Integrand size = 23, antiderivative size = 272 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]
1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/ 2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*( a-b)*(a^2+4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1 /4*(a-b)*(a^2+4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/ 2)+2*b*(3*a^2-b^2)*tan(d*x+c)^(1/2)/d+8/5*a*b^2*tan(d*x+c)^(3/2)/d+2/5*b^2 *tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))/d
Result contains complex when optimal does not.
Time = 0.84 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.44 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\frac {-5 \sqrt [4]{-1} (i a+b)^3 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-5 (-1)^{3/4} (a+i b)^3 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 a^2-5 b^2+5 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{5 d} \]
(-5*(-1)^(1/4)*(I*a + b)^3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 5*(-1)^ (3/4)*(a + I*b)^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c + d*x]]*(15*a^2 - 5*b^2 + 5*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2))/(5*d)
Time = 0.87 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.92, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4049, 27, 3042, 4113, 3042, 4011, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\tan (c+d x)} \left (12 a b^2 \tan ^2(c+d x)+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (12 a b^2 \tan ^2(c+d x)+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (12 a b^2 \tan (c+d x)^2+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 a \left (a^2-3 b^2\right )+5 b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 a \left (a^2-3 b^2\right )+5 b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2-3 b^2\right ) \tan (c+d x)-5 b \left (3 a^2-b^2\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2-3 b^2\right ) \tan (c+d x)-5 b \left (3 a^2-b^2\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 \int -\frac {5 \left (b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
(2*b^2*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))/(5*d) + ((-10*(-1/2*((a + b)*(a^2 - 4*a*b + b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((a - b)*(a^2 + 4*a*b + b^2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/d + (10*b*(3*a^2 - b^2)*Sqrt[Tan[c + d*x]])/d + (8*a*b^2*Tan[c + d*x]^(3/2))/d )/5
3.6.71.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
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[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Time = 0.05 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(250\) |
| default | \(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(250\) |
| parts | \(\frac {a^{3} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {b^{3} \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {3 a^{2} b \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(416\) |
1/d*(2/5*b^3*tan(d*x+c)^(5/2)+2*a*b^2*tan(d*x+c)^(3/2)+6*a^2*b*tan(d*x+c)^ (1/2)-2*b^3*tan(d*x+c)^(1/2)+1/4*(-3*a^2*b+b^3)*2^(1/2)*(ln((1+2^(1/2)*tan (d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arcta n(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*( a^3-3*a*b^2)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2 )*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arc tan(-1+2^(1/2)*tan(d*x+c)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 1388 vs. \(2 (234) = 468\).
Time = 0.29 (sec) , antiderivative size = 1388, normalized size of antiderivative = 5.10 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]
-1/10*(5*d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^1 0*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) )/d^2)*log(((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b ^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^ 5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^ 8 - 30*a^2*b^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27 *a^4*b^8 + 12*a^2*b^10 - b^12)*sqrt(tan(d*x + c))) - 5*d*sqrt((6*a^5*b - 2 0*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^ 6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(-((a^3 - 3*a*b^2) *d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^ 7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^ 10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4 ))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^ 12)*sqrt(tan(d*x + c))) - 5*d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 - d^2*s qrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^ 2*b^10 + b^12)/d^4))/d^2)*log(((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b ^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) + (3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b...
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sqrt {\tan {\left (c + d x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.88 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\frac {8 \, b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 40 \, a b^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (3 \, a^{2} b - b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{20 \, d} \]
1/20*(8*b^3*tan(d*x + c)^(5/2) + 40*a*b^2*tan(d*x + c)^(3/2) + 10*sqrt(2)* (a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d *x + c)))) + 10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2 )*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 40*(3*a^2*b - b^3)*sqrt(tan(d*x + c)))/d
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\text {Timed out} \]
Time = 7.79 (sec) , antiderivative size = 1742, normalized size of antiderivative = 6.40 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]
(2*b^3*tan(c + d*x)^(5/2))/(5*d) - atan((((8*(4*b^3*d^2 - 12*a^2*b*d^2)*(( 6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*1 5i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 2 0*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i - ((8*(4*b^3*d^2 - 12*a^2*b*d^2 )*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b ^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b ^4 - 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i)/(((8*(4*b^3*d^2 - 12*a^2*b *d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a ^4*b^2*15i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a ^2*b^4 - 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4 *15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2) - (16*(3*a*b^8 - a^9 + 8*a ^3*b^6 + 6*a^5*b^4))/d^3 + ((8*(4*b^3*d^2 - 12*a^2*b*d^2)*((6*a*b^5 + 6*a^ 5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^( 1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/ d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^ 4*b^2*15i)/(4*d^2))^(1/2)))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^ 4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*2i - atan((((8*(4*b^3*d^2 - 12*a^2*b*d^2)*((6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - ...