\(\int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [480]

Optimal result

Integrand size = 21, antiderivative size = 149 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {a \left (a^2-3 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^2 \tan (c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^2+5 b^2\right )}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3646, 3709, 3612, 3611} \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {a^2 \left (a^2+5 b^2\right )}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {a^2 \tan (c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a \left (a^2-3 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3} \]

[In]

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

[Out]

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

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \tan (c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {a^2-2 a b \tan (c+d x)+\left (a^2+2 b^2\right ) \tan ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )} \\ & = -\frac {a^2 \tan (c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^2+5 b^2\right )}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-4 a b^2-2 b \left (a^2-b^2\right ) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{2 b \left (a^2+b^2\right )^2} \\ & = -\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {a^2 \tan (c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^2+5 b^2\right )}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (a \left (a^2-3 b^2\right )\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3} \\ & = -\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {a \left (a^2-3 b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^2 \tan (c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^2+5 b^2\right )}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.80 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.81 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {i \log (i-\tan (c+d x))}{(a+i b)^2}+\frac {a \log (i-\tan (c+d x))}{(-i a+b)^3}-\frac {i \log (i+\tan (c+d x))}{(a-i b)^2}+\frac {a \log (i+\tan (c+d x))}{(i a+b)^3}-\frac {4 a b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}-\frac {a}{b (a+b \tan (c+d x))^2}-\frac {2 \tan (c+d x)}{(a+b \tan (c+d x))^2}+\frac {2 b}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {a b \left (\left (-6 a^2+2 b^2\right ) \log (a+b \tan (c+d x))+\frac {\left (a^2+b^2\right ) \left (5 a^2+b^2+4 a b \tan (c+d x)\right )}{(a+b \tan (c+d x))^2}\right )}{\left (a^2+b^2\right )^3}}{2 b d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3}}{2 b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\frac {\left (-a^{3}+3 a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-3 a^{2} b +b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(158\)
default	\(\frac {\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}-\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a^{2}+b^{2}\right )^{2} b^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3}}{2 b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\frac {\left (-a^{3}+3 a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-3 a^{2} b +b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(158\)
norman	\(\frac {\frac {\left (a^{4}+3 a^{2} b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {a^{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {b^{3} \left (3 a^{2}-b^{2}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {\left (3 a^{2}-b^{2}\right ) a^{2} b x}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}-\frac {2 b^{2} \left (3 a^{2}-b^{2}\right ) a x \tan \left (d x +c \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(332\)
risch	\(-\frac {i x}{3 i b \,a^{2}-i b^{3}-a^{3}+3 a \,b^{2}}-\frac {2 i a^{3} x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}+\frac {6 i a x \,b^{2}}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {2 i a^{3} c}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {6 i a \,b^{2} c}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {2 a^{2} \left (2 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a b -3 b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )^{2} \left (i b +a \right )^{2} d \left (-i b +a \right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) b^{2}}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}\)	\(401\)
parallelrisch	\(-\frac {a^{7}+\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{5} b^{2}+6 a^{5} b^{2}+6 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{3} b^{4}-3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} b^{4}-2 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{5} b^{2}+2 \tan \left (d x +c \right ) a^{6} b +8 \tan \left (d x +c \right ) a^{4} b^{3}+6 \tan \left (d x +c \right ) a^{2} b^{5}+5 a^{3} b^{4}-2 a^{2} b^{5} x d -6 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{2} b^{5}-4 \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{4} b^{3}+6 x \,a^{4} b^{3} d +\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a^{3} b^{4}-3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a \,b^{6}+2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{4} b^{3}-2 x \left (\tan ^{2}\left (d x +c \right )\right ) b^{7} d +6 x \left (\tan ^{2}\left (d x +c \right )\right ) a^{2} b^{5} d +12 x \tan \left (d x +c \right ) a^{3} b^{4} d -2 \ln \left (a +b \tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a^{3} b^{4}-4 b^{6} a \tan \left (d x +c \right ) x d +6 \ln \left (a +b \tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a \,b^{6}+12 \tan \left (d x +c \right ) \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2} b^{5}}{2 \left (a +b \tan \left (d x +c \right )\right )^{2} \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right ) b^{2} d}\)	\(467\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (145) = 290\).

Time = 0.25 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {a^{5} - 5 \, a^{3} b^{2} - 2 \, {\left (3 \, a^{4} b - a^{2} b^{3}\right )} d x + {\left (a^{5} + 7 \, a^{3} b^{2} - 2 \, {\left (3 \, a^{2} b^{3} - b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (a^{5} - 3 \, a^{3} b^{2} + {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 3 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (3 \, a^{4} b - 3 \, a^{2} b^{3} - 2 \, {\left (3 \, a^{3} b^{2} - a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \]

[In]

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

[Out]

1/2*(a^5 - 5*a^3*b^2 - 2*(3*a^4*b - a^2*b^3)*d*x + (a^5 + 7*a^3*b^2 - 2*(3*a^2*b^3 - b^5)*d*x)*tan(d*x + c)^2
+ (a^5 - 3*a^3*b^2 + (a^3*b^2 - 3*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b - 3*a^2*b^3)*tan(d*x + c))*log((b^2*tan(d*x
+ c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + 2*(3*a^4*b - 3*a^2*b^3 - 2*(3*a^3*b^2 - a*b^4)*d*x
)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 +
a*b^7)*d*tan(d*x + c) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*d)

Sympy [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Exception raised: AttributeError} \]

[In]

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

[Out]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

Maxima [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.76 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {a^{5} + 5 \, a^{3} b^{2} + 2 \, {\left (a^{4} b + 3 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3 - 3*a*b^2)*log(b*tan(d*x + c) +
a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4
+ b^6) + (a^5 + 5*a^3*b^2 + 2*(a^4*b + 3*a^2*b^3)*tan(d*x + c))/(a^6*b^2 + 2*a^4*b^4 + a^2*b^6 + (a^4*b^4 + 2
*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(a^5*b^3 + 2*a^3*b^5 + a*b^7)*tan(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.82 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (a^{3} b - 3 \, a b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} + \frac {3 \, a^{3} b^{4} \tan \left (d x + c\right )^{2} - 9 \, a b^{6} \tan \left (d x + c\right )^{2} + 2 \, a^{6} b \tan \left (d x + c\right ) + 14 \, a^{4} b^{3} \tan \left (d x + c\right ) - 12 \, a^{2} b^{5} \tan \left (d x + c\right ) + a^{7} + 9 \, a^{5} b^{2} - 4 \, a^{3} b^{4}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1
)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3*b - 3*a*b^3)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 +
3*a^2*b^5 + b^7) + (3*a^3*b^4*tan(d*x + c)^2 - 9*a*b^6*tan(d*x + c)^2 + 2*a^6*b*tan(d*x + c) + 14*a^4*b^3*tan(
d*x + c) - 12*a^2*b^5*tan(d*x + c) + a^7 + 9*a^5*b^2 - 4*a^3*b^4)/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*(b*
tan(d*x + c) + a)^2))/d

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {a}{{\left (a^2+b^2\right )}^2}-\frac {4\,a\,b^2}{{\left (a^2+b^2\right )}^3}\right )}{d}-\frac {\frac {a\,\left (a^4+5\,a^2\,b^2\right )}{2\,b^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+3\,a^2\,b^2\right )}{b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )} \]

[In]

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

[Out]

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