\(\int (a+b \tan ^4(c+d x))^3 \, dx\) [382]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 144 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=(a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)}{3 d}-\frac {b^2 (3 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 (3 a+b) \tan ^7(c+d x)}{7 d}-\frac {b^3 \tan ^9(c+d x)}{9 d}+\frac {b^3 \tan ^{11}(c+d x)}{11 d} \] Output:

(a+b)^3*x-b*(3*a^2+3*a*b+b^2)*tan(d*x+c)/d+1/3*b*(3*a^2+3*a*b+b^2)*tan(d*x 
+c)^3/d-1/5*b^2*(3*a+b)*tan(d*x+c)^5/d+1/7*b^2*(3*a+b)*tan(d*x+c)^7/d-1/9* 
b^3*tan(d*x+c)^9/d+1/11*b^3*tan(d*x+c)^11/d
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.89 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\frac {(a+b)^3 \arctan (\tan (c+d x))}{d}+\frac {b \tan (c+d x) \left (-3465 \left (3 a^2+3 a b+b^2\right )+1155 \left (3 a^2+3 a b+b^2\right ) \tan ^2(c+d x)-693 b (3 a+b) \tan ^4(c+d x)+495 b (3 a+b) \tan ^6(c+d x)-385 b^2 \tan ^8(c+d x)+315 b^2 \tan ^{10}(c+d x)\right )}{3465 d} \] Input:

Integrate[(a + b*Tan[c + d*x]^4)^3,x]
 

Output:

((a + b)^3*ArcTan[Tan[c + d*x]])/d + (b*Tan[c + d*x]*(-3465*(3*a^2 + 3*a*b 
 + b^2) + 1155*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x]^2 - 693*b*(3*a + b)*Tan[ 
c + d*x]^4 + 495*b*(3*a + b)*Tan[c + d*x]^6 - 385*b^2*Tan[c + d*x]^8 + 315 
*b^2*Tan[c + d*x]^10))/(3465*d)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4144, 1468, 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.

\(\displaystyle \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \tan (c+d x)^4\right )^3dx\)

\(\Big \downarrow \) 4144

\(\displaystyle \frac {\int \frac {\left (b \tan ^4(c+d x)+a\right )^3}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 1468

\(\displaystyle \frac {\int \left (b^3 \tan ^{10}(c+d x)-b^3 \tan ^8(c+d x)+b^2 (3 a+b) \tan ^6(c+d x)-b^2 (3 a+b) \tan ^4(c+d x)+b \left (3 a^2+3 b a+b^2\right ) \tan ^2(c+d x)-b \left (3 a^2+3 b a+b^2\right )+\frac {(a+b)^3}{\tan ^2(c+d x)+1}\right )d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{3} b \left (3 a^2+3 a b+b^2\right ) \tan ^3(c+d x)-b \left (3 a^2+3 a b+b^2\right ) \tan (c+d x)+(a+b)^3 \arctan (\tan (c+d x))+\frac {1}{7} b^2 (3 a+b) \tan ^7(c+d x)-\frac {1}{5} b^2 (3 a+b) \tan ^5(c+d x)+\frac {1}{11} b^3 \tan ^{11}(c+d x)-\frac {1}{9} b^3 \tan ^9(c+d x)}{d}\)

Input:

Int[(a + b*Tan[c + d*x]^4)^3,x]
 

Output:

((a + b)^3*ArcTan[Tan[c + d*x]] - b*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x] + ( 
b*(3*a^2 + 3*a*b + b^2)*Tan[c + d*x]^3)/3 - (b^2*(3*a + b)*Tan[c + d*x]^5) 
/5 + (b^2*(3*a + b)*Tan[c + d*x]^7)/7 - (b^3*Tan[c + d*x]^9)/9 + (b^3*Tan[ 
c + d*x]^11)/11)/d
 

Defintions of rubi rules used

rule 1468
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> I 
nt[ExpandIntegrand[(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e 
}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

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 4144
Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 
Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03

method result size
norman \(\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) x -\frac {b^{3} \tan \left (d x +c \right )^{9}}{9 d}+\frac {b^{3} \tan \left (d x +c \right )^{11}}{11 d}-\frac {b \left (3 a^{2}+3 a b +b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (3 a^{2}+3 a b +b^{2}\right ) \tan \left (d x +c \right )^{3}}{3 d}-\frac {b^{2} \left (3 a +b \right ) \tan \left (d x +c \right )^{5}}{5 d}+\frac {b^{2} \left (3 a +b \right ) \tan \left (d x +c \right )^{7}}{7 d}\) \(149\)
parts \(a^{3} x +\frac {b^{3} \left (\frac {\tan \left (d x +c \right )^{11}}{11}-\frac {\tan \left (d x +c \right )^{9}}{9}+\frac {\tan \left (d x +c \right )^{7}}{7}-\frac {\tan \left (d x +c \right )^{5}}{5}+\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\tan \left (d x +c \right )^{7}}{7}-\frac {\tan \left (d x +c \right )^{5}}{5}+\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{2} b \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) \(170\)
derivativedivides \(\frac {\frac {b^{3} \tan \left (d x +c \right )^{11}}{11}-\frac {\tan \left (d x +c \right )^{9} b^{3}}{9}+\frac {3 a \,b^{2} \tan \left (d x +c \right )^{7}}{7}+\frac {b^{3} \tan \left (d x +c \right )^{7}}{7}-\frac {3 \tan \left (d x +c \right )^{5} a \,b^{2}}{5}-\frac {b^{3} \tan \left (d x +c \right )^{5}}{5}+\tan \left (d x +c \right )^{3} a^{2} b +a \,b^{2} \tan \left (d x +c \right )^{3}+\frac {b^{3} \tan \left (d x +c \right )^{3}}{3}-3 a^{2} b \tan \left (d x +c \right )-3 \tan \left (d x +c \right ) a \,b^{2}-\tan \left (d x +c \right ) b^{3}+\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) \(187\)
default \(\frac {\frac {b^{3} \tan \left (d x +c \right )^{11}}{11}-\frac {\tan \left (d x +c \right )^{9} b^{3}}{9}+\frac {3 a \,b^{2} \tan \left (d x +c \right )^{7}}{7}+\frac {b^{3} \tan \left (d x +c \right )^{7}}{7}-\frac {3 \tan \left (d x +c \right )^{5} a \,b^{2}}{5}-\frac {b^{3} \tan \left (d x +c \right )^{5}}{5}+\tan \left (d x +c \right )^{3} a^{2} b +a \,b^{2} \tan \left (d x +c \right )^{3}+\frac {b^{3} \tan \left (d x +c \right )^{3}}{3}-3 a^{2} b \tan \left (d x +c \right )-3 \tan \left (d x +c \right ) a \,b^{2}-\tan \left (d x +c \right ) b^{3}+\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) \(187\)
parallelrisch \(\frac {315 b^{3} \tan \left (d x +c \right )^{11}-385 \tan \left (d x +c \right )^{9} b^{3}+1485 a \,b^{2} \tan \left (d x +c \right )^{7}+495 b^{3} \tan \left (d x +c \right )^{7}-2079 \tan \left (d x +c \right )^{5} a \,b^{2}-693 b^{3} \tan \left (d x +c \right )^{5}+3465 \tan \left (d x +c \right )^{3} a^{2} b +3465 a \,b^{2} \tan \left (d x +c \right )^{3}+1155 b^{3} \tan \left (d x +c \right )^{3}+3465 a^{3} d x +10395 a^{2} b d x +10395 a \,b^{2} d x +3465 b^{3} d x -10395 a^{2} b \tan \left (d x +c \right )-10395 \tan \left (d x +c \right ) a \,b^{2}-3465 \tan \left (d x +c \right ) b^{3}}{3465 d}\) \(193\)
risch \(a^{3} x +3 a^{2} b x +3 a \,b^{2} x +b^{3} x -\frac {4 i b \left (6930 a^{2}+3254 b^{2}+8712 a b +928620 a^{2} {\mathrm e}^{14 i \left (d x +c \right )}+438900 b^{2} {\mathrm e}^{14 i \left (d x +c \right )}+1503810 a^{2} {\mathrm e}^{12 i \left (d x +c \right )}+751674 b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+1697850 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+751674 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+1358280 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+634920 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+762300 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+317460 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+287595 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+126995 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+65835 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+25399 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+219450 b^{2} {\mathrm e}^{16 i \left (d x +c \right )}+381150 a^{2} {\mathrm e}^{16 i \left (d x +c \right )}+51975 b^{2} {\mathrm e}^{18 i \left (d x +c \right )}+93555 a^{2} {\mathrm e}^{18 i \left (d x +c \right )}+10395 b^{2} {\mathrm e}^{20 i \left (d x +c \right )}+10395 a^{2} {\mathrm e}^{20 i \left (d x +c \right )}+910800 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+333630 \,{\mathrm e}^{4 i \left (d x +c \right )} a b +20790 a b \,{\mathrm e}^{20 i \left (d x +c \right )}+1219680 a b \,{\mathrm e}^{14 i \left (d x +c \right )}+1655280 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+2109492 a b \,{\mathrm e}^{10 i \left (d x +c \right )}+526680 a b \,{\mathrm e}^{16 i \left (d x +c \right )}+145530 a b \,{\mathrm e}^{18 i \left (d x +c \right )}+1915452 a b \,{\mathrm e}^{12 i \left (d x +c \right )}+75042 a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{3465 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{11}}\) \(471\)

Input:

int((a+b*tan(d*x+c)^4)^3,x,method=_RETURNVERBOSE)
 

Output:

(a^3+3*a^2*b+3*a*b^2+b^3)*x-1/9*b^3*tan(d*x+c)^9/d+1/11*b^3*tan(d*x+c)^11/ 
d-b*(3*a^2+3*a*b+b^2)*tan(d*x+c)/d+1/3*b*(3*a^2+3*a*b+b^2)*tan(d*x+c)^3/d- 
1/5*b^2*(3*a+b)*tan(d*x+c)^5/d+1/7*b^2*(3*a+b)*tan(d*x+c)^7/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\frac {315 \, b^{3} \tan \left (d x + c\right )^{11} - 385 \, b^{3} \tan \left (d x + c\right )^{9} + 495 \, {\left (3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{7} - 693 \, {\left (3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{5} + 1155 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{3} + 3465 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3465 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )}{3465 \, d} \] Input:

integrate((a+b*tan(d*x+c)^4)^3,x, algorithm="fricas")
 

Output:

1/3465*(315*b^3*tan(d*x + c)^11 - 385*b^3*tan(d*x + c)^9 + 495*(3*a*b^2 + 
b^3)*tan(d*x + c)^7 - 693*(3*a*b^2 + b^3)*tan(d*x + c)^5 + 1155*(3*a^2*b + 
 3*a*b^2 + b^3)*tan(d*x + c)^3 + 3465*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x 
- 3465*(3*a^2*b + 3*a*b^2 + b^3)*tan(d*x + c))/d
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.56 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x + 3 a^{2} b x + \frac {a^{2} b \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} + 3 a b^{2} x + \frac {3 a b^{2} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {3 a b^{2} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tan {\left (c + d x \right )}}{d} + b^{3} x + \frac {b^{3} \tan ^{11}{\left (c + d x \right )}}{11 d} - \frac {b^{3} \tan ^{9}{\left (c + d x \right )}}{9 d} + \frac {b^{3} \tan ^{7}{\left (c + d x \right )}}{7 d} - \frac {b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} + \frac {b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{4}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*tan(d*x+c)**4)**3,x)
 

Output:

Piecewise((a**3*x + 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*x + 3*a*b**2*tan(c + d*x)**7/(7*d) - 3*a*b**2*tan(c + 
 d*x)**5/(5*d) + a*b**2*tan(c + d*x)**3/d - 3*a*b**2*tan(c + d*x)/d + b**3 
*x + b**3*tan(c + d*x)**11/(11*d) - b**3*tan(c + d*x)**9/(9*d) + b**3*tan( 
c + d*x)**7/(7*d) - 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)**4)**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.16 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=a^{3} x + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} b}{d} + \frac {{\left (15 \, \tan \left (d x + c\right )^{7} - 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - 105 \, \tan \left (d x + c\right )\right )} a b^{2}}{35 \, d} + \frac {{\left (315 \, \tan \left (d x + c\right )^{11} - 385 \, \tan \left (d x + c\right )^{9} + 495 \, \tan \left (d x + c\right )^{7} - 693 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3} + 3465 \, d x + 3465 \, c - 3465 \, \tan \left (d x + c\right )\right )} b^{3}}{3465 \, d} \] Input:

integrate((a+b*tan(d*x+c)^4)^3,x, algorithm="maxima")
 

Output:

a^3*x + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2*b/d + 1/35*(15 
*tan(d*x + c)^7 - 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 105*d*x + 105*c 
- 105*tan(d*x + c))*a*b^2/d + 1/3465*(315*tan(d*x + c)^11 - 385*tan(d*x + 
c)^9 + 495*tan(d*x + c)^7 - 693*tan(d*x + c)^5 + 1155*tan(d*x + c)^3 + 346 
5*d*x + 3465*c - 3465*tan(d*x + c))*b^3/d
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.58 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )}}{d} + \frac {315 \, b^{3} d^{10} \tan \left (d x + c\right )^{11} - 385 \, b^{3} d^{10} \tan \left (d x + c\right )^{9} + 1485 \, a b^{2} d^{10} \tan \left (d x + c\right )^{7} + 495 \, b^{3} d^{10} \tan \left (d x + c\right )^{7} - 2079 \, a b^{2} d^{10} \tan \left (d x + c\right )^{5} - 693 \, b^{3} d^{10} \tan \left (d x + c\right )^{5} + 3465 \, a^{2} b d^{10} \tan \left (d x + c\right )^{3} + 3465 \, a b^{2} d^{10} \tan \left (d x + c\right )^{3} + 1155 \, b^{3} d^{10} \tan \left (d x + c\right )^{3} - 10395 \, a^{2} b d^{10} \tan \left (d x + c\right ) - 10395 \, a b^{2} d^{10} \tan \left (d x + c\right ) - 3465 \, b^{3} d^{10} \tan \left (d x + c\right )}{3465 \, d^{11}} \] Input:

integrate((a+b*tan(d*x+c)^4)^3,x, algorithm="giac")
 

Output:

(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c)/d + 1/3465*(315*b^3*d^10*tan(d*x 
 + c)^11 - 385*b^3*d^10*tan(d*x + c)^9 + 1485*a*b^2*d^10*tan(d*x + c)^7 + 
495*b^3*d^10*tan(d*x + c)^7 - 2079*a*b^2*d^10*tan(d*x + c)^5 - 693*b^3*d^1 
0*tan(d*x + c)^5 + 3465*a^2*b*d^10*tan(d*x + c)^3 + 3465*a*b^2*d^10*tan(d* 
x + c)^3 + 1155*b^3*d^10*tan(d*x + c)^3 - 10395*a^2*b*d^10*tan(d*x + c) - 
10395*a*b^2*d^10*tan(d*x + c) - 3465*b^3*d^10*tan(d*x + c))/d^11
 

Mupad [B] (verification not implemented)

Time = 7.65 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.25 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2\,b+a\,b^2+\frac {b^3}{3}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a+b\right )}^3}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}\right )\,{\left (a+b\right )}^3}{d}-\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^9}{9\,d}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{11}}{11\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {b^3}{5}+\frac {3\,a\,b^2}{5}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {b^3}{7}+\frac {3\,a\,b^2}{7}\right )}{d} \] Input:

int((a + b*tan(c + d*x)^4)^3,x)
 

Output:

(tan(c + d*x)^3*(a*b^2 + a^2*b + b^3/3))/d + (atan((tan(c + d*x)*(a + b)^3 
)/(3*a*b^2 + 3*a^2*b + a^3 + b^3))*(a + b)^3)/d - (b^3*tan(c + d*x)^9)/(9* 
d) + (b^3*tan(c + d*x)^11)/(11*d) - (tan(c + d*x)*(3*a*b^2 + 3*a^2*b + b^3 
))/d - (tan(c + d*x)^5*((3*a*b^2)/5 + b^3/5))/d + (tan(c + d*x)^7*((3*a*b^ 
2)/7 + b^3/7))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.33 \[ \int \left (a+b \tan ^4(c+d x)\right )^3 \, dx=\frac {315 \tan \left (d x +c \right )^{11} b^{3}-385 \tan \left (d x +c \right )^{9} b^{3}+1485 \tan \left (d x +c \right )^{7} a \,b^{2}+495 \tan \left (d x +c \right )^{7} b^{3}-2079 \tan \left (d x +c \right )^{5} a \,b^{2}-693 \tan \left (d x +c \right )^{5} b^{3}+3465 \tan \left (d x +c \right )^{3} a^{2} b +3465 \tan \left (d x +c \right )^{3} a \,b^{2}+1155 \tan \left (d x +c \right )^{3} b^{3}-10395 \tan \left (d x +c \right ) a^{2} b -10395 \tan \left (d x +c \right ) a \,b^{2}-3465 \tan \left (d x +c \right ) b^{3}+3465 a^{3} d x +10395 a^{2} b d x +10395 a \,b^{2} d x +3465 b^{3} d x}{3465 d} \] Input:

int((a+b*tan(d*x+c)^4)^3,x)
 

Output:

(315*tan(c + d*x)**11*b**3 - 385*tan(c + d*x)**9*b**3 + 1485*tan(c + d*x)* 
*7*a*b**2 + 495*tan(c + d*x)**7*b**3 - 2079*tan(c + d*x)**5*a*b**2 - 693*t 
an(c + d*x)**5*b**3 + 3465*tan(c + d*x)**3*a**2*b + 3465*tan(c + d*x)**3*a 
*b**2 + 1155*tan(c + d*x)**3*b**3 - 10395*tan(c + d*x)*a**2*b - 10395*tan( 
c + d*x)*a*b**2 - 3465*tan(c + d*x)*b**3 + 3465*a**3*d*x + 10395*a**2*b*d* 
x + 10395*a*b**2*d*x + 3465*b**3*d*x)/(3465*d)