3.90 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=167 \[ \frac{2 a^2 c^3 (11 A-B) \cos ^5(e+f x)}{99 f \sqrt{c-c \sin (e+f x)}}+\frac{16 a^2 c^4 (11 A-B) \cos ^5(e+f x)}{693 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a^2 c^5 (11 A-B) \cos ^5(e+f x)}{3465 f (c-c \sin (e+f x))^{5/2}}-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f} \]
[Out]
(64*a^2*(11*A - B)*c^5*Cos[e + f*x]^5)/(3465*f*(c - c*Sin[e + f*x])^(5/2)) + (16*a^2*(11*A - B)*c^4*Cos[e + f*
x]^5)/(693*f*(c - c*Sin[e + f*x])^(3/2)) + (2*a^2*(11*A - B)*c^3*Cos[e + f*x]^5)/(99*f*Sqrt[c - c*Sin[e + f*x]
]) - (2*a^2*B*c^2*Cos[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(11*f)
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Rubi [A] time = 0.452875, antiderivative size = 167, 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} \[ \frac{2 a^2 c^3 (11 A-B) \cos ^5(e+f x)}{99 f \sqrt{c-c \sin (e+f x)}}+\frac{16 a^2 c^4 (11 A-B) \cos ^5(e+f x)}{693 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a^2 c^5 (11 A-B) \cos ^5(e+f x)}{3465 f (c-c \sin (e+f x))^{5/2}}-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]
[Out]
(64*a^2*(11*A - B)*c^5*Cos[e + f*x]^5)/(3465*f*(c - c*Sin[e + f*x])^(5/2)) + (16*a^2*(11*A - B)*c^4*Cos[e + f*
x]^5)/(693*f*(c - c*Sin[e + f*x])^(3/2)) + (2*a^2*(11*A - B)*c^3*Cos[e + f*x]^5)/(99*f*Sqrt[c - c*Sin[e + f*x]
]) - (2*a^2*B*c^2*Cos[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(11*f)
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))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx\\ &=-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{11} \left (a^2 (11 A-B) c^2\right ) \int \cos ^4(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{2 a^2 (11 A-B) c^3 \cos ^5(e+f x)}{99 f \sqrt{c-c \sin (e+f x)}}-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{99} \left (8 a^2 (11 A-B) c^3\right ) \int \frac{\cos ^4(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{16 a^2 (11 A-B) c^4 \cos ^5(e+f x)}{693 f (c-c \sin (e+f x))^{3/2}}+\frac{2 a^2 (11 A-B) c^3 \cos ^5(e+f x)}{99 f \sqrt{c-c \sin (e+f x)}}-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{693} \left (32 a^2 (11 A-B) c^4\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{64 a^2 (11 A-B) c^5 \cos ^5(e+f x)}{3465 f (c-c \sin (e+f x))^{5/2}}+\frac{16 a^2 (11 A-B) c^4 \cos ^5(e+f x)}{693 f (c-c \sin (e+f x))^{3/2}}+\frac{2 a^2 (11 A-B) c^3 \cos ^5(e+f x)}{99 f \sqrt{c-c \sin (e+f x)}}-\frac{2 a^2 B c^2 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}\\ \end{align*}
Mathematica [B] time = 6.55698, size = 1173, normalized size = 7.02 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x]
[Out]
((6*A - B)*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(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])^4) - ((4*A + B)*Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^
2*(c - c*Sin[e + f*x])^(5/2))/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2])^4) + ((8*A - 3*B)*Cos[(5*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2))/(80*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) - ((2*A + 3*B)*Cos[(7*(e + f*x))/2]*(a
+ a*Sin[e + f*x])^2*(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] + Sin[(e + f*x)/2])^4) + ((2*A - B)*Cos[(9*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(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])^4) - (B*Cos[(11*(e + f*x
))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2))/(176*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])^4) + ((6*A - B)*Sin[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(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])^4) + ((4*A + B)*(a +
a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)*Sin[(3*(e + f*x))/2])/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + ((8*A - 3*B)*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)*S
in[(5*(e + f*x))/2])/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) +
((2*A + 3*B)*(a + a*Sin[e + f*x])^2*(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])^4) + ((2*A - B)*(a + a*Sin[e + f*x])^2*(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])^4) + (B*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2)*Sin[(11*(e + f*x))/2])/(176*f*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)
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Maple [A] time = 0.951, size = 105, normalized size = 0.6 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{3}{a}^{2} \left ( -315\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( -1210\,A+1370\,B \right ) \sin \left ( fx+e \right ) + \left ( -385\,A+980\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+1562\,A-1402\,B \right ) }{3465\,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))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x)
[Out]
-2/3465*(-1+sin(f*x+e))*c^3*(1+sin(f*x+e))^3*a^2*(-315*B*cos(f*x+e)^2*sin(f*x+e)+(-1210*A+1370*B)*sin(f*x+e)+(
-385*A+980*B)*cos(f*x+e)^2+1562*A-1402*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 )}^{2}{\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))^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)^2*(-c*sin(f*x + e) + c)^(5/2), x)
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Fricas [B] time = 1.53081, size = 740, normalized size = 4.43 \begin{align*} -\frac{2 \,{\left (315 \, B a^{2} c^{2} \cos \left (f x + e\right )^{6} - 35 \,{\left (11 \, A - 10 \, B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{5} + 5 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{4} - 8 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{3} + 16 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{2} - 64 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right ) - 128 \,{\left (11 \, A - B\right )} a^{2} c^{2} -{\left (315 \, B a^{2} c^{2} \cos \left (f x + e\right )^{5} + 35 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{4} + 40 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{3} + 48 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right )^{2} + 64 \,{\left (11 \, A - B\right )} a^{2} c^{2} \cos \left (f x + e\right ) + 128 \,{\left (11 \, A - B\right )} a^{2} c^{2}\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
-2/3465*(315*B*a^2*c^2*cos(f*x + e)^6 - 35*(11*A - 10*B)*a^2*c^2*cos(f*x + e)^5 + 5*(11*A - B)*a^2*c^2*cos(f*x
+ e)^4 - 8*(11*A - B)*a^2*c^2*cos(f*x + e)^3 + 16*(11*A - B)*a^2*c^2*cos(f*x + e)^2 - 64*(11*A - B)*a^2*c^2*c
os(f*x + e) - 128*(11*A - B)*a^2*c^2 - (315*B*a^2*c^2*cos(f*x + e)^5 + 35*(11*A - B)*a^2*c^2*cos(f*x + e)^4 +
40*(11*A - B)*a^2*c^2*cos(f*x + e)^3 + 48*(11*A - B)*a^2*c^2*cos(f*x + e)^2 + 64*(11*A - B)*a^2*c^2*cos(f*x +
e) + 128*(11*A - B)*a^2*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))**2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [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 )}^{2}{\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))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2*(-c*sin(f*x + e) + c)^(5/2), x)