Integrand size = 25, antiderivative size = 377 \[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {1}{4} \sqrt [3]{a-i b} (A-i B) x-\frac {1}{4} \sqrt [3]{a+i b} (A+i B) x-\frac {\sqrt {3} \sqrt [3]{a-i b} (i A+B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}+\frac {\sqrt {3} \sqrt [3]{a+i b} (i A-B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}-\frac {\sqrt [3]{a+i b} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {\sqrt [3]{a-i b} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 \sqrt [3]{a-i b} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 \sqrt [3]{a+i b} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d} \]
-1/4*(a-I*b)^(1/3)*(A-I*B)*x-1/4*(a+I*b)^(1/3)*(A+I*B)*x-1/4*(a+I*b)^(1/3) *(I*A-B)*ln(cos(d*x+c))/d+1/4*(a-I*b)^(1/3)*(I*A+B)*ln(cos(d*x+c))/d+3/4*( a-I*b)^(1/3)*(I*A+B)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d-3/4*(a+I*b )^(1/3)*(I*A-B)*ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d-1/2*(a-I*b)^(1/ 3)*(I*A+B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))* 3^(1/2)/d+1/2*(a+I*b)^(1/3)*(I*A-B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3) /(a+I*b)^(1/3))*3^(1/2))*3^(1/2)/d+3*B*(a+b*tan(d*x+c))^(1/3)/d
Time = 0.99 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.92 \[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {i \left ((A-i B) \left (-\frac {1}{2} \sqrt [3]{a-i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )+3 \sqrt [3]{a+b \tan (c+d x)}\right )-(A+i B) \left (-\frac {1}{2} \sqrt [3]{a+i b} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )+3 \sqrt [3]{a+b \tan (c+d x)}\right )\right )}{2 d} \]
((I/2)*((A - I*B)*(-1/2*((a - I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b* Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] - 2*Log[(a - I*b)^(1/3) - ( a + b*Tan[c + d*x])^(1/3)] + 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)])) + 3*(a + b*Tan[c + d*x ])^(1/3)) - (A + I*B)*(-1/2*((a + I*b)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - 2*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] + 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)])) + 3*(a + b*Tan[c + d*x])^(1/3))))/d
Time = 0.73 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.71, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4011, 3042, 4022, 3042, 4020, 25, 69, 16, 1082, 217}
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.
| \(\displaystyle \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {a A-b B+(A b+a B) \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a A-b B+(A b+a B) \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {1}{2} (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{(a+b \tan (c+d x))^{2/3}}dx+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {1}{2} (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{(a+b \tan (c+d x))^{2/3}}dx+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (a-i b) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) (a+b \tan (c+d x))^{2/3}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) (a+b \tan (c+d x))^{2/3}}d(-i \tan (c+d x))}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (a-i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) (a+b \tan (c+d x))^{2/3}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) (a+b \tan (c+d x))^{2/3}}d(-i \tan (c+d x))}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {i (a-i b) (A-i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a-i b}}-\frac {3 \int \frac {1}{\sqrt [3]{a-i b}-i \tan (c+d x)}d\sqrt [3]{a+b \tan (c+d x)}}{2 (a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (a+i b) (A+i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a+i b}}-\frac {3 \int \frac {1}{i \tan (c+d x)+\sqrt [3]{a+i b}}d\sqrt [3]{a+b \tan (c+d x)}}{2 (a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}\right )}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i (a-i b) (A-i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (a+i b) (A+i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i (a-i b) (A-i B) \left (\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (\frac {2 i \tan (c+d x)}{\sqrt [3]{a-i b}}+1\right )}{(a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (a+i b) (A+i B) \left (\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (1-\frac {2 i \tan (c+d x)}{\sqrt [3]{a+i b}}\right )}{(a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i (a-i b) (A-i B) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{(a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (a+i b) (A+i B) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{(a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}+\frac {3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\) |
((I/2)*(a - I*b)*(A - I*B)*(((-I)*Sqrt[3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/( a - I*b)^(2/3) - Log[1 - I*Tan[c + d*x]]/(2*(a - I*b)^(2/3)) + (3*Log[(a - I*b)^(1/3) - I*Tan[c + d*x]])/(2*(a - I*b)^(2/3))))/d - ((I/2)*(a + I*b)* (A + I*B)*((I*Sqrt[3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a + I*b)^(2/3) - Log [1 + I*Tan[c + d*x]]/(2*(a + I*b)^(2/3)) + (3*Log[(a + I*b)^(1/3) + I*Tan[ c + d*x]])/(2*(a + I*b)^(2/3))))/d + (3*B*(a + b*Tan[c + d*x])^(1/3))/d
3.5.74.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.61 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.26
| method | result | size |
| derivativedivides | \(\frac {3 B \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{3}-B \,a^{2}-B \,b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\) | \(97\) |
| default | \(\frac {3 B \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{3}-B \,a^{2}-B \,b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\) | \(97\) |
| parts | \(\frac {A b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{3}-a}\right )}{2 d}+\frac {B \left (3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3} a +a^{2}+b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{2} \left (-\textit {\_R}^{3}+a \right )}\right )}{2}\right )}{d}\) | \(142\) |
1/d*(3*B*(a+b*tan(d*x+c))^(1/3)+1/2*sum(((A*b+B*a)*_R^3-B*a^2-B*b^2)/(_R^5 -_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2)))
Leaf count of result is larger than twice the leaf count of optimal. 2785 vs. \(2 (281) = 562\).
Time = 0.37 (sec) , antiderivative size = 2785, normalized size of antiderivative = 7.39 \[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
1/4*(2*d*(-(d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B - 10 *A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6)*b^2)/d^6) + (3*A^2 *B - B^3)*a + (A^3 - 3*A*B^2)*b)/d^3)^(1/3)*log(((A^5 - 2*A^3*B^2 - 3*A*B^ 4)*a - (3*A^4*B + 2*A^2*B^3 - B^5)*b)*(b*tan(d*x + c) + a)^(1/3) + (A*d^4* sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B ^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6)*b^2)/d^6) - ((A^3*B - 3*A*B^3)*a - (3*A^2*B^2 - B^4)*b)*d)*(-(d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6)*b^ 2)/d^6) + (3*A^2*B - B^3)*a + (A^3 - 3*A*B^2)*b)/d^3)^(1/3)) - (sqrt(-3)*d + d)*(-(d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B - 10*A^ 3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6)*b^2)/d^6) + (3*A^2*B - B^3)*a + (A^3 - 3*A*B^2)*b)/d^3)^(1/3)*log(((A^5 - 2*A^3*B^2 - 3*A*B^4)* a - (3*A^4*B + 2*A^2*B^3 - B^5)*b)*(b*tan(d*x + c) + a)^(1/3) + 1/2*(sqrt( -3)*((A^3*B - 3*A*B^3)*a - (3*A^2*B^2 - B^4)*b)*d + ((A^3*B - 3*A*B^3)*a - (3*A^2*B^2 - B^4)*b)*d - (sqrt(-3)*A*d^4 + A*d^4)*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6)*b^2)/d^6))*(-(d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^ 2 - 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6) *b^2)/d^6) + (3*A^2*B - B^3)*a + (A^3 - 3*A*B^2)*b)/d^3)^(1/3)) + (sqrt(-3 )*d - d)*(-(d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^2 - 2*(3*A^5*B -...
\[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt [3]{a + b \tan {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
Time = 20.45 (sec) , antiderivative size = 2537, normalized size of antiderivative = 6.73 \[ \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
log(a*d^7*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(4/3) + A*b*(a + b*tan( c + d*x))^(1/3)*(-A^6*a^2*d^6)^(1/2) - A^4*a^2*d^3*(a + b*tan(c + d*x))^(1 /3) + 2*A^3*a*b*d^4*(((-A^6*a^2*d^6)^(1/2) - A^3*b*d^3)/d^6)^(1/3))*(((-A^ 6*a^2*d^6)^(1/2) - A^3*b*d^3)/(8*d^6))^(1/3) + log(A*b*(a + b*tan(c + d*x) )^(1/3)*(-A^6*a^2*d^6)^(1/2) + A^4*a^2*d^3*(a + b*tan(c + d*x))^(1/3) + a* d*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3)*(-A^6*a^2*d^6)^(1/2) - A ^3*a*b*d^4*(-((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/d^6)^(1/3))*(-((-A^6*a^2*d ^6)^(1/2) + A^3*b*d^3)/(8*d^6))^(1/3) + log(d*(((-B^6*b^2*d^6)^(1/2) + B^3 *a*d^3)/d^6)^(1/3) - B*(a + b*tan(c + d*x))^(1/3))*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(8*d^6))^(1/3) + log(d*(-((-B^6*b^2*d^6)^(1/2) - B^3*a*d^3)/d ^6)^(1/3) - B*(a + b*tan(c + d*x))^(1/3))*(-((-B^6*b^2*d^6)^(1/2) - B^3*a* d^3)/(8*d^6))^(1/3) + (log(- ((3^(1/2)*1i - 1)*(((3^(1/2)*1i - 1)^2*(1944* a*b^4*(3^(1/2)*1i - 1)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3)*(a^2 + b^2) - (3888*B*a*b^4*(a^2 + b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(((-B^6 *b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(2/3))/16 - (972*B^3*b^4*(a^4 - b^4))/d^ 3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/d^6)^(1/3))/4 - (486*B^4*b^4*(a^4 - b^4)*(a + b*tan(c + d*x))^(1/3))/d^4)*(3^(1/2)*1i - 1)*((-B^6*b^2*d^6)^(1 /2)/(8*d^6) + (B^3*a)/(8*d^3))^(1/3))/2 - (log(- ((3^(1/2)*1i + 1)*(((3^(1 /2)*1i + 1)^2*(1944*a*b^4*(3^(1/2)*1i + 1)*(((-B^6*b^2*d^6)^(1/2) + B^3*a* d^3)/d^6)^(1/3)*(a^2 + b^2) + (3888*B*a*b^4*(a^2 + b^2)*(a + b*tan(c + ...