Integrand size = 14, antiderivative size = 256 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\frac {a x}{a^2+b^2}+\frac {\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a^2+b^2\right ) d}-\frac {b \log \left (a \cos ^3(c+d x)+b \sin ^3(c+d x)\right )}{3 \left (a^2+b^2\right ) d}+\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )}{3 a^{2/3} \left (a^2+b^2\right ) d}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )}{6 a^{2/3} \left (a^2+b^2\right ) d} \] Output:
a*x/(a^2+b^2)+1/3*b^(1/3)*(a^(4/3)-b^(4/3))*arctan(1/3*(a^(1/3)-2*b^(1/3)* tan(d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/a^(2/3)/(a^2+b^2)/d-1/3*b*ln(a*cos(d* x+c)^3+b*sin(d*x+c)^3)/(a^2+b^2)/d+1/3*b^(1/3)*(a^(4/3)+b^(4/3))*ln(a^(1/3 )+b^(1/3)*tan(d*x+c))/a^(2/3)/(a^2+b^2)/d-1/6*b^(1/3)*(a^(4/3)+b^(4/3))*ln (a^(2/3)-a^(1/3)*b^(1/3)*tan(d*x+c)+b^(2/3)*tan(d*x+c)^2)/a^(2/3)/(a^2+b^2 )/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.42 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.09 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\frac {-2 \sqrt {3} b^{5/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )-3 i a^{5/3} \log (i-\tan (c+d x))+3 a^{2/3} b \log (i-\tan (c+d x))+3 i a^{5/3} \log (i+\tan (c+d x))+3 a^{2/3} b \log (i+\tan (c+d x))+2 b^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )-b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )-2 a^{2/3} b \log \left (a+b \tan ^3(c+d x)\right )-3 a^{2/3} b \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b \tan ^3(c+d x)}{a}\right ) \tan ^2(c+d x)}{6 a^{2/3} \left (a^2+b^2\right ) d} \] Input:
Integrate[(a + b*Tan[c + d*x]^3)^(-1),x]
Output:
(-2*Sqrt[3]*b^(5/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x])/(Sqrt[3]*a^( 1/3))] - (3*I)*a^(5/3)*Log[I - Tan[c + d*x]] + 3*a^(2/3)*b*Log[I - Tan[c + d*x]] + (3*I)*a^(5/3)*Log[I + Tan[c + d*x]] + 3*a^(2/3)*b*Log[I + Tan[c + d*x]] + 2*b^(5/3)*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]] - b^(5/3)*Log[a^(2/ 3) - a^(1/3)*b^(1/3)*Tan[c + d*x] + b^(2/3)*Tan[c + d*x]^2] - 2*a^(2/3)*b* Log[a + b*Tan[c + d*x]^3] - 3*a^(2/3)*b*Hypergeometric2F1[2/3, 1, 5/3, -(( b*Tan[c + d*x]^3)/a)]*Tan[c + d*x]^2)/(6*a^(2/3)*(a^2 + b^2)*d)
Time = 0.58 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4144, 7276, 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.
| \(\displaystyle \int \frac {1}{a+b \tan ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a+b \tan (c+d x)^3}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^3(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {\int \left (\frac {a+b \tan (c+d x)}{\left (a^2+b^2\right ) \left (\tan ^2(c+d x)+1\right )}-\frac {b \left (b \tan ^2(c+d x)+a \tan (c+d x)-b\right )}{\left (a^2+b^2\right ) \left (b \tan ^3(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a \arctan (\tan (c+d x))}{a^2+b^2}-\frac {b \log \left (a+b \tan ^3(c+d x)\right )}{3 \left (a^2+b^2\right )}+\frac {b \log \left (\tan ^2(c+d x)+1\right )}{2 \left (a^2+b^2\right )}+\frac {\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \left (a^2+b^2\right )}-\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )}{6 a^{2/3} \left (a^2+b^2\right )}+\frac {\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )}{3 a^{2/3} \left (a^2+b^2\right )}}{d}\) |
Input:
Int[(a + b*Tan[c + d*x]^3)^(-1),x]
Output:
((a*ArcTan[Tan[c + d*x]])/(a^2 + b^2) + (b^(1/3)*(a^(4/3) - b^(4/3))*ArcTa n[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)* (a^2 + b^2)) + (b^(1/3)*(a^(4/3) + b^(4/3))*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]])/(3*a^(2/3)*(a^2 + b^2)) + (b*Log[1 + Tan[c + d*x]^2])/(2*(a^2 + b^2 )) - (b^(1/3)*(a^(4/3) + b^(4/3))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d* x] + b^(2/3)*Tan[c + d*x]^2])/(6*a^(2/3)*(a^2 + b^2)) - (b*Log[a + b*Tan[c + d*x]^3])/(3*(a^2 + b^2)))/d
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])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.57 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.04
| method | result | size |
| risch | \(\frac {x}{i b +a}+\frac {2 i b \,a^{2} d^{3} x}{a^{4} d^{3}+a^{2} b^{2} d^{3}}+\frac {2 i b \,a^{2} d^{2} c}{a^{4} d^{3}+a^{2} b^{2} d^{3}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (27 a^{4} d^{3}+27 a^{2} b^{2} d^{3}\right ) \textit {\_Z}^{3}+27 \textit {\_Z}^{2} a^{2} b \,d^{2}-b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (-\frac {18 d^{2} a^{4}}{a^{2}-b^{2}}-\frac {18 d^{2} b^{2} a^{2}}{a^{2}-b^{2}}\right ) \textit {\_R}^{2}+\left (\frac {6 i d \,a^{3}}{a^{2}-b^{2}}-\frac {6 i d \,b^{2} a}{a^{2}-b^{2}}-\frac {6 d b \,a^{2}}{a^{2}-b^{2}}\right ) \textit {\_R} +\frac {a^{2}}{a^{2}-b^{2}}+\frac {b^{2}}{a^{2}-b^{2}}\right )\right )\) | \(265\) |
| derivativedivides | \(\frac {\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {\left (-b \left (\frac {\ln \left (\tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+a \left (-\frac {\ln \left (\tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\ln \left (a +b \tan \left (d x +c \right )^{3}\right )}{3}\right ) b}{a^{2}+b^{2}}}{d}\) | \(293\) |
| default | \(\frac {\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {\left (-b \left (\frac {\ln \left (\tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+a \left (-\frac {\ln \left (\tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\tan \left (d x +c \right )^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \tan \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\ln \left (a +b \tan \left (d x +c \right )^{3}\right )}{3}\right ) b}{a^{2}+b^{2}}}{d}\) | \(293\) |
Input:
int(1/(a+b*tan(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
x/(a+I*b)+2*I*b*a^2*d^3/(a^4*d^3+a^2*b^2*d^3)*x+2*I*b*a^2*d^2/(a^4*d^3+a^2 *b^2*d^3)*c+sum(_R*ln(exp(2*I*(d*x+c))+(-18/(a^2-b^2)*d^2*a^4-18/(a^2-b^2) *d^2*b^2*a^2)*_R^2+(6*I/(a^2-b^2)*d*a^3-6*I/(a^2-b^2)*d*b^2*a-6/(a^2-b^2)* d*b*a^2)*_R+1/(a^2-b^2)*a^2+1/(a^2-b^2)*b^2),_R=RootOf((27*a^4*d^3+27*a^2* b^2*d^3)*_Z^3+27*_Z^2*a^2*b*d^2-b))
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 4817, normalized size of antiderivative = 18.82 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*tan(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\int \frac {1}{a + b \tan ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(1/(a+b*tan(d*x+c)**3),x)
Output:
Integral(1/(a + b*tan(c + d*x)**3), x)
Time = 0.11 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.14 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=-\frac {\frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2\right )} - b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {2 \, a}{b}\right )}\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \tan \left (d x + c\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {18 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac {3 \, {\left (b {\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} + a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\tan \left (d x + c\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x + c\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {9 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {6 \, {\left (b {\left (\left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} - a \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \tan \left (d x + c\right )\right )}{a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{18 \, d} \] Input:
integrate(1/(a+b*tan(d*x+c)^3),x, algorithm="maxima")
Output:
-1/18*(2*sqrt(3)*(a*(3*(a/b)^(2/3) - 2) - b*(3*(a/b)^(1/3) - 2*a/b))*arcta n(-1/3*sqrt(3)*((a/b)^(1/3) - 2*tan(d*x + c))/(a/b)^(1/3))/((a^2*(a/b)^(2/ 3) + b^2*(a/b)^(2/3))*(a/b)^(1/3)) - 18*(d*x + c)*a/(a^2 + b^2) + 3*(b*(2* (a/b)^(2/3) + 1) + a*(a/b)^(1/3))*log(tan(d*x + c)^2 - (a/b)^(1/3)*tan(d*x + c) + (a/b)^(2/3))/(a^2*(a/b)^(2/3) + b^2*(a/b)^(2/3)) - 9*b*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) + 6*(b*((a/b)^(2/3) - 1) - a*(a/b)^(1/3))*log((a/ b)^(1/3) + tan(d*x + c))/(a^2*(a/b)^(2/3) + b^2*(a/b)^(2/3)))/d
Time = 0.22 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.31 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\frac {{\left (a^{3} b^{2} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a b^{4} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b^{3} d - b^{5} d\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \tan \left (d x + c\right ) \right |}\right )}{3 \, {\left (a^{5} b d^{2} + 2 \, a^{3} b^{3} d^{2} + a b^{5} d^{2}\right )}} + \frac {{\left (d x + c\right )} a}{a^{2} d + b^{2} d} + \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {b \log \left ({\left | b \tan \left (d x + c\right )^{3} + a \right |}\right )}{3 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} + \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \tan \left (d x + c\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (\sqrt {3} a^{3} b + \sqrt {3} a b^{3}\right )} d} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac {2}{3}} a\right )} \log \left (\tan \left (d x + c\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x + c\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{3} b + a b^{3}\right )} d} \] Input:
integrate(1/(a+b*tan(d*x+c)^3),x, algorithm="giac")
Output:
1/3*(a^3*b^2*d*(-a/b)^(1/3) + a*b^4*d*(-a/b)^(1/3) - a^2*b^3*d - b^5*d)*(- a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + tan(d*x + c)))/(a^5*b*d^2 + 2*a^3*b^3*d ^2 + a*b^5*d^2) + (d*x + c)*a/(a^2*d + b^2*d) + 1/2*b*log(tan(d*x + c)^2 + 1)/(a^2*d + b^2*d) - 1/3*b*log(abs(b*tan(d*x + c)^3 + a))/(a^2*d + b^2*d) + ((-a*b^2)^(1/3)*b^2 + (-a*b^2)^(2/3)*a)*arctan(1/3*sqrt(3)*((-a/b)^(1/3 ) + 2*tan(d*x + c))/(-a/b)^(1/3))/((sqrt(3)*a^3*b + sqrt(3)*a*b^3)*d) + 1/ 6*((-a*b^2)^(1/3)*b^2 - (-a*b^2)^(2/3)*a)*log(tan(d*x + c)^2 + (-a/b)^(1/3 )*tan(d*x + c) + (-a/b)^(2/3))/((a^3*b + a*b^3)*d)
Time = 8.42 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.34 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\frac {\sum _{k=1}^3\ln \left (\mathrm {root}\left (27\,a^2\,b^2\,z^3+27\,a^4\,z^3+27\,a^2\,b\,z^2-b,z,k\right )\,\left (\mathrm {root}\left (27\,a^2\,b^2\,z^3+27\,a^4\,z^3+27\,a^2\,b\,z^2-b,z,k\right )\,\left (\mathrm {root}\left (27\,a^2\,b^2\,z^3+27\,a^4\,z^3+27\,a^2\,b\,z^2-b,z,k\right )\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (12\,b^6-69\,a^2\,b^4\right )+\mathrm {root}\left (27\,a^2\,b^2\,z^3+27\,a^4\,z^3+27\,a^2\,b\,z^2-b,z,k\right )\,\left (36\,a\,b^6-180\,a^3\,b^4+\mathrm {tan}\left (c+d\,x\right )\,\left (162\,a^2\,b^5-54\,a^4\,b^3\right )\right )-36\,a\,b^5+27\,a^3\,b^3\right )+13\,a\,b^4-16\,b^5\,\mathrm {tan}\left (c+d\,x\right )\right )+5\,b^4\,\mathrm {tan}\left (c+d\,x\right )\right )\right )\,\mathrm {root}\left (27\,a^2\,b^2\,z^3+27\,a^4\,z^3+27\,a^2\,b\,z^2-b,z,k\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \] Input:
int(1/(a + b*tan(c + d*x)^3),x)
Output:
symsum(log(root(27*a^2*b^2*z^3 + 27*a^4*z^3 + 27*a^2*b*z^2 - b, z, k)*(roo t(27*a^2*b^2*z^3 + 27*a^4*z^3 + 27*a^2*b*z^2 - b, z, k)*(root(27*a^2*b^2*z ^3 + 27*a^4*z^3 + 27*a^2*b*z^2 - b, z, k)*(tan(c + d*x)*(12*b^6 - 69*a^2*b ^4) + root(27*a^2*b^2*z^3 + 27*a^4*z^3 + 27*a^2*b*z^2 - b, z, k)*(36*a*b^6 - 180*a^3*b^4 + tan(c + d*x)*(162*a^2*b^5 - 54*a^4*b^3)) - 36*a*b^5 + 27* a^3*b^3) + 13*a*b^4 - 16*b^5*tan(c + d*x)) + 5*b^4*tan(c + d*x)))*root(27* a^2*b^2*z^3 + 27*a^4*z^3 + 27*a^2*b*z^2 - b, z, k), k, 1, 3)/d + log(tan(c + d*x) - 1i)/(2*d*(a*1i + b)) + (log(tan(c + d*x) + 1i)*1i)/(2*d*(a + b*1 i))
Time = 0.18 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.16 \[ \int \frac {1}{a+b \tan ^3(c+d x)} \, dx=\frac {-2 b^{\frac {7}{3}} a^{\frac {2}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} \tan \left (d x +c \right )}{a^{\frac {1}{3}} \sqrt {3}}\right )+2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} \tan \left (d x +c \right )}{a^{\frac {1}{3}} \sqrt {3}}\right ) a^{2} b -b^{\frac {7}{3}} a^{\frac {2}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} \tan \left (d x +c \right )+b^{\frac {2}{3}} \tan \left (d x +c \right )^{2}\right )+2 b^{\frac {7}{3}} a^{\frac {2}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} \tan \left (d x +c \right )\right )+3 b^{\frac {5}{3}} a^{\frac {4}{3}} \mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-2 b^{\frac {5}{3}} a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} \tan \left (d x +c \right )+b^{\frac {2}{3}} \tan \left (d x +c \right )^{2}\right )-2 b^{\frac {5}{3}} a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} \tan \left (d x +c \right )\right )+6 b^{\frac {2}{3}} a^{\frac {7}{3}} d x -\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} \tan \left (d x +c \right )+b^{\frac {2}{3}} \tan \left (d x +c \right )^{2}\right ) a^{2} b +2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} \tan \left (d x +c \right )\right ) a^{2} b}{6 b^{\frac {2}{3}} a^{\frac {4}{3}} d \left (a^{2}+b^{2}\right )} \] Input:
int(1/(a+b*tan(d*x+c)^3),x)
Output:
( - 2*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*tan(c + d*x))/ (a**(1/3)*sqrt(3)))*b**2 + 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*tan(c + d *x))/(a**(1/3)*sqrt(3)))*a**2*b - b**(1/3)*a**(2/3)*log(a**(2/3) - b**(1/3 )*a**(1/3)*tan(c + d*x) + b**(2/3)*tan(c + d*x)**2)*b**2 + 2*b**(1/3)*a**( 2/3)*log(a**(1/3) + b**(1/3)*tan(c + d*x))*b**2 + 3*b**(2/3)*a**(1/3)*log( tan(c + d*x)**2 + 1)*a*b - 2*b**(2/3)*a**(1/3)*log(a**(2/3) - b**(1/3)*a** (1/3)*tan(c + d*x) + b**(2/3)*tan(c + d*x)**2)*a*b - 2*b**(2/3)*a**(1/3)*l og(a**(1/3) + b**(1/3)*tan(c + d*x))*a*b + 6*b**(2/3)*a**(1/3)*a**2*d*x - log(a**(2/3) - b**(1/3)*a**(1/3)*tan(c + d*x) + b**(2/3)*tan(c + d*x)**2)* a**2*b + 2*log(a**(1/3) + b**(1/3)*tan(c + d*x))*a**2*b)/(6*b**(2/3)*a**(1 /3)*a*d*(a**2 + b**2))