Integrand size = 23, antiderivative size = 198 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d} \] Output:
-1/2*(a^2+2*a*b-b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d-1/2*(a^ 2+2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d-1/2*(a^2-2*a*b-b ^2)*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/d+2*(a^2-b^2) *tan(d*x+c)^(1/2)/d+4/3*a*b*tan(d*x+c)^(3/2)/d+2/5*b^2*tan(d*x+c)^(5/2)/d
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.61 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \sqrt [4]{-1} (a-i b)^2 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+15 \sqrt [4]{-1} (a+i b)^2 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (15 a^2-15 b^2+10 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{15 d} \] Input:
Integrate[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2,x]
Output:
(15*(-1)^(1/4)*(a - I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 15*(-1) ^(1/4)*(a + I*b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(15*a^2 - 15*b^2 + 10*a*b*Tan[c + d*x] + 3*b^2*Tan[c + d*x]^2))/(15 *d)
Time = 0.72 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.15, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 3042, 4017, 25, 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.
| \(\displaystyle \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^{3/2} (a+b \tan (c+d x))^2dx\) |
| \(\Big \downarrow \) 4026 |
| \(\displaystyle \int \tan ^{\frac {3}{2}}(c+d x) \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^{3/2} \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {-a^2-2 b \tan (c+d x) a+b^2}{\sqrt {\tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {-a^2-2 b \tan (c+d x) a+b^2}{\sqrt {\tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {2 \int -\frac {a^2+2 b \tan (c+d x) a-b^2}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int \frac {a^2+2 b \tan (c+d x) a-b^2}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 \left (-\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {4 a b \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b^2 \tan ^{\frac {5}{2}}(c+d x)}{5 d}\) |
Input:
Int[Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2,x]
Output:
(2*(-1/2*((a^2 + 2*a*b - b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sq rt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) - ((a^2 - 2*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 + ( 2*(a^2 - b^2)*Sqrt[Tan[c + d*x]])/d + (4*a*b*Tan[c + d*x]^(3/2))/(3*d) + ( 2*b^2*Tan[c + d*x]^(5/2))/(5*d)
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_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Time = 0.13 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 a b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {\tan \left (d x +c \right )}-2 b^{2} \sqrt {\tan \left (d x +c \right )}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}}{d}\) | \(238\) |
| default | \(\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 a b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {\tan \left (d x +c \right )}-2 b^{2} \sqrt {\tan \left (d x +c \right )}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}}{d}\) | \(238\) |
| parts | \(\frac {a^{2} \left (2 \sqrt {\tan \left (d x +c \right )}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {b^{2} \left (\frac {2 \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (d x +c \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {2 a b \left (\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\) | \(321\) |
Input:
int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(2/5*b^2*tan(d*x+c)^(5/2)+4/3*a*b*tan(d*x+c)^(3/2)+2*a^2*tan(d*x+c)^(1 /2)-2*b^2*tan(d*x+c)^(1/2)+1/4*(-a^2+b^2)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)* tan(d*x+c)^(1/2)+1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^ (1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))-1/2*a*b*2^( 1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)+2^(1/2)*tan(d *x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*t an(d*x+c)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (170) = 340\).
Time = 0.09 (sec) , antiderivative size = 673, normalized size of antiderivative = 3.40 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {30 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \arctan \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a^{2} - 2 \, a b - b^{2}\right )} d \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} + d^{2} \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}}}{a^{4} - 6 \, a^{2} b^{2} + b^{4}}\right ) + 30 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \arctan \left (-\frac {2 \, \sqrt {\frac {1}{2}} {\left (a^{2} - 2 \, a b - b^{2}\right )} d \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} - d^{2} \sqrt {\frac {a^{4} + 4 \, a^{3} b + 2 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}}}{a^{4} - 6 \, a^{2} b^{2} + b^{4}}\right ) + 15 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}} \log \left (2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} - a^{2} + 2 \, a b + b^{2} - {\left (a^{2} - 2 \, a b - b^{2}\right )} \tan \left (d x + c\right )\right ) - 15 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}} \log \left (-2 \, \sqrt {\frac {1}{2}} d \sqrt {\frac {a^{4} - 4 \, a^{3} b + 2 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}}{d^{2}}} \sqrt {\tan \left (d x + c\right )} - a^{2} + 2 \, a b + b^{2} - {\left (a^{2} - 2 \, a b - b^{2}\right )} \tan \left (d x + c\right )\right ) + 4 \, {\left (3 \, b^{2} \tan \left (d x + c\right )^{2} + 10 \, a b \tan \left (d x + c\right ) + 15 \, a^{2} - 15 \, b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{30 \, d} \] Input:
integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/30*(30*sqrt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2) *arctan(-(2*sqrt(1/2)*(a^2 - 2*a*b - b^2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^ 2 - 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c)) + d^2*sqrt((a^4 + 4*a^3*b + 2*a ^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b ^4)/d^2))/(a^4 - 6*a^2*b^2 + b^4)) + 30*sqrt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*arctan(-(2*sqrt(1/2)*(a^2 - 2*a*b - b^2)*d *sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c)) - d^2*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt((a^4 - 4* a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2))/(a^4 - 6*a^2*b^2 + b^4)) + 15*sqr t(1/2)*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*log(2*sqrt( 1/2)*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c)) - a^2 + 2*a*b + b^2 - (a^2 - 2*a*b - b^2)*tan(d*x + c)) - 15*sqrt(1/ 2)*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*log(-2*sqrt(1/2 )*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c )) - a^2 + 2*a*b + b^2 - (a^2 - 2*a*b - b^2)*tan(d*x + c)) + 4*(3*b^2*tan( d*x + c)^2 + 10*a*b*tan(d*x + c) + 15*a^2 - 15*b^2)*sqrt(tan(d*x + c)))/d
\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \] Input:
integrate(tan(d*x+c)**(3/2)*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*tan(c + d*x)**(3/2), x)
Time = 0.14 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {24 \, b^{2} \tan \left (d x + c\right )^{\frac {5}{2}} + 80 \, a b \tan \left (d x + c\right )^{\frac {3}{2}} - 30 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 30 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 120 \, {\left (a^{2} - b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{60 \, d} \] Input:
integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/60*(24*b^2*tan(d*x + c)^(5/2) + 80*a*b*tan(d*x + c)^(3/2) - 30*sqrt(2)*( a^2 + 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 30*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d *x + c)))) - 15*sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 15*sqrt(2)*(a^2 - 2*a*b - b^2)*log(-sqrt(2)*sqrt(ta n(d*x + c)) + tan(d*x + c) + 1) + 120*(a^2 - b^2)*sqrt(tan(d*x + c)))/d
Exception generated. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 3.11 (sec) , antiderivative size = 986, normalized size of antiderivative = 4.98 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Too large to display} \] Input:
int(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^2,x)
Output:
tan(c + d*x)^(1/2)*((2*a^2)/d - (2*b^2)/d) - atan((a^4*tan(c + d*x)^(1/2)* ((a*b^3)/d^2 - (b^4*1i)/(4*d^2) - (a^4*1i)/(4*d^2) - (a^3*b)/d^2 + (a^2*b^ 2*3i)/(2*d^2))^(1/2)*32i)/((a^6*16i)/d - (b^6*16i)/d + (32*a*b^5)/d + (32* a^5*b)/d + (a^2*b^4*112i)/d - (192*a^3*b^3)/d - (a^4*b^2*112i)/d) + (b^4*t an(c + d*x)^(1/2)*((a*b^3)/d^2 - (b^4*1i)/(4*d^2) - (a^4*1i)/(4*d^2) - (a^ 3*b)/d^2 + (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((a^6*16i)/d - (b^6*16i)/d + ( 32*a*b^5)/d + (32*a^5*b)/d + (a^2*b^4*112i)/d - (192*a^3*b^3)/d - (a^4*b^2 *112i)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a*b^3)/d^2 - (b^4*1i)/(4*d^2) - (a^4*1i)/(4*d^2) - (a^3*b)/d^2 + (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((a^6*1 6i)/d - (b^6*16i)/d + (32*a*b^5)/d + (32*a^5*b)/d + (a^2*b^4*112i)/d - (19 2*a^3*b^3)/d - (a^4*b^2*112i)/d))*(-(4*a^3*b - 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)/(4*d^2))^(1/2)*2i - atan((a^4*tan(c + d*x)^(1/2)*((a^4*1i)/(4 *d^2) + (b^4*1i)/(4*d^2) + (a*b^3)/d^2 - (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2 ))^(1/2)*32i)/((b^6*16i)/d - (a^6*16i)/d + (32*a*b^5)/d + (32*a^5*b)/d - ( a^2*b^4*112i)/d - (192*a^3*b^3)/d + (a^4*b^2*112i)/d) + (b^4*tan(c + d*x)^ (1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) + (a*b^3)/d^2 - (a^3*b)/d^2 - ( a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((b^6*16i)/d - (a^6*16i)/d + (32*a*b^5)/d + (32*a^5*b)/d - (a^2*b^4*112i)/d - (192*a^3*b^3)/d + (a^4*b^2*112i)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) + (a*b^3) /d^2 - (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((b^6*16i)/d - (...
\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} b^{2}+10 \sqrt {\tan \left (d x +c \right )}\, a^{2}-10 \sqrt {\tan \left (d x +c \right )}\, b^{2}-5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a^{2} d +5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b^{2} d +10 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a b d}{5 d} \] Input:
int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2,x)
Output:
(2*sqrt(tan(c + d*x))*tan(c + d*x)**2*b**2 + 10*sqrt(tan(c + d*x))*a**2 - 10*sqrt(tan(c + d*x))*b**2 - 5*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a**2 *d + 5*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*b**2*d + 10*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*a*b*d)/(5*d)