3.3.50 \(\int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^3 \, dx\) [250]

Optimal. Leaf size=91 \[ \frac {5}{16} a^3 c^3 x+\frac {5 a^3 c^3 \cos (e+f x) \sin (e+f x)}{16 f}+\frac {5 a^3 c^3 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^3 c^3 \cos ^5(e+f x) \sin (e+f x)}{6 f} \]

[Out]

5/16*a^3*c^3*x+5/16*a^3*c^3*cos(f*x+e)*sin(f*x+e)/f+5/24*a^3*c^3*cos(f*x+e)^3*sin(f*x+e)/f+1/6*a^3*c^3*cos(f*x
+e)^5*sin(f*x+e)/f

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Rubi [A]
time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2815, 2715, 8} \begin {gather*} \frac {a^3 c^3 \sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac {5 a^3 c^3 \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac {5 a^3 c^3 \sin (e+f x) \cos (e+f x)}{16 f}+\frac {5}{16} a^3 c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*c^3*x)/16 + (5*a^3*c^3*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (5*a^3*c^3*Cos[e + f*x]^3*Sin[e + f*x])/(24*
f) + (a^3*c^3*Cos[e + f*x]^5*Sin[e + f*x])/(6*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2815

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))^3 \, dx &=\left (a^3 c^3\right ) \int \cos ^6(e+f x) \, dx\\ &=\frac {a^3 c^3 \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {1}{6} \left (5 a^3 c^3\right ) \int \cos ^4(e+f x) \, dx\\ &=\frac {5 a^3 c^3 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^3 c^3 \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {1}{8} \left (5 a^3 c^3\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac {5 a^3 c^3 \cos (e+f x) \sin (e+f x)}{16 f}+\frac {5 a^3 c^3 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^3 c^3 \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {1}{16} \left (5 a^3 c^3\right ) \int 1 \, dx\\ &=\frac {5}{16} a^3 c^3 x+\frac {5 a^3 c^3 \cos (e+f x) \sin (e+f x)}{16 f}+\frac {5 a^3 c^3 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a^3 c^3 \cos ^5(e+f x) \sin (e+f x)}{6 f}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 49, normalized size = 0.54 \begin {gather*} \frac {a^3 c^3 (60 e+60 f x+45 \sin (2 (e+f x))+9 \sin (4 (e+f x))+\sin (6 (e+f x)))}{192 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^3*(60*e + 60*f*x + 45*Sin[2*(e + f*x)] + 9*Sin[4*(e + f*x)] + Sin[6*(e + f*x)]))/(192*f)

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Maple [A]
time = 0.26, size = 140, normalized size = 1.54

method	result	size
risch	\(\frac {5 a^{3} c^{3} x}{16}+\frac {c^{3} a^{3} \sin \left (6 f x +6 e \right )}{192 f}+\frac {3 c^{3} a^{3} \sin \left (4 f x +4 e \right )}{64 f}+\frac {15 c^{3} a^{3} \sin \left (2 f x +2 e \right )}{64 f}\)	\(71\)
derivativedivides	\(\frac {-c^{3} a^{3} \left (-\frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+3 c^{3} a^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-3 c^{3} a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+c^{3} a^{3} \left (f x +e \right )}{f}\)	\(140\)
default	\(\frac {-c^{3} a^{3} \left (-\frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+3 c^{3} a^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-3 c^{3} a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+c^{3} a^{3} \left (f x +e \right )}{f}\)	\(140\)
norman	\(\frac {\frac {5 a^{3} c^{3} x}{16}+\frac {15 a^{3} c^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {75 a^{3} c^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{16}+\frac {25 a^{3} c^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {75 a^{3} c^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{16}+\frac {15 a^{3} c^{3} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {5 a^{3} c^{3} x \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{16}+\frac {11 c^{3} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}-\frac {5 c^{3} a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}+\frac {15 c^{3} a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {15 c^{3} a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {5 c^{3} a^{3} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}-\frac {11 c^{3} a^{3} \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{6}}\)	\(277\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(-c^3*a^3*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)+3*c^3*a^3*(-1/
4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-3*c^3*a^3*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e)
+c^3*a^3*(f*x+e))

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Maxima [A]
time = 0.28, size = 142, normalized size = 1.56 \begin {gather*} -\frac {{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{3} - 18 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{3} + 144 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{3} - 192 \, {\left (f x + e\right )} a^{3} c^{3}}{192 \, f} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

-1/192*((4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*a^3*c^3 - 18*(12*f*x
+ 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^3*c^3 + 144*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3*c^3 - 192*
(f*x + e)*a^3*c^3)/f

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Fricas [A]
time = 0.32, size = 74, normalized size = 0.81 \begin {gather*} \frac {15 \, a^{3} c^{3} f x + {\left (8 \, a^{3} c^{3} \cos \left (f x + e\right )^{5} + 10 \, a^{3} c^{3} \cos \left (f x + e\right )^{3} + 15 \, a^{3} c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

1/48*(15*a^3*c^3*f*x + (8*a^3*c^3*cos(f*x + e)^5 + 10*a^3*c^3*cos(f*x + e)^3 + 15*a^3*c^3*cos(f*x + e))*sin(f*
x + e))/f

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (88) = 176\).
time = 0.57, size = 398, normalized size = 4.37 \begin {gather*} \begin {cases} - \frac {5 a^{3} c^{3} x \sin ^{6}{\left (e + f x \right )}}{16} - \frac {15 a^{3} c^{3} x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac {9 a^{3} c^{3} x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {15 a^{3} c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac {9 a^{3} c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {3 a^{3} c^{3} x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {5 a^{3} c^{3} x \cos ^{6}{\left (e + f x \right )}}{16} + \frac {9 a^{3} c^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {3 a^{3} c^{3} x \cos ^{2}{\left (e + f x \right )}}{2} + a^{3} c^{3} x + \frac {11 a^{3} c^{3} \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{16 f} + \frac {5 a^{3} c^{3} \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac {15 a^{3} c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 a^{3} c^{3} \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} - \frac {9 a^{3} c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {3 a^{3} c^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right )^{3} \left (- c \sin {\left (e \right )} + c\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

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))**3,x)

[Out]

Piecewise((-5*a**3*c**3*x*sin(e + f*x)**6/16 - 15*a**3*c**3*x*sin(e + f*x)**4*cos(e + f*x)**2/16 + 9*a**3*c**3
*x*sin(e + f*x)**4/8 - 15*a**3*c**3*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + 9*a**3*c**3*x*sin(e + f*x)**2*cos(e
+ f*x)**2/4 - 3*a**3*c**3*x*sin(e + f*x)**2/2 - 5*a**3*c**3*x*cos(e + f*x)**6/16 + 9*a**3*c**3*x*cos(e + f*x)
**4/8 - 3*a**3*c**3*x*cos(e + f*x)**2/2 + a**3*c**3*x + 11*a**3*c**3*sin(e + f*x)**5*cos(e + f*x)/(16*f) + 5*a
**3*c**3*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - 15*a**3*c**3*sin(e + f*x)**3*cos(e + f*x)/(8*f) + 5*a**3*c**3
*sin(e + f*x)*cos(e + f*x)**5/(16*f) - 9*a**3*c**3*sin(e + f*x)*cos(e + f*x)**3/(8*f) + 3*a**3*c**3*sin(e + f*
x)*cos(e + f*x)/(2*f), Ne(f, 0)), (x*(a*sin(e) + a)**3*(-c*sin(e) + c)**3, True))

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Giac [A]
time = 0.45, size = 73, normalized size = 0.80 \begin {gather*} \frac {5}{16} \, a^{3} c^{3} x + \frac {a^{3} c^{3} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {3 \, a^{3} c^{3} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {15 \, a^{3} c^{3} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

5/16*a^3*c^3*x + 1/192*a^3*c^3*sin(6*f*x + 6*e)/f + 3/64*a^3*c^3*sin(4*f*x + 4*e)/f + 15/64*a^3*c^3*sin(2*f*x
+ 2*e)/f

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Mupad [B]
time = 10.14, size = 143, normalized size = 1.57 \begin {gather*} \frac {5\,a^3\,c^3\,x}{16}-\frac {\frac {11\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}-\frac {5\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}+\frac {15\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {15\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{4}+\frac {5\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24}-\frac {11\,a^3\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^6} \end {gather*}

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))^3,x)

[Out]

(5*a^3*c^3*x)/16 - ((5*a^3*c^3*tan(e/2 + (f*x)/2)^3)/24 - (15*a^3*c^3*tan(e/2 + (f*x)/2)^5)/4 + (15*a^3*c^3*ta
n(e/2 + (f*x)/2)^7)/4 - (5*a^3*c^3*tan(e/2 + (f*x)/2)^9)/24 + (11*a^3*c^3*tan(e/2 + (f*x)/2)^11)/8 - (11*a^3*c
^3*tan(e/2 + (f*x)/2))/8)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^6)

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