\(\int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx\) [459]

Optimal result

Integrand size = 20, antiderivative size = 118 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a b^{5/2}} \]

[Out]

2/3*d*(d*x+c)^(3/2)/b-2*c^(5/2)*arctanh((d*x+c)^(1/2)/c^(1/2))/a+2*(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x+c)^(1
/2)/(-a*d+b*c)^(1/2))/a/b^(5/2)+2*d*(-a*d+2*b*c)*(d*x+c)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {86, 159, 162, 65, 214} \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a b^{5/2}}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {2 d \sqrt {c+d x} (2 b c-a d)}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b} \]

[In]

Int[(c + d*x)^(5/2)/(x*(a + b*x)),x]

[Out]

(2*d*(2*b*c - a*d)*Sqrt[c + d*x])/b^2 + (2*d*(c + d*x)^(3/2))/(3*b) - (2*c^(5/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]
])/a + (2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 d (c+d x)^{3/2}}{3 b}+\frac {\int \frac {\sqrt {c+d x} \left (b c^2+d (2 b c-a d) x\right )}{x (a+b x)} \, dx}{b} \\ & = \frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {2 \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}{b^2} \\ & = \frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {c^3 \int \frac {1}{x \sqrt {c+d x}} \, dx}{a}-\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a b^2} \\ & = \frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}+\frac {\left (2 c^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a d}-\frac {\left (2 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a b^2 d} \\ & = \frac {2 d (2 b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 d (c+d x)^{3/2}}{3 b}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}+\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\frac {2 d \sqrt {c+d x} (7 b c-3 a d+b d x)}{3 b^2}+\frac {2 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{a b^{5/2}}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a} \]

[In]

Integrate[(c + d*x)^(5/2)/(x*(a + b*x)),x]

[Out]

(2*d*Sqrt[c + d*x]*(7*b*c - 3*a*d + b*d*x))/(3*b^2) + (2*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/S
qrt[-(b*c) + a*d]])/(a*b^(5/2)) - (2*c^(5/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\left (a d -b c \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) c^{\frac {5}{2}} b^{2}+\left (\frac {\left (-d x -7 c \right ) b}{3}+a d \right ) d a \sqrt {d x +c}\right ) \sqrt {\left (a d -b c \right ) b}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2} a}\)	\(114\)
derivativedivides	\(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a d -2 \sqrt {d x +c}\, b c}{b^{2}}-\frac {c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} d a \sqrt {\left (a d -b c \right ) b}}\right )\)	\(145\)
default	\(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a d -2 \sqrt {d x +c}\, b c}{b^{2}}-\frac {c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} d a \sqrt {\left (a d -b c \right ) b}}\right )\)	\(145\)

[In]

int((d*x+c)^(5/2)/x/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-2/((a*d-b*c)*b)^(1/2)*(-(a*d-b*c)^3*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))+(arctanh((d*x+c)^(1/2)/c^(1/2
))*c^(5/2)*b^2+(1/3*(-d*x-7*c)*b+a*d)*d*a*(d*x+c)^(1/2))*((a*d-b*c)*b)^(1/2))/b^2/a

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 598, normalized size of antiderivative = 5.07 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\left [\frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 6 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {6 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}}{3 \, a b^{2}}, \frac {2 \, {\left (3 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (a b d^{2} x + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x + c}\right )}}{3 \, a b^{2}}\right ] \]

[In]

integrate((d*x+c)^(5/2)/x/(b*x+a),x, algorithm="fricas")

[Out]

[1/3*(3*b^2*c^(5/2)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c
- a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(a*b*d^2*x + 7*a*b
*c*d - 3*a^2*d^2)*sqrt(d*x + c))/(a*b^2), 1/3*(3*b^2*c^(5/2)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 6*
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d))
+ 2*(a*b*d^2*x + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x + c))/(a*b^2), 1/3*(6*b^2*sqrt(-c)*c^2*arctan(sqrt(d*x + c)*
sqrt(-c)/c) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)
*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(a*b*d^2*x + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x + c))/(a*b^2), 2/3*(3*b^2*
sqrt(-c)*c^2*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan(
-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (a*b*d^2*x + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x + c))/(a*b^2
)]

Sympy [A] (verification not implemented)

Time = 2.91 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\begin {cases} \frac {2 d \left (c + d x\right )^{\frac {3}{2}}}{3 b} + \frac {2 \sqrt {c + d x} \left (- a d^{2} + 2 b c d\right )}{b^{2}} + \frac {2 c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} + \frac {2 \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{a b^{3} \sqrt {\frac {a d - b c}{b}}} & \text {for}\: d \neq 0 \\- c^{\frac {5}{2}} \left (\begin {cases} \frac {1}{b x} & \text {for}\: a = 0 \\\frac {\log {\left (\frac {a}{x} + b \right )}}{a} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((d*x+c)**(5/2)/x/(b*x+a),x)

[Out]

Piecewise((2*d*(c + d*x)**(3/2)/(3*b) + 2*sqrt(c + d*x)*(-a*d**2 + 2*b*c*d)/b**2 + 2*c**3*atan(sqrt(c + d*x)/s
qrt(-c))/(a*sqrt(-c)) + 2*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(a*b**3*sqrt((a*d - b*c)/b)),
Ne(d, 0)), (-c**(5/2)*Piecewise((1/(b*x), Eq(a, 0)), (log(a/x + b)/a, True)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(5/2)/x/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\frac {2 \, c^{3} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a b^{2}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} d + 6 \, \sqrt {d x + c} b^{2} c d - 3 \, \sqrt {d x + c} a b d^{2}\right )}}{3 \, b^{3}} \]

[In]

integrate((d*x+c)^(5/2)/x/(b*x+a),x, algorithm="giac")

[Out]

2*c^3*arctan(sqrt(d*x + c)/sqrt(-c))/(a*sqrt(-c)) - 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arct
an(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a*b^2) + 2/3*((d*x + c)^(3/2)*b^2*d + 6*sqrt(d*
x + c)*b^2*c*d - 3*sqrt(d*x + c)*a*b*d^2)/b^3

Mupad [B] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 2048, normalized size of antiderivative = 17.36 \[ \int \frac {(c+d x)^{5/2}}{x (a+b x)} \, dx=\text {Too large to display} \]

[In]

int((c + d*x)^(5/2)/(x*(a + b*x)),x)

[Out]

(atan(((((8*(c + d*x)^(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 + (((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))/
b^3 + (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(c^5)^(1/2)*(c + d*x)^(1/2))/(a*b^3))*(c^5)^(1/2))/a)*(c^5)^(1/2)*1i)
/a + (((8*(c + d*x)^(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 - (((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))/b^
3 - (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(c^5)^(1/2)*(c + d*x)^(1/2))/(a*b^3))*(c^5)^(1/2))/a)*(c^5)^(1/2)*1i)/a
)/((16*(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 - (((8*(c + d*x)^(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 + (((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*
d^4))/b^3 + (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(c^5)^(1/2)*(c + d*x)^(1/2))/(a*b^3))*(c^5)^(1/2))/a)*(c^5)^(1/
2))/a + (((8*(c + d*x)^(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 - (((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))
/b^3 - (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(c^5)^(1/2)*(c + d*x)^(1/2))/(a*b^3))*(c^5)^(1/2))/a)*(c^5)^(1/2))/a
))*(c^5)^(1/2)*2i)/a + (2*d*(c + d*x)^(3/2))/(3*b) + (atan((((-b^5*(a*d - b*c)^5)^(1/2)*((8*(c + d*x)^(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)*((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))/b^3
+ (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x)^(1/2))/(a*b^8)))/(a*b^5))*1i)/(a*b^
5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((8*(c + d*x)^(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)*((8*(a^4*
b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))/b^3 - (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)
^5)^(1/2)*(c + d*x)^(1/2))/(a*b^8)))/(a*b^5))*1i)/(a*b^5))/((16*(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)*((8*(c + d*x
)^(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)*((8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2
*d^4))/b^3 + (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x)^(1/2))/(a*b^8)))/(a*b^5))
)/(a*b^5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((8*(c + d*x)^(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)*((
8*(a^4*b^3*c*d^5 + 2*a^2*b^5*c^3*d^3 - 3*a^3*b^4*c^2*d^4))/b^3 - (8*(a^3*b^5*d^3 - 2*a^2*b^6*c*d^2)*(-b^5*(a*d
- b*c)^5)^(1/2)*(c + d*x)^(1/2))/(a*b^8)))/(a*b^5)))/(a*b^5)))*(-b^5*(a*d - b*c)^5)^(1/2)*2i)/(a*b^5) - (2*d*
(a*d - 2*b*c)*(c + d*x)^(1/2))/b^2