^7)/(9009*f*(c - c*Sin[e + f*x])^(7/2)) + (16*a^3*(13*A + B)*c^5*Cos[e + f*
x]^7)/(1287*f*(c - c*Sin[e + f*x])^(5/2)) + (2*a^3*(13*A + B)*c^4*Cos[e + f*x]^7)/(143*f*(c - c*Sin[e + f*x])^
(3/2)) - (2*a^3*B*c^3*Cos[e + f*x]^7)/(13*f*Sqrt[c - c*Sin[e + f*x]])

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))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{13} \left (a^3 (13 A+B) c^3\right ) \int \frac {\cos ^6(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {2 a^3 (13 A+B) c^4 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{143} \left (8 a^3 (13 A+B) c^4\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {16 a^3 (13 A+B) c^5 \cos ^7(e+f x)}{1287 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 (13 A+B) c^4 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (32 a^3 (13 A+B) c^5\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{1287}\\ &=\frac {64 a^3 (13 A+B) c^6 \cos ^7(e+f x)}{9009 f (c-c \sin (e+f x))^{7/2}}+\frac {16 a^3 (13 A+B) c^5 \cos ^7(e+f x)}{1287 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 (13 A+B) c^4 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 6.71, size = 1351, normalized size = 8.39 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(5*A*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x
)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (5*(4*A + B)*Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c
 - c*Sin[e + f*x])^(5/2))/(96*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^
6) + ((2*A - B)*Cos[(5*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(32*f*(Cos[(e + f*x)/2
] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - ((5*A + 2*B)*Cos[(7*(e + f*x))/2]*(a + a*Si
n[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^6) + ((A - 2*B)*Cos[(9*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(144*
f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - ((2*A + B)*Cos[(11*(e + f
*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(352*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (B*Cos[(13*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5
/2))/(416*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*A*Sin[(e + f
*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e
 + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*(4*A + B)*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(3*(e +
 f*x))/2])/(96*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((2*A - B)
*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(5*(e + f*x))/2])/(32*f*(Cos[(e + f*x)/2] - Sin[(e + f*
x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((5*A + 2*B)*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(
5/2)*Sin[(7*(e + f*x))/2])/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]
)^6) + ((A - 2*B)*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(9*(e + f*x))/2])/(144*f*(Cos[(e + f*x
)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((2*A + B)*(a + a*Sin[e + f*x])^3*(c - c
*Sin[e + f*x])^(5/2)*Sin[(11*(e + f*x))/2])/(352*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] +
 Sin[(e + f*x)/2])^6) - (B*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(13*(e + f*x))/2])/(416*f*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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fricas [B]  time = 0.44, size = 334, normalized size = 2.07 \[ -\frac {2 \, {\left (693 \, B a^{3} c^{2} \cos \left (f x + e\right )^{7} + 63 \, {\left (13 \, A + 12 \, B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{6} - 7 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{5} + 10 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{4} - 16 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{3} + 32 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{2} - 128 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right ) - 256 \, {\left (13 \, A + B\right )} a^{3} c^{2} + {\left (693 \, B a^{3} c^{2} \cos \left (f x + e\right )^{6} - 63 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{5} - 70 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{4} - 80 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{3} - 96 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{2} - 128 \, {\left (13 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right ) - 256 \, {\left (13 \, A + B\right )} a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{9009 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/9009*(693*B*a^3*c^2*cos(f*x + e)^7 + 63*(13*A + 12*B)*a^3*c^2*cos(f*x + e)^6 - 7*(13*A + B)*a^3*c^2*cos(f*x
 + e)^5 + 10*(13*A + B)*a^3*c^2*cos(f*x + e)^4 - 16*(13*A + B)*a^3*c^2*cos(f*x + e)^3 + 32*(13*A + B)*a^3*c^2*
cos(f*x + e)^2 - 128*(13*A + B)*a^3*c^2*cos(f*x + e) - 256*(13*A + B)*a^3*c^2 + (693*B*a^3*c^2*cos(f*x + e)^6
- 63*(13*A + B)*a^3*c^2*cos(f*x + e)^5 - 70*(13*A + B)*a^3*c^2*cos(f*x + e)^4 - 80*(13*A + B)*a^3*c^2*cos(f*x
+ e)^3 - 96*(13*A + B)*a^3*c^2*cos(f*x + e)^2 - 128*(13*A + B)*a^3*c^2*cos(f*x + e) - 256*(13*A + B)*a^3*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))^3*(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)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)*(16*f*(2*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a^3*c^2*sign(sin
(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(2*f*x+2*exp(1)+pi))/(16*f)^2+288*f*(2*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))
-1/4*pi))+2*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(18*f*x+18*exp(1)+pi))/(288*f)^2-288*f*(4*A*
a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(18*f*x+18*
exp(1)-pi))/(288*f)^2+160*f*(6*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+6*B*a^3*c^2*sign(sin(1/2*(f*x+exp(
1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(1)+pi))/(160*f)^2-16*f*(12*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2
*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x-pi)+1/2*exp(1))/(16*f)^2-320*f*(32*A*a^3*c^2*sig
n(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(10*f*x+10*exp(1)-pi))
/(320*f)^2+48*f*(-2*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)
))*sin(1/4*(6*f*x+6*exp(1)-pi))/(48*f)^2+352*f*(-2*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a^3*c^2*si
gn(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(22*f*x+22*exp(1)-pi))/(352*f)^2-224*f*(-4*A*a^3*c^2*sign(sin(1/2*(f
*x+exp(1))-1/4*pi))+2*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(14*f*x+14*exp(1)+pi))/(224*f)^2+2
24*f*(-6*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-6*B*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*
(14*f*x+14*exp(1)-pi))/(224*f)^2-192*f*(-32*A*a^3*c^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a^3*c^2*sign(sin(
1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(6*f*x+6*exp(1)+pi))/(192*f)^2-1408*B*a^3*c^2*f*sign(sin(1/2*(f*x+exp(1))-1
/4*pi))*cos(1/4*(22*f*x+22*exp(1)+pi))/(704*f)^2+1664*B*a^3*c^2*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(
26*f*x+26*exp(1)-pi))/(832*f)^2)

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maple [A]  time = 1.34, size = 105, normalized size = 0.65 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{4} a^{3} \left (-693 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-2366 A +2590 B \right ) \sin \left (f x +e \right )+\left (-819 A +2016 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2782 A -2558 B \right )}{9009 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9009*(sin(f*x+e)-1)*c^3*(1+sin(f*x+e))^4*a^3*(-693*B*cos(f*x+e)^2*sin(f*x+e)+(-2366*A+2590*B)*sin(f*x+e)+(-
819*A+2016*B)*cos(f*x+e)^2+2782*A-2558*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 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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