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 \[ \frac{2 a^3 c^4 (13 A+B) \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a^3 c^5 (13 A+B) \cos ^7(e+f x)}{1287 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^3 c^6 (13 A+B) \cos ^7(e+f x)}{9009 f (c-c \sin (e+f x))^{7/2}}-\frac{2 a^3 B c^3 \cos ^7(e+f x)}{13 f \sqrt{c-c \sin (e+f x)}} \]
[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]])
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Rubi [A] time = 0.471644, antiderivative size = 161, normalized size of antiderivative = 1.,
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 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]))
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 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 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]
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*}
Mathematica [B] time = 6.71864, 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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Maple [A] time = 0.887, size = 105, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{4}{a}^{3} \left ( -693\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( -2366\,A+2590\,B \right ) \sin \left ( fx+e \right ) + \left ( -819\,A+2016\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2782\,A-2558\,B \right ) }{9009\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}
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*(-1+sin(f*x+e))*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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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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Fricas [B] time = 1.53978, size = 849, normalized size = 5.27 \begin{align*} -\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 )}} \end{align*}
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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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out