Integrand size = 14, antiderivative size = 648 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=\frac {x}{(a+b)^2}+\frac {\left (\sqrt {a}-\sqrt {b}\right ) \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} (a+b)^2 d}+\frac {\left (\sqrt {a}-3 \sqrt {b}\right ) \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} (a+b) d}-\frac {\left (\sqrt {a}-\sqrt {b}\right ) \sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} (a+b)^2 d}-\frac {\left (\sqrt {a}-3 \sqrt {b}\right ) \sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} (a+b) d}-\frac {\left (\sqrt {a}+\sqrt {b}\right ) \sqrt [4]{b} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{4 \sqrt {2} a^{3/4} (a+b)^2 d}-\frac {\left (\sqrt {a}+3 \sqrt {b}\right ) \sqrt [4]{b} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b) d}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \sqrt [4]{b} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{4 \sqrt {2} a^{3/4} (a+b)^2 d}+\frac {\left (\sqrt {a}+3 \sqrt {b}\right ) \sqrt [4]{b} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b) d}+\frac {b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )} \]
x/(a+b)^2+1/16*b^(1/4)*arctan(1-b^(1/4)*2^(1/2)*tan(d*x+c)/a^(1/4))*(a^(1/ 2)-3*b^(1/2))/a^(7/4)/(a+b)/d*2^(1/2)-1/16*b^(1/4)*arctan(1+b^(1/4)*2^(1/2 )*tan(d*x+c)/a^(1/4))*(a^(1/2)-3*b^(1/2))/a^(7/4)/(a+b)/d*2^(1/2)+1/4*b^(1 /4)*arctan(1-b^(1/4)*2^(1/2)*tan(d*x+c)/a^(1/4))*(a^(1/2)-b^(1/2))/a^(3/4) /(a+b)^2/d*2^(1/2)-1/4*b^(1/4)*arctan(1+b^(1/4)*2^(1/2)*tan(d*x+c)/a^(1/4) )*(a^(1/2)-b^(1/2))/a^(3/4)/(a+b)^2/d*2^(1/2)-1/8*b^(1/4)*ln(a^(1/2)-a^(1/ 4)*b^(1/4)*2^(1/2)*tan(d*x+c)+b^(1/2)*tan(d*x+c)^2)*(a^(1/2)+b^(1/2))/a^(3 /4)/(a+b)^2/d*2^(1/2)+1/8*b^(1/4)*ln(a^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*tan(d *x+c)+b^(1/2)*tan(d*x+c)^2)*(a^(1/2)+b^(1/2))/a^(3/4)/(a+b)^2/d*2^(1/2)-1/ 32*b^(1/4)*ln(a^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*tan(d*x+c)+b^(1/2)*tan(d*x+c )^2)*(a^(1/2)+3*b^(1/2))/a^(7/4)/(a+b)/d*2^(1/2)+1/32*b^(1/4)*ln(a^(1/2)+a ^(1/4)*b^(1/4)*2^(1/2)*tan(d*x+c)+b^(1/2)*tan(d*x+c)^2)*(a^(1/2)+3*b^(1/2) )/a^(7/4)/(a+b)/d*2^(1/2)+1/4*b*tan(d*x+c)*(1-tan(d*x+c)^2)/a/(a+b)/d/(a+t an(d*x+c)^4*b)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.34 (sec) , antiderivative size = 609, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=\frac {\arctan (\tan (c+d x))}{(a+b)^2 d}-\frac {3 b^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} (a+b) d}+\frac {3 b^{3/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} (a+b) d}+\frac {\left (\sqrt {a}-\sqrt {b}\right ) \left (\frac {\sqrt {2} \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}-\frac {\sqrt {2} \sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}\right )}{4 \sqrt {a} (a+b)^2 d}-\frac {3 b^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b) d}+\frac {3 b^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b) d}-\frac {\left (\sqrt {a}+\sqrt {b}\right ) \left (\frac {\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}-\frac {\sqrt {2} \sqrt [4]{b} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {b} \tan ^2(c+d x)\right )}{\sqrt [4]{a}}\right )}{8 \sqrt {a} (a+b)^2 d}-\frac {b \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {7}{4},-\frac {b \tan ^4(c+d x)}{a}\right ) \tan ^3(c+d x)}{3 a^2 (a+b) d}+\frac {b \tan (c+d x)}{4 a (a+b) d \left (a+b \tan ^4(c+d x)\right )} \]
ArcTan[Tan[c + d*x]]/((a + b)^2*d) - (3*b^(3/4)*ArcTan[1 - (Sqrt[2]*b^(1/4 )*Tan[c + d*x])/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(a + b)*d) + (3*b^(3/4)*ArcTa n[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(a + b)* d) + ((Sqrt[a] - Sqrt[b])*((Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Ta n[c + d*x])/a^(1/4)])/a^(1/4) - (Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/ 4)*Tan[c + d*x])/a^(1/4)])/a^(1/4)))/(4*Sqrt[a]*(a + b)^2*d) - (3*b^(3/4)* Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^ 2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) + (3*b^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1 /4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)*d) - ((Sqrt[a] + Sqrt[b])*((Sqrt[2]*b^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^ (1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/a^(1/4) - (Sqrt[2]*b ^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/a^(1/4)))/(8*Sqrt[a]*(a + b)^2*d) - (b*Hypergeometric2F1[3/4, 2 , 7/4, -((b*Tan[c + d*x]^4)/a)]*Tan[c + d*x]^3)/(3*a^2*(a + b)*d) + (b*Tan [c + d*x])/(4*a*(a + b)*d*(a + b*Tan[c + d*x]^4))
Time = 0.70 (sec) , antiderivative size = 631, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4144, 1568, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \tan (c+d x)^4\right )^2}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^4(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1568 |
| \(\displaystyle \frac {\int \left (\frac {b-b \tan ^2(c+d x)}{(a+b)^2 \left (b \tan ^4(c+d x)+a\right )}+\frac {b-b \tan ^2(c+d x)}{(a+b) \left (b \tan ^4(c+d x)+a\right )^2}+\frac {1}{(a+b)^2 \left (\tan ^2(c+d x)+1\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sqrt [4]{b} \left (\sqrt {a}-3 \sqrt {b}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} (a+b)}+\frac {\sqrt [4]{b} \left (\sqrt {a}-\sqrt {b}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{3/4} (a+b)^2}-\frac {\sqrt [4]{b} \left (\sqrt {a}-3 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} (a+b)}-\frac {\sqrt [4]{b} \left (\sqrt {a}-\sqrt {b}\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} a^{3/4} (a+b)^2}-\frac {\sqrt [4]{b} \left (\sqrt {a}+3 \sqrt {b}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b)}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )}{4 \sqrt {2} a^{3/4} (a+b)^2}+\frac {\sqrt [4]{b} \left (\sqrt {a}+3 \sqrt {b}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )}{16 \sqrt {2} a^{7/4} (a+b)}+\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \tan (c+d x)+\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )}{4 \sqrt {2} a^{3/4} (a+b)^2}+\frac {\arctan (\tan (c+d x))}{(a+b)^2}+\frac {b \tan (c+d x) \left (1-\tan ^2(c+d x)\right )}{4 a (a+b) \left (a+b \tan ^4(c+d x)\right )}}{d}\) |
(ArcTan[Tan[c + d*x]]/(a + b)^2 + ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*(a + b)^2) + ( (Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^ (1/4)])/(8*Sqrt[2]*a^(7/4)*(a + b)) - ((Sqrt[a] - Sqrt[b])*b^(1/4)*ArcTan[ 1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*(a + b)^2) - ((Sqrt[a] - 3*Sqrt[b])*b^(1/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Tan[c + d*x] )/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(a + b)) - ((Sqrt[a] + Sqrt[b])*b^(1/4)*Log [Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2]) /(4*Sqrt[2]*a^(3/4)*(a + b)^2) - ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a ] - Sqrt[2]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(16*Sq rt[2]*a^(7/4)*(a + b)) + ((Sqrt[a] + Sqrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2 ]*a^(1/4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(4*Sqrt[2]*a^(3/ 4)*(a + b)^2) + ((Sqrt[a] + 3*Sqrt[b])*b^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/ 4)*b^(1/4)*Tan[c + d*x] + Sqrt[b]*Tan[c + d*x]^2])/(16*Sqrt[2]*a^(7/4)*(a + b)) + (b*Tan[c + d*x]*(1 - Tan[c + d*x]^2))/(4*a*(a + b)*(a + b*Tan[c + d*x]^4)))/d
3.4.86.3.1 Defintions of rubi rules used
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int [ExpandIntegrand[(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, p, q}, x] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0])
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 = 0.37 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.55
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 a}-\frac {\left (a +b \right ) \tan \left (d x +c \right )}{4 a}}{a +\tan \left (d x +c \right )^{4} b}+\frac {\frac {\left (-7 a -3 b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}{\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )\right )}{8 a}+\frac {\left (5 a +b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}{\tan \left (d x +c \right )^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{4 a}\right )}{\left (a +b \right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a +b \right )^{2}}}{d}\) | \(356\) |
| default | \(\frac {-\frac {b \left (\frac {\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 a}-\frac {\left (a +b \right ) \tan \left (d x +c \right )}{4 a}}{a +\tan \left (d x +c \right )^{4} b}+\frac {\frac {\left (-7 a -3 b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}{\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )\right )}{8 a}+\frac {\left (5 a +b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}{\tan \left (d x +c \right )^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x +c \right ) \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{4 a}\right )}{\left (a +b \right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a +b \right )^{2}}}{d}\) | \(356\) |
| risch | \(\text {Expression too large to display}\) | \(1197\) |
1/d*(-1/(a+b)^2*b*((1/4*(a+b)/a*tan(d*x+c)^3-1/4*(a+b)/a*tan(d*x+c))/(a+ta n(d*x+c)^4*b)+1/4/a*(1/8*(-7*a-3*b)*(a/b)^(1/4)/a*2^(1/2)*(ln((tan(d*x+c)^ 2+(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(tan(d*x+c)^2-(a/b)^(1/4)*ta n(d*x+c)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1)- 2*arctan(-2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1))+1/8*(5*a+b)/b/(a/b)^(1/4)*2^( 1/2)*(ln((tan(d*x+c)^2-(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2))/(tan(d* x+c)^2+(a/b)^(1/4)*tan(d*x+c)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b) ^(1/4)*tan(d*x+c)+1)-2*arctan(-2^(1/2)/(a/b)^(1/4)*tan(d*x+c)+1))))+1/(a+b )^2*arctan(tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 4291 vs. \(2 (484) = 968\).
Time = 0.53 (sec) , antiderivative size = 4291, normalized size of antiderivative = 6.62 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
1/32*(32*a*b*d*x*tan(d*x + c)^4 + 32*a^2*d*x - 8*(a*b + b^2)*tan(d*x + c)^ 3 + ((a^3*b + 2*a^2*b^2 + a*b^3)*d*tan(d*x + c)^4 + (a^4 + 2*a^3*b + a^2*b ^2)*d)*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2*sqrt(-( 625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738 *a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a^11*b ^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) + 70*a^2*b + 44 *a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d^2))*l og((625*a^5 - 750*a^4*b - 1376*a^3*b^2 - 594*a^2*b^3 - 81*a*b^4 + (625*a^4 *b - 750*a^3*b^2 - 1376*a^2*b^3 - 594*a*b^4 - 81*b^5)*tan(d*x + c)^2 + 2*( 2*(a^11 + 5*a^10*b + 10*a^9*b^2 + 10*a^8*b^3 + 5*a^7*b^4 + a^6*b^5)*d^3*sq rt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 2383*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12*b^3 + 70*a ^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4))*tan(d*x + c) + (125*a^7 + 5*a^6*b - 442*a^5*b^2 - 490*a^4*b^3 - 195*a^3*b^4 - 27*a^2 *b^5)*d*tan(d*x + c))*sqrt(((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b ^4)*d^2*sqrt(-(625*a^6*b - 1950*a^5*b^2 - 529*a^4*b^3 + 2748*a^3*b^4 + 238 3*a^2*b^5 + 738*a*b^6 + 81*b^7)/((a^15 + 8*a^14*b + 28*a^13*b^2 + 56*a^12* b^3 + 70*a^11*b^4 + 56*a^10*b^5 + 28*a^9*b^6 + 8*a^8*b^7 + a^7*b^8)*d^4)) + 70*a^2*b + 44*a*b^2 + 6*b^3)/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a ^3*b^4)*d^2)) + ((25*a^9 + 109*a^8*b + 186*a^7*b^2 + 154*a^6*b^3 + 61*a...
Timed out. \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 394, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=-\frac {\frac {b {\left (\frac {2 \, \sqrt {2} {\left (b {\left (\sqrt {a} - 3 \, \sqrt {b}\right )} + 5 \, a^{\frac {3}{2}} - 7 \, a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} \tan \left (d x + c\right ) + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} {\left (b {\left (\sqrt {a} - 3 \, \sqrt {b}\right )} + 5 \, a^{\frac {3}{2}} - 7 \, a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} \tan \left (d x + c\right ) - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} {\left (b {\left (\sqrt {a} + 3 \, \sqrt {b}\right )} + 5 \, a^{\frac {3}{2}} + 7 \, a \sqrt {b}\right )} \log \left (\sqrt {b} \tan \left (d x + c\right )^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \tan \left (d x + c\right ) + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (b {\left (\sqrt {a} + 3 \, \sqrt {b}\right )} + 5 \, a^{\frac {3}{2}} + 7 \, a \sqrt {b}\right )} \log \left (\sqrt {b} \tan \left (d x + c\right )^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \tan \left (d x + c\right ) + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}}\right )}}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac {8 \, {\left (b \tan \left (d x + c\right )^{3} - b \tan \left (d x + c\right )\right )}}{{\left (a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{4} + a^{3} + a^{2} b} - \frac {32 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}}}{32 \, d} \]
-1/32*(b*(2*sqrt(2)*(b*(sqrt(a) - 3*sqrt(b)) + 5*a^(3/2) - 7*a*sqrt(b))*ar ctan(1/2*sqrt(2)*(2*sqrt(b)*tan(d*x + c) + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(s qrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*(b*(s qrt(a) - 3*sqrt(b)) + 5*a^(3/2) - 7*a*sqrt(b))*arctan(1/2*sqrt(2)*(2*sqrt( b)*tan(d*x + c) - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a) *sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*(b*(sqrt(a) + 3*sqrt(b)) + 5*a^( 3/2) + 7*a*sqrt(b))*log(sqrt(b)*tan(d*x + c)^2 + sqrt(2)*a^(1/4)*b^(1/4)*t an(d*x + c) + sqrt(a))/(a^(3/4)*b^(3/4)) + sqrt(2)*(b*(sqrt(a) + 3*sqrt(b) ) + 5*a^(3/2) + 7*a*sqrt(b))*log(sqrt(b)*tan(d*x + c)^2 - sqrt(2)*a^(1/4)* b^(1/4)*tan(d*x + c) + sqrt(a))/(a^(3/4)*b^(3/4)))/(a^3 + 2*a^2*b + a*b^2) + 8*(b*tan(d*x + c)^3 - b*tan(d*x + c))/((a^2*b + a*b^2)*tan(d*x + c)^4 + a^3 + a^2*b) - 32*(d*x + c)/(a^2 + 2*a*b + b^2))/d
Time = 0.87 (sec) , antiderivative size = 517, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \tan \left (d x + c\right )\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )} {\left (\left (a b^{3}\right )^{\frac {3}{4}} {\left (5 \, a + b\right )} - \left (a b^{3}\right )^{\frac {1}{4}} {\left (7 \, a b^{2} + 3 \, b^{3}\right )}\right )}}{\sqrt {2} a^{4} b^{2} + 2 \, \sqrt {2} a^{3} b^{3} + \sqrt {2} a^{2} b^{4}} + \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \tan \left (d x + c\right )\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )} {\left (\left (a b^{3}\right )^{\frac {3}{4}} {\left (5 \, a + b\right )} - \left (a b^{3}\right )^{\frac {1}{4}} {\left (7 \, a b^{2} + 3 \, b^{3}\right )}\right )}}{\sqrt {2} a^{4} b^{2} + 2 \, \sqrt {2} a^{3} b^{3} + \sqrt {2} a^{2} b^{4}} - \frac {{\left (\left (a b^{3}\right )^{\frac {3}{4}} {\left (5 \, a + b\right )} + \left (a b^{3}\right )^{\frac {1}{4}} {\left (7 \, a b^{2} + 3 \, b^{3}\right )}\right )} \log \left (\tan \left (d x + c\right )^{2} + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x + c\right ) + \sqrt {\frac {a}{b}}\right )}{\sqrt {2} a^{4} b^{2} + 2 \, \sqrt {2} a^{3} b^{3} + \sqrt {2} a^{2} b^{4}} + \frac {{\left (\left (a b^{3}\right )^{\frac {3}{4}} {\left (5 \, a + b\right )} + \left (a b^{3}\right )^{\frac {1}{4}} {\left (7 \, a b^{2} + 3 \, b^{3}\right )}\right )} \log \left (\tan \left (d x + c\right )^{2} - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} \tan \left (d x + c\right ) + \sqrt {\frac {a}{b}}\right )}{\sqrt {2} a^{4} b^{2} + 2 \, \sqrt {2} a^{3} b^{3} + \sqrt {2} a^{2} b^{4}} - \frac {16 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {4 \, {\left (b \tan \left (d x + c\right )^{3} - b \tan \left (d x + c\right )\right )}}{{\left (b \tan \left (d x + c\right )^{4} + a\right )} {\left (a^{2} + a b\right )}}}{16 \, d} \]
-1/16*(2*(pi*floor((d*x + c)/pi + 1/2) + arctan(1/2*sqrt(2)*(sqrt(2)*(a/b) ^(1/4) + 2*tan(d*x + c))/(a/b)^(1/4)))*((a*b^3)^(3/4)*(5*a + b) - (a*b^3)^ (1/4)*(7*a*b^2 + 3*b^3))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2)*a^ 2*b^4) + 2*(pi*floor((d*x + c)/pi + 1/2) + arctan(-1/2*sqrt(2)*(sqrt(2)*(a /b)^(1/4) - 2*tan(d*x + c))/(a/b)^(1/4)))*((a*b^3)^(3/4)*(5*a + b) - (a*b^ 3)^(1/4)*(7*a*b^2 + 3*b^3))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2) *a^2*b^4) - ((a*b^3)^(3/4)*(5*a + b) + (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))*lo g(tan(d*x + c)^2 + sqrt(2)*(a/b)^(1/4)*tan(d*x + c) + sqrt(a/b))/(sqrt(2)* a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2)*a^2*b^4) + ((a*b^3)^(3/4)*(5*a + b) + (a*b^3)^(1/4)*(7*a*b^2 + 3*b^3))*log(tan(d*x + c)^2 - sqrt(2)*(a/b)^(1/4 )*tan(d*x + c) + sqrt(a/b))/(sqrt(2)*a^4*b^2 + 2*sqrt(2)*a^3*b^3 + sqrt(2) *a^2*b^4) - 16*(d*x + c)/(a^2 + 2*a*b + b^2) + 4*(b*tan(d*x + c)^3 - b*tan (d*x + c))/((b*tan(d*x + c)^4 + a)*(a^2 + a*b)))/d
Time = 15.58 (sec) , antiderivative size = 11516, normalized size of antiderivative = 17.77 \[ \int \frac {1}{\left (a+b \tan ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
((b*tan(c + d*x))/(4*a*(a + b)) - (b*tan(c + d*x)^3)/(4*a*(a + b)))/(d*(a + b*tan(c + d*x)^4)) - (2*atan((((((((((960*a^7*b^8 - 224*a^5*b^10 - 144*a ^6*b^9 - 48*a^4*b^11 + 2480*a^8*b^7 + 2592*a^9*b^6 + 1296*a^10*b^5 + 256*a ^11*b^4)*1i)/(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2) - (tan(c + d*x)*(65536*a^6*b^11 + 327680*a^7*b^10 + 589824*a^8*b^9 + 327680*a^9*b^8 - 327680*a^10*b^7 - 589824*a^11*b^6 - 327680*a^12*b^5 - 65536*a^13*b^4))/(1 28*(4*a*b + 2*a^2 + 2*b^2)*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^ 2)))*1i)/(4*a*b + 2*a^2 + 2*b^2) - (tan(c + d*x)*(1152*a^2*b^11 + 7936*a^3 *b^10 + 20352*a^4*b^9 + 8704*a^5*b^8 - 66688*a^6*b^7 - 110848*a^7*b^6 - 49 024*a^8*b^5)*1i)/(128*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2)))* 1i)/(4*a*b + 2*a^2 + 2*b^2) - (((45*a*b^10)/16 + (305*a^2*b^9)/16 + (385*a ^3*b^8)/8 + (657*a^4*b^7)/8 + (2081*a^5*b^6)/16 + (1277*a^6*b^5)/16)*1i)/( 4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2))/(4*a*b + 2*a^2 + 2*b^2) - (tan(c + d*x)*(612*a*b^8 + 81*b^9 + 1894*a^2*b^7 + 2532*a^3*b^6 + 1425*a ^4*b^5))/(128*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2)))/(4*a*b + 2*a^2 + 2*b^2) - ((((((((960*a^7*b^8 - 224*a^5*b^10 - 144*a^6*b^9 - 48*a^ 4*b^11 + 2480*a^8*b^7 + 2592*a^9*b^6 + 1296*a^10*b^5 + 256*a^11*b^4)*1i)/( 4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2) + (tan(c + d*x)*(65536*a^ 6*b^11 + 327680*a^7*b^10 + 589824*a^8*b^9 + 327680*a^9*b^8 - 327680*a^10*b ^7 - 589824*a^11*b^6 - 327680*a^12*b^5 - 65536*a^13*b^4))/(128*(4*a*b +...