Integrand size = 14, antiderivative size = 415 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=-\frac {x}{4 \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {x}{4 \left (a+\sqrt {-b^2}\right )^{2/3}}+\frac {\sqrt {3} b \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3} d}-\frac {\sqrt {3} b \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3} d}-\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3} d}+\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3} d}-\frac {3 b \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3} d}+\frac {3 b \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3} d} \]
-1/4*x/(a-(-b^2)^(1/2))^(2/3)-1/4*b*ln(cos(d*x+c))/d/(a-(-b^2)^(1/2))^(2/3 )/(-b^2)^(1/2)-3/4*b*ln((a-(-b^2)^(1/2))^(1/3)-(a+b*tan(d*x+c))^(1/3))/d/( a-(-b^2)^(1/2))^(2/3)/(-b^2)^(1/2)+1/2*b*arctan(1/3*(1+2*(a+b*tan(d*x+c))^ (1/3)/(a-(-b^2)^(1/2))^(1/3))*3^(1/2))*3^(1/2)/d/(a-(-b^2)^(1/2))^(2/3)/(- b^2)^(1/2)-1/4*x/(a+(-b^2)^(1/2))^(2/3)+1/4*b*ln(cos(d*x+c))/d/(-b^2)^(1/2 )/(a+(-b^2)^(1/2))^(2/3)+3/4*b*ln((a+(-b^2)^(1/2))^(1/3)-(a+b*tan(d*x+c))^ (1/3))/d/(-b^2)^(1/2)/(a+(-b^2)^(1/2))^(2/3)-1/2*b*arctan(1/3*(1+2*(a+b*ta n(d*x+c))^(1/3)/(a+(-b^2)^(1/2))^(1/3))*3^(1/2))*3^(1/2)/d/(-b^2)^(1/2)/(a +(-b^2)^(1/2))^(2/3)
Result contains complex when optimal does not.
Time = 0.41 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\frac {i \left (\frac {2 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{(a-i b)^{2/3}}-\frac {2 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{(a+i b)^{2/3}}-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )}{(a-i b)^{2/3}}+\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )}{(a+i b)^{2/3}}\right )}{4 d} \]
((I/4)*((2*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(a - I*b)^(2 /3) - (2*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(a + I*b)^(2/3 ) - (2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3)) /Sqrt[3]] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/ 3) + (a + b*Tan[c + d*x])^(2/3)])/(a - I*b)^(2/3) + (2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] + Log[(a + I*b)^ (2/3) + (a + I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^ (2/3)])/(a + I*b)^(2/3)))/d
Time = 0.49 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3966, 485, 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.
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{2/3}}dx\) |
\(\Big \downarrow \) 3966 |
\(\displaystyle \frac {b \int \frac {1}{(a+b \tan (c+d x))^{2/3} \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 485 |
\(\displaystyle \frac {b \int \left (\frac {\sqrt {-b^2}}{2 b^2 \left (\sqrt {-b^2}-b \tan (c+d x)\right ) (a+b \tan (c+d x))^{2/3}}+\frac {\sqrt {-b^2}}{2 b^2 (a+b \tan (c+d x))^{2/3} \left (b \tan (c+d x)+\sqrt {-b^2}\right )}\right )d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3}}-\frac {\log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{4 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3}}+\frac {\log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{4 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3}}-\frac {3 \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right )^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right )^{2/3}}\right )}{d}\) |
(b*((Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - Sqrt[-b^2])^( 1/3))/Sqrt[3]])/(2*Sqrt[-b^2]*(a - Sqrt[-b^2])^(2/3)) - (Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + Sqrt[-b^2])^(1/3))/Sqrt[3]])/(2*Sqr t[-b^2]*(a + Sqrt[-b^2])^(2/3)) - Log[Sqrt[-b^2] - b*Tan[c + d*x]]/(4*Sqrt [-b^2]*(a + Sqrt[-b^2])^(2/3)) + Log[Sqrt[-b^2] + b*Tan[c + d*x]]/(4*Sqrt[ -b^2]*(a - Sqrt[-b^2])^(2/3)) - (3*Log[(a - Sqrt[-b^2])^(1/3) - (a + b*Tan [c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a - Sqrt[-b^2])^(2/3)) + (3*Log[(a + Sqr t[-b^2])^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*Sqrt[-b^2]*(a + Sqrt[-b^2 ])^(2/3))))/d
3.7.93.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & & !IntegerQ[2*n]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Su bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && NeQ[a^2 + b^2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.66 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.14
method | result | size |
derivativedivides | \(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\) | \(57\) |
default | \(\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\) | \(57\) |
Leaf count of result is larger than twice the leaf count of optimal. 2139 vs. \(2 (325) = 650\).
Time = 0.32 (sec) , antiderivative size = 2139, normalized size of antiderivative = 5.15 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \]
-1/4*(sqrt(-3) + 1)*(((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^ 4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d*x + c) + a)^(1/ 3) + 1/2*(sqrt(-3)*(a^2*b - b^3)*d + (a^2*b - b^3)*d - (sqrt(-3)*(a^5 + 2* a^3*b^2 + a*b^4)*d^4 + (a^5 + 2*a^3*b^2 + a*b^4)*d^4)*sqrt(-(a^4 - 2*a^2*b ^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)))*(((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6* a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^( 1/3)) + 1/4*(sqrt(-3) - 1)*(((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^ 2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a* b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d*x + c) + a)^(1/3) - 1/2*(sqrt(-3)*(a^2*b - b^3)*d - (a^2*b - b^3)*d - (sqrt(-3)*(a ^5 + 2*a^3*b^2 + a*b^4)*d^4 - (a^5 + 2*a^3*b^2 + a*b^4)*d^4)*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)))*( ((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b ^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)* d^3))^(1/3)) - 1/4*(sqrt(-3) + 1)*(-((a^4 + 2*a^2*b^2 + b^4)*d^3*sqrt(-(a^ 4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^6) ) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^3))^(1/3)*log(-(a^2 - b^2)*(b*tan(d* x + c) + a)^(1/3) + 1/2*(sqrt(-3)*(a^2*b - b^3)*d + (a^2*b - b^3)*d + (...
\[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
Time = 9.43 (sec) , antiderivative size = 1048, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \]
(log(((-1i/(d^3*(a*1i - b)^2))^(4/3)*(a^2*b^5*d*486i - b^7*d*486i + 486*a^ 3*b^4*d - 486*a*b^6*d + (972*a*b^5*(a + b*tan(c + d*x))^(1/3))/(-1i/(d^3*( a*1i - b)^2))^(1/3)))/d - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1/(b^ 2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3))^(1/3))/2 + log((((7776*a*b^5*(a + b*ta n(c + d*x))^(1/3))/d + 7776*a*b^4*(a^2 + b^2)*(1i/(8*d^3*(a*1i + b)^2))^(1 /3))*(1i/(8*d^3*(a*1i + b)^2))^(2/3) - (972*b^5)/d^3)*(1i/(8*d^3*(a*1i + b )^2))^(1/3) - (486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(1i/(8*(b^2*d^3 - a^2*d^3 + a*b*d^3*2i)))^(1/3) + (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/ d^4 + ((3^(1/2)*1i - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*a*b^5* (a + b*tan(c + d*x))^(1/3))/d + 1944*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*(- 1i/(d^3*(a*1i - b)^2))^(1/3))*(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^ 3*(a*1i - b)^2))^(1/3))/4)*(3^(1/2)*1i - 1)*(1/(b^2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3))^(1/3))/4 - (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((3 ^(1/2)*1i + 1)*((972*b^5)/d^3 - ((3^(1/2)*1i + 1)^2*((7776*a*b^5*(a + b*ta n(c + d*x))^(1/3))/d - 1944*a*b^4*(3^(1/2)*1i + 1)*(a^2 + b^2)*(-1i/(d^3*( a*1i - b)^2))^(1/3))*(-1i/(d^3*(a*1i - b)^2))^(2/3))/16)*(-1i/(d^3*(a*1i - b)^2))^(1/3))/4)*(3^(1/2)*1i + 1)*(1/(b^2*d^3*1i - a^2*d^3*1i + 2*a*b*d^3 ))^(1/3))/4 + (log((486*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 + ((3^(1/2)*1i - 1)*((972*b^5)/d^3 - ((3^(1/2)*1i - 1)^2*((7776*a*b^5*(a + b*tan(c + d*x ))^(1/3))/d + 3888*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*(1i/(8*d^3*(a*1i ...