Integrand size = 21, antiderivative size = 87 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d} \]
-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d/(a-I*b)^(1/2)-arctanh((a+ b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d/(a+I*b)^(1/2)
Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(-a+i b) d}+\frac {\sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(-a-i b) d} \]
(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((-a + I*b )*d) + (Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/((- a - I*b)*d)
Time = 0.41 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4022, 3042, 4020, 25, 73, 221}
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.
\(\displaystyle \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} i \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} i \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {\int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {\int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {\int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {i \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {i \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {i \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}-\frac {i \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}\) |
((-I)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (I*ArcTan[Ta n[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d)
3.6.32.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(338\) vs. \(2(71)=142\).
Time = 0.98 (sec) , antiderivative size = 339, normalized size of antiderivative = 3.90
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{2 \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{2 \sqrt {a^{2}+b^{2}}}}{d}\) | \(339\) |
default | \(\frac {\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{2 \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{2 \sqrt {a^{2}+b^{2}}}}{d}\) | \(339\) |
1/d*(1/2/(a^2+b^2)^(1/2)*(1/2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln((a+b*tan(d* x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+ 2*(a-(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^( 1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)))+ 1/2/(a^2+b^2)^(1/2)*(-1/2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+ (a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+2*(( a^2+b^2)^(1/2)-a)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c)) ^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (67) = 134\).
Time = 0.27 (sec) , antiderivative size = 693, normalized size of antiderivative = 7.97 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {1}{2} \, \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left ({\left ({\left (a^{2} + b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a d\right )} \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - \frac {1}{2} \, \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} + b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a d\right )} \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - \frac {1}{2} \, \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left ({\left ({\left (a^{2} + b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a d\right )} \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) + \frac {1}{2} \, \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} + b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a d\right )} \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) \]
1/2*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/(( a^2 + b^2)*d^2))*log(((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d ^4)) - a*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) - 1/2*sqrt(((a^2 + b^ 2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2))*lo g(-((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a*d)*sqrt(( (a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2) *d^2)) + sqrt(b*tan(d*x + c) + a)) - 1/2*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/ ((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*log(((a^2 + b^2)*d^ 3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a*d)*sqrt(-((a^2 + b^2)*d^2*s qrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2)) + sqrt(b*t an(d*x + c) + a)) + 1/2*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*log(-((a^2 + b^2)*d^3*sqrt(-b^2/((a^ 4 + 2*a^2*b^2 + b^4)*d^4)) + a*d)*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a) )
\[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
Exception generated. \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Time = 5.48 (sec) , antiderivative size = 781, normalized size of antiderivative = 8.98 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-2\,\mathrm {atanh}\left (\frac {32\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,b^2}{d}-\frac {64\,a^2\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^3\,d^2\,64{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}-\frac {128\,a^2\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {a\,b^3\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,128{}\mathrm {i}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}-\mathrm {atanh}\left (\frac {16\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,b^2}{d}-\frac {16\,a\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}+\frac {16\,a\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^3\,16{}\mathrm {i}}{d}-\frac {16\,a\,b^2}{d}-\frac {a\,b^3\,d^2\,16{}\mathrm {i}}{a\,d^3-b\,d^3\,1{}\mathrm {i}}+\frac {16\,a^2\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}\right )\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}} \]
- 2*atanh((32*b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d ^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16*b^2)/d - (64*a^2*b^2*d^2)/(4*a ^2*d^3 + 4*b^2*d^3) + (a*b^3*d^2*64i)/(4*a^2*d^3 + 4*b^2*d^3)) - (128*a^2* b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((64*b^4)/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/ (4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (25 6*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b ^2*d^3)) + (a*b^3*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d ^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*128i)/((64*b^4)/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256 i)/(4*a^2*d^3 + 4*b^2*d^3)))*((a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2) - atanh((16*b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((1 6*b^2)/d - (16*a*b^2*d^2)/(a*d^3 - b*d^3*1i)) + (16*a*b^2*(1/(a*d^2 - b*d^ 2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^3*16i)/d - (16*a*b^2)/d - (a* b^3*d^2*16i)/(a*d^3 - b*d^3*1i) + (16*a^2*b^2*d^2)/(a*d^3 - b*d^3*1i)))*(1 /(a*d^2 - b*d^2*1i))^(1/2)