\(\int \frac {(d+e x)^{3/2} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\) [245]

Optimal result

Integrand size = 46, antiderivative size = 221 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 (c e f+c d g-b e g) (d+e x)^{3/2}}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {2 (e f-d g) \sqrt {d+e x}}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 (e f-d g) \text {arctanh}\left (\frac {\sqrt {2 c d-b e} \sqrt {d+e x}}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e^2 (2 c d-b e)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^{5/2} \left (-\frac {(-b e+c (d-e x)) \left (4 b c e^2 f-b^2 e^2 g+c^2 \left (d^2 g+3 e^2 f x-d e (5 f+3 g x)\right )\right )}{c (-2 c d+b e)^2}+\frac {3 (e f-d g) (-b e+c (d-e x))^{5/2} \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )}{(-2 c d+b e)^{5/2}}\right )}{3 e^2 ((d+e x) (-b e+c (d-e x)))^{5/2}} \] Input:

Integrate[((d + e*x)^(3/2)*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2 
)^(5/2),x]
 

Output:

(2*(d + e*x)^(5/2)*(-(((-(b*e) + c*(d - e*x))*(4*b*c*e^2*f - b^2*e^2*g + c 
^2*(d^2*g + 3*e^2*f*x - d*e*(5*f + 3*g*x))))/(c*(-2*c*d + b*e)^2)) + (3*(e 
*f - d*g)*(-(b*e) + c*(d - e*x))^(5/2)*ArcTan[Sqrt[c*d - b*e - c*e*x]/Sqrt 
[-2*c*d + b*e]])/(-2*c*d + b*e)^(5/2)))/(3*e^2*((d + e*x)*(-(b*e) + c*(d - 
 e*x)))^(5/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1218, 1132, 1136, 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.

\(\displaystyle \int \frac {(d+e x)^{3/2} (f+g x)}{\left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1218

\(\displaystyle \frac {(e f-d g) \int \frac {\sqrt {d+e x}}{\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}dx}{e (2 c d-b e)}+\frac {2 (d+e x)^{3/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle \frac {(e f-d g) \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{2 c d-b e}+\frac {2 \sqrt {d+e x}}{e (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e (2 c d-b e)}+\frac {2 (d+e x)^{3/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(e f-d g) \left (\frac {2 e \int \frac {1}{\frac {e^2 \left (-c x^2 e^2-b x e^2+d (c d-b e)\right )}{d+e x}-e^2 (2 c d-b e)}d\frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}}{2 c d-b e}+\frac {2 \sqrt {d+e x}}{e (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e (2 c d-b e)}+\frac {2 (d+e x)^{3/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(e f-d g) \left (\frac {2 \sqrt {d+e x}}{e (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e (2 c d-b e)^{3/2}}\right )}{e (2 c d-b e)}+\frac {2 (d+e x)^{3/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}\)

Input:

Int[((d + e*x)^(3/2)*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2 
),x]
 

Output:

(2*(c*e*f + c*d*g - b*e*g)*(d + e*x)^(3/2))/(3*c*e^2*(2*c*d - b*e)*(d*(c*d 
 - b*e) - b*e^2*x - c*e^2*x^2)^(3/2)) + ((e*f - d*g)*((2*Sqrt[d + e*x])/(e 
*(2*c*d - b*e)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]) - (2*ArcTanh[Sqr 
t[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]/(Sqrt[2*c*d - b*e]*Sqrt[d + e*x])]) 
/(e*(2*c*d - b*e)^(3/2))))/(e*(2*c*d - b*e))
 

Defintions of rubi rules used

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]
 

rule 1132

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 
 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 
 4*a*c)))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ 
[0, m, 1] && IntegerQ[2*p]
 

rule 1136

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 

rule 1218

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(476\) vs. \(2(203)=406\).

Time = 1.54 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.16

method	result	size
default	\(\frac {2 \left (3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e g x \sqrt {-c e x -b e +c d}-3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} e^{2} f x \sqrt {-c e x -b e +c d}+3 \sqrt {-c e x -b e +c d}\, \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d e g -3 \sqrt {-c e x -b e +c d}\, \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{2} f -3 \sqrt {-c e x -b e +c d}\, \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} g +3 \sqrt {-c e x -b e +c d}\, \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e f +3 \sqrt {b e -2 c d}\, c^{2} d e g x -3 \sqrt {b e -2 c d}\, c^{2} e^{2} f x +\sqrt {b e -2 c d}\, b^{2} e^{2} g -4 \sqrt {b e -2 c d}\, b c \,e^{2} f -\sqrt {b e -2 c d}\, c^{2} d^{2} g +5 \sqrt {b e -2 c d}\, c^{2} d e f \right ) \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}}{3 \left (b e -2 c d \right )^{\frac {5}{2}} c \,e^{2} \left (c e x +b e -c d \right )^{2} \sqrt {e x +d}}\)	\(477\)

Input:

int((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x,method= 
_RETURNVERBOSE)
 

Output:

2/3*(3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e*g*x*(-c*e* 
x-b*e+c*d)^(1/2)-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*e^ 
2*f*x*(-c*e*x-b*e+c*d)^(1/2)+3*(-c*e*x-b*e+c*d)^(1/2)*arctan((-c*e*x-b*e+c 
*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d*e*g-3*(-c*e*x-b*e+c*d)^(1/2)*arctan((-c 
*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*e^2*f-3*(-c*e*x-b*e+c*d)^(1/2)* 
arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^2*g+3*(-c*e*x-b*e+c 
*d)^(1/2)*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e*f+3*(b* 
e-2*c*d)^(1/2)*c^2*d*e*g*x-3*(b*e-2*c*d)^(1/2)*c^2*e^2*f*x+(b*e-2*c*d)^(1/ 
2)*b^2*e^2*g-4*(b*e-2*c*d)^(1/2)*b*c*e^2*f-(b*e-2*c*d)^(1/2)*c^2*d^2*g+5*( 
b*e-2*c*d)^(1/2)*c^2*d*e*f)*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)/(b*e-2*c*d)^( 
5/2)/c/e^2/(c*e*x+b*e-c*d)^2/(e*x+d)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (203) = 406\).

Time = 0.21 (sec) , antiderivative size = 1439, normalized size of antiderivative = 6.51 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, 
algorithm="fricas")
 

Output:

[-1/3*(3*((c^3*e^4*f - c^3*d*e^3*g)*x^3 - ((c^3*d*e^3 - 2*b*c^2*e^4)*f - ( 
c^3*d^2*e^2 - 2*b*c^2*d*e^3)*g)*x^2 + (c^3*d^3*e - 2*b*c^2*d^2*e^2 + b^2*c 
*d*e^3)*f - (c^3*d^4 - 2*b*c^2*d^3*e + b^2*c*d^2*e^2)*g - ((c^3*d^2*e^2 - 
b^2*c*e^4)*f - (c^3*d^3*e - b^2*c*d*e^3)*g)*x)*sqrt(2*c*d - b*e)*log(-(c*e 
^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x - 2*sqrt(-c*e^2*x^2 - b*e 
^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x 
+ d^2)) - 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((10*c^3*d^2*e - 13 
*b*c^2*d*e^2 + 4*b^2*c*e^3)*f - (2*c^3*d^3 - b*c^2*d^2*e - 2*b^2*c*d*e^2 + 
 b^3*e^3)*g - 3*((2*c^3*d*e^2 - b*c^2*e^3)*f - (2*c^3*d^2*e - b*c^2*d*e^2) 
*g)*x)*sqrt(e*x + d))/(8*c^6*d^6*e^2 - 28*b*c^5*d^5*e^3 + 38*b^2*c^4*d^4*e 
^4 - 25*b^3*c^3*d^3*e^5 + 8*b^4*c^2*d^2*e^6 - b^5*c*d*e^7 + (8*c^6*d^3*e^5 
 - 12*b*c^5*d^2*e^6 + 6*b^2*c^4*d*e^7 - b^3*c^3*e^8)*x^3 - (8*c^6*d^4*e^4 
- 28*b*c^5*d^3*e^5 + 30*b^2*c^4*d^2*e^6 - 13*b^3*c^3*d*e^7 + 2*b^4*c^2*e^8 
)*x^2 - (8*c^6*d^5*e^3 - 12*b*c^5*d^4*e^4 - 2*b^2*c^4*d^3*e^5 + 11*b^3*c^3 
*d^2*e^6 - 6*b^4*c^2*d*e^7 + b^5*c*e^8)*x), -2/3*(3*((c^3*e^4*f - c^3*d*e^ 
3*g)*x^3 - ((c^3*d*e^3 - 2*b*c^2*e^4)*f - (c^3*d^2*e^2 - 2*b*c^2*d*e^3)*g) 
*x^2 + (c^3*d^3*e - 2*b*c^2*d^2*e^2 + b^2*c*d*e^3)*f - (c^3*d^4 - 2*b*c^2* 
d^3*e + b^2*c*d^2*e^2)*g - ((c^3*d^2*e^2 - b^2*c*e^4)*f - (c^3*d^3*e - b^2 
*c*d*e^3)*g)*x)*sqrt(-2*c*d + b*e)*arctan(-sqrt(-c*e^2*x^2 - b*e^2*x + c*d 
^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(2*c*d^2 - b*d*e + (2*c*d*...
 

Sympy [F]

\[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x+d)**(3/2)*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/ 
2),x)
 

Output:

Integral((d + e*x)**(3/2)*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(5/2 
), x)
 

Maxima [F]

\[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (g x + f\right )}}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, 
algorithm="maxima")
 

Output:

integrate((e*x + d)^(3/2)*(g*x + f)/(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e) 
^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e f - d g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{{\left (4 \, c^{2} d^{2} e - 4 \, b c d e^{2} + b^{2} e^{3}\right )} \sqrt {-2 \, c d + b e}} - \frac {2 \, c^{2} d e f - b c e^{2} f + 2 \, c^{2} d^{2} g - 3 \, b c d e g + b^{2} e^{2} g - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )} c e f + 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )} c d g}{{\left (4 \, c^{3} d^{2} e - 4 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}\right )}}{3 \, e} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, 
algorithm="giac")
 

Output:

2/3*(3*(e*f - d*g)*arctan(sqrt(-(e*x + d)*c + 2*c*d - b*e)/sqrt(-2*c*d + b 
*e))/((4*c^2*d^2*e - 4*b*c*d*e^2 + b^2*e^3)*sqrt(-2*c*d + b*e)) - (2*c^2*d 
*e*f - b*c*e^2*f + 2*c^2*d^2*g - 3*b*c*d*e*g + b^2*e^2*g - 3*((e*x + d)*c 
- 2*c*d + b*e)*c*e*f + 3*((e*x + d)*c - 2*c*d + b*e)*c*d*g)/((4*c^3*d^2*e 
- 4*b*c^2*d*e^2 + b^2*c*e^3)*((e*x + d)*c - 2*c*d + b*e)*sqrt(-(e*x + d)*c 
 + 2*c*d - b*e)))/e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{5/2}} \,d x \] Input:

int(((f + g*x)*(d + e*x)^(3/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2 
),x)
 

Output:

int(((f + g*x)*(d + e*x)^(3/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 588, normalized size of antiderivative = 2.66 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx=\frac {-2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d e g +2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{2} f +2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} g -2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e f -2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e g x +2 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} e^{2} f x -\frac {2 b^{3} e^{3} g}{3}+\frac {4 b^{2} c d \,e^{2} g}{3}+\frac {8 b^{2} c \,e^{3} f}{3}+\frac {2 b \,c^{2} d^{2} e g}{3}-\frac {26 b \,c^{2} d \,e^{2} f}{3}-2 b \,c^{2} d \,e^{2} g x +2 b \,c^{2} e^{3} f x -\frac {4 c^{3} d^{3} g}{3}+\frac {20 c^{3} d^{2} e f}{3}+4 c^{3} d^{2} e g x -4 c^{3} d \,e^{2} f x}{\sqrt {-c e x -b e +c d}\, c \,e^{2} \left (b^{3} c \,e^{4} x -6 b^{2} c^{2} d \,e^{3} x +12 b \,c^{3} d^{2} e^{2} x -8 c^{4} d^{3} e x +b^{4} e^{4}-7 b^{3} c d \,e^{3}+18 b^{2} c^{2} d^{2} e^{2}-20 b \,c^{3} d^{3} e +8 c^{4} d^{4}\right )} \] Input:

int((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x)
 

Output:

(2*( - 3*sqrt(b*e - 2*c*d)*sqrt( - b*e + c*d - c*e*x)*atan(sqrt( - b*e + c 
*d - c*e*x)/sqrt(b*e - 2*c*d))*b*c*d*e*g + 3*sqrt(b*e - 2*c*d)*sqrt( - b*e 
 + c*d - c*e*x)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*b*c*e** 
2*f + 3*sqrt(b*e - 2*c*d)*sqrt( - b*e + c*d - c*e*x)*atan(sqrt( - b*e + c* 
d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d**2*g - 3*sqrt(b*e - 2*c*d)*sqrt( - b* 
e + c*d - c*e*x)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d 
*e*f - 3*sqrt(b*e - 2*c*d)*sqrt( - b*e + c*d - c*e*x)*atan(sqrt( - b*e + c 
*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d*e*g*x + 3*sqrt(b*e - 2*c*d)*sqrt( - 
b*e + c*d - c*e*x)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2 
*e**2*f*x - b**3*e**3*g + 2*b**2*c*d*e**2*g + 4*b**2*c*e**3*f + b*c**2*d** 
2*e*g - 13*b*c**2*d*e**2*f - 3*b*c**2*d*e**2*g*x + 3*b*c**2*e**3*f*x - 2*c 
**3*d**3*g + 10*c**3*d**2*e*f + 6*c**3*d**2*e*g*x - 6*c**3*d*e**2*f*x))/(3 
*sqrt( - b*e + c*d - c*e*x)*c*e**2*(b**4*e**4 - 7*b**3*c*d*e**3 + b**3*c*e 
**4*x + 18*b**2*c**2*d**2*e**2 - 6*b**2*c**2*d*e**3*x - 20*b*c**3*d**3*e + 
 12*b*c**3*d**2*e**2*x + 8*c**4*d**4 - 8*c**4*d**3*e*x))