Integrand size = 27, antiderivative size = 151 \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\frac {(i a-b) (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (i a+b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 b \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d} \]
(I*a-b)*(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+I *b)^(5/2)*(I*a+b)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*b*(a^2 +b^2)*(a+b*tan(d*x+c))^(1/2)/d+2/5*b*(a+b*tan(d*x+c))^(5/2)/d
Time = 1.42 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.28 \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\frac {\cos (c+d x) (a-b \tan (c+d x)) \left (5 i (a-i b)^{5/2} (a+i b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-5 i (a-i b) (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 b \sqrt {a+b \tan (c+d x)} \left (-4 a^2-5 b^2+2 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )\right )}{5 d (a \cos (c+d x)-b \sin (c+d x))} \]
(Cos[c + d*x]*(a - b*Tan[c + d*x])*((5*I)*(a - I*b)^(5/2)*(a + I*b)*ArcTan h[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - (5*I)*(a - I*b)*(a + I*b)^(5/2 )*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*b*Sqrt[a + b*Tan[c + d*x]]*(-4*a^2 - 5*b^2 + 2*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2)))/(5*d*( a*Cos[c + d*x] - b*Sin[c + d*x]))
Time = 0.69 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.80, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4011, 27, 3042, 3963, 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.
\(\displaystyle \int (b \tan (c+d x)-a) (a+b \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (b \tan (c+d x)-a) (a+b \tan (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \left (-a^2-b^2\right ) (a+b \tan (c+d x))^{3/2}dx+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \int (a+b \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \int (a+b \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3963 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\int \frac {a^2+2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\int \frac {a^2+2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\frac {1}{2} (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\frac {1}{2} (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\frac {i (a-i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (-\frac {i (a-i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\frac {(a-i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}-\left (a^2+b^2\right ) \left (\frac {(a-i b)^{3/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )\) |
(2*b*(a + b*Tan[c + d*x])^(5/2))/(5*d) - (a^2 + b^2)*(((a - I*b)^(3/2)*Arc Tan[Tan[c + d*x]/Sqrt[a - I*b]])/d + ((a + I*b)^(3/2)*ArcTan[Tan[c + d*x]/ Sqrt[a + I*b]])/d + (2*b*Sqrt[a + b*Tan[c + d*x]])/d)
3.4.40.3.1 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_))^(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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d *x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[n, 1]
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. \(1374\) vs. \(2(127)=254\).
Time = 0.08 (sec) , antiderivative size = 1375, normalized size of antiderivative = 9.11
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1375\) |
default | \(\text {Expression too large to display}\) | \(1375\) |
parts | \(\text {Expression too large to display}\) | \(2381\) |
2/5*b*(a+b*tan(d*x+c))^(5/2)/d-2/d*b*(a+b*tan(d*x+c))^(1/2)*a^2-2/d*b^3*(a +b*tan(d*x+c))^(1/2)+1/4/d/b*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))*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1 /2)+2*a)^(1/2)*a^3+1/4/d*b*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))*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2 )+2*a)^(1/2)*a-1/4/d/b*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)+2*a)^(1/2)*a^4+1/4 /d*b^3*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)+2*a)^(1/2)+1/d*b/(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))*(a^2+b^2)^(1/2)*a^2+1/d*b^3/(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))*(a^2+b^2)^(1/2)-2/d*b/(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))*a^3-2/d*b^3/(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))*a-1/4/d/b*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))*(a^2+b^2)^(1/2)*(2*(a^2 +b^2)^(1/2)+2*a)^(1/2)*a^3-1/4/d*b*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))*(a^2+b^2)^(1/2)*(2*(a^...
Leaf count of result is larger than twice the leaf count of optimal. 1472 vs. \(2 (121) = 242\).
Time = 0.29 (sec) , antiderivative size = 1472, normalized size of antiderivative = 9.75 \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
-1/10*(5*d*sqrt(-(a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6 + d^2*sqrt(-(9*a^12* b^2 + 30*a^10*b^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^1 4)/d^4))/d^2)*log(-(3*a^10*b + 11*a^8*b^3 + 14*a^6*b^5 + 6*a^4*b^7 - a^2*b ^9 - b^11)*sqrt(b*tan(d*x + c) + a) + (a*d^3*sqrt(-(9*a^12*b^2 + 30*a^10*b ^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^14)/d^4) + (3*a^ 6*b^2 + 5*a^4*b^4 + a^2*b^6 - b^8)*d)*sqrt(-(a^7 - a^5*b^2 - 5*a^3*b^4 - 3 *a*b^6 + d^2*sqrt(-(9*a^12*b^2 + 30*a^10*b^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9* a^4*b^10 - 2*a^2*b^12 + b^14)/d^4))/d^2)) - 5*d*sqrt(-(a^7 - a^5*b^2 - 5*a ^3*b^4 - 3*a*b^6 + d^2*sqrt(-(9*a^12*b^2 + 30*a^10*b^4 + 31*a^8*b^6 + 4*a^ 6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^14)/d^4))/d^2)*log(-(3*a^10*b + 11*a^8 *b^3 + 14*a^6*b^5 + 6*a^4*b^7 - a^2*b^9 - b^11)*sqrt(b*tan(d*x + c) + a) - (a*d^3*sqrt(-(9*a^12*b^2 + 30*a^10*b^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b ^10 - 2*a^2*b^12 + b^14)/d^4) + (3*a^6*b^2 + 5*a^4*b^4 + a^2*b^6 - b^8)*d) *sqrt(-(a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6 + d^2*sqrt(-(9*a^12*b^2 + 30*a ^10*b^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^14)/d^4))/d ^2)) - 5*d*sqrt(-(a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6 - d^2*sqrt(-(9*a^12* b^2 + 30*a^10*b^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^1 4)/d^4))/d^2)*log(-(3*a^10*b + 11*a^8*b^3 + 14*a^6*b^5 + 6*a^4*b^7 - a^2*b ^9 - b^11)*sqrt(b*tan(d*x + c) + a) + (a*d^3*sqrt(-(9*a^12*b^2 + 30*a^10*b ^4 + 31*a^8*b^6 + 4*a^6*b^8 - 9*a^4*b^10 - 2*a^2*b^12 + b^14)/d^4) - (3...
\[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=- \int a^{3} \sqrt {a + b \tan {\left (c + d x \right )}}\, dx - \int \left (- b^{3} \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\right )\, dx - \int \left (- a b^{2} \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\right )\, dx - \int a^{2} b \sqrt {a + b \tan {\left (c + d x \right )}} \tan {\left (c + d x \right )}\, dx \]
-Integral(a**3*sqrt(a + b*tan(c + d*x)), x) - Integral(-b**3*sqrt(a + b*ta n(c + d*x))*tan(c + d*x)**3, x) - Integral(-a*b**2*sqrt(a + b*tan(c + d*x) )*tan(c + d*x)**2, x) - Integral(a**2*b*sqrt(a + b*tan(c + d*x))*tan(c + d *x), x)
Exception generated. \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 32.67 (sec) , antiderivative size = 3441, normalized size of antiderivative = 22.79 \[ \int (-a+b \tan (c+d x)) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]
log((8*a^3*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3 - ((((((-a^4*b^2*d^4*(5*a^ 4 + b^4 - 10*a^2*b^2)^2)^(1/2) - a^7*d^2 - 5*a^3*b^4*d^2 + 10*a^5*b^2*d^2) /d^4)^(1/2)*(64*a^2*b^5 + 64*a^4*b^3 + 32*a*b^2*d*(((-a^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - a^7*d^2 - 5*a^3*b^4*d^2 + 10*a^5*b^2*d^2)/d^ 4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) + (16*a^2*b^2*(a + b*tan(c + d *x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-a^4*b^2*d^4*(5* a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) - a^7*d^2 - 5*a^3*b^4*d^2 + 10*a^5*b^2*d^ 2)/d^4)^(1/2))/2)*((20*a^6*b^8*d^4 - a^4*b^10*d^4 - 110*a^8*b^6*d^4 + 100* a^10*b^4*d^4 - 25*a^12*b^2*d^4)^(1/2)/(4*d^4) - a^7/(4*d^2) - (5*a^3*b^4)/ (4*d^2) + (5*a^5*b^2)/(2*d^2))^(1/2) - log((8*a^3*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3 - ((((-((-a^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a ^7*d^2 + 5*a^3*b^4*d^2 - 10*a^5*b^2*d^2)/d^4)^(1/2)*(64*a^2*b^5 + 64*a^4*b ^3 - 32*a*b^2*d*(-((-a^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^7 *d^2 + 5*a^3*b^4*d^2 - 10*a^5*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/ 2)))/(2*d) - (16*a^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 15*a^2*b^ 4 - 15*a^4*b^2))/d^2)*(-((-a^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^7*d^2 + 5*a^3*b^4*d^2 - 10*a^5*b^2*d^2)/d^4)^(1/2))/2)*(-(a^7*d^2 + ( 20*a^6*b^8*d^4 - a^4*b^10*d^4 - 110*a^8*b^6*d^4 + 100*a^10*b^4*d^4 - 25*a^ 12*b^2*d^4)^(1/2) + 5*a^3*b^4*d^2 - 10*a^5*b^2*d^2)/(4*d^4))^(1/2) - log(( 8*a^3*b^3*(3*a^2 - b^2)*(a^2 + b^2)^3)/d^3 - ((((((-a^4*b^2*d^4*(5*a^4 ...