3.194 \(\int \frac {\sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 \cos (c+d x)}{a d}+\frac {\sin ^2(c+d x) \cos (c+d x)}{d (a \sin (c+d x)+a)}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {3 x}{2 a} \]

[Out]

3/2*x/a+2*cos(d*x+c)/a/d-3/2*cos(d*x+c)*sin(d*x+c)/a/d+cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2767, 2734} \[ \frac {2 \cos (c+d x)}{a d}+\frac {\sin ^2(c+d x) \cos (c+d x)}{d (a \sin (c+d x)+a)}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

(3*x)/(2*a) + (2*Cos[c + d*x])/(a*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) + (Cos[c + d*x]*Sin[c + d*x]^2)/(
d*(a + a*Sin[c + d*x]))

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2767

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(a + b*Sin[e + f*x])), x] - Dist[d/(a*b), Int[(c +
d*Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*n)*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && (IntegerQ[2
*n] || EqQ[c, 0])

Rubi steps

\begin {align*} \int \frac {\sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\cos (c+d x) \sin ^2(c+d x)}{d (a+a \sin (c+d x))}-\frac {\int \sin (c+d x) (2 a-3 a \sin (c+d x)) \, dx}{a^2}\\ &=\frac {3 x}{2 a}+\frac {2 \cos (c+d x)}{a d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 117, normalized size = 1.56 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right ) (-\sin (2 (c+d x))+4 \cos (c+d x)+6 c+6 d x-8)+\cos \left (\frac {1}{2} (c+d x)\right ) (-\sin (2 (c+d x))+4 \cos (c+d x)+6 c+6 d x)\right )}{4 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Sin[(c + d*x)/2]*(-8 + 6*c + 6*d*x + 4*Cos[c + d*x] - Sin[2*(c + d*x)]
) + Cos[(c + d*x)/2]*(6*c + 6*d*x + 4*Cos[c + d*x] - Sin[2*(c + d*x)])))/(4*a*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.47, size = 92, normalized size = 1.23 \[ \frac {\cos \left (d x + c\right )^{3} + 3 \, d x + 3 \, {\left (d x + 1\right )} \cos \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )^{2} + {\left (3 \, d x - \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 2}{2 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(cos(d*x + c)^3 + 3*d*x + 3*(d*x + 1)*cos(d*x + c) + 2*cos(d*x + c)^2 + (3*d*x - cos(d*x + c)^2 + cos(d*x
+ c) - 2)*sin(d*x + c) + 2)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

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giac [A]  time = 0.40, size = 91, normalized size = 1.21 \[ \frac {\frac {3 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a} + \frac {4}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/2*(3*(d*x + c)/a + 2*(tan(1/2*d*x + 1/2*c)^3 + 2*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 2)/((tan(1/
2*d*x + 1/2*c)^2 + 1)^2*a) + 4/(a*(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [B]  time = 0.06, size = 163, normalized size = 2.17 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2-1/
a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2+3/a/d*arctan(tan(1/2*d*x+1/
2*c))+2/a/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.67, size = 212, normalized size = 2.83 \[ \frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 4}{a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

((sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4)/(a + a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 2*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(
d*x + c)^5/(cos(d*x + c) + 1)^5) + 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 3.29, size = 92, normalized size = 1.23 \[ \frac {3\,x}{2\,a}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4}{a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3/(a + a*sin(c + d*x)),x)

[Out]

(3*x)/(2*a) + (tan(c/2 + (d*x)/2) + 5*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^3 + 3*tan(c/2 + (d*x)/2)^4 +
4)/(a*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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sympy [A]  time = 6.52, size = 1127, normalized size = 15.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 +
d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 3*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan
(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*t
an(c/2 + d*x/2) + 2*a*d) + 6*d*x*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 +
4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 6*d*x*tan(c/2 + d*x/
2)**2/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x
/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 3*d*x*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 +
d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 3*d*x/(
2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2
+ 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 6*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)*
*4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 6*tan(c/2 + d*x
/2)**3/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*
x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 10*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2
+ d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 2*tan(
c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/
2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d) + 8/(2*a*d*tan(c/2 + d*x/2)**5 + 2*a*d*tan(c/2 + d*x/2)**4 + 4
*a*d*tan(c/2 + d*x/2)**3 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d*tan(c/2 + d*x/2) + 2*a*d), Ne(d, 0)), (x*sin(c)**
3/(a*sin(c) + a), True))

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