Integrand size = 25, antiderivative size = 188 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=-\frac {(i a+b) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (b c+a d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f} \]
-(I*a+b)*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+(I* a-b)*(c+I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(2*a* c*d+b*(c^2-d^2))*(c+d*tan(f*x+e))^(1/2)/f+2/3*(a*d+b*c)*(c+d*tan(f*x+e))^( 3/2)/f+2/5*b*(c+d*tan(f*x+e))^(5/2)/f
Time = 1.17 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.24 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\frac {i \left ((a-i b) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c-i d) \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )\right )-(a+i b) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c+i d) \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )\right )\right )}{2 f} \]
((I/2)*((a - I*b)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d* Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/3) - (a + I*b)*((2*(c + d *Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3)))/f
Time = 1.00 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.90, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 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.
| \(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} \left (-2 b c d+a \left (c^2-d^2\right )+\left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} \left (-2 b c d+a \left (c^2-d^2\right )+\left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {-b d \left (3 c^2-d^2\right )+a \left (c^3-3 c d^2\right )+\left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {-b d \left (3 c^2-d^2\right )+a \left (c^3-3 c d^2\right )+\left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (a-i b) (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (a-i b) (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (a-i b) (c-i d)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c+i d)^3 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (a-i b) (c-i d)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (a+i b) (c+i d)^3 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a-i b) (c-i d)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b) (c+i d)^3 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a-i b) (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b) (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a c d+b \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (a d+b c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b (c+d \tan (e+f x))^{5/2}}{5 f}\) |
((a - I*b)*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a + I *b)*(c + I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (2*(2*a*c*d + b*(c^2 - d^2))*Sqrt[c + d*Tan[e + f*x]])/f + (2*(b*c + a*d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*b*(c + d*Tan[e + f*x])^(5/2))/(5*f)
3.13.43.3.1 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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(2391\) vs. \(2(160)=320\).
Time = 0.89 (sec) , antiderivative size = 2392, normalized size of antiderivative = 12.72
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2392\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2405\) |
| default | \(\text {Expression too large to display}\) | \(2405\) |
a*(2/3/f*d*(c+d*tan(f*x+e))^(3/2)+4/f*d*(c+d*tan(f*x+e))^(1/2)*c-1/4/f/d*l n(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 +d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2+1/4/f*d*ln( d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d ^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+1/4/f/d*ln(d*tan( f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1 /2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3/4/f*d*ln(d*tan(f*x+e)+c+(c+d*tan( f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^ (1/2)+2*c)^(1/2)*c+3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan( f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2) )*c^2-1/f*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/ 2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d/(2* (c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^( 1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*c+1/4/f/d* ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^ 2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2-1/4/f*d*ln ((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+ d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)-1/4/f/d*ln((c+d* tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^( 1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+3/4/f*d*ln((c+d*tan(f*x+e))^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 4855 vs. \(2 (155) = 310\).
Time = 0.85 (sec) , antiderivative size = 4855, normalized size of antiderivative = 25.82 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
-1/30*(15*f*sqrt((10*a*b*c^4*d - 20*a*b*c^2*d^3 + 2*a*b*d^5 - (a^2 - b^2)* c^5 + 10*(a^2 - b^2)*c^3*d^2 - 5*(a^2 - b^2)*c*d^4 + f^2*sqrt(-(4*a^2*b^2* c^10 + 20*(a^3*b - a*b^3)*c^9*d + 5*(5*a^4 - 26*a^2*b^2 + 5*b^4)*c^8*d^2 - 240*(a^3*b - a*b^3)*c^7*d^3 - 20*(5*a^4 - 32*a^2*b^2 + 5*b^4)*c^6*d^4 + 5 04*(a^3*b - a*b^3)*c^5*d^5 + 10*(11*a^4 - 62*a^2*b^2 + 11*b^4)*c^4*d^6 - 2 40*(a^3*b - a*b^3)*c^3*d^7 - 20*(a^4 - 7*a^2*b^2 + b^4)*c^2*d^8 + 20*(a^3* b - a*b^3)*c*d^9 + (a^4 - 2*a^2*b^2 + b^4)*d^10)/f^4))/f^2)*log(-(2*(a^3*b + a*b^3)*c^9 + 5*(a^4 - b^4)*c^8*d - 16*(a^3*b + a*b^3)*c^7*d^2 - 28*(a^3 *b + a*b^3)*c^5*d^4 - 14*(a^4 - b^4)*c^4*d^5 - 8*(a^4 - b^4)*c^2*d^7 + 10* (a^3*b + a*b^3)*c*d^8 + (a^4 - b^4)*d^9)*sqrt(d*tan(f*x + e) + c) + ((a*c^ 2 - 2*b*c*d - a*d^2)*f^3*sqrt(-(4*a^2*b^2*c^10 + 20*(a^3*b - a*b^3)*c^9*d + 5*(5*a^4 - 26*a^2*b^2 + 5*b^4)*c^8*d^2 - 240*(a^3*b - a*b^3)*c^7*d^3 - 2 0*(5*a^4 - 32*a^2*b^2 + 5*b^4)*c^6*d^4 + 504*(a^3*b - a*b^3)*c^5*d^5 + 10* (11*a^4 - 62*a^2*b^2 + 11*b^4)*c^4*d^6 - 240*(a^3*b - a*b^3)*c^3*d^7 - 20* (a^4 - 7*a^2*b^2 + b^4)*c^2*d^8 + 20*(a^3*b - a*b^3)*c*d^9 + (a^4 - 2*a^2* b^2 + b^4)*d^10)/f^4) - (2*a*b^2*c^7 + (9*a^2*b - 5*b^3)*c^6*d + 2*(5*a^3 - 16*a*b^2)*c^5*d^2 - 5*(11*a^2*b - 3*b^3)*c^4*d^3 - 10*(2*a^3 - 5*a*b^2)* c^3*d^4 + (31*a^2*b - 11*b^3)*c^2*d^5 + 2*(a^3 - 6*a*b^2)*c*d^6 - (a^2*b - b^3)*d^7)*f)*sqrt((10*a*b*c^4*d - 20*a*b*c^2*d^3 + 2*a*b*d^5 - (a^2 - b^2 )*c^5 + 10*(a^2 - b^2)*c^3*d^2 - 5*(a^2 - b^2)*c*d^4 + f^2*sqrt(-(4*a^2...
\[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
Exception generated. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for more details)Is
Timed out. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 39.96 (sec) , antiderivative size = 3863, normalized size of antiderivative = 20.55 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
log(- ((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3 *d^2*f^2)/f^4)^(1/2)*(32*b*d^6 - 32*b*c^4*d^2 + 32*c*d^2*f*(((-b^4*d^2*f^4 *(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10* b^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*b^2*d ^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2)) /2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3)*((20*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6*d^4*f^4 - 25*b^4*c^8*d^2 *f^4)^(1/2)/(4*f^4) + (b^2*c^5)/(4*f^2) + (5*b^2*c*d^4)/(4*f^2) - (5*b^2*c ^3*d^2)/(2*f^2))^(1/2) - log(((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2 )^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*( ((((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2 *c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(32*b*c^4*d^2 - 32*b*d^6 + 32* c*d^2*f*(((-b^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 10*b^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1 /2)))/(2*f) - (16*b^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d ^4 - 15*c^4*d^2))/f^2))/2 - (8*b^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3) *(((20*b^4*c^2*d^8*f^4 - b^4*d^10*f^4 - 110*b^4*c^4*d^6*f^4 + 100*b^4*c^6* d^4*f^4 - 25*b^4*c^8*d^2*f^4)^(1/2) + b^2*c^5*f^2 + 5*b^2*c*d^4*f^2 - 1...