\(\int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx\) [125]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {5 a^2 (8 b c-7 a d) \sqrt {a x+b x^2}}{64 b^4}-\frac {5 a (8 b c-7 a d) x \sqrt {a x+b x^2}}{96 b^3}+\frac {(8 b c-7 a d) x^2 \sqrt {a x+b x^2}}{24 b^2}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}-\frac {5 a^3 (8 b c-7 a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{9/2}} \] Output:

5/64*a^2*(-7*a*d+8*b*c)*(b*x^2+a*x)^(1/2)/b^4-5/96*a*(-7*a*d+8*b*c)*x*(b*x 
^2+a*x)^(1/2)/b^3+1/24*(-7*a*d+8*b*c)*x^2*(b*x^2+a*x)^(1/2)/b^2+1/4*d*x^3* 
(b*x^2+a*x)^(1/2)/b-5/64*a^3*(-7*a*d+8*b*c)*arctanh(b^(1/2)*x/(b*x^2+a*x)^ 
(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {\sqrt {x} (a+b x) \left (120 a^2 b c \sqrt {x}-105 a^3 d \sqrt {x}-80 a b^2 c x^{3/2}+70 a^2 b d x^{3/2}+64 b^3 c x^{5/2}-56 a b^2 d x^{5/2}+48 b^3 d x^{7/2}\right )}{192 b^4 \sqrt {x (a+b x)}}+\frac {5 a^3 (-8 b c+7 a d) \sqrt {x} \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{32 b^{9/2} \sqrt {x (a+b x)}} \] Input:

Integrate[(x^3*(c + d*x))/Sqrt[a*x + b*x^2],x]
 

Output:

(Sqrt[x]*(a + b*x)*(120*a^2*b*c*Sqrt[x] - 105*a^3*d*Sqrt[x] - 80*a*b^2*c*x 
^(3/2) + 70*a^2*b*d*x^(3/2) + 64*b^3*c*x^(5/2) - 56*a*b^2*d*x^(5/2) + 48*b 
^3*d*x^(7/2)))/(192*b^4*Sqrt[x*(a + b*x)]) + (5*a^3*(-8*b*c + 7*a*d)*Sqrt[ 
x]*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(3 
2*b^(9/2)*Sqrt[x*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1221, 1134, 1134, 1160, 1091, 219}

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.

\(\displaystyle \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {(8 b c-7 a d) \int \frac {x^3}{\sqrt {b x^2+a x}}dx}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {x^2 \sqrt {a x+b x^2}}{3 b}-\frac {5 a \int \frac {x^2}{\sqrt {b x^2+a x}}dx}{6 b}\right )}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {x^2 \sqrt {a x+b x^2}}{3 b}-\frac {5 a \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \int \frac {x}{\sqrt {b x^2+a x}}dx}{4 b}\right )}{6 b}\right )}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {x^2 \sqrt {a x+b x^2}}{3 b}-\frac {5 a \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \int \frac {1}{\sqrt {b x^2+a x}}dx}{2 b}\right )}{4 b}\right )}{6 b}\right )}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {(8 b c-7 a d) \left (\frac {x^2 \sqrt {a x+b x^2}}{3 b}-\frac {5 a \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{b}\right )}{4 b}\right )}{6 b}\right )}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (\frac {x^2 \sqrt {a x+b x^2}}{3 b}-\frac {5 a \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}\right ) (8 b c-7 a d)}{8 b}+\frac {d x^3 \sqrt {a x+b x^2}}{4 b}\)

Input:

Int[(x^3*(c + d*x))/Sqrt[a*x + b*x^2],x]
 

Output:

(d*x^3*Sqrt[a*x + b*x^2])/(4*b) + ((8*b*c - 7*a*d)*((x^2*Sqrt[a*x + b*x^2] 
)/(3*b) - (5*a*((x*Sqrt[a*x + b*x^2])/(2*b) - (3*a*(Sqrt[a*x + b*x^2]/b - 
(a*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/b^(3/2)))/(4*b)))/(6*b)))/(8*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1134

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d + e*x)^ 
(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ 
c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 
*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {\frac {35 a^{3} \left (a d -\frac {8 b c}{7}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )}{64}-\frac {35 \left (-\frac {8 \left (\frac {7 d x}{12}+c \right ) a^{2} b^{\frac {3}{2}}}{7}+\frac {16 \left (\frac {7 d x}{10}+c \right ) x a \,b^{\frac {5}{2}}}{21}-\frac {64 x^{2} \left (\frac {3 d x}{4}+c \right ) b^{\frac {7}{2}}}{105}+\sqrt {b}\, a^{3} d \right ) \sqrt {x \left (b x +a \right )}}{64}}{b^{\frac {9}{2}}}\)	\(97\)
risch	\(-\frac {\left (-48 b^{3} d \,x^{3}+56 a \,b^{2} d \,x^{2}-64 b^{3} c \,x^{2}-70 a^{2} b d x +80 a \,b^{2} c x +105 a^{3} d -120 c \,a^{2} b \right ) x \left (b x +a \right )}{192 b^{4} \sqrt {x \left (b x +a \right )}}+\frac {5 a^{3} \left (7 a d -8 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {9}{2}}}\)	\(121\)
default	\(c \left (\frac {x^{2} \sqrt {b \,x^{2}+a x}}{3 b}-\frac {5 a \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+d \left (\frac {x^{3} \sqrt {b \,x^{2}+a x}}{4 b}-\frac {7 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a x}}{3 b}-\frac {5 a \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )\)	\(224\)

Input:

int(x^3*(d*x+c)/(b*x^2+a*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

35/64/b^(9/2)*(a^3*(a*d-8/7*b*c)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))-(-8/ 
7*(7/12*d*x+c)*a^2*b^(3/2)+16/21*(7/10*d*x+c)*x*a*b^(5/2)-64/105*x^2*(3/4* 
d*x+c)*b^(7/2)+b^(1/2)*a^3*d)*(x*(b*x+a))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.58 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\left [-\frac {15 \, {\left (8 \, a^{3} b c - 7 \, a^{4} d\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (48 \, b^{4} d x^{3} + 120 \, a^{2} b^{2} c - 105 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c - 7 \, a b^{3} d\right )} x^{2} - 10 \, {\left (8 \, a b^{3} c - 7 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{384 \, b^{5}}, \frac {15 \, {\left (8 \, a^{3} b c - 7 \, a^{4} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (48 \, b^{4} d x^{3} + 120 \, a^{2} b^{2} c - 105 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c - 7 \, a b^{3} d\right )} x^{2} - 10 \, {\left (8 \, a b^{3} c - 7 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{192 \, b^{5}}\right ] \] Input:

integrate(x^3*(d*x+c)/(b*x^2+a*x)^(1/2),x, algorithm="fricas")
 

Output:

[-1/384*(15*(8*a^3*b*c - 7*a^4*d)*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a 
*x)*sqrt(b)) - 2*(48*b^4*d*x^3 + 120*a^2*b^2*c - 105*a^3*b*d + 8*(8*b^4*c 
- 7*a*b^3*d)*x^2 - 10*(8*a*b^3*c - 7*a^2*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^5, 
 1/192*(15*(8*a^3*b*c - 7*a^4*d)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b 
)/(b*x + a)) + (48*b^4*d*x^3 + 120*a^2*b^2*c - 105*a^3*b*d + 8*(8*b^4*c - 
7*a*b^3*d)*x^2 - 10*(8*a*b^3*c - 7*a^2*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^5]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\begin {cases} - \frac {5 a^{3} \left (- \frac {7 a d}{8 b} + c\right ) \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{3}} + \sqrt {a x + b x^{2}} \cdot \left (\frac {5 a^{2} \left (- \frac {7 a d}{8 b} + c\right )}{8 b^{3}} - \frac {5 a x \left (- \frac {7 a d}{8 b} + c\right )}{12 b^{2}} + \frac {d x^{3}}{4 b} + \frac {x^{2} \left (- \frac {7 a d}{8 b} + c\right )}{3 b}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (\frac {c \left (a x\right )^{\frac {7}{2}}}{7} + \frac {d \left (a x\right )^{\frac {9}{2}}}{9 a}\right )}{a^{4}} & \text {for}\: a \neq 0 \\\tilde {\infty } \left (\frac {c x^{4}}{4} + \frac {d x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(x**3*(d*x+c)/(b*x**2+a*x)**(1/2),x)
 

Output:

Piecewise((-5*a**3*(-7*a*d/(8*b) + c)*Piecewise((log(a + 2*sqrt(b)*sqrt(a* 
x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + 
 x)/sqrt(b*(a/(2*b) + x)**2), True))/(16*b**3) + sqrt(a*x + b*x**2)*(5*a** 
2*(-7*a*d/(8*b) + c)/(8*b**3) - 5*a*x*(-7*a*d/(8*b) + c)/(12*b**2) + d*x** 
3/(4*b) + x**2*(-7*a*d/(8*b) + c)/(3*b)), Ne(b, 0)), (2*(c*(a*x)**(7/2)/7 
+ d*(a*x)**(9/2)/(9*a))/a**4, Ne(a, 0)), (zoo*(c*x**4/4 + d*x**5/5), True) 
)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a x} d x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a x} c x^{2}}{3 \, b} - \frac {7 \, \sqrt {b x^{2} + a x} a d x^{2}}{24 \, b^{2}} - \frac {5 \, \sqrt {b x^{2} + a x} a c x}{12 \, b^{2}} + \frac {35 \, \sqrt {b x^{2} + a x} a^{2} d x}{96 \, b^{3}} - \frac {5 \, a^{3} c \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {7}{2}}} + \frac {35 \, a^{4} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {9}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} a^{2} c}{8 \, b^{3}} - \frac {35 \, \sqrt {b x^{2} + a x} a^{3} d}{64 \, b^{4}} \] Input:

integrate(x^3*(d*x+c)/(b*x^2+a*x)^(1/2),x, algorithm="maxima")
 

Output:

1/4*sqrt(b*x^2 + a*x)*d*x^3/b + 1/3*sqrt(b*x^2 + a*x)*c*x^2/b - 7/24*sqrt( 
b*x^2 + a*x)*a*d*x^2/b^2 - 5/12*sqrt(b*x^2 + a*x)*a*c*x/b^2 + 35/96*sqrt(b 
*x^2 + a*x)*a^2*d*x/b^3 - 5/16*a^3*c*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s 
qrt(b))/b^(7/2) + 35/128*a^4*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b) 
)/b^(9/2) + 5/8*sqrt(b*x^2 + a*x)*a^2*c/b^3 - 35/64*sqrt(b*x^2 + a*x)*a^3* 
d/b^4
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (\frac {6 \, d x}{b} + \frac {8 \, b^{3} c - 7 \, a b^{2} d}{b^{4}}\right )} x - \frac {5 \, {\left (8 \, a b^{2} c - 7 \, a^{2} b d\right )}}{b^{4}}\right )} x + \frac {15 \, {\left (8 \, a^{2} b c - 7 \, a^{3} d\right )}}{b^{4}}\right )} + \frac {5 \, {\left (8 \, a^{3} b c - 7 \, a^{4} d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{128 \, b^{\frac {9}{2}}} \] Input:

integrate(x^3*(d*x+c)/(b*x^2+a*x)^(1/2),x, algorithm="giac")
 

Output:

1/192*sqrt(b*x^2 + a*x)*(2*(4*(6*d*x/b + (8*b^3*c - 7*a*b^2*d)/b^4)*x - 5* 
(8*a*b^2*c - 7*a^2*b*d)/b^4)*x + 15*(8*a^2*b*c - 7*a^3*d)/b^4) + 5/128*(8* 
a^3*b*c - 7*a^4*d)*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a)) 
/b^(9/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\int \frac {x^3\,\left (c+d\,x\right )}{\sqrt {b\,x^2+a\,x}} \,d x \] Input:

int((x^3*(c + d*x))/(a*x + b*x^2)^(1/2),x)
 

Output:

int((x^3*(c + d*x))/(a*x + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.09 \[ \int \frac {x^3 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {-105 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b d +120 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} c +70 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} d x -80 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c x -56 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} d \,x^{2}+64 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c \,x^{2}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{4} d \,x^{3}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} d -120 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} b c}{192 b^{5}} \] Input:

int(x^3*(d*x+c)/(b*x^2+a*x)^(1/2),x)
 

Output:

( - 105*sqrt(x)*sqrt(a + b*x)*a**3*b*d + 120*sqrt(x)*sqrt(a + b*x)*a**2*b* 
*2*c + 70*sqrt(x)*sqrt(a + b*x)*a**2*b**2*d*x - 80*sqrt(x)*sqrt(a + b*x)*a 
*b**3*c*x - 56*sqrt(x)*sqrt(a + b*x)*a*b**3*d*x**2 + 64*sqrt(x)*sqrt(a + b 
*x)*b**4*c*x**2 + 48*sqrt(x)*sqrt(a + b*x)*b**4*d*x**3 + 105*sqrt(b)*log(( 
sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*d - 120*sqrt(b)*log((sqrt(a 
 + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*b*c)/(192*b**5)