Integrand size = 14, antiderivative size = 77 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=(a-b)^3 x+\frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {(3 a-b) b^2 \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d} \]
(a-b)^3*x+b*(3*a^2-3*a*b+b^2)*tan(d*x+c)/d+1/3*(3*a-b)*b^2*tan(d*x+c)^3/d+ 1/5*b^3*tan(d*x+c)^5/d
Time = 1.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=\frac {\tan (c+d x) \left (\frac {15 (a-b)^3 \text {arctanh}\left (\sqrt {-\tan ^2(c+d x)}\right )}{\sqrt {-\tan ^2(c+d x)}}+b \left (45 a^2-15 a b \left (3-\tan ^2(c+d x)\right )+b^2 \left (15-5 \tan ^2(c+d x)+3 \tan ^4(c+d x)\right )\right )\right )}{15 d} \]
(Tan[c + d*x]*((15*(a - b)^3*ArcTanh[Sqrt[-Tan[c + d*x]^2]])/Sqrt[-Tan[c + d*x]^2] + b*(45*a^2 - 15*a*b*(3 - Tan[c + d*x]^2) + b^2*(15 - 5*Tan[c + d *x]^2 + 3*Tan[c + d*x]^4))))/(15*d)
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4144, 300, 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.
\(\displaystyle \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \tan (c+d x)^2\right )^3dx\) |
\(\Big \downarrow \) 4144 |
\(\displaystyle \frac {\int \frac {\left (b \tan ^2(c+d x)+a\right )^3}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (b^3 \tan ^4(c+d x)+(3 a-b) b^2 \tan ^2(c+d x)+b \left (3 a^2-3 b a+b^2\right )+\frac {(a-b)^3}{\tan ^2(c+d x)+1}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)+(a-b)^3 \arctan (\tan (c+d x))+\frac {1}{3} b^2 (3 a-b) \tan ^3(c+d x)+\frac {1}{5} b^3 \tan ^5(c+d x)}{d}\) |
((a - b)^3*ArcTan[Tan[c + d*x]] + b*(3*a^2 - 3*a*b + b^2)*Tan[c + d*x] + ( (3*a - b)*b^2*Tan[c + d*x]^3)/3 + (b^3*Tan[c + d*x]^5)/5)/d
3.3.51.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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])
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14
method | result | size |
norman | \(\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) x +\frac {b \left (3 a^{2}-3 a b +b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \tan \left (d x +c \right )^{5}}{5 d}+\frac {\left (3 a -b \right ) b^{2} \tan \left (d x +c \right )^{3}}{3 d}\) | \(88\) |
derivativedivides | \(\frac {\frac {b^{3} \tan \left (d x +c \right )^{5}}{5}+a \,b^{2} \tan \left (d x +c \right )^{3}-\frac {b^{3} \tan \left (d x +c \right )^{3}}{3}+3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(108\) |
default | \(\frac {\frac {b^{3} \tan \left (d x +c \right )^{5}}{5}+a \,b^{2} \tan \left (d x +c \right )^{3}-\frac {b^{3} \tan \left (d x +c \right )^{3}}{3}+3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(108\) |
parts | \(x \,a^{3}+\frac {b^{3} \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{2} b \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(110\) |
parallelrisch | \(\frac {3 b^{3} \tan \left (d x +c \right )^{5}+15 a \,b^{2} \tan \left (d x +c \right )^{3}-5 b^{3} \tan \left (d x +c \right )^{3}+15 a^{3} d x -45 a^{2} b d x +45 a \,b^{2} d x -15 b^{3} d x +45 a^{2} b \tan \left (d x +c \right )-45 a \,b^{2} \tan \left (d x +c \right )+15 b^{3} \tan \left (d x +c \right )}{15 d}\) | \(112\) |
risch | \(x \,a^{3}-3 a^{2} b x +3 a \,b^{2} x -b^{3} x +\frac {2 i b \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-90 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+180 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-270 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-330 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+180 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+70 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{2}-60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(226\) |
(a^3-3*a^2*b+3*a*b^2-b^3)*x+b*(3*a^2-3*a*b+b^2)*tan(d*x+c)/d+1/5*b^3*tan(d *x+c)^5/d+1/3*(3*a-b)*b^2*tan(d*x+c)^3/d
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} \tan \left (d x + c\right )^{5} + 5 \, {\left (3 \, a b^{2} - b^{3}\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x + 15 \, {\left (3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )}{15 \, d} \]
1/15*(3*b^3*tan(d*x + c)^5 + 5*(3*a*b^2 - b^3)*tan(d*x + c)^3 + 15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*x + 15*(3*a^2*b - 3*a*b^2 + b^3)*tan(d*x + c))/ d
Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.64 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x - 3 a^{2} b x + \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} + 3 a b^{2} x + \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tan {\left (c + d x \right )}}{d} - b^{3} x + \frac {b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{2}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \]
Piecewise((a**3*x - 3*a**2*b*x + 3*a**2*b*tan(c + d*x)/d + 3*a*b**2*x + 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)**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)**2)**3, True))
Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.35 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=a^{3} x - \frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b}{d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b^{2}}{d} + \frac {{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} b^{3}}{15 \, d} \]
a^3*x - 3*(d*x + c - tan(d*x + c))*a^2*b/d + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a*b^2/d + 1/15*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 1 5*d*x - 15*c + 15*tan(d*x + c))*b^3/d
Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (73) = 146\).
Time = 0.95 (sec) , antiderivative size = 1027, normalized size of antiderivative = 13.34 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
1/15*(15*a^3*d*x*tan(d*x)^5*tan(c)^5 - 45*a^2*b*d*x*tan(d*x)^5*tan(c)^5 + 45*a*b^2*d*x*tan(d*x)^5*tan(c)^5 - 15*b^3*d*x*tan(d*x)^5*tan(c)^5 - 75*a^3 *d*x*tan(d*x)^4*tan(c)^4 + 225*a^2*b*d*x*tan(d*x)^4*tan(c)^4 - 225*a*b^2*d *x*tan(d*x)^4*tan(c)^4 + 75*b^3*d*x*tan(d*x)^4*tan(c)^4 - 45*a^2*b*tan(d*x )^5*tan(c)^4 + 45*a*b^2*tan(d*x)^5*tan(c)^4 - 15*b^3*tan(d*x)^5*tan(c)^4 - 45*a^2*b*tan(d*x)^4*tan(c)^5 + 45*a*b^2*tan(d*x)^4*tan(c)^5 - 15*b^3*tan( d*x)^4*tan(c)^5 + 150*a^3*d*x*tan(d*x)^3*tan(c)^3 - 450*a^2*b*d*x*tan(d*x) ^3*tan(c)^3 + 450*a*b^2*d*x*tan(d*x)^3*tan(c)^3 - 150*b^3*d*x*tan(d*x)^3*t an(c)^3 - 15*a*b^2*tan(d*x)^5*tan(c)^2 + 5*b^3*tan(d*x)^5*tan(c)^2 + 180*a ^2*b*tan(d*x)^4*tan(c)^3 - 225*a*b^2*tan(d*x)^4*tan(c)^3 + 75*b^3*tan(d*x) ^4*tan(c)^3 + 180*a^2*b*tan(d*x)^3*tan(c)^4 - 225*a*b^2*tan(d*x)^3*tan(c)^ 4 + 75*b^3*tan(d*x)^3*tan(c)^4 - 15*a*b^2*tan(d*x)^2*tan(c)^5 + 5*b^3*tan( d*x)^2*tan(c)^5 - 150*a^3*d*x*tan(d*x)^2*tan(c)^2 + 450*a^2*b*d*x*tan(d*x) ^2*tan(c)^2 - 450*a*b^2*d*x*tan(d*x)^2*tan(c)^2 + 150*b^3*d*x*tan(d*x)^2*t an(c)^2 - 3*b^3*tan(d*x)^5 + 30*a*b^2*tan(d*x)^4*tan(c) - 25*b^3*tan(d*x)^ 4*tan(c) - 270*a^2*b*tan(d*x)^3*tan(c)^2 + 360*a*b^2*tan(d*x)^3*tan(c)^2 - 150*b^3*tan(d*x)^3*tan(c)^2 - 270*a^2*b*tan(d*x)^2*tan(c)^3 + 360*a*b^2*t an(d*x)^2*tan(c)^3 - 150*b^3*tan(d*x)^2*tan(c)^3 + 30*a*b^2*tan(d*x)*tan(c )^4 - 25*b^3*tan(d*x)*tan(c)^4 - 3*b^3*tan(c)^5 + 75*a^3*d*x*tan(d*x)*tan( c) - 225*a^2*b*d*x*tan(d*x)*tan(c) + 225*a*b^2*d*x*tan(d*x)*tan(c) - 75...
Time = 10.47 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.49 \[ \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx=\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b-3\,a\,b^2+b^3\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^3}{a^3-3\,a^2\,b+3\,a\,b^2-b^3}\right )\,{\left (a-b\right )}^3}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a\,b^2-\frac {b^3}{3}\right )}{d} \]