\(\int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx\) [224]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {\cos (c+d x)}{a^3 d}-\frac {b^3}{2 a^4 d (b+a \cos (c+d x))^2}+\frac {3 b^2}{a^4 d (b+a \cos (c+d x))}+\frac {3 b \log (b+a \cos (c+d x))}{a^4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2912, 12, 45} \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {b^3}{2 a^4 d (a \cos (c+d x)+b)^2}+\frac {3 b^2}{a^4 d (a \cos (c+d x)+b)}+\frac {3 b \log (a \cos (c+d x)+b)}{a^4 d}-\frac {\cos (c+d x)}{a^3 d} \]

[In]

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

[Out]

-(Cos[c + d*x]/(a^3*d)) - b^3/(2*a^4*d*(b + a*Cos[c + d*x])^2) + (3*b^2)/(a^4*d*(b + a*Cos[c + d*x])) + (3*b*L
og[b + a*Cos[c + d*x]])/(a^4*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

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

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*} \text {integral}& = -\int \frac {\cos ^3(c+d x) \sin (c+d x)}{(-b-a \cos (c+d x))^3} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{a^3 (-b+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{(-b+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (1-\frac {b^3}{(b-x)^3}+\frac {3 b^2}{(b-x)^2}-\frac {3 b}{b-x}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = -\frac {\cos (c+d x)}{a^3 d}-\frac {b^3}{2 a^4 d (b+a \cos (c+d x))^2}+\frac {3 b^2}{a^4 d (b+a \cos (c+d x))}+\frac {3 b \log (b+a \cos (c+d x))}{a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {-2 a^3 \cos ^3(c+d x)+2 a^2 b \cos ^2(c+d x) (-2+3 \log (b+a \cos (c+d x)))+4 a b^2 \cos (c+d x) (1+3 \log (b+a \cos (c+d x)))+b^3 (5+6 \log (b+a \cos (c+d x)))}{2 a^4 d (b+a \cos (c+d x))^2} \]

[In]

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

[Out]

(-2*a^3*Cos[c + d*x]^3 + 2*a^2*b*Cos[c + d*x]^2*(-2 + 3*Log[b + a*Cos[c + d*x]]) + 4*a*b^2*Cos[c + d*x]*(1 + 3
*Log[b + a*Cos[c + d*x]]) + b^3*(5 + 6*Log[b + a*Cos[c + d*x]]))/(2*a^4*d*(b + a*Cos[c + d*x])^2)

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{3} \sec \left (d x +c \right )}-\frac {3 b \ln \left (\sec \left (d x +c \right )\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \sec \left (d x +c \right )\right )^{2}}+\frac {3 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \sec \left (d x +c \right )\right )}}{d}\)	\(85\)
default	\(\frac {-\frac {1}{a^{3} \sec \left (d x +c \right )}-\frac {3 b \ln \left (\sec \left (d x +c \right )\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \sec \left (d x +c \right )\right )^{2}}+\frac {3 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \sec \left (d x +c \right )\right )}}{d}\)	\(85\)
risch	\(-\frac {3 i x b}{a^{4}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}-\frac {6 i b c}{a^{4} d}+\frac {2 b^{2} \left (3 a \,{\mathrm e}^{3 i \left (d x +c \right )}+5 b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} a \right )}{a^{4} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2} d}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{4} d}\)	\(166\)
norman	\(\frac {\frac {-2 a^{4}+10 a^{2} b^{2}-6 b^{4}}{d \,a^{3} \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3}-6 a^{2} b +12 a \,b^{2}-6 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,a^{3} \left (a -b \right )}-\frac {\left (-4 a^{4}+8 a^{3} b -18 a \,b^{3}+12 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \,a^{3} \left (a^{2}-2 a b +b^{2}\right )}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {3 b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a^{4} d}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{a^{4} d}\)	\(265\)
parallelrisch	\(\frac {6 b \left (a -b \right )^{2} \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right ) \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )-6 b \left (a -b \right )^{2} \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-3 a \left (\left (\frac {2}{3} a^{4}-4 a^{2} b^{2}+3 a \,b^{3}\right ) \cos \left (2 d x +2 c \right )+\frac {a^{2} \left (a -b \right )^{2} \cos \left (3 d x +3 c \right )}{3}+\left (a^{4}+\frac {2}{3} a^{3} b -7 a^{2} b^{2}+4 b^{4}\right ) \cos \left (d x +c \right )+\frac {2 a^{4}}{3}-\frac {8 a^{2} b^{2}}{3}-3 a \,b^{3}+4 b^{4}\right )}{2 a^{4} d \left (a -b \right )^{2} \left (\cos \left (2 d x +2 c \right ) a^{2}+4 \cos \left (d x +c \right ) a b +a^{2}+2 b^{2}\right )}\)	\(278\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right )^{3} + 4 \, a^{2} b \cos \left (d x + c\right )^{2} - 4 \, a b^{2} \cos \left (d x + c\right ) - 5 \, b^{3} - 6 \, {\left (a^{2} b \cos \left (d x + c\right )^{2} + 2 \, a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{2 \, {\left (a^{6} d \cos \left (d x + c\right )^{2} + 2 \, a^{5} b d \cos \left (d x + c\right ) + a^{4} b^{2} d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sin {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {6 \, a b^{2} \cos \left (d x + c\right ) + 5 \, b^{3}}{a^{6} \cos \left (d x + c\right )^{2} + 2 \, a^{5} b \cos \left (d x + c\right ) + a^{4} b^{2}} - \frac {2 \, \cos \left (d x + c\right )}{a^{3}} + \frac {6 \, b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {\cos \left (d x + c\right )}{a^{3} d} + \frac {3 \, b \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{4} d} + \frac {6 \, a b^{2} \cos \left (d x + c\right ) + 5 \, b^{3}}{2 \, {\left (a \cos \left (d x + c\right ) + b\right )}^{2} a^{4} d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {3\,b^2\,\cos \left (c+d\,x\right )+\frac {5\,b^3}{2\,a}}{d\,\left (a^5\,{\cos \left (c+d\,x\right )}^2+2\,a^4\,b\,\cos \left (c+d\,x\right )+a^3\,b^2\right )}-\frac {\cos \left (c+d\,x\right )}{a^3\,d}+\frac {3\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{a^4\,d} \]

[In]

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

[Out]

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