\(\int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx\) [435]

Optimal result

Integrand size = 21, antiderivative size = 147 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=b \left (3 a^2-b^2\right ) x+\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a (a+b \tan (c+d x))^2}{2 d}-\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d} \]

[Out]

b*(3*a^2-b^2)*x+a*(a^2-3*b^2)*ln(cos(d*x+c))/d-b*(a^2-b^2)*tan(d*x+c)/d-1/2*a*(a+b*tan(d*x+c))^2/d-1/3*(a+b*ta
n(d*x+c))^3/d-1/20*a*(a+b*tan(d*x+c))^4/b^2/d+1/5*tan(d*x+c)*(a+b*tan(d*x+c))^4/b/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3647, 3711, 12, 3609, 3606, 3556} \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+b x \left (3 a^2-b^2\right )-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^2}{2 d} \]

[In]

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

[Out]

b*(3*a^2 - b^2)*x + (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d - (b*(a^2 - b^2)*Tan[c + d*x])/d - (a*(a + b*Tan[c +
d*x])^2)/(2*d) - (a + b*Tan[c + d*x])^3/(3*d) - (a*(a + b*Tan[c + d*x])^4)/(20*b^2*d) + (Tan[c + d*x]*(a + b*
Tan[c + d*x])^4)/(5*b*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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

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

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

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

Rubi steps \begin{align*} \text {integral}& = \frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int (a+b \tan (c+d x))^3 \left (-a-5 b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{5 b} \\ & = -\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac {\int -5 b \tan (c+d x) (a+b \tan (c+d x))^3 \, dx}{5 b} \\ & = -\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx \\ & = -\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int (-b+a \tan (c+d x)) (a+b \tan (c+d x))^2 \, dx \\ & = -\frac {a (a+b \tan (c+d x))^2}{2 d}-\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int (a+b \tan (c+d x)) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx \\ & = b \left (3 a^2-b^2\right ) x-\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a (a+b \tan (c+d x))^2}{2 d}-\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\left (a \left (a^2-3 b^2\right )\right ) \int \tan (c+d x) \, dx \\ & = b \left (3 a^2-b^2\right ) x+\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a (a+b \tan (c+d x))^2}{2 d}-\frac {(a+b \tan (c+d x))^3}{3 d}-\frac {a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac {\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.86 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.10 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-3 \left (a^5+10 (a+i b)^3 b^2 \log (i-\tan (c+d x))+10 (a-i b)^3 b^2 \log (i+\tan (c+d x))\right )+60 b^3 \left (-3 a^2+b^2\right ) \tan (c+d x)+30 a b^2 \left (a^2-3 b^2\right ) \tan ^2(c+d x)-20 b^3 \left (-3 a^2+b^2\right ) \tan ^3(c+d x)+45 a b^4 \tan ^4(c+d x)+12 b^5 \tan ^5(c+d x)}{60 b^2 d} \]

[In]

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

[Out]

(-3*(a^5 + 10*(a + I*b)^3*b^2*Log[I - Tan[c + d*x]] + 10*(a - I*b)^3*b^2*Log[I + Tan[c + d*x]]) + 60*b^3*(-3*a
^2 + b^2)*Tan[c + d*x] + 30*a*b^2*(a^2 - 3*b^2)*Tan[c + d*x]^2 - 20*b^3*(-3*a^2 + b^2)*Tan[c + d*x]^3 + 45*a*b
^4*Tan[c + d*x]^4 + 12*b^5*Tan[c + d*x]^5)/(60*b^2*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99

method	result	size
norman	\(b \left (3 a^{2}-b^{2}\right ) x +\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {3 a \,b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a \left (a^{2}-3 b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \left (3 a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (3 a^{2}-b^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(146\)
derivativedivides	\(\frac {\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a \,b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4}+a^{2} b \left (\tan ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {3 a \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-3 a^{2} b \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (3 a^{2} b -b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(153\)
default	\(\frac {\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {3 a \,b^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4}+a^{2} b \left (\tan ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {3 a \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}-3 a^{2} b \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (3 a^{2} b -b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(153\)
parts	\(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{3} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(154\)
parallelrisch	\(-\frac {-12 b^{3} \left (\tan ^{5}\left (d x +c \right )\right )-45 a \,b^{2} \left (\tan ^{4}\left (d x +c \right )\right )-60 a^{2} b \left (\tan ^{3}\left (d x +c \right )\right )+20 b^{3} \left (\tan ^{3}\left (d x +c \right )\right )-180 a^{2} b d x +60 b^{3} d x -30 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )+90 a \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )+30 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}-90 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a \,b^{2}+180 a^{2} b \tan \left (d x +c \right )-60 b^{3} \tan \left (d x +c \right )}{60 d}\)	\(159\)
risch	\(3 a^{2} b x -b^{3} x -i a^{3} x +3 i a \,b^{2} x -\frac {2 i a^{3} c}{d}+\frac {6 i a \,b^{2} c}{d}+\frac {2 i \left (-15 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+90 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-90 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-45 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+180 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-270 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+90 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-45 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-330 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-15 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+180 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+70 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 a^{2} b +23 b^{3}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\)	\(362\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.93 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {12 \, b^{3} \tan \left (d x + c\right )^{5} + 45 \, a b^{2} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} + 60 \, {\left (3 \, a^{2} b - b^{3}\right )} d x + 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2} + 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 60 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

1/60*(12*b^3*tan(d*x + c)^5 + 45*a*b^2*tan(d*x + c)^4 + 20*(3*a^2*b - b^3)*tan(d*x + c)^3 + 60*(3*a^2*b - b^3)
*d*x + 30*(a^3 - 3*a*b^2)*tan(d*x + c)^2 + 30*(a^3 - 3*a*b^2)*log(1/(tan(d*x + c)^2 + 1)) - 60*(3*a^2*b - b^3)
*tan(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.32 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\begin {cases} - \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 a^{2} b x + \frac {a^{2} b \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} + \frac {3 a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 a b^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {3 a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} - b^{3} x + \frac {b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a**3*log(tan(c + d*x)**2 + 1)/(2*d) + a**3*tan(c + d*x)**2/(2*d) + 3*a**2*b*x + a**2*b*tan(c + d*x
)**3/d - 3*a**2*b*tan(c + d*x)/d + 3*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*a*b**2*tan(c + d*x)**4/(4*d) -
3*a*b**2*tan(c + d*x)**2/(2*d) - b**3*x + b**3*tan(c + d*x)**5/(5*d) - b**3*tan(c + d*x)**3/(3*d) + b**3*tan(c
+ d*x)/d, Ne(d, 0)), (x*(a + b*tan(c))**3*tan(c)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {12 \, b^{3} \tan \left (d x + c\right )^{5} + 45 \, a b^{2} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} + 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2} + 60 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} - 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

1/60*(12*b^3*tan(d*x + c)^5 + 45*a*b^2*tan(d*x + c)^4 + 20*(3*a^2*b - b^3)*tan(d*x + c)^3 + 30*(a^3 - 3*a*b^2)
*tan(d*x + c)^2 + 60*(3*a^2*b - b^3)*(d*x + c) - 30*(a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1) - 60*(3*a^2*b - b^
3)*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1877 vs. \(2 (139) = 278\).

Time = 2.80 (sec) , antiderivative size = 1877, normalized size of antiderivative = 12.77 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/60*(180*a^2*b*d*x*tan(d*x)^5*tan(c)^5 - 60*b^3*d*x*tan(d*x)^5*tan(c)^5 + 30*a^3*log(4*(tan(d*x)^2*tan(c)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^5*tan(c)^5 - 90*a*b^2*log(
4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^5*
tan(c)^5 - 900*a^2*b*d*x*tan(d*x)^4*tan(c)^4 + 300*b^3*d*x*tan(d*x)^4*tan(c)^4 + 30*a^3*tan(d*x)^5*tan(c)^5 -
135*a*b^2*tan(d*x)^5*tan(c)^5 - 150*a^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)
^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 450*a*b^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c
) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 180*a^2*b*tan(d*x)^5*tan(c)^4
- 60*b^3*tan(d*x)^5*tan(c)^4 + 180*a^2*b*tan(d*x)^4*tan(c)^5 - 60*b^3*tan(d*x)^4*tan(c)^5 + 1800*a^2*b*d*x*tan
(d*x)^3*tan(c)^3 - 600*b^3*d*x*tan(d*x)^3*tan(c)^3 + 30*a^3*tan(d*x)^5*tan(c)^3 - 90*a*b^2*tan(d*x)^5*tan(c)^3
- 90*a^3*tan(d*x)^4*tan(c)^4 + 495*a*b^2*tan(d*x)^4*tan(c)^4 + 30*a^3*tan(d*x)^3*tan(c)^5 - 90*a*b^2*tan(d*x)
^3*tan(c)^5 - 60*a^2*b*tan(d*x)^5*tan(c)^2 + 20*b^3*tan(d*x)^5*tan(c)^2 + 300*a^3*log(4*(tan(d*x)^2*tan(c)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 900*a*b^2*log
(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3
*tan(c)^3 - 900*a^2*b*tan(d*x)^4*tan(c)^3 + 300*b^3*tan(d*x)^4*tan(c)^3 - 900*a^2*b*tan(d*x)^3*tan(c)^4 + 300*
b^3*tan(d*x)^3*tan(c)^4 - 60*a^2*b*tan(d*x)^2*tan(c)^5 + 20*b^3*tan(d*x)^2*tan(c)^5 + 45*a*b^2*tan(d*x)^5*tan(
c) - 1800*a^2*b*d*x*tan(d*x)^2*tan(c)^2 + 600*b^3*d*x*tan(d*x)^2*tan(c)^2 - 90*a^3*tan(d*x)^4*tan(c)^2 + 450*a
*b^2*tan(d*x)^4*tan(c)^2 + 120*a^3*tan(d*x)^3*tan(c)^3 - 540*a*b^2*tan(d*x)^3*tan(c)^3 - 90*a^3*tan(d*x)^2*tan
(c)^4 + 450*a*b^2*tan(d*x)^2*tan(c)^4 + 45*a*b^2*tan(d*x)*tan(c)^5 - 12*b^3*tan(d*x)^5 + 120*a^2*b*tan(d*x)^4*
tan(c) - 100*b^3*tan(d*x)^4*tan(c) - 300*a^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*t
an(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 900*a*b^2*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*
tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 1440*a^2*b*tan(d*x)^3*tan
(c)^2 - 600*b^3*tan(d*x)^3*tan(c)^2 + 1440*a^2*b*tan(d*x)^2*tan(c)^3 - 600*b^3*tan(d*x)^2*tan(c)^3 + 120*a^2*b
*tan(d*x)*tan(c)^4 - 100*b^3*tan(d*x)*tan(c)^4 - 12*b^3*tan(c)^5 - 45*a*b^2*tan(d*x)^4 + 900*a^2*b*d*x*tan(d*x
)*tan(c) - 300*b^3*d*x*tan(d*x)*tan(c) + 90*a^3*tan(d*x)^3*tan(c) - 450*a*b^2*tan(d*x)^3*tan(c) - 120*a^3*tan(
d*x)^2*tan(c)^2 + 540*a*b^2*tan(d*x)^2*tan(c)^2 + 90*a^3*tan(d*x)*tan(c)^3 - 450*a*b^2*tan(d*x)*tan(c)^3 - 45*
a*b^2*tan(c)^4 - 60*a^2*b*tan(d*x)^3 + 20*b^3*tan(d*x)^3 + 150*a^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan
(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 450*a*b^2*log(4*(tan(d*x)^2*tan(
c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 900*a^2*b*t
an(d*x)^2*tan(c) + 300*b^3*tan(d*x)^2*tan(c) - 900*a^2*b*tan(d*x)*tan(c)^2 + 300*b^3*tan(d*x)*tan(c)^2 - 60*a^
2*b*tan(c)^3 + 20*b^3*tan(c)^3 - 180*a^2*b*d*x + 60*b^3*d*x - 30*a^3*tan(d*x)^2 + 90*a*b^2*tan(d*x)^2 + 90*a^3
*tan(d*x)*tan(c) - 495*a*b^2*tan(d*x)*tan(c) - 30*a^3*tan(c)^2 + 90*a*b^2*tan(c)^2 - 30*a^3*log(4*(tan(d*x)^2*
tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 90*a*b^2*log(4*(tan(d*x
)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 180*a^2*b*tan(d*x)
- 60*b^3*tan(d*x) + 180*a^2*b*tan(c) - 60*b^3*tan(c) - 30*a^3 + 135*a*b^2)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*
x)^4*tan(c)^4 + 10*d*tan(d*x)^3*tan(c)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)

Mupad [B] (verification not implemented)

Time = 4.93 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.24 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b-b^3\right )}{d}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {3\,a\,b^2}{2}-\frac {a^3}{2}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,a\,b^2}{2}-\frac {a^3}{2}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2\,b-\frac {b^3}{3}\right )}{d}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2-b^2\right )}{3\,a^2\,b-b^3}\right )\,\left (3\,a^2-b^2\right )}{d} \]

[In]

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

[Out]

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