\(\int \frac {(e+f x)^{5/2}}{(a c+(b c+a d) x+b d x^2)^2} \, dx\) [396]

Optimal result

Integrand size = 31, antiderivative size = 261 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {f (2 b d e-b c f-a d f) \sqrt {e+f x}}{b d (b c-a d)^2}-\frac {(e+f x)^{3/2} (b c e+a d e-2 a c f+(2 b d e-(b c+a d) f) x)}{(b c-a d)^2 \left (a c+(b c+a d) x+b d x^2\right )}+\frac {(b e-a f)^{3/2} (4 b d e-5 b c f+a d f) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{3/2} (b c-a d)^3}-\frac {(d e-c f)^{3/2} (4 b d e+b c f-5 a d f) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (b c-a d)^3} \] Output:

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

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.00 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {\sqrt {e+f x} \left (a^2 d f^2 (c+d x)+b^2 \left (2 d^2 e^2 x+c^2 f^2 x+c d e (e-2 f x)\right )+a b \left (-4 c d e f+c^2 f^2+d^2 e (e-2 f x)\right )\right )}{b d (b c-a d)^2 (a+b x) (c+d x)}-\frac {(-b e+a f)^{3/2} (4 b d e-5 b c f+a d f) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{b^{3/2} (b c-a d)^3}-\frac {(-d e+c f)^{3/2} (4 b d e+b c f-5 a d f) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{3/2} (-b c+a d)^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {1164, 27, 1196, 25, 1197, 27, 1480, 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.

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

\(\Big \downarrow \) 1164

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1196

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1196

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

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(2 f^{3} \left (-\frac {\left (a f -b e \right )^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 b \left (b \left (f x +e \right )+a f -b e \right )}-\frac {\left (a d f -5 b c f +4 b d e \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 b \sqrt {\left (a f -b e \right ) b}}\right )}{f^{3} \left (a d -b c \right )^{3}}+\frac {\left (c f -d e \right )^{2} \left (-\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 d \left (d \left (f x +e \right )+c f -d e \right )}+\frac {\left (5 a d f -b c f -4 b d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 d \sqrt {\left (c f -d e \right ) d}}\right )}{f^{3} \left (a d -b c \right )^{3}}\right )\)	\(243\)
default	\(2 f^{3} \left (-\frac {\left (a f -b e \right )^{2} \left (\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 b \left (b \left (f x +e \right )+a f -b e \right )}-\frac {\left (a d f -5 b c f +4 b d e \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{2 b \sqrt {\left (a f -b e \right ) b}}\right )}{f^{3} \left (a d -b c \right )^{3}}+\frac {\left (c f -d e \right )^{2} \left (-\frac {f \left (a d -b c \right ) \sqrt {f x +e}}{2 d \left (d \left (f x +e \right )+c f -d e \right )}+\frac {\left (5 a d f -b c f -4 b d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{2 d \sqrt {\left (c f -d e \right ) d}}\right )}{f^{3} \left (a d -b c \right )^{3}}\right )\)	\(243\)
pseudoelliptic	\(-\frac {-d \left (a f -b e \right )^{2} \left (d x +c \right ) \left (\left (-5 c f +4 d e \right ) b +a d f \right ) \sqrt {\left (c f -d e \right ) d}\, \left (b x +a \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )+\sqrt {\left (a f -b e \right ) b}\, \left (-5 \left (\frac {\left (-c f -4 d e \right ) b}{5}+a d f \right ) b \left (c f -d e \right )^{2} \left (d x +c \right ) \left (b x +a \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+\left (\left (2 d^{2} e^{2} x +c e \left (-2 f x +e \right ) d +c^{2} f^{2} x \right ) b^{2}+\left (e \left (-2 f x +e \right ) d^{2}-4 e f d c +c^{2} f^{2}\right ) a b +a^{2} d \,f^{2} \left (d x +c \right )\right ) \sqrt {f x +e}\, \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}\right )}{\sqrt {\left (c f -d e \right ) d}\, \sqrt {\left (a f -b e \right ) b}\, b \left (b x +a \right ) \left (a d -b c \right )^{3} \left (d x +c \right ) d}\)	\(309\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 850 vs. \(2 (237) = 474\).

Time = 3.34 (sec) , antiderivative size = 3442, normalized size of antiderivative = 13.19 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((f*x+e)^(5/2)/(a*c+(a*d+b*c)*x+b*d*x^2)^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(c*f-d*e>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. 632 vs. \(2 (237) = 474\).

Time = 0.40 (sec) , antiderivative size = 632, normalized size of antiderivative = 2.42 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {{\left (4 \, b^{3} d e^{3} - 5 \, b^{3} c e^{2} f - 7 \, a b^{2} d e^{2} f + 10 \, a b^{2} c e f^{2} + 2 \, a^{2} b d e f^{2} - 5 \, a^{2} b c f^{3} + a^{3} d f^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt {-b^{2} e + a b f}} + \frac {{\left (4 \, b d^{3} e^{3} - 7 \, b c d^{2} e^{2} f - 5 \, a d^{3} e^{2} f + 2 \, b c^{2} d e f^{2} + 10 \, a c d^{2} e f^{2} + b c^{3} f^{3} - 5 \, a c^{2} d f^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} \sqrt {-d^{2} e + c d f}} - \frac {2 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} d^{2} e^{2} f - 2 \, \sqrt {f x + e} b^{2} d^{2} e^{3} f - 2 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c d e f^{2} - 2 \, {\left (f x + e\right )}^{\frac {3}{2}} a b d^{2} e f^{2} + 3 \, \sqrt {f x + e} b^{2} c d e^{2} f^{2} + 3 \, \sqrt {f x + e} a b d^{2} e^{2} f^{2} + {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c^{2} f^{3} + {\left (f x + e\right )}^{\frac {3}{2}} a^{2} d^{2} f^{3} - \sqrt {f x + e} b^{2} c^{2} e f^{3} - 4 \, \sqrt {f x + e} a b c d e f^{3} - \sqrt {f x + e} a^{2} d^{2} e f^{3} + \sqrt {f x + e} a b c^{2} f^{4} + \sqrt {f x + e} a^{2} c d f^{4}}{{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} {\left ({\left (f x + e\right )}^{2} b d - 2 \, {\left (f x + e\right )} b d e + b d e^{2} + {\left (f x + e\right )} b c f + {\left (f x + e\right )} a d f - b c e f - a d e f + a c f^{2}\right )}} \] Input:

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

Output:

-(4*b^3*d*e^3 - 5*b^3*c*e^2*f - 7*a*b^2*d*e^2*f + 10*a*b^2*c*e*f^2 + 2*a^2 
*b*d*e*f^2 - 5*a^2*b*c*f^3 + a^3*d*f^3)*arctan(sqrt(f*x + e)*b/sqrt(-b^2*e 
 + a*b*f))/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*sqrt(- 
b^2*e + a*b*f)) + (4*b*d^3*e^3 - 7*b*c*d^2*e^2*f - 5*a*d^3*e^2*f + 2*b*c^2 
*d*e*f^2 + 10*a*c*d^2*e*f^2 + b*c^3*f^3 - 5*a*c^2*d*f^3)*arctan(sqrt(f*x + 
 e)*d/sqrt(-d^2*e + c*d*f))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 
- a^3*d^4)*sqrt(-d^2*e + c*d*f)) - (2*(f*x + e)^(3/2)*b^2*d^2*e^2*f - 2*sq 
rt(f*x + e)*b^2*d^2*e^3*f - 2*(f*x + e)^(3/2)*b^2*c*d*e*f^2 - 2*(f*x + e)^ 
(3/2)*a*b*d^2*e*f^2 + 3*sqrt(f*x + e)*b^2*c*d*e^2*f^2 + 3*sqrt(f*x + e)*a* 
b*d^2*e^2*f^2 + (f*x + e)^(3/2)*b^2*c^2*f^3 + (f*x + e)^(3/2)*a^2*d^2*f^3 
- sqrt(f*x + e)*b^2*c^2*e*f^3 - 4*sqrt(f*x + e)*a*b*c*d*e*f^3 - sqrt(f*x + 
 e)*a^2*d^2*e*f^3 + sqrt(f*x + e)*a*b*c^2*f^4 + sqrt(f*x + e)*a^2*c*d*f^4) 
/((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*((f*x + e)^2*b*d - 2*(f*x + e)*b 
*d*e + b*d*e^2 + (f*x + e)*b*c*f + (f*x + e)*a*d*f - b*c*e*f - a*d*e*f + a 
*c*f^2))
 

Mupad [B] (verification not implemented)

Time = 8.04 (sec) , antiderivative size = 15161, normalized size of antiderivative = 58.09 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(((e + f*x)^(1/2)*(a^2*d^2*e*f^3 + b^2*c^2*e*f^3 + 2*b^2*d^2*e^3*f - a*b*c 
^2*f^4 - a^2*c*d*f^4 - 3*a*b*d^2*e^2*f^2 - 3*b^2*c*d*e^2*f^2 + 4*a*b*c*d*e 
*f^3))/(b*d*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (f*(e + f*x)^(3/2)*(a^2*d^2 
*f^2 + b^2*c^2*f^2 + 2*b^2*d^2*e^2 - 2*a*b*d^2*e*f - 2*b^2*c*d*e*f))/(b*d* 
(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/((e + f*x)*(a*d*f + b*c*f - 2*b*d*e) + b 
*d*(e + f*x)^2 + a*c*f^2 + b*d*e^2 - a*d*e*f - b*c*e*f) + (atan((((-b^3*(a 
*f - b*e)^3)^(1/2)*((2*(e + f*x)^(1/2)*(a^6*d^6*f^8 + b^6*c^6*f^8 + 32*b^6 
*d^6*e^6*f^2 + 25*a^2*b^4*c^4*d^2*f^8 + 25*a^4*b^2*c^2*d^4*f^8 + 90*a^2*b^ 
4*d^6*e^4*f^4 - 20*a^3*b^3*d^6*e^3*f^5 - 10*a^4*b^2*d^6*e^2*f^6 + 90*b^6*c 
^2*d^4*e^4*f^4 - 20*b^6*c^3*d^3*e^3*f^5 - 10*b^6*c^4*d^2*e^2*f^6 - 10*a*b^ 
5*c^5*d*f^8 - 10*a^5*b*c*d^5*f^8 + 4*a^5*b*d^6*e*f^7 + 4*b^6*c^5*d*e*f^7 - 
 96*a*b^5*d^6*e^5*f^3 - 96*b^6*c*d^5*e^5*f^3 + 300*a*b^5*c*d^5*e^4*f^4 - 3 
00*a*b^5*c^2*d^4*e^3*f^5 + 100*a*b^5*c^3*d^3*e^2*f^6 - 300*a^2*b^4*c*d^5*e 
^3*f^5 - 100*a^2*b^4*c^3*d^3*e*f^7 + 100*a^3*b^3*c*d^5*e^2*f^6 - 100*a^3*b 
^3*c^2*d^4*e*f^7 + 300*a^2*b^4*c^2*d^4*e^2*f^6))/(a^4*b*d^5 + b^5*c^4*d - 
4*a*b^4*c^3*d^2 - 4*a^3*b^2*c*d^4 + 6*a^2*b^3*c^2*d^3) - (((20*a^2*b^8*c^7 
*d^3*f^6 - 36*a^3*b^7*c^6*d^4*f^6 + 20*a^4*b^6*c^5*d^5*f^6 + 20*a^5*b^5*c^ 
4*d^6*f^6 - 36*a^6*b^4*c^3*d^7*f^6 + 20*a^7*b^3*c^2*d^8*f^6 + 8*a^6*b^4*d^ 
10*e^3*f^3 - 12*a^7*b^3*d^10*e^2*f^4 + 8*b^10*c^6*d^4*e^3*f^3 - 12*b^10*c^ 
7*d^3*e^2*f^4 - 4*a*b^9*c^8*d^2*f^6 - 4*a^8*b^2*c*d^9*f^6 + 4*a^8*b^2*d...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 2167, normalized size of antiderivative = 8.30 \[ \int \frac {(e+f x)^{5/2}}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e))) 
*a**3*c*d**3*f**2 + sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b 
)*sqrt(a*f - b*e)))*a**3*d**4*f**2*x - 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqr 
t(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*c**2*d**2*f**2 + 3*sqrt(b) 
*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b* 
c*d**3*e*f - 4*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqr 
t(a*f - b*e)))*a**2*b*c*d**3*f**2*x + 3*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt 
(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*d**4*e*f*x + sqrt(b)*sqrt(a 
*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a**2*b*d**4*f* 
*2*x**2 + 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a 
*f - b*e)))*a*b**2*c**2*d**2*e*f - 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e 
+ f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c**2*d**2*f**2*x - 4*sqrt(b)*s 
qrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c* 
d**3*e**2 + 8*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt 
(a*f - b*e)))*a*b**2*c*d**3*e*f*x - 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e 
 + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*c*d**3*f**2*x**2 - 4*sqrt(b)* 
sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b**2*d 
**4*e**2*x + 3*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqr 
t(a*f - b*e)))*a*b**2*d**4*e*f*x**2 + 5*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt 
(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*b**3*c**2*d**2*e*f*x - 4*sqrt(b...