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

Optimal. Leaf size=112 \[ \frac {\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac {\left (a^2+b^2\right ) \cos (e+f x)}{f}-\frac {a b \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac {3 a b \sin (e+f x) \cos (e+f x)}{4 f}+\frac {3 a b x}{4}-\frac {b^2 \cos ^5(e+f x)}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2789, 2635, 8, 3013, 373} \[ \frac {\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac {\left (a^2+b^2\right ) \cos (e+f x)}{f}-\frac {a b \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac {3 a b \sin (e+f x) \cos (e+f x)}{4 f}+\frac {3 a b x}{4}-\frac {b^2 \cos ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 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 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 91, normalized size = 0.81 \[ \frac {-30 \left (6 a^2+5 b^2\right ) \cos (e+f x)+5 \left (4 a^2+5 b^2\right ) \cos (3 (e+f x))-3 b (b \cos (5 (e+f x))-5 a (12 (e+f x)-8 \sin (2 (e+f x))+\sin (4 (e+f x))))}{240 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*(6*a^2 + 5*b^2)*Cos[e + f*x] + 5*(4*a^2 + 5*b^2)*Cos[3*(e + f*x)] - 3*b*(b*Cos[5*(e + f*x)] - 5*a*(12*(e
+ f*x) - 8*Sin[2*(e + f*x)] + Sin[4*(e + f*x)])))/(240*f)

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fricas [A]  time = 0.49, size = 90, normalized size = 0.80 \[ -\frac {12 \, b^{2} \cos \left (f x + e\right )^{5} - 45 \, a b f x - 20 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 60 \, {\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right ) - 15 \, {\left (2 \, a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{60 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*b^2*cos(f*x + e)^5 - 45*a*b*f*x - 20*(a^2 + 2*b^2)*cos(f*x + e)^3 + 60*(a^2 + b^2)*cos(f*x + e) - 15
*(2*a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e))*sin(f*x + e))/f

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giac [A]  time = 0.22, size = 128, normalized size = 1.14 \[ \frac {3}{4} \, a b x - \frac {b^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {a b \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac {a b \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} + \frac {{\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*a*b*x - 1/80*b^2*cos(5*f*x + 5*e)/f + 1/16*a*b*sin(4*f*x + 4*e)/f - 1/2*a*b*sin(2*f*x + 2*e)/f + 1/48*(4*a
^2 + 5*b^2)*cos(3*f*x + 3*e)/f - 1/8*(2*a^2 + 3*b^2)*cos(f*x + e)/f - 1/4*(2*a^2 + b^2)*cos(f*x + e)/f

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maple [A]  time = 0.28, size = 95, normalized size = 0.85 \[ \frac {-\frac {b^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/5*b^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*a*b*(-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 = 0.72, size = 94, normalized size = 0.84 \[ \frac {80 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} + 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a b - 16 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{2}}{240 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(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
*b - 16*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*b^2)/f

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mupad [B]  time = 10.37, size = 157, normalized size = 1.40 \[ \frac {3\,a\,b\,x}{4}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {20\,a^2}{3}+\frac {16\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {28\,a^2}{3}+\frac {32\,b^2}{3}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\frac {4\,a^2}{3}+\frac {16\,b^2}{15}+7\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-7\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7-\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*b*x)/4 - (tan(e/2 + (f*x)/2)^2*((20*a^2)/3 + (16*b^2)/3) + tan(e/2 + (f*x)/2)^4*((28*a^2)/3 + (32*b^2)/3)
+ 4*a^2*tan(e/2 + (f*x)/2)^6 + (4*a^2)/3 + (16*b^2)/15 + 7*a*b*tan(e/2 + (f*x)/2)^3 - 7*a*b*tan(e/2 + (f*x)/2
)^7 - (3*a*b*tan(e/2 + (f*x)/2)^9)/2 + (3*a*b*tan(e/2 + (f*x)/2))/2)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^5)

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sympy [A]  time = 3.10, size = 221, normalized size = 1.97 \[ \begin {cases} - \frac {a^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {3 a b x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 a b x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac {3 a b x \cos ^{4}{\left (e + f x \right )}}{4} - \frac {5 a b \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} - \frac {3 a b \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac {b^{2} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 b^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 b^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\relax (e )}\right )^{2} \sin ^{3}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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