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

3.3.11.1 Optimal result
3.3.11.2 Mathematica [A] (verified)
3.3.11.3 Rubi [A] (verified)
3.3.11.4 Maple [A] (verified)
3.3.11.5 Fricas [A] (verification not implemented)
3.3.11.6 Sympy [F(-1)]
3.3.11.7 Maxima [A] (verification not implemented)
3.3.11.8 Giac [A] (verification not implemented)
3.3.11.9 Mupad [B] (verification not implemented)

3.3.11.1 Optimal result

Integrand size = 21, antiderivative size = 119 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{a^4 d}-\frac {b \cos ^2(c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {b^2 \left (a^2-b^2\right )}{a^5 d (b+a \cos (c+d x))}+\frac {2 b \left (a^2-2 b^2\right ) \log (b+a \cos (c+d x))}{a^5 d} \]

output
-(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
 
3.3.11.2 Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.40 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-9 a^4+60 a^2 b^2-24 b^4-8 \left (a^4-3 a^2 b^2\right ) \cos (2 (c+d x))-4 a^3 b \cos (3 (c+d x))+a^4 \cos (4 (c+d x))+48 a^2 b^2 \log (b+a \cos (c+d x))-96 b^4 \log (b+a \cos (c+d x))+24 a b \cos (c+d x) \left (-a^2+3 b^2+2 \left (a^2-2 b^2\right ) \log (b+a \cos (c+d x))\right )}{24 a^5 d (b+a \cos (c+d x))} \]

input
Integrate[Sin[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
 
output
(-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]))
 
3.3.11.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 3042, 25, 3316, 27, 522, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^3}{\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^2(c+d x)}{(-a \cos (c+d x)-b)^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^3}{\left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}{\left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a^3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {a^2 \cos ^2(c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2}d(a \cos (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 522

\(\displaystyle -\frac {\int \left (-\cos ^2(c+d x) a^2+\left (1-\frac {3 b^2}{a^2}\right ) a^2+2 b \cos (c+d x) a+\frac {2 b \left (2 b^2-a^2\right )}{b+a \cos (c+d x)}-\frac {b^2 \left (b^2-a^2\right )}{(b+a \cos (c+d x))^2}\right )d(a \cos (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {1}{3} a^3 \cos ^3(c+d x)+a \left (a^2-3 b^2\right ) \cos (c+d x)-\frac {b^2 \left (a^2-b^2\right )}{a \cos (c+d x)+b}-2 b \left (a^2-2 b^2\right ) \log (a \cos (c+d x)+b)+a^2 b \cos ^2(c+d x)}{a^5 d}\)

input
Int[Sin[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
 
output
-((a*(a^2 - 3*b^2)*Cos[c + d*x] + a^2*b*Cos[c + d*x]^2 - (a^3*Cos[c + d*x] 
^3)/3 - (b^2*(a^2 - b^2))/(b + a*Cos[c + d*x]) - 2*b*(a^2 - 2*b^2)*Log[b + 
 a*Cos[c + d*x]])/(a^5*d))
 

3.3.11.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 522
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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 4360
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 
3.3.11.4 Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.94

method result size
derivativedivides \(\frac {\frac {\frac {\cos \left (d x +c \right )^{3} a^{2}}{3}-a b \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right ) a^{2}+3 \cos \left (d x +c \right ) b^{2}}{a^{4}}+\frac {b^{2} \left (a^{2}-b^{2}\right )}{a^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{5}}}{d}\) \(112\)
default \(\frac {\frac {\frac {\cos \left (d x +c \right )^{3} a^{2}}{3}-a b \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right ) a^{2}+3 \cos \left (d x +c \right ) b^{2}}{a^{4}}+\frac {b^{2} \left (a^{2}-b^{2}\right )}{a^{5} \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{5}}}{d}\) \(112\)
parallelrisch \(\frac {48 b^{2} \left (a^{2}-2 b^{2}\right ) \left (b +a \cos \left (d x +c \right )\right ) \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )-48 b^{2} \left (a^{2}-2 b^{2}\right ) \left (b +a \cos \left (d x +c \right )\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-8 a^{4} b +24 a^{2} b^{3}\right ) \cos \left (2 d x +2 c \right )-4 \cos \left (3 d x +3 c \right ) a^{3} b^{2}+b \cos \left (4 d x +4 c \right ) a^{4}+\left (-16 a^{5}-44 a^{3} b^{2}+96 a \,b^{4}\right ) \cos \left (d x +c \right )-25 a^{4} b +40 a^{2} b^{3}}{24 d \,a^{5} b \left (b +a \cos \left (d x +c \right )\right )}\) \(204\)
risch \(-\frac {2 i b x}{a^{3}}+\frac {4 i b^{3} x}{a^{5}}-\frac {b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{3} d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 a^{4} d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 a^{4} d}-\frac {b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{3} d}+\frac {8 i b^{3} c}{a^{5} d}-\frac {4 i b c}{a^{3} d}-\frac {2 b^{2} \left (-a^{2}+b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{a^{5} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d}-\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{5} d}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,a^{2}}\) \(301\)
norman \(\frac {-\frac {\left (4 a^{4}-8 a^{3} b +12 a^{2} b^{2}+16 a \,b^{3}-24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{6 d \,a^{4} b}+\frac {\left (4 a^{3}+4 a^{2} b +8 a \,b^{2}-24 b^{3}\right ) \left (a +b \right )}{6 a^{4} b d}-\frac {\left (4 a^{4}-12 a^{3} b +4 a^{2} b^{2}+8 a \,b^{3}-24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 a^{4} b d}+\frac {\left (4 a^{4}+12 a^{3} b +4 a^{2} b^{2}-8 a \,b^{3}-24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d \,a^{4} b}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{5}}+\frac {2 b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \,a^{5}}\) \(336\)

input
int(sin(d*x+c)^3/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
1/d*(1/a^4*(1/3*cos(d*x+c)^3*a^2-a*b*cos(d*x+c)^2-cos(d*x+c)*a^2+3*cos(d*x 
+c)*b^2)+b^2*(a^2-b^2)/a^5/(b+a*cos(d*x+c))+2/a^5*b*(a^2-2*b^2)*ln(b+a*cos 
(d*x+c)))
 
3.3.11.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, a^{4} \cos \left (d x + c\right )^{4} - 4 \, a^{3} b \cos \left (d x + c\right )^{3} + 9 \, a^{2} b^{2} - 6 \, b^{4} - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right ) + 12 \, {\left (a^{2} b^{2} - 2 \, b^{4} + {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{6 \, {\left (a^{6} d \cos \left (d x + c\right ) + a^{5} b d\right )}} \]

input
integrate(sin(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
 
output
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)
 
3.3.11.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]

input
integrate(sin(d*x+c)**3/(a+b*sec(d*x+c))**2,x)
 
output
Timed out
 
3.3.11.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (a^{2} b^{2} - b^{4}\right )}}{a^{6} \cos \left (d x + c\right ) + a^{5} b} + \frac {a^{2} \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )}{a^{4}} + \frac {6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{5}}}{3 \, d} \]

input
integrate(sin(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
 
output
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
 
3.3.11.8 Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{5} d} + \frac {a^{2} b^{2} - b^{4}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{5} d} + \frac {a^{4} d^{5} \cos \left (d x + c\right )^{3} - 3 \, a^{3} b d^{5} \cos \left (d x + c\right )^{2} - 3 \, a^{4} d^{5} \cos \left (d x + c\right ) + 9 \, a^{2} b^{2} d^{5} \cos \left (d x + c\right )}{3 \, a^{6} d^{6}} \]

input
integrate(sin(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")
 
output
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*c 
os(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)
 
3.3.11.9 Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a^2}-\frac {3\,b^2}{a^4}\right )-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}+\frac {b\,{\cos \left (c+d\,x\right )}^2}{a^3}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (2\,a^2\,b-4\,b^3\right )}{a^5}+\frac {b^4-a^2\,b^2}{a\,\left (\cos \left (c+d\,x\right )\,a^5+b\,a^4\right )}}{d} \]

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