3.298 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx\)
Optimal. Leaf size=145 \[ \frac {256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac {64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f} \]
[Out]
256/1155*a^2*c^6*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+64/231*a^2*c^5*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(3/2)+8/
33*a^2*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(1/2)+2/11*a^2*c^3*cos(f*x+e)^5*(c-c*sin(f*x+e))^(1/2)/f
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Rubi [A] time = 0.33, antiderivative size = 145, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used =
{2736, 2674, 2673} \[ \frac {256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac {64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2),x]
[Out]
(256*a^2*c^6*Cos[e + f*x]^5)/(1155*f*(c - c*Sin[e + f*x])^(5/2)) + (64*a^2*c^5*Cos[e + f*x]^5)/(231*f*(c - c*S
in[e + f*x])^(3/2)) + (8*a^2*c^4*Cos[e + f*x]^5)/(33*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a^2*c^3*Cos[e + f*x]^5*S
qrt[c - c*Sin[e + f*x]])/(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 2736
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[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))^2 (c-c \sin (e+f x))^{7/2} \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}+\frac {1}{11} \left (12 a^2 c^3\right ) \int \cos ^4(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}+\frac {1}{33} \left (32 a^2 c^4\right ) \int \frac {\cos ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}+\frac {1}{231} \left (128 a^2 c^5\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac {64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\\ \end {align*}
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Mathematica [B] time = 6.44, size = 1105, normalized size = 7.62 \[ \text {result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2),x]
[Out]
(7*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) - (Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f
*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*Cos[(5
*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(7*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])
^(7/2))/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(9*(e +
f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(48*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(11*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/
2))/(176*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (7*Sin[(e + f*x)
/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^4) + ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(3*(e + f*x))/2])/(8*f
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*(a + a*Sin[e + f*x])^2
*(c - c*Sin[e + f*x])^(7/2)*Sin[(5*(e + f*x))/2])/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^4) - ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(7*(e + f*x))/2])/(112*f*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + ((a + a*Sin[e + f*x])^2*(c -
c*Sin[e + f*x])^(7/2)*Sin[(9*(e + f*x))/2])/(48*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^4) - ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(11*(e + f*x))/2])/(176*f*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)
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fricas [A] time = 0.48, size = 234, normalized size = 1.61 \[ \frac {2 \, {\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 245 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 20 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 32 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 64 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 256 \, a^{2} c^{3} \cos \left (f x + e\right ) + 512 \, a^{2} c^{3} - {\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 140 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} - 160 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 192 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 256 \, a^{2} c^{3} \cos \left (f x + e\right ) - 512 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1155 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")
[Out]
2/1155*(105*a^2*c^3*cos(f*x + e)^6 + 245*a^2*c^3*cos(f*x + e)^5 - 20*a^2*c^3*cos(f*x + e)^4 + 32*a^2*c^3*cos(f
*x + e)^3 - 64*a^2*c^3*cos(f*x + e)^2 + 256*a^2*c^3*cos(f*x + e) + 512*a^2*c^3 - (105*a^2*c^3*cos(f*x + e)^5 -
140*a^2*c^3*cos(f*x + e)^4 - 160*a^2*c^3*cos(f*x + e)^3 - 192*a^2*c^3*cos(f*x + e)^2 - 256*a^2*c^3*cos(f*x +
e) - 512*a^2*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))^2*(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)sqrt(2*c)*(-2*a^2*c^3*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1
/4*(2*f*x-pi)+1/2*exp(1))/f-32*a^2*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(2*f*x+2*exp(1)+pi))/(16*f
)^2+96*a^2*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(6*f*x+6*exp(1)-pi))/(48*f)^2-960*a^2*c^3*f*sign(s
in(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(10*f*x+10*exp(1)+pi))/(160*f)^2+1344*a^2*c^3*f*sign(sin(1/2*(f*x+exp(1))
-1/4*pi))*sin(1/4*(14*f*x+14*exp(1)-pi))/(224*f)^2-576*a^2*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sin(1/4*(1
8*f*x+18*exp(1)+pi))/(288*f)^2+704*a^2*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+24*a^2*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^2*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+224*a^2*c^3*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*a^2*c^3*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+80*a^2*c^3*f*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)
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maple [A] time = 0.87, size = 81, normalized size = 0.56 \[ \frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{3} a^{2} \left (105 \left (\sin ^{3}\left (f x +e \right )\right )-455 \left (\sin ^{2}\left (f x +e \right )\right )+755 \sin \left (f x +e \right )-533\right )}{1155 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^(7/2),x)
[Out]
2/1155*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^3*a^2*(105*sin(f*x+e)^3-455*sin(f*x+e)^2+755*sin(f*x+e)-533)/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^2*(-c*sin(f*x + e) + c)^(7/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^(7/2),x)
[Out]
int((a + a*sin(e + f*x))^2*(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))**2*(c-c*sin(f*x+e))**(7/2),x)
[Out]
Timed out
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