3.1.26 \(\int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx\) [26]

3.1.26.1 Optimal result
3.1.26.2 Mathematica [A] (verified)
3.1.26.3 Rubi [A] (verified)
3.1.26.4 Maple [C] (verified)
3.1.26.5 Fricas [A] (verification not implemented)
3.1.26.6 Sympy [F]
3.1.26.7 Maxima [A] (verification not implemented)
3.1.26.8 Giac [F(-1)]
3.1.26.9 Mupad [B] (verification not implemented)

3.1.26.1 Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {7 a^3 \log (1-\sin (c+d x))}{d}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {2 a^4}{d (a-a \sin (c+d x))} \]

output
7*a^3*ln(1-sin(d*x+c))/d+5*a^3*sin(d*x+c)/d+3/2*a^3*sin(d*x+c)^2/d+1/3*a^3 
*sin(d*x+c)^3/d+2*a^4/d/(a-a*sin(d*x+c))
 
3.1.26.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {a^3 \left (42 \log (1-\sin (c+d x))+\frac {12}{1-\sin (c+d x)}+30 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \]

input
Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
 
output
(a^3*(42*Log[1 - Sin[c + d*x]] + 12/(1 - Sin[c + d*x]) + 30*Sin[c + d*x] + 
 9*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3))/(6*d)
 
3.1.26.3 Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 86, 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.

\(\displaystyle \int \tan ^3(c+d x) (a \sin (c+d x)+a)^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 (a \sin (c+d x)+a)^3dx\)

\(\Big \downarrow \) 3186

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

\(\Big \downarrow \) 86

\(\displaystyle \frac {\int \left (\frac {2 a^4}{(a-a \sin (c+d x))^2}-\frac {7 a^3}{a-a \sin (c+d x)}+\sin ^2(c+d x) a^2+3 \sin (c+d x) a^2+5 a^2\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 a^4}{a-a \sin (c+d x)}+\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {3}{2} a^3 \sin ^2(c+d x)+5 a^3 \sin (c+d x)+7 a^3 \log (a-a \sin (c+d x))}{d}\)

input
Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
 
output
(7*a^3*Log[a - a*Sin[c + d*x]] + 5*a^3*Sin[c + d*x] + (3*a^3*Sin[c + d*x]^ 
2)/2 + (a^3*Sin[c + d*x]^3)/3 + (2*a^4)/(a - a*Sin[c + d*x]))/d
 

3.1.26.3.1 Defintions of rubi rules used

rule 86
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((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 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 3186
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) 
^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
 
3.1.26.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.38 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56

method result size
risch \(-7 i a^{3} x -\frac {21 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {21 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {14 i a^{3} c}{d}-\frac {4 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{\left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d}+\frac {14 a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{4 d}\) \(142\)
derivativedivides \(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(204\)
default \(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(204\)
parts \(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) \(218\)

input
int((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
 
output
-7*I*a^3*x-21/8*I/d*a^3*exp(I*(d*x+c))+21/8*I/d*a^3*exp(-I*(d*x+c))-14*I/d 
*a^3*c-4*I*a^3*exp(I*(d*x+c))/(-I+exp(I*(d*x+c)))^2/d+14*a^3/d*ln(-I+exp(I 
*(d*x+c)))-1/12*a^3/d*sin(3*d*x+3*c)-3/4/d*a^3*cos(2*d*x+2*c)
 
3.1.26.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right )^{4} - 50 \, a^{3} \cos \left (d x + c\right )^{2} + 31 \, a^{3} + 84 \, {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 55 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

input
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="fricas")
 
output
1/12*(4*a^3*cos(d*x + c)^4 - 50*a^3*cos(d*x + c)^2 + 31*a^3 + 84*(a^3*sin( 
d*x + c) - a^3)*log(-sin(d*x + c) + 1) - (14*a^3*cos(d*x + c)^2 + 55*a^3)* 
sin(d*x + c))/(d*sin(d*x + c) - d)
 
3.1.26.6 Sympy [F]

\[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \]

input
integrate((a+a*sin(d*x+c))**3*tan(d*x+c)**3,x)
 
output
a**3*(Integral(3*sin(c + d*x)*tan(c + d*x)**3, x) + Integral(3*sin(c + d*x 
)**2*tan(c + d*x)**3, x) + Integral(sin(c + d*x)**3*tan(c + d*x)**3, x) + 
Integral(tan(c + d*x)**3, x))
 
3.1.26.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 42 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, a^{3} \sin \left (d x + c\right ) - \frac {12 \, a^{3}}{\sin \left (d x + c\right ) - 1}}{6 \, d} \]

input
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="maxima")
 
output
1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 42*a^3*log(sin(d*x + c) 
 - 1) + 30*a^3*sin(d*x + c) - 12*a^3/(sin(d*x + c) - 1))/d
 
3.1.26.8 Giac [F(-1)]

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

input
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="giac")
 
output
Timed out
 
3.1.26.9 Mupad [B] (verification not implemented)

Time = 6.95 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.88 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {14\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}+\frac {14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {100\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+14\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {7\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

input
int(tan(c + d*x)^3*(a + a*sin(c + d*x))^3,x)
 
output
(14*a^3*log(tan(c/2 + (d*x)/2) - 1))/d + ((98*a^3*tan(c/2 + (d*x)/2)^3)/3 
- 14*a^3*tan(c/2 + (d*x)/2)^2 - (100*a^3*tan(c/2 + (d*x)/2)^4)/3 + (98*a^3 
*tan(c/2 + (d*x)/2)^5)/3 - 14*a^3*tan(c/2 + (d*x)/2)^6 + 14*a^3*tan(c/2 + 
(d*x)/2)^7 + 14*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 - 2*tan 
(c/2 + (d*x)/2) - 6*tan(c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)/2)^4 - 6*tan( 
c/2 + (d*x)/2)^5 + 4*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + tan(c 
/2 + (d*x)/2)^8 + 1)) - (7*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d