\(\int \frac {(e x)^{9/2} (c+d x)}{(a x^2+b x^3)^{3/2}} \, dx\) [340]

Optimal result

Integrand size = 28, antiderivative size = 171 \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 (b c-a d) e^2 (e x)^{5/2}}{b^2 \sqrt {a x^2+b x^3}}+\frac {3 (4 b c-5 a d) e^5 \sqrt {a x^2+b x^3}}{4 b^3 \sqrt {e x}}+\frac {d e^4 \sqrt {e x} \sqrt {a x^2+b x^3}}{2 b^2}-\frac {3 a (4 b c-5 a d) e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{4 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00 \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {e^4 \sqrt {x} \sqrt {e x} \left (\sqrt {b} \sqrt {x} \left (-15 a^2 d+a b (12 c-5 d x)+2 b^2 x (2 c+d x)\right )+24 a b c \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )+30 a^2 d \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{4 b^{7/2} \sqrt {x^2 (a+b x)}} \] Input:

Integrate[((e*x)^(9/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
 

Output:

(e^4*Sqrt[x]*Sqrt[e*x]*(Sqrt[b]*Sqrt[x]*(-15*a^2*d + a*b*(12*c - 5*d*x) + 
2*b^2*x*(2*c + d*x)) + 24*a*b*c*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(S 
qrt[a] - Sqrt[a + b*x])] + 30*a^2*d*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt[x] 
)/(-Sqrt[a] + Sqrt[a + b*x])]))/(4*b^(7/2)*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1943, 1930, 1930, 1937, 1935, 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.

\(\displaystyle \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1937

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

\(\Big \downarrow \) 1935

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

\(\Big \downarrow \) 219

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

Input:

Int[((e*x)^(9/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
 

Output:

(2*(b*c - a*d)*e*(e*x)^(7/2))/(a*b*Sqrt[a*x^2 + b*x^3]) - ((4*b*c - 5*a*d) 
*e^2*((e^2*Sqrt[e*x]*Sqrt[a*x^2 + b*x^3])/(2*b) - (3*a*e*((e^2*Sqrt[a*x^2 
+ b*x^3])/(b*Sqrt[e*x]) - (a*e*Sqrt[e*x]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a* 
x^2 + b*x^3]])/(b^(3/2)*Sqrt[x])))/(4*b)))/(a*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 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1935

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 

rule 1937

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
:> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a*x^j + b 
*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2] && 
NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]
 

rule 1943

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j 
+ 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( 
m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1)))   Int[(e*x)^(m 
 - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, 
n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 
] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {\left (-2 b d x +7 a d -4 b c \right ) x^{2} \left (b x +a \right ) e^{5}}{4 b^{3} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {a \left (\frac {15 a d \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right )}{\sqrt {b e}}-\frac {12 b c \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right )}{\sqrt {b e}}-\frac {16 \left (a d -b c \right ) \sqrt {b e \left (x +\frac {a}{b}\right )^{2}-a e \left (x +\frac {a}{b}\right )}}{b e \left (x +\frac {a}{b}\right )}\right ) e^{5} x \sqrt {e x \left (b x +a \right )}}{8 b^{3} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\)	\(214\)
default	\(-\frac {x^{3} \left (b x +a \right ) \left (-4 b^{2} d \,x^{2} \sqrt {b e}\, \sqrt {e x \left (b x +a \right )}-15 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{2} b d e x +12 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a \,b^{2} c e x +10 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a b d x -8 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b^{2} c x -15 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{3} d e +12 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{2} b c e +30 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{2} d -24 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a b c \right ) \sqrt {e x}\, e^{4}}{8 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \sqrt {b e}\, \sqrt {e x \left (b x +a \right )}\, b^{3}}\)	\(328\)

Input:

int((e*x)^(9/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-2*b*d*x+7*a*d-4*b*c)*x^2*(b*x+a)/b^3*e^5/(x^2*(b*x+a))^(1/2)/(e*x)^ 
(1/2)+1/8*a/b^3*(15*a*d*ln((1/2*a*e+b*e*x)/(b*e)^(1/2)+(b*e*x^2+a*e*x)^(1/ 
2))/(b*e)^(1/2)-12*b*c*ln((1/2*a*e+b*e*x)/(b*e)^(1/2)+(b*e*x^2+a*e*x)^(1/2 
))/(b*e)^(1/2)-16*(a*d-b*c)/b/e/(x+a/b)*(b*e*(x+a/b)^2-a*e*(x+a/b))^(1/2)) 
*e^5/(x^2*(b*x+a))^(1/2)*x*(e*x*(b*x+a))^(1/2)/(e*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.12 \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} e^{4} x^{2} + {\left (4 \, a^{2} b c - 5 \, a^{3} d\right )} e^{4} x\right )} \sqrt {\frac {e}{b}} \log \left (\frac {2 \, b e x^{2} + a e x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {\frac {e}{b}}}{x}\right ) - 2 \, {\left (2 \, b^{2} d e^{4} x^{2} + {\left (4 \, b^{2} c - 5 \, a b d\right )} e^{4} x + 3 \, {\left (4 \, a b c - 5 \, a^{2} d\right )} e^{4}\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{8 \, {\left (b^{4} x^{2} + a b^{3} x\right )}}, \frac {3 \, {\left ({\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} e^{4} x^{2} + {\left (4 \, a^{2} b c - 5 \, a^{3} d\right )} e^{4} x\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {-\frac {e}{b}}}{b e x^{2} + a e x}\right ) + {\left (2 \, b^{2} d e^{4} x^{2} + {\left (4 \, b^{2} c - 5 \, a b d\right )} e^{4} x + 3 \, {\left (4 \, a b c - 5 \, a^{2} d\right )} e^{4}\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{4 \, {\left (b^{4} x^{2} + a b^{3} x\right )}}\right ] \] Input:

integrate((e*x)^(9/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/8*(3*((4*a*b^2*c - 5*a^2*b*d)*e^4*x^2 + (4*a^2*b*c - 5*a^3*d)*e^4*x)*s 
qrt(e/b)*log((2*b*e*x^2 + a*e*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*sqrt(e 
/b))/x) - 2*(2*b^2*d*e^4*x^2 + (4*b^2*c - 5*a*b*d)*e^4*x + 3*(4*a*b*c - 5* 
a^2*d)*e^4)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^4*x^2 + a*b^3*x), 1/4*(3*((4 
*a*b^2*c - 5*a^2*b*d)*e^4*x^2 + (4*a^2*b*c - 5*a^3*d)*e^4*x)*sqrt(-e/b)*ar 
ctan(sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*sqrt(-e/b)/(b*e*x^2 + a*e*x)) + (2*b^ 
2*d*e^4*x^2 + (4*b^2*c - 5*a*b*d)*e^4*x + 3*(4*a*b*c - 5*a^2*d)*e^4)*sqrt( 
b*x^3 + a*x^2)*sqrt(e*x))/(b^4*x^2 + a*b^3*x)]
 

Sympy [F]

\[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {\left (e x\right )^{\frac {9}{2}} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((e*x)**(9/2)*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
 

Output:

Integral((e*x)**(9/2)*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(9/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)*(e*x)^(9/2)/(b*x^3 + a*x^2)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.35 \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, d e^{3} x {\left | e \right |}}{b \mathrm {sgn}\left (x\right )} + \frac {4 \, b^{4} c e^{5} {\left | e \right |} \mathrm {sgn}\left (x\right ) - 5 \, a b^{3} d e^{5} {\left | e \right |} \mathrm {sgn}\left (x\right )}{b^{5} e^{2}}\right )} e x + \frac {3 \, {\left (4 \, a b^{3} c e^{6} {\left | e \right |} \mathrm {sgn}\left (x\right ) - 5 \, a^{2} b^{2} d e^{6} {\left | e \right |} \mathrm {sgn}\left (x\right )\right )}}{b^{5} e^{2}}\right )} \sqrt {e x}}{4 \, \sqrt {b e^{2} x + a e^{2}}} - \frac {3 \, {\left (4 \, a b c e^{4} {\left | e \right |} \log \left (e^{2} {\left | a \right |}\right ) - 5 \, a^{2} d e^{4} {\left | e \right |} \log \left (e^{2} {\left | a \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{8 \, \sqrt {b e} b^{3}} + \frac {3 \, {\left (4 \, a b c e^{4} {\left | e \right |} - 5 \, a^{2} d e^{4} {\left | e \right |}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {e x} + \sqrt {b e^{2} x + a e^{2}} \right |}\right )}{4 \, \sqrt {b e} b^{3} \mathrm {sgn}\left (x\right )} \] Input:

integrate((e*x)^(9/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
 

Output:

1/4*((2*d*e^3*x*abs(e)/(b*sgn(x)) + (4*b^4*c*e^5*abs(e)*sgn(x) - 5*a*b^3*d 
*e^5*abs(e)*sgn(x))/(b^5*e^2))*e*x + 3*(4*a*b^3*c*e^6*abs(e)*sgn(x) - 5*a^ 
2*b^2*d*e^6*abs(e)*sgn(x))/(b^5*e^2))*sqrt(e*x)/sqrt(b*e^2*x + a*e^2) - 3/ 
8*(4*a*b*c*e^4*abs(e)*log(e^2*abs(a)) - 5*a^2*d*e^4*abs(e)*log(e^2*abs(a)) 
)*sgn(x)/(sqrt(b*e)*b^3) + 3/4*(4*a*b*c*e^4*abs(e) - 5*a^2*d*e^4*abs(e))*l 
og(abs(-sqrt(b*e)*sqrt(e*x) + sqrt(b*e^2*x + a*e^2)))/(sqrt(b*e)*b^3*sgn(x 
))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^{9/2}\,\left (c+d\,x\right )}{{\left (b\,x^3+a\,x^2\right )}^{3/2}} \,d x \] Input:

int(((e*x)^(9/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
 

Output:

int(((e*x)^(9/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {(e x)^{9/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {e}\, e^{4} \left (15 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} d -12 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a b c -10 \sqrt {b}\, \sqrt {b x +a}\, a^{2} d +9 \sqrt {b}\, \sqrt {b x +a}\, a b c -15 \sqrt {x}\, a^{2} b d +12 \sqrt {x}\, a \,b^{2} c -5 \sqrt {x}\, a \,b^{2} d x +4 \sqrt {x}\, b^{3} c x +2 \sqrt {x}\, b^{3} d \,x^{2}\right )}{4 \sqrt {b x +a}\, b^{4}} \] Input:

int((e*x)^(9/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
 

Output:

(sqrt(e)*e**4*(15*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt( 
b))/sqrt(a))*a**2*d - 12*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x 
)*sqrt(b))/sqrt(a))*a*b*c - 10*sqrt(b)*sqrt(a + b*x)*a**2*d + 9*sqrt(b)*sq 
rt(a + b*x)*a*b*c - 15*sqrt(x)*a**2*b*d + 12*sqrt(x)*a*b**2*c - 5*sqrt(x)* 
a*b**2*d*x + 4*sqrt(x)*b**3*c*x + 2*sqrt(x)*b**3*d*x**2))/(4*sqrt(a + b*x) 
*b**4)