3.2 \(\int \sin ^3(e+f x) (a+a \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=129 \[ -\frac {3 a^3 \cos ^5(e+f x)}{5 f}+\frac {7 a^3 \cos ^3(e+f x)}{3 f}-\frac {4 a^3 \cos (e+f x)}{f}-\frac {a^3 \sin ^5(e+f x) \cos (e+f x)}{6 f}-\frac {23 a^3 \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac {23 a^3 \sin (e+f x) \cos (e+f x)}{16 f}+\frac {23 a^3 x}{16} \]

[Out]

23/16*a^3*x-4*a^3*cos(f*x+e)/f+7/3*a^3*cos(f*x+e)^3/f-3/5*a^3*cos(f*x+e)^5/f-23/16*a^3*cos(f*x+e)*sin(f*x+e)/f
-23/24*a^3*cos(f*x+e)*sin(f*x+e)^3/f-1/6*a^3*cos(f*x+e)*sin(f*x+e)^5/f

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2633, 2635, 8} \[ -\frac {3 a^3 \cos ^5(e+f x)}{5 f}+\frac {7 a^3 \cos ^3(e+f x)}{3 f}-\frac {4 a^3 \cos (e+f x)}{f}-\frac {a^3 \sin ^5(e+f x) \cos (e+f x)}{6 f}-\frac {23 a^3 \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac {23 a^3 \sin (e+f x) \cos (e+f x)}{16 f}+\frac {23 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*a^3*x)/16 - (4*a^3*Cos[e + f*x])/f + (7*a^3*Cos[e + f*x]^3)/(3*f) - (3*a^3*Cos[e + f*x]^5)/(5*f) - (23*a^3
*Cos[e + f*x]*Sin[e + f*x])/(16*f) - (23*a^3*Cos[e + f*x]*Sin[e + f*x]^3)/(24*f) - (a^3*Cos[e + f*x]*Sin[e + f
*x]^5)/(6*f)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

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] && Integer
Q[2*n]

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \sin ^3(e+f x) (a+a \sin (e+f x))^3 \, dx &=\int \left (a^3 \sin ^3(e+f x)+3 a^3 \sin ^4(e+f x)+3 a^3 \sin ^5(e+f x)+a^3 \sin ^6(e+f x)\right ) \, dx\\ &=a^3 \int \sin ^3(e+f x) \, dx+a^3 \int \sin ^6(e+f x) \, dx+\left (3 a^3\right ) \int \sin ^4(e+f x) \, dx+\left (3 a^3\right ) \int \sin ^5(e+f x) \, dx\\ &=-\frac {3 a^3 \cos (e+f x) \sin ^3(e+f x)}{4 f}-\frac {a^3 \cos (e+f x) \sin ^5(e+f x)}{6 f}+\frac {1}{6} \left (5 a^3\right ) \int \sin ^4(e+f x) \, dx+\frac {1}{4} \left (9 a^3\right ) \int \sin ^2(e+f x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {4 a^3 \cos (e+f x)}{f}+\frac {7 a^3 \cos ^3(e+f x)}{3 f}-\frac {3 a^3 \cos ^5(e+f x)}{5 f}-\frac {9 a^3 \cos (e+f x) \sin (e+f x)}{8 f}-\frac {23 a^3 \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac {a^3 \cos (e+f x) \sin ^5(e+f x)}{6 f}+\frac {1}{8} \left (5 a^3\right ) \int \sin ^2(e+f x) \, dx+\frac {1}{8} \left (9 a^3\right ) \int 1 \, dx\\ &=\frac {9 a^3 x}{8}-\frac {4 a^3 \cos (e+f x)}{f}+\frac {7 a^3 \cos ^3(e+f x)}{3 f}-\frac {3 a^3 \cos ^5(e+f x)}{5 f}-\frac {23 a^3 \cos (e+f x) \sin (e+f x)}{16 f}-\frac {23 a^3 \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac {a^3 \cos (e+f x) \sin ^5(e+f x)}{6 f}+\frac {1}{16} \left (5 a^3\right ) \int 1 \, dx\\ &=\frac {23 a^3 x}{16}-\frac {4 a^3 \cos (e+f x)}{f}+\frac {7 a^3 \cos ^3(e+f x)}{3 f}-\frac {3 a^3 \cos ^5(e+f x)}{5 f}-\frac {23 a^3 \cos (e+f x) \sin (e+f x)}{16 f}-\frac {23 a^3 \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac {a^3 \cos (e+f x) \sin ^5(e+f x)}{6 f}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 115, normalized size = 0.89 \[ -\frac {a^3 \cos (e+f x) \left (690 \sin ^{-1}\left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\left (40 \sin ^5(e+f x)+144 \sin ^4(e+f x)+230 \sin ^3(e+f x)+272 \sin ^2(e+f x)+345 \sin (e+f x)+544\right ) \sqrt {\cos ^2(e+f x)}\right )}{240 f \sqrt {\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a^3*Cos[e + f*x]*(690*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2]*(544 + 345*Sin[e +
f*x] + 272*Sin[e + f*x]^2 + 230*Sin[e + f*x]^3 + 144*Sin[e + f*x]^4 + 40*Sin[e + f*x]^5)))/(f*Sqrt[Cos[e + f*
x]^2])

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fricas [A]  time = 0.47, size = 96, normalized size = 0.74 \[ -\frac {144 \, a^{3} \cos \left (f x + e\right )^{5} - 560 \, a^{3} \cos \left (f x + e\right )^{3} - 345 \, a^{3} f x + 960 \, a^{3} \cos \left (f x + e\right ) + 5 \, {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 62 \, a^{3} \cos \left (f x + e\right )^{3} + 123 \, a^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/240*(144*a^3*cos(f*x + e)^5 - 560*a^3*cos(f*x + e)^3 - 345*a^3*f*x + 960*a^3*cos(f*x + e) + 5*(8*a^3*cos(f*
x + e)^5 - 62*a^3*cos(f*x + e)^3 + 123*a^3*cos(f*x + e))*sin(f*x + e))/f

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giac [A]  time = 1.93, size = 112, normalized size = 0.87 \[ \frac {23}{16} \, a^{3} x - \frac {3 \, a^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {19 \, a^{3} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {21 \, a^{3} \cos \left (f x + e\right )}{8 \, f} - \frac {a^{3} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {9 \, a^{3} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} - \frac {63 \, a^{3} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

23/16*a^3*x - 3/80*a^3*cos(5*f*x + 5*e)/f + 19/48*a^3*cos(3*f*x + 3*e)/f - 21/8*a^3*cos(f*x + e)/f - 1/192*a^3
*sin(6*f*x + 6*e)/f + 9/64*a^3*sin(4*f*x + 4*e)/f - 63/64*a^3*sin(2*f*x + 2*e)/f

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maple [A]  time = 0.34, size = 143, normalized size = 1.11 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )-\frac {3 a^{3} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+3 a^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{3} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(a+a*sin(f*x+e))^3,x)

[Out]

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

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maxima [A]  time = 0.43, size = 143, normalized size = 1.11 \[ -\frac {192 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{3} - 320 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} - 5 \, {\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} - 90 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3}}{960 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/960*(192*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^3 - 320*(cos(f*x + e)^3 - 3*cos(f*x + e
))*a^3 - 5*(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 - 90*(12*f*x
+ 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^3)/f

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mupad [B]  time = 10.33, size = 294, normalized size = 2.28 \[ \frac {23\,a^3\,x}{16}-\frac {\frac {23\,a^3\,\left (e+f\,x\right )}{16}+\frac {391\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24}+\frac {75\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{4}-\frac {75\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {391\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{24}-\frac {23\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}}{8}-\frac {a^3\,\left (345\,e+345\,f\,x-1088\right )}{240}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {69\,a^3\,\left (e+f\,x\right )}{8}-\frac {a^3\,\left (2070\,e+2070\,f\,x-6528\right )}{240}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (\frac {345\,a^3\,\left (e+f\,x\right )}{16}-\frac {a^3\,\left (5175\,e+5175\,f\,x-960\right )}{240}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {115\,a^3\,\left (e+f\,x\right )}{4}-\frac {a^3\,\left (6900\,e+6900\,f\,x-10880\right )}{240}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {345\,a^3\,\left (e+f\,x\right )}{16}-\frac {a^3\,\left (5175\,e+5175\,f\,x-15360\right )}{240}\right )+\frac {23\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^3*(a + a*sin(e + f*x))^3,x)

[Out]

(23*a^3*x)/16 - ((23*a^3*(e + f*x))/16 + (391*a^3*tan(e/2 + (f*x)/2)^3)/24 + (75*a^3*tan(e/2 + (f*x)/2)^5)/4 -
(75*a^3*tan(e/2 + (f*x)/2)^7)/4 - (391*a^3*tan(e/2 + (f*x)/2)^9)/24 - (23*a^3*tan(e/2 + (f*x)/2)^11)/8 - (a^3
*(345*e + 345*f*x - 1088))/240 + tan(e/2 + (f*x)/2)^2*((69*a^3*(e + f*x))/8 - (a^3*(2070*e + 2070*f*x - 6528))
/240) + tan(e/2 + (f*x)/2)^8*((345*a^3*(e + f*x))/16 - (a^3*(5175*e + 5175*f*x - 960))/240) + tan(e/2 + (f*x)/
2)^6*((115*a^3*(e + f*x))/4 - (a^3*(6900*e + 6900*f*x - 10880))/240) + tan(e/2 + (f*x)/2)^4*((345*a^3*(e + f*x
))/16 - (a^3*(5175*e + 5175*f*x - 15360))/240) + (23*a^3*tan(e/2 + (f*x)/2))/8)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^
6)

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sympy [A]  time = 6.82, size = 379, normalized size = 2.94 \[ \begin {cases} \frac {5 a^{3} x \sin ^{6}{\left (e + f x \right )}}{16} + \frac {15 a^{3} x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac {9 a^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {15 a^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac {9 a^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {5 a^{3} x \cos ^{6}{\left (e + f x \right )}}{16} + \frac {9 a^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {11 a^{3} \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{16 f} - \frac {3 a^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 a^{3} \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac {15 a^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {4 a^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {a^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 a^{3} \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} - \frac {9 a^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {8 a^{3} \cos ^{5}{\left (e + f x \right )}}{5 f} - \frac {2 a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\relax (e )} + a\right )^{3} \sin ^{3}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3*(a+a*sin(f*x+e))**3,x)

[Out]

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

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