3.1 \(\int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=102 \[ -\frac{a^2 \cos ^5(e+f x)}{5 f}+\frac{a^2 \cos ^3(e+f x)}{f}-\frac{2 a^2 \cos (e+f x)}{f}-\frac{a^2 \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac{3 a^2 \sin (e+f x) \cos (e+f x)}{4 f}+\frac{3 a^2 x}{4} \]

[Out]

(3*a^2*x)/4 - (2*a^2*Cos[e + f*x])/f + (a^2*Cos[e + f*x]^3)/f - (a^2*Cos[e + f*x]^5)/(5*f) - (3*a^2*Cos[e + f*
x]*Sin[e + f*x])/(4*f) - (a^2*Cos[e + f*x]*Sin[e + f*x]^3)/(2*f)

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Rubi [A]  time = 0.101512, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 2633, 2635, 8} \[ -\frac{a^2 \cos ^5(e+f x)}{5 f}+\frac{a^2 \cos ^3(e+f x)}{f}-\frac{2 a^2 \cos (e+f x)}{f}-\frac{a^2 \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac{3 a^2 \sin (e+f x) \cos (e+f x)}{4 f}+\frac{3 a^2 x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*x)/4 - (2*a^2*Cos[e + f*x])/f + (a^2*Cos[e + f*x]^3)/f - (a^2*Cos[e + f*x]^5)/(5*f) - (3*a^2*Cos[e + f*
x]*Sin[e + f*x])/(4*f) - (a^2*Cos[e + f*x]*Sin[e + f*x]^3)/(2*f)

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]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.428919, size = 105, normalized size = 1.03 \[ -\frac{a^2 \cos (e+f x) \left (30 \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\left (4 \sin ^4(e+f x)+10 \sin ^3(e+f x)+12 \sin ^2(e+f x)+15 \sin (e+f x)+24\right ) \sqrt{\cos ^2(e+f x)}\right )}{20 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cos[e + f*x]*(30*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2]*(24 + 15*Sin[e + f*x] + 1
2*Sin[e + f*x]^2 + 10*Sin[e + f*x]^3 + 4*Sin[e + f*x]^4)))/(20*f*Sqrt[Cos[e + f*x]^2])

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Maple [A]  time = 0.031, size = 96, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{2}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+2\,{a}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{{a}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x)

[Out]

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

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Maxima [A]  time = 1.72938, size = 128, normalized size = 1.25 \begin{align*} -\frac{16 \,{\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^{2} - 80 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} - 15 \,{\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^{2}}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(16*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2 - 80*(cos(f*x + e)^3 - 3*cos(f*x + e))
*a^2 - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2)/f

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Fricas [A]  time = 1.8163, size = 205, normalized size = 2.01 \begin{align*} -\frac{4 \, a^{2} \cos \left (f x + e\right )^{5} - 20 \, a^{2} \cos \left (f x + e\right )^{3} - 15 \, a^{2} f x + 40 \, a^{2} \cos \left (f x + e\right ) - 5 \,{\left (2 \, a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{20 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(4*a^2*cos(f*x + e)^5 - 20*a^2*cos(f*x + e)^3 - 15*a^2*f*x + 40*a^2*cos(f*x + e) - 5*(2*a^2*cos(f*x + e)
^3 - 5*a^2*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 2.24003, size = 221, normalized size = 2.17 \begin{align*} \begin{cases} \frac{3 a^{2} x \sin ^{4}{\left (e + f x \right )}}{4} + \frac{3 a^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac{3 a^{2} x \cos ^{4}{\left (e + f x \right )}}{4} - \frac{a^{2} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{5 a^{2} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{4 f} - \frac{4 a^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a^{2} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac{8 a^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{2} \sin ^{3}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3*(a+a*sin(f*x+e))**2,x)

[Out]

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

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Giac [A]  time = 2.08549, size = 127, normalized size = 1.25 \begin{align*} \frac{3}{4} \, a^{2} x - \frac{a^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{3 \, a^{2} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac{11 \, a^{2} \cos \left (f x + e\right )}{8 \, f} + \frac{a^{2} \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac{a^{2} \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

3/4*a^2*x - 1/80*a^2*cos(5*f*x + 5*e)/f + 3/16*a^2*cos(3*f*x + 3*e)/f - 11/8*a^2*cos(f*x + e)/f + 1/16*a^2*sin
(4*f*x + 4*e)/f - 1/2*a^2*sin(2*f*x + 2*e)/f