3.1.78 \(\int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [78]

Optimal. Leaf size=66 \[ \frac {2 \cos (c+d x)}{a^2 d}-\frac {\cos ^2(c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {2 \log (1+\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 78} \begin {gather*} \frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {\cos ^2(c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {2 \log (\cos (c+d x)+1)}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2915

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {(-a-x) x^2}{a^2 (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \frac {(-a-x) x^2}{-a+x} \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac {\text {Subst}\left (\int \left (-2 a^2+\frac {2 a^3}{a-x}-2 a x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac {2 \cos (c+d x)}{a^2 d}-\frac {\cos ^2(c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {2 \log (1+\cos (c+d x))}{a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 51, normalized size = 0.77 \begin {gather*} \frac {-22+27 \cos (c+d x)-6 \cos (2 (c+d x))+\cos (3 (c+d x))-48 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{12 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

(-22 + 27*Cos[c + d*x] - 6*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] - 48*Log[Cos[(c + d*x)/2]])/(12*a^2*d)

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Maple [A]
time = 0.12, size = 58, normalized size = 0.88

method	result	size
derivativedivides	\(-\frac {-\frac {1}{3 \sec \left (d x +c \right )^{3}}+\frac {1}{\sec \left (d x +c \right )^{2}}-\frac {2}{\sec \left (d x +c \right )}-2 \ln \left (\sec \left (d x +c \right )\right )+2 \ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{2}}\)	\(58\)
default	\(-\frac {-\frac {1}{3 \sec \left (d x +c \right )^{3}}+\frac {1}{\sec \left (d x +c \right )^{2}}-\frac {2}{\sec \left (d x +c \right )}-2 \ln \left (\sec \left (d x +c \right )\right )+2 \ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{2}}\)	\(58\)
norman	\(\frac {\frac {10 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {14}{3 a d}+\frac {12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a}+\frac {2 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(90\)
risch	\(\frac {2 i x}{a^{2}}+\frac {9 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {9 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}+\frac {4 i c}{a^{2} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,a^{2}}-\frac {\cos \left (2 d x +2 c \right )}{2 d \,a^{2}}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 51, normalized size = 0.77 \begin {gather*} \frac {\frac {\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right )}{a^{2}} - \frac {6 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.60, size = 48, normalized size = 0.73 \begin {gather*} \frac {\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) - 6 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{3 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.50, size = 75, normalized size = 1.14 \begin {gather*} -\frac {2 \, \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a^{2} d} + \frac {a^{4} d^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{4} d^{2} \cos \left (d x + c\right )^{2} + 6 \, a^{4} d^{2} \cos \left (d x + c\right )}{3 \, a^{6} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-2*log(abs(-cos(d*x + c) - 1))/(a^2*d) + 1/3*(a^4*d^2*cos(d*x + c)^3 - 3*a^4*d^2*cos(d*x + c)^2 + 6*a^4*d^2*co
s(d*x + c))/(a^6*d^3)

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Mupad [B]
time = 0.06, size = 56, normalized size = 0.85 \begin {gather*} -\frac {\frac {2\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^2}-\frac {2\,\cos \left (c+d\,x\right )}{a^2}+\frac {{\cos \left (c+d\,x\right )}^2}{a^2}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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