Integrand size = 12, antiderivative size = 210 \[ \int (d \tan (a+b x))^{3/2} \, dx=\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} b}-\frac {d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} b}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{2 \sqrt {2} b}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{2 \sqrt {2} b}+\frac {2 d \sqrt {d \tan (a+b x)}}{b} \]
1/2*d^(3/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-1/2*d ^(3/2)*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)+1/4*d^(3/2 )*ln(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)-1/ 4*d^(3/2)*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^ (1/2)+2*d*(d*tan(b*x+a))^(1/2)/b
Time = 0.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.76 \[ \int (d \tan (a+b x))^{3/2} \, dx=\frac {\left (\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right )}{2 \sqrt {2}}+2 \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{3/2}}{b \tan ^{\frac {3}{2}}(a+b x)} \]
((ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Sqrt [Tan[a + b*x]]]/Sqrt[2] + Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x ]]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]/(2*Sqr t[2]) + 2*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(b*Tan[a + b*x]^(3/2 ))
Time = 0.46 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 3954, 3042, 3957, 266, 755, 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.
\(\displaystyle \int (d \tan (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \tan (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-d^2 \int \frac {1}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-d^2 \int \frac {1}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {d^3 \int \frac {1}{\sqrt {d \tan (a+b x)} \left (\tan ^2(a+b x) d^2+d^2\right )}d(d \tan (a+b x))}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \int \frac {1}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{b}\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(a+b x)+d}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2 d \sqrt {d \tan (a+b x)}}{b}-\frac {2 d^3 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{b}\) |
(-2*d^3*((-(ArcTan[1 - Sqrt[2]*Sqrt[d]*Tan[a + b*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sqrt[d]*Tan[a + b*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2* Log[d - Sqrt[2]*d^(3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(Sqrt[2]*Sqrt[d ]) + Log[d + Sqrt[2]*d^(3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(2*Sqrt[2] *Sqrt[d]))/(2*d)))/b + (2*d*Sqrt[d*Tan[a + b*x]])/b
3.3.39.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Time = 0.18 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {2 d \left (\sqrt {d \tan \left (b x +a \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (b x +a \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (b x +a \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{b}\) | \(149\) |
default | \(\frac {2 d \left (\sqrt {d \tan \left (b x +a \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (b x +a \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (b x +a \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{b}\) | \(149\) |
2/b*d*((d*tan(b*x+a))^(1/2)-1/8*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(b*x+a)+(d^2 )^(1/4)*(d*tan(b*x+a))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(b*x+a)-(d^2)^(1/4 )*(d*tan(b*x+a))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)* (d*tan(b*x+a))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(b*x+a))^(1/2) +1)))
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int (d \tan (a+b x))^{3/2} \, dx=-\frac {\left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d \tan \left (b x + a\right )} d + \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b\right ) + i \, \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d \tan \left (b x + a\right )} d + i \, \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b\right ) - i \, \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d \tan \left (b x + a\right )} d - i \, \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b\right ) - \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (\sqrt {d \tan \left (b x + a\right )} d - \left (-\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b\right ) - 4 \, \sqrt {d \tan \left (b x + a\right )} d}{2 \, b} \]
-1/2*((-d^6/b^4)^(1/4)*b*log(sqrt(d*tan(b*x + a))*d + (-d^6/b^4)^(1/4)*b) + I*(-d^6/b^4)^(1/4)*b*log(sqrt(d*tan(b*x + a))*d + I*(-d^6/b^4)^(1/4)*b) - I*(-d^6/b^4)^(1/4)*b*log(sqrt(d*tan(b*x + a))*d - I*(-d^6/b^4)^(1/4)*b) - (-d^6/b^4)^(1/4)*b*log(sqrt(d*tan(b*x + a))*d - (-d^6/b^4)^(1/4)*b) - 4* sqrt(d*tan(b*x + a))*d)/b
\[ \int (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.81 \[ \int (d \tan (a+b x))^{3/2} \, dx=-\frac {2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 8 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{4 \, b d} \]
-1/4*(2*sqrt(2)*d^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan (b*x + a)))/sqrt(d)) + 2*sqrt(2)*d^(5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt (d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + sqrt(2)*d^(5/2)*log(d*tan(b*x + a ) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - sqrt(2)*d^(5/2)*log(d*tan( b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 8*sqrt(d*tan(b*x + a))*d^2)/(b*d)
\[ \int (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
Time = 3.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int (d \tan (a+b x))^{3/2} \, dx=\frac {2\,d\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{b}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{b}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{b} \]