\(\int \frac {(c+d x^2)^{5/2}}{x (a+b x^2)} \, dx\) [699]
Optimal result
Integrand size = 24, antiderivative size = 124 \[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\frac {d (2 b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}-\frac {c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {(b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{5/2}}
\]
[Out]
1/3*d*(d*x^2+c)^(3/2)/b-c^(5/2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))/a+(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x^2+c)^
(1/2)/(-a*d+b*c)^(1/2))/a/b^(5/2)+d*(-a*d+2*b*c)*(d*x^2+c)^(1/2)/b^2
Rubi [A] (verified)
Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of
steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 86, 159, 162, 65, 214}
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\frac {(b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{5/2}}-\frac {c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \sqrt {c+d x^2} (2 b c-a d)}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}
\]
[In]
Int[(c + d*x^2)^(5/2)/(x*(a + b*x^2)),x]
[Out]
(d*(2*b*c - a*d)*Sqrt[c + d*x^2])/b^2 + (d*(c + d*x^2)^(3/2))/(3*b) - (c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]
])/a + ((b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*b^(5/2))
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 86
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[f*((e + f*x)^(p -
1)/(b*d*(p - 1))), x] + Dist[1/(b*d), Int[(b*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p -
2)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
Rule 159
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]
Rule 162
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Rule 214
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]
Rule 457
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rubi steps \begin{align*}
\text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {d \left (c+d x^2\right )^{3/2}}{3 b}+\frac {\text {Subst}\left (\int \frac {\sqrt {c+d x} \left (b c^2+d (2 b c-a d) x\right )}{x (a+b x)} \, dx,x,x^2\right )}{2 b} \\ & = \frac {d (2 b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}+\frac {\text {Subst}\left (\int \frac {\frac {b^2 c^3}{2}+\frac {1}{2} d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{b^2} \\ & = \frac {d (2 b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}+\frac {c^3 \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a}-\frac {(b c-a d)^3 \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a b^2} \\ & = \frac {d (2 b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}+\frac {c^3 \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a d}-\frac {(b c-a d)^3 \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{a b^2 d} \\ & = \frac {d (2 b c-a d) \sqrt {c+d x^2}}{b^2}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{5/2}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\frac {d \sqrt {c+d x^2} \left (7 b c-3 a d+b d x^2\right )}{3 b^2}+\frac {(-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{a b^{5/2}}-\frac {c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}
\]
[In]
Integrate[(c + d*x^2)^(5/2)/(x*(a + b*x^2)),x]
[Out]
(d*Sqrt[c + d*x^2]*(7*b*c - 3*a*d + b*d*x^2))/(3*b^2) + ((-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])
/Sqrt[-(b*c) + a*d]])/(a*b^(5/2)) - (c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/a
Maple [A] (verified)
Time = 3.02 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98
| | |
| method | result | size |
| | |
| pseudoelliptic |
\(-\frac {-\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) b^{2}+\left (\frac {\left (-d \,x^{2}-7 c \right ) b}{3}+a d \right ) d a \sqrt {d \,x^{2}+c}\right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{2} a}\) |
\(122\) |
| default |
\(\text {Expression too large to display}\) |
\(2138\) |
| | |
|
|
|
|
[In]
int((d*x^2+c)^(5/2)/x/(b*x^2+a),x,method=_RETURNVERBOSE)
[Out]
-(-(a*d-b*c)^3*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))+(c^(5/2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))*b^2+(1/
3*(-d*x^2-7*c)*b+a*d)*d*a*(d*x^2+c)^(1/2))*((a*d-b*c)*b)^(1/2))/((a*d-b*c)*b)^(1/2)/b^2/a
Fricas [A] (verification not implemented)
none
Time = 0.85 (sec) , antiderivative size = 837, normalized size of antiderivative = 6.75
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\left [\frac {6 \, b^{2} c^{\frac {5}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a b^{2}}, \frac {12 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a b^{2}}, \frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a b^{2}}, \frac {6 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a b^{2}}\right ]
\]
[In]
integrate((d*x^2+c)^(5/2)/x/(b*x^2+a),x, algorithm="fricas")
[Out]
[1/12*(6*b^2*c^(5/2)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*s
qrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(b^2
*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(a*b*d^2*x^2 +
7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/12*(12*b^2*sqrt(-c)*c^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) +
3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*
(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 +
2*a*b*x^2 + a^2)) + 4*(a*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/6*(3*b^2*c^(5/2)*log(-
(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan
(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) + 2*
(a*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/6*(6*b^2*sqrt(-c)*c^2*arctan(sqrt(-c)/sqrt(d
*x^2 + c)) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d
*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) + 2*(a*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2
)*sqrt(d*x^2 + c))/(a*b^2)]
Sympy [A] (verification not implemented)
Time = 8.98 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.62
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {d^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}{6 b} + \frac {\sqrt {c + d x^{2}} \left (- a d^{3} + 2 b c d^{2}\right )}{2 b^{2}} + \frac {c^{3} d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}} + \frac {d \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a b^{3} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (- \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\- \frac {\log {\left (a - 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a} - \frac {b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{2}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + 2 b \left (\frac {a}{2 b} + x^{2}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{a}\right ) & \text {otherwise} \end {cases}
\]
[In]
integrate((d*x**2+c)**(5/2)/x/(b*x**2+a),x)
[Out]
Piecewise((2*(d**2*(c + d*x**2)**(3/2)/(6*b) + sqrt(c + d*x**2)*(-a*d**3 + 2*b*c*d**2)/(2*b**2) + c**3*d*atan(
sqrt(c + d*x**2)/sqrt(-c))/(2*a*sqrt(-c)) + d*(a*d - b*c)**3*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*a*b
**3*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(5/2)*(-b*Piecewise(((a/(2*b) + x**2)/a, Eq(b, 0)), (-log(a - 2*b*
(a/(2*b) + x**2))/(2*b), True))/a - b*Piecewise(((a/(2*b) + x**2)/a, Eq(b, 0)), (log(a + 2*b*(a/(2*b) + x**2))
/(2*b), True))/a), True))
Maxima [F]
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )} x} \,d x }
\]
[In]
integrate((d*x^2+c)^(5/2)/x/(b*x^2+a),x, algorithm="maxima")
[Out]
integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)*x), x)
Giac [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.31
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\frac {c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + 6 \, \sqrt {d x^{2} + c} b^{2} c d - 3 \, \sqrt {d x^{2} + c} a b d^{2}}{3 \, b^{3}}
\]
[In]
integrate((d*x^2+c)^(5/2)/x/(b*x^2+a),x, algorithm="giac")
[Out]
c^3*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a*sqrt(-c)) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan
(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a*b^2) + 1/3*((d*x^2 + c)^(3/2)*b^2*d + 6*sqrt(
d*x^2 + c)*b^2*c*d - 3*sqrt(d*x^2 + c)*a*b*d^2)/b^3
Mupad [B] (verification not implemented)
Time = 5.61 (sec) , antiderivative size = 2094, normalized size of antiderivative = 16.89
\[
\int \frac {\left (c+d x^2\right )^{5/2}}{x \left (a+b x^2\right )} \, dx=\text {Too large to display}
\]
[In]
int((c + d*x^2)^(5/2)/(x*(a + b*x^2)),x)
[Out]
(atan((((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20
*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + (((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*
b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2))/(a*b^3))*(c^5)^(1/2))/(2*
a))*1i)/(2*a) + ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4
*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - (((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3
- 12*a^3*b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2))/(a*b^3))*(c^5)^(
1/2))/(2*a))*1i)/(2*a))/((2*(a^5*c^3*d^8 - 3*b^5*c^8*d^3 + 12*a*b^4*c^7*d^4 - 6*a^4*b*c^4*d^7 - 19*a^2*b^3*c^6
*d^5 + 15*a^3*b^2*c^5*d^6))/b^3 - ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^
3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + (((4*a^4*b^3*c*d^5 +
8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2)
)/(a*b^3))*(c^5)^(1/2))/(2*a)))/(2*a) + ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*
c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - (((4*a^4*b^3*c*
d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)
^(1/2))/(a*b^3))*(c^5)^(1/2))/(2*a)))/(2*a)))*(c^5)^(1/2)*1i)/a + (d*(c + d*x^2)^(3/2))/(3*b) + (atan((((-b^5*
(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 2
0*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 +
8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(
c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5))*1i)/(2*a*b^5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*
d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b
*c*d^7))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 - (
(4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5))*1i)/(2*a*
b^5))/((2*(a^5*c^3*d^8 - 3*b^5*c^8*d^3 + 12*a*b^4*c^7*d^4 - 6*a^4*b*c^4*d^7 - 19*a^2*b^3*c^6*d^5 + 15*a^3*b^2*
c^5*d^6))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 +
15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + ((-b^5*(a*d - b*c)^5)^(1
/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5
*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5)))/(2*a*b^5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c
+ d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b
^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*
b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/
(2*a*b^5)))/(2*a*b^5)))*(-b^5*(a*d - b*c)^5)^(1/2)*1i)/(a*b^5) - (d*(c + d*x^2)^(1/2)*(a*d - 2*b*c))/b^2