\(\int (a+b x)^{5/2} \sqrt {c+d x} \, dx\) [311]

Optimal result

Integrand size = 19, antiderivative size = 192 \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}-\frac {5 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87 \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^3 d^3+a^2 b d^2 (73 c+118 d x)+a b^2 d \left (-55 c^2+36 c d x+136 d^2 x^2\right )+b^3 \left (15 c^3-10 c^2 d x+8 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b d^3}-\frac {5 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{3/2} d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^3*d^3 + a^2*b*d^2*(73*c + 118*d*x) + a* 
b^2*d*(-55*c^2 + 36*c*d*x + 136*d^2*x^2) + b^3*(15*c^3 - 10*c^2*d*x + 8*c* 
d^2*x^2 + 48*d^3*x^3)))/(192*b*d^3) - (5*(b*c - a*d)^4*ArcTanh[(Sqrt[b]*Sq 
rt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(64*b^(3/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {60, 60, 60, 60, 66, 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.

\(\displaystyle \int (a+b x)^{5/2} \sqrt {c+d x} \, dx\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 221

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]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {3}{2}}}{4 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}{2 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {x d +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {d b}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {x d +c}\, \sqrt {b x +a}\, \sqrt {d b}}\right )}{4 d}\right )}{2 d}\right )}{8 d}\)	\(206\)

Input:

int((b*x+a)^(5/2)*(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*(b*x+a)^(5/2)*(d*x+c)^(3/2)-5/8*(-a*d+b*c)/d*(1/3/d*(b*x+a)^(3/2)*(d 
*x+c)^(3/2)-1/2*(-a*d+b*c)/d*(1/2/d*(b*x+a)^(1/2)*(d*x+c)^(3/2)-1/4*(-a*d+ 
b*c)/d*((b*x+a)^(1/2)*(d*x+c)^(1/2)/b-1/2*(a*d-b*c)/b*((b*x+a)*(d*x+c))^(1 
/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(d*b)^(1/2)+(b* 
d*x^2+(a*d+b*c)*x+a*c)^(1/2))/(d*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.81 \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\left [\frac {15 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{2} d^{4}}, \frac {15 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{2} d^{4}}\right ] \] Input:

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/768*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 
a^4*d^4)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*( 
2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + 
a*b*d^2)*x) + 4*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 55*a*b^3*c^2*d^2 + 73*a^2 
*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4*c*d^3 + 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^ 
2*d^2 - 18*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/( 
b^2*d^4), 1/384*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b 
*c*d^3 + a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*s 
qrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x) 
) + 2*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 55*a*b^3*c^2*d^2 + 73*a^2*b^2*c*d^3 
 + 15*a^3*b*d^4 + 8*(b^4*c*d^3 + 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 18 
*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^4)]
 

Sympy [F]

\[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\int \left (a + b x\right )^{\frac {5}{2}} \sqrt {c + d x}\, dx \] Input:

integrate((b*x+a)**(5/2)*(d*x+c)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (154) = 308\).

Time = 0.22 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.70 \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx =\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

-1/192*(192*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c 
 + (b*x + a)*b*d - a*b*d)))/sqrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d 
)*sqrt(b*x + a))*a^3*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2* 
(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^ 
14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6 
)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^1 
1*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^ 
2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) 
+ sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*b*abs(b) - 14 
4*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^ 
2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b*d 
)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a^2 
*abs(b)/b^2 - 24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*b*x + 4*a + (b 
*c*d^3 - 13*a*d^4)/d^4)*(b*x + a) - 3*(b^2*c^2*d^2 + 2*a*b*c*d^3 - 11*a^2* 
d^4)/d^4)*sqrt(b*x + a) - 3*(b^4*c^3 + a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - 5*a 
^3*b*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - 
a*b*d)))/(sqrt(b*d)*d^2))*a*abs(b)/b^2)/b
 

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\int {\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x} \,d x \] Input:

int((a + b*x)^(5/2)*(c + d*x)^(1/2),x)
 

Output:

int((a + b*x)^(5/2)*(c + d*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.45 \[ \int (a+b x)^{5/2} \sqrt {c+d x} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} b \,d^{4}+73 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} c \,d^{3}+118 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} d^{4} x -55 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c^{2} d^{2}+36 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c \,d^{3} x +136 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} d^{4} x^{2}+15 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{3} d -10 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{2} d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c \,d^{3} x^{2}+48 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} d^{4} x^{3}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{4} d^{4}+60 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} b c \,d^{3}-90 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b^{2} c^{2} d^{2}+60 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{3} c^{3} d -15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{4} c^{4}}{192 b^{2} d^{4}} \] Input:

int((b*x+a)^(5/2)*(d*x+c)^(1/2),x)
 

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**4 + 73*sqrt(c + d*x)*sqrt(a + b* 
x)*a**2*b**2*c*d**3 + 118*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*d**4*x - 5 
5*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**2*d**2 + 36*sqrt(c + d*x)*sqrt(a + 
 b*x)*a*b**3*c*d**3*x + 136*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*d**4*x**2 + 
 15*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**3*d - 10*sqrt(c + d*x)*sqrt(a + b* 
x)*b**4*c**2*d**2*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c*d**3*x**2 + 48* 
sqrt(c + d*x)*sqrt(a + b*x)*b**4*d**4*x**3 - 15*sqrt(d)*sqrt(b)*log((sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*d**4 + 60* 
sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a 
*d - b*c))*a**3*b*c*d**3 - 90*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + 
 sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b**2*c**2*d**2 + 60*sqrt(d)* 
sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c 
))*a*b**3*c**3*d - 15*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b) 
*sqrt(c + d*x))/sqrt(a*d - b*c))*b**4*c**4)/(192*b**2*d**4)