3.3.11 \(\int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [211]

Optimal. Leaf size=119 \[ -\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{a^4 d}-\frac {b \cos ^2(c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {b^2 \left (a^2-b^2\right )}{a^5 d (b+a \cos (c+d x))}+\frac {2 b \left (a^2-2 b^2\right ) \log (b+a \cos (c+d x))}{a^5 d} \]

[Out]

-(a^2-3*b^2)*cos(d*x+c)/a^4/d-b*cos(d*x+c)^2/a^3/d+1/3*cos(d*x+c)^3/a^2/d+b^2*(a^2-b^2)/a^5/d/(b+a*cos(d*x+c))
+2*b*(a^2-2*b^2)*ln(b+a*cos(d*x+c))/a^5/d

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Rubi [A]
time = 0.16, antiderivative size = 119, 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, 2916, 12, 908} \begin {gather*} -\frac {b \cos ^2(c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {b^2 \left (a^2-b^2\right )}{a^5 d (a \cos (c+d x)+b)}+\frac {2 b \left (a^2-2 b^2\right ) \log (a \cos (c+d x)+b)}{a^5 d}-\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 3*b^2)*Cos[c + d*x])/(a^4*d)) - (b*Cos[c + d*x]^2)/(a^3*d) + Cos[c + d*x]^3/(3*a^2*d) + (b^2*(a^2 -
b^2))/(a^5*d*(b + a*Cos[c + d*x])) + (2*b*(a^2 - 2*b^2)*Log[b + a*Cos[c + d*x]])/(a^5*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 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x^2 \left (a^2-x^2\right )}{a^2 (-b+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \frac {x^2 \left (a^2-x^2\right )}{(-b+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac {\text {Subst}\left (\int \left (a^2 \left (1-\frac {3 b^2}{a^2}\right )-\frac {b^2 \left (-a^2+b^2\right )}{(b-x)^2}+\frac {2 b \left (-a^2+2 b^2\right )}{b-x}-2 b x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=-\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{a^4 d}-\frac {b \cos ^2(c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {b^2 \left (a^2-b^2\right )}{a^5 d (b+a \cos (c+d x))}+\frac {2 b \left (a^2-2 b^2\right ) \log (b+a \cos (c+d x))}{a^5 d}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 167, normalized size = 1.40 \begin {gather*} \frac {-9 a^4+60 a^2 b^2-24 b^4-8 \left (a^4-3 a^2 b^2\right ) \cos (2 (c+d x))-4 a^3 b \cos (3 (c+d x))+a^4 \cos (4 (c+d x))+48 a^2 b^2 \log (b+a \cos (c+d x))-96 b^4 \log (b+a \cos (c+d x))+24 a b \cos (c+d x) \left (-a^2+3 b^2+2 \left (a^2-2 b^2\right ) \log (b+a \cos (c+d x))\right )}{24 a^5 d (b+a \cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^4 + 60*a^2*b^2 - 24*b^4 - 8*(a^4 - 3*a^2*b^2)*Cos[2*(c + d*x)] - 4*a^3*b*Cos[3*(c + d*x)] + a^4*Cos[4*(c
+ d*x)] + 48*a^2*b^2*Log[b + a*Cos[c + d*x]] - 96*b^4*Log[b + a*Cos[c + d*x]] + 24*a*b*Cos[c + d*x]*(-a^2 + 3
*b^2 + 2*(a^2 - 2*b^2)*Log[b + a*Cos[c + d*x]]))/(24*a^5*d*(b + a*Cos[c + d*x]))

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Maple [A]
time = 0.13, size = 112, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{3}-b \left (\cos ^{2}\left (d x +c \right )\right ) a -a^{2} \cos \left (d x +c \right )+3 b^{2} \cos \left (d x +c \right )}{a^{4}}+\frac {b^{2} \left (a^{2}-b^{2}\right )}{a^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{5}}}{d}\)	\(112\)
default	\(\frac {\frac {\frac {\left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{3}-b \left (\cos ^{2}\left (d x +c \right )\right ) a -a^{2} \cos \left (d x +c \right )+3 b^{2} \cos \left (d x +c \right )}{a^{4}}+\frac {b^{2} \left (a^{2}-b^{2}\right )}{a^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{5}}}{d}\)	\(112\)
risch	\(\frac {8 i b^{3} c}{a^{5} d}+\frac {4 i b^{3} x}{a^{5}}-\frac {b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{3} d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 a^{4} d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 a^{4} d}-\frac {b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{3} d}-\frac {4 i b c}{a^{3} d}-\frac {2 i b x}{a^{3}}-\frac {2 b^{2} \left (-a^{2}+b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{a^{5} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{5} d}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,a^{2}}\)	\(301\)
norman	\(\frac {-\frac {\left (4 a^{4}-8 b \,a^{3}+12 b^{2} a^{2}+16 b^{3} a -24 b^{4}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a^{4} d b}+\frac {\left (4 a^{3}+4 b \,a^{2}+8 b^{2} a -24 b^{3}\right ) \left (a +b \right )}{6 a^{4} b d}-\frac {\left (4 a^{4}-12 b \,a^{3}+4 b^{2} a^{2}+8 b^{3} a -24 b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{4} b d}+\frac {\left (4 a^{4}+12 b \,a^{3}+4 b^{2} a^{2}-8 b^{3} a -24 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{4} d b}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{5} d}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}{a^{5} d}\)	\(336\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/a^4*(1/3*cos(d*x+c)^3*a^2-b*cos(d*x+c)^2*a-a^2*cos(d*x+c)+3*b^2*cos(d*x+c))+b^2*(a^2-b^2)/a^5/(b+a*cos(
d*x+c))+2*b/a^5*(a^2-2*b^2)*ln(b+a*cos(d*x+c)))

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Maxima [A]
time = 0.26, size = 112, normalized size = 0.94 \begin {gather*} \frac {\frac {3 \, {\left (a^{2} b^{2} - b^{4}\right )}}{a^{6} \cos \left (d x + c\right ) + a^{5} b} + \frac {a^{2} \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )}{a^{4}} + \frac {6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{5}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(a^2*b^2 - b^4)/(a^6*cos(d*x + c) + a^5*b) + (a^2*cos(d*x + c)^3 - 3*a*b*cos(d*x + c)^2 - 3*(a^2 - 3*b^
2)*cos(d*x + c))/a^4 + 6*(a^2*b - 2*b^3)*log(a*cos(d*x + c) + b)/a^5)/d

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Fricas [A]
time = 2.70, size = 150, normalized size = 1.26 \begin {gather*} \frac {2 \, a^{4} \cos \left (d x + c\right )^{4} - 4 \, a^{3} b \cos \left (d x + c\right )^{3} + 9 \, a^{2} b^{2} - 6 \, b^{4} - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right ) + 12 \, {\left (a^{2} b^{2} - 2 \, b^{4} + {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{6 \, {\left (a^{6} d \cos \left (d x + c\right ) + a^{5} b d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^4*cos(d*x + c)^4 - 4*a^3*b*cos(d*x + c)^3 + 9*a^2*b^2 - 6*b^4 - 6*(a^4 - 2*a^2*b^2)*cos(d*x + c)^2 -
3*(a^3*b - 6*a*b^3)*cos(d*x + c) + 12*(a^2*b^2 - 2*b^4 + (a^3*b - 2*a*b^3)*cos(d*x + c))*log(a*cos(d*x + c) +
b))/(a^6*d*cos(d*x + c) + a^5*b*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+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 139, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{5} d} + \frac {a^{2} b^{2} - b^{4}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{5} d} + \frac {a^{4} d^{5} \cos \left (d x + c\right )^{3} - 3 \, a^{3} b d^{5} \cos \left (d x + c\right )^{2} - 3 \, a^{4} d^{5} \cos \left (d x + c\right ) + 9 \, a^{2} b^{2} d^{5} \cos \left (d x + c\right )}{3 \, a^{6} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^2*b - 2*b^3)*log(abs(-a*cos(d*x + c) - b))/(a^5*d) + (a^2*b^2 - b^4)/((a*cos(d*x + c) + b)*a^5*d) + 1/3*(
a^4*d^5*cos(d*x + c)^3 - 3*a^3*b*d^5*cos(d*x + c)^2 - 3*a^4*d^5*cos(d*x + c) + 9*a^2*b^2*d^5*cos(d*x + c))/(a^
6*d^6)

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Mupad [B]
time = 0.09, size = 113, normalized size = 0.95 \begin {gather*} -\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a^2}-\frac {3\,b^2}{a^4}\right )-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}+\frac {b\,{\cos \left (c+d\,x\right )}^2}{a^3}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (2\,a^2\,b-4\,b^3\right )}{a^5}+\frac {b^4-a^2\,b^2}{a\,\left (\cos \left (c+d\,x\right )\,a^5+b\,a^4\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(c + d*x)*(1/a^2 - (3*b^2)/a^4) - cos(c + d*x)^3/(3*a^2) + (b*cos(c + d*x)^2)/a^3 - (log(b + a*cos(c + d*
x))*(2*a^2*b - 4*b^3))/a^5 + (b^4 - a^2*b^2)/(a*(a^5*cos(c + d*x) + a^4*b)))/d

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