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

Optimal. Leaf size=37 \[ \frac {\sin ^2(c+d x)}{2 a d}+\frac {\cos ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3872, 2835, 2564, 30, 2565} \[ \frac {\sin ^2(c+d x)}{2 a d}+\frac {\cos ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 32, normalized size = 0.86 \[ \frac {2 \sin ^4\left (\frac {1}{2} (c+d x)\right ) (2 \cos (c+d x)+1)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 + 2*Cos[c + d*x])*Sin[(c + d*x)/2]^4)/(3*a*d)

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fricas [A]  time = 0.67, size = 29, normalized size = 0.78 \[ \frac {2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2}}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3 - 3*cos(d*x + c)^2)/(a*d)

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giac [A]  time = 1.00, size = 32, normalized size = 0.86 \[ \frac {\frac {2 \, \cos \left (d x + c\right )^{3}}{d} - \frac {3 \, \cos \left (d x + c\right )^{2}}{d}}{6 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3/d - 3*cos(d*x + c)^2/d)/a

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maple [A]  time = 0.43, size = 30, normalized size = 0.81 \[ -\frac {\frac {1}{2 \sec \left (d x +c \right )^{2}}-\frac {1}{3 \sec \left (d x +c \right )^{3}}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 29, normalized size = 0.78 \[ \frac {2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2}}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3 - 3*cos(d*x + c)^2)/(a*d)

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mupad [B]  time = 0.88, size = 26, normalized size = 0.70 \[ \frac {{\cos \left (c+d\,x\right )}^2\,\left (2\,\cos \left (c+d\,x\right )-3\right )}{6\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(c + d*x)^2*(2*cos(c + d*x) - 3))/(6*a*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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