3.307 \(\int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=145 \[ \frac {256 a^3 c^7 \cos ^7(e+f x)}{3003 f (c-c \sin (e+f x))^{7/2}}+\frac {64 a^3 c^6 \cos ^7(e+f x)}{429 f (c-c \sin (e+f x))^{5/2}}+\frac {24 a^3 c^5 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

256/3003*a^3*c^7*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+64/429*a^3*c^6*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(5/2)+24
/143*a^3*c^5*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(3/2)+2/13*a^3*c^4*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(1/2)

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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} \[ \frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {24 a^3 c^5 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {64 a^3 c^6 \cos ^7(e+f x)}{429 f (c-c \sin (e+f x))^{5/2}}+\frac {256 a^3 c^7 \cos ^7(e+f x)}{3003 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(256*a^3*c^7*Cos[e + f*x]^7)/(3003*f*(c - c*Sin[e + f*x])^(7/2)) + (64*a^3*c^6*Cos[e + f*x]^7)/(429*f*(c - c*S
in[e + f*x])^(5/2)) + (24*a^3*c^5*Cos[e + f*x]^7)/(143*f*(c - c*Sin[e + f*x])^(3/2)) + (2*a^3*c^4*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 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))^3 (c-c \sin (e+f x))^{7/2} \, dx &=\left (a^3 c^3\right ) \int \cos ^6(e+f x) \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{13} \left (12 a^3 c^4\right ) \int \frac {\cos ^6(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {24 a^3 c^5 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{143} \left (96 a^3 c^5\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {64 a^3 c^6 \cos ^7(e+f x)}{429 f (c-c \sin (e+f x))^{5/2}}+\frac {24 a^3 c^5 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{429} \left (128 a^3 c^6\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac {256 a^3 c^7 \cos ^7(e+f x)}{3003 f (c-c \sin (e+f x))^{7/2}}+\frac {64 a^3 c^6 \cos ^7(e+f x)}{429 f (c-c \sin (e+f x))^{5/2}}+\frac {24 a^3 c^5 \cos ^7(e+f x)}{143 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{13 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 6.27, size = 112, normalized size = 0.77 \[ \frac {a^3 c^3 \cos ^6(e+f x) \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (-6377 \sin (e+f x)+231 \sin (3 (e+f x))-1890 \cos (2 (e+f x))+5230)}{6006 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(a^3*c^3*Cos[e + f*x]^6*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(5230 - 1890*Cos[2*(e +
f*x)] - 6377*Sin[e + f*x] + 231*Sin[3*(e + f*x)]))/(6006*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7)

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fricas [B]  time = 0.45, size = 265, normalized size = 1.83 \[ \frac {2 \, {\left (231 \, a^{3} c^{3} \cos \left (f x + e\right )^{7} - 21 \, a^{3} c^{3} \cos \left (f x + e\right )^{6} + 28 \, a^{3} c^{3} \cos \left (f x + e\right )^{5} - 40 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} + 64 \, a^{3} c^{3} \cos \left (f x + e\right )^{3} - 128 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 512 \, a^{3} c^{3} \cos \left (f x + e\right ) + 1024 \, a^{3} c^{3} + {\left (231 \, a^{3} c^{3} \cos \left (f x + e\right )^{6} + 252 \, a^{3} c^{3} \cos \left (f x + e\right )^{5} + 280 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} + 320 \, a^{3} c^{3} \cos \left (f x + e\right )^{3} + 384 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 512 \, a^{3} c^{3} \cos \left (f x + e\right ) + 1024 \, a^{3} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3003 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/3003*(231*a^3*c^3*cos(f*x + e)^7 - 21*a^3*c^3*cos(f*x + e)^6 + 28*a^3*c^3*cos(f*x + e)^5 - 40*a^3*c^3*cos(f*
x + e)^4 + 64*a^3*c^3*cos(f*x + e)^3 - 128*a^3*c^3*cos(f*x + e)^2 + 512*a^3*c^3*cos(f*x + e) + 1024*a^3*c^3 +
(231*a^3*c^3*cos(f*x + e)^6 + 252*a^3*c^3*cos(f*x + e)^5 + 280*a^3*c^3*cos(f*x + e)^4 + 320*a^3*c^3*cos(f*x +
e)^3 + 384*a^3*c^3*cos(f*x + e)^2 + 512*a^3*c^3*cos(f*x + e) + 1024*a^3*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))^3*(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)sqrt(2*c)*(-2*a^3*
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+5760*a^3*c^3*f*sign(sin(1/2*(f*x+exp(1
))-1/4*pi))*cos(1/4*(6*f*x+6*exp(1)+pi))/(192*f)^2-9600*a^3*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(
10*f*x+10*exp(1)-pi))/(320*f)^2+1344*a^3*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(14*f*x+14*exp(1)+pi
))/(224*f)^2-1728*a^3*c^3*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(18*f*x+18*exp(1)-pi))/(288*f)^2+1408*a
^3*c^3*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*a^3*c^3*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+352*a^3*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))/(16*f)^2)

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maple [A]  time = 0.89, size = 81, normalized size = 0.56 \[ \frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (1+\sin \left (f x +e \right )\right )^{4} a^{3} \left (231 \left (\sin ^{3}\left (f x +e \right )\right )-945 \left (\sin ^{2}\left (f x +e \right )\right )+1421 \sin \left (f x +e \right )-835\right )}{3003 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(7/2),x)

[Out]

2/3003*(sin(f*x+e)-1)*c^4*(1+sin(f*x+e))^4*a^3*(231*sin(f*x+e)^3-945*sin(f*x+e)^2+1421*sin(f*x+e)-835)/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3*(-c*sin(f*x + e) + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(7/2),x)

[Out]

int((a + a*sin(e + f*x))^3*(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))**3*(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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