\(\int \frac {(d+e x)^{5/2}}{(f+g x)^2 (a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [54]

Optimal result

Integrand size = 46, antiderivative size = 252 \[ \int \frac {(d+e x)^{5/2}}{(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {5 c d (d+e x)^{3/2}}{3 (c d f-a e g)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {(d+e x)^{3/2}}{(c d f-a e g) (f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {5 c d g \sqrt {d+e x}}{(c d f-a e g)^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {5 c d g^{3/2} \arctan \left (\frac {\sqrt {c d f-a e g} \sqrt {d+e x}}{\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{(c d f-a e g)^{7/2}} \] Output:

-5/3*c*d*(e*x+d)^(3/2)/(-a*e*g+c*d*f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^ 
(3/2)+(e*x+d)^(3/2)/(-a*e*g+c*d*f)/(g*x+f)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^ 
2)^(3/2)+5*c*d*g*(e*x+d)^(1/2)/(-a*e*g+c*d*f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d 
*e*x^2)^(1/2)-5*c*d*g^(3/2)*arctan(1/g^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e* 
x^2)^(1/2)*(-a*e*g+c*d*f)^(1/2)*(e*x+d)^(1/2))/(-a*e*g+c*d*f)^(7/2)
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.71 \[ \int \frac {(d+e x)^{5/2}}{(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {(d+e x)^{3/2} \left (\sqrt {c d f-a e g} \left (3 a^2 e^2 g^2+2 a c d e g (7 f+10 g x)+c^2 d^2 \left (-2 f^2+10 f g x+15 g^2 x^2\right )\right )+15 c d g^{3/2} (a e+c d x)^{3/2} (f+g x) \arctan \left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{3 (c d f-a e g)^{7/2} ((a e+c d x) (d+e x))^{3/2} (f+g x)} \] Input:

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

Output:

((d + e*x)^(3/2)*(Sqrt[c*d*f - a*e*g]*(3*a^2*e^2*g^2 + 2*a*c*d*e*g*(7*f + 
10*g*x) + c^2*d^2*(-2*f^2 + 10*f*g*x + 15*g^2*x^2)) + 15*c*d*g^(3/2)*(a*e 
+ c*d*x)^(3/2)*(f + g*x)*ArcTan[(Sqrt[g]*Sqrt[a*e + c*d*x])/Sqrt[c*d*f - a 
*e*g]]))/(3*(c*d*f - a*e*g)^(7/2)*((a*e + c*d*x)*(d + e*x))^(3/2)*(f + g*x 
))
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {1252, 1252, 1254, 1255, 218}

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.

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

\(\Big \downarrow \) 1252

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

\(\Big \downarrow \) 1252

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

\(\Big \downarrow \) 1254

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

\(\Big \downarrow \) 1255

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

\(\Big \downarrow \) 218

\(\displaystyle -\frac {5 g \left (-\frac {3 g \left (\frac {c d \arctan \left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{\sqrt {g} (c d f-a e g)^{3/2}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} (f+g x) (c d f-a e g)}\right )}{c d f-a e g}-\frac {2 \sqrt {d+e x}}{(f+g x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}\right )}{3 (c d f-a e g)}-\frac {2 (d+e x)^{3/2}}{3 (f+g x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1252

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

rule 1254

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

rule 1255

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

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.64

method	result	size
default	\(\frac {\sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) \sqrt {c d x +a e}\, c^{2} d^{2} g^{3} x^{2}+15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) a c d e \,g^{3} x \sqrt {c d x +a e}+15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) \sqrt {c d x +a e}\, c^{2} d^{2} f \,g^{2} x +15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) a c d e f \,g^{2} \sqrt {c d x +a e}-15 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} g^{2} x^{2}-20 \sqrt {\left (a e g -d f c \right ) g}\, a c d e \,g^{2} x -10 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} f g x -3 \sqrt {\left (a e g -d f c \right ) g}\, a^{2} e^{2} g^{2}-14 \sqrt {\left (a e g -d f c \right ) g}\, a c d e f g +2 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} f^{2}\right )}{3 \sqrt {e x +d}\, \left (c d x +a e \right )^{2} \left (a e g -d f c \right )^{3} \left (g x +f \right ) \sqrt {\left (a e g -d f c \right ) g}}\)	\(414\)

Input:

int((e*x+d)^(5/2)/(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2),x,meth 
od=_RETURNVERBOSE)
 

Output:

1/3*((e*x+d)*(c*d*x+a*e))^(1/2)*(15*arctanh(g*(c*d*x+a*e)^(1/2)/((a*e*g-c* 
d*f)*g)^(1/2))*(c*d*x+a*e)^(1/2)*c^2*d^2*g^3*x^2+15*arctanh(g*(c*d*x+a*e)^ 
(1/2)/((a*e*g-c*d*f)*g)^(1/2))*a*c*d*e*g^3*x*(c*d*x+a*e)^(1/2)+15*arctanh( 
g*(c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2))*(c*d*x+a*e)^(1/2)*c^2*d^2*f*g 
^2*x+15*arctanh(g*(c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2))*a*c*d*e*f*g^2 
*(c*d*x+a*e)^(1/2)-15*((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*g^2*x^2-20*((a*e*g-c 
*d*f)*g)^(1/2)*a*c*d*e*g^2*x-10*((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*f*g*x-3*(( 
a*e*g-c*d*f)*g)^(1/2)*a^2*e^2*g^2-14*((a*e*g-c*d*f)*g)^(1/2)*a*c*d*e*f*g+2 
*((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*f^2)/(e*x+d)^(1/2)/(c*d*x+a*e)^2/(a*e*g-c 
*d*f)^3/(g*x+f)/((a*e*g-c*d*f)*g)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 932 vs. \(2 (226) = 452\).

Time = 0.42 (sec) , antiderivative size = 1907, normalized size of antiderivative = 7.57 \[ \int \frac {(d+e x)^{5/2}}{(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/6*(15*(c^3*d^3*e*g^2*x^4 + a^2*c*d^2*e^2*f*g + (c^3*d^3*e*f*g + (c^3*d 
^4 + 2*a*c^2*d^2*e^2)*g^2)*x^3 + ((c^3*d^4 + 2*a*c^2*d^2*e^2)*f*g + (2*a*c 
^2*d^3*e + a^2*c*d*e^3)*g^2)*x^2 + (a^2*c*d^2*e^2*g^2 + (2*a*c^2*d^3*e + a 
^2*c*d*e^3)*f*g)*x)*sqrt(-g/(c*d*f - a*e*g))*log(-(c*d*e*g*x^2 - c*d^2*f + 
 2*a*d*e*g - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d*f - a*e*g) 
*sqrt(e*x + d)*sqrt(-g/(c*d*f - a*e*g)) - (c*d*e*f - (c*d^2 + 2*a*e^2)*g)* 
x)/(e*g*x^2 + d*f + (e*f + d*g)*x)) - 2*(15*c^2*d^2*g^2*x^2 - 2*c^2*d^2*f^ 
2 + 14*a*c*d*e*f*g + 3*a^2*e^2*g^2 + 10*(c^2*d^2*f*g + 2*a*c*d*e*g^2)*x)*s 
qrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a^2*c^3*d^4*e^2 
*f^4 - 3*a^3*c^2*d^3*e^3*f^3*g + 3*a^4*c*d^2*e^4*f^2*g^2 - a^5*d*e^5*f*g^3 
 + (c^5*d^5*e*f^3*g - 3*a*c^4*d^4*e^2*f^2*g^2 + 3*a^2*c^3*d^3*e^3*f*g^3 - 
a^3*c^2*d^2*e^4*g^4)*x^4 + (c^5*d^5*e*f^4 + (c^5*d^6 - a*c^4*d^4*e^2)*f^3* 
g - 3*(a*c^4*d^5*e + a^2*c^3*d^3*e^3)*f^2*g^2 + (3*a^2*c^3*d^4*e^2 + 5*a^3 
*c^2*d^2*e^4)*f*g^3 - (a^3*c^2*d^3*e^3 + 2*a^4*c*d*e^5)*g^4)*x^3 + ((c^5*d 
^6 + 2*a*c^4*d^4*e^2)*f^4 - (a*c^4*d^5*e + 5*a^2*c^3*d^3*e^3)*f^3*g - 3*(a 
^2*c^3*d^4*e^2 - a^3*c^2*d^2*e^4)*f^2*g^2 + (5*a^3*c^2*d^3*e^3 + a^4*c*d*e 
^5)*f*g^3 - (2*a^4*c*d^2*e^4 + a^5*e^6)*g^4)*x^2 - (a^5*d*e^5*g^4 - (2*a*c 
^4*d^5*e + a^2*c^3*d^3*e^3)*f^4 + (5*a^2*c^3*d^4*e^2 + 3*a^3*c^2*d^2*e^4)* 
f^3*g - 3*(a^3*c^2*d^3*e^3 + a^4*c*d*e^5)*f^2*g^2 - (a^4*c*d^2*e^4 - a^5*e 
^6)*f*g^3)*x), 1/3*(15*(c^3*d^3*e*g^2*x^4 + a^2*c*d^2*e^2*f*g + (c^3*d^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.65 \[ \int \frac {(d+e x)^{5/2}}{(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {1}{3} \, {\left (\frac {3 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c d g^{2}}{{\left (c^{3} d^{3} e^{2} f^{3} {\left | e \right |} - 3 \, a c^{2} d^{2} e^{3} f^{2} g {\left | e \right |} + 3 \, a^{2} c d e^{4} f g^{2} {\left | e \right |} - a^{3} e^{5} g^{3} {\left | e \right |}\right )} {\left (c d e^{2} f - a e^{3} g + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} g\right )}} + \frac {15 \, c d g^{2} \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} g}{\sqrt {c d f g - a e g^{2}} e}\right )}{{\left (c^{3} d^{3} e^{2} f^{3} {\left | e \right |} - 3 \, a c^{2} d^{2} e^{3} f^{2} g {\left | e \right |} + 3 \, a^{2} c d e^{4} f g^{2} {\left | e \right |} - a^{3} e^{5} g^{3} {\left | e \right |}\right )} \sqrt {c d f g - a e g^{2}} e} - \frac {2 \, {\left (c^{2} d^{2} e^{2} f - a c d e^{3} g - 6 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c d g\right )}}{{\left (c^{3} d^{3} e^{2} f^{3} {\left | e \right |} - 3 \, a c^{2} d^{2} e^{3} f^{2} g {\left | e \right |} + 3 \, a^{2} c d e^{4} f g^{2} {\left | e \right |} - a^{3} e^{5} g^{3} {\left | e \right |}\right )} {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}}\right )} e^{4} \] Input:

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

Output:

1/3*(3*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c*d*g^2/((c^3*d^3*e^2*f^3*a 
bs(e) - 3*a*c^2*d^2*e^3*f^2*g*abs(e) + 3*a^2*c*d*e^4*f*g^2*abs(e) - a^3*e^ 
5*g^3*abs(e))*(c*d*e^2*f - a*e^3*g + ((e*x + d)*c*d*e - c*d^2*e + a*e^3)*g 
)) + 15*c*d*g^2*arctan(sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*g/(sqrt(c*d 
*f*g - a*e*g^2)*e))/((c^3*d^3*e^2*f^3*abs(e) - 3*a*c^2*d^2*e^3*f^2*g*abs(e 
) + 3*a^2*c*d*e^4*f*g^2*abs(e) - a^3*e^5*g^3*abs(e))*sqrt(c*d*f*g - a*e*g^ 
2)*e) - 2*(c^2*d^2*e^2*f - a*c*d*e^3*g - 6*((e*x + d)*c*d*e - c*d^2*e + a* 
e^3)*c*d*g)/((c^3*d^3*e^2*f^3*abs(e) - 3*a*c^2*d^2*e^3*f^2*g*abs(e) + 3*a^ 
2*c*d*e^4*f*g^2*abs(e) - a^3*e^5*g^3*abs(e))*((e*x + d)*c*d*e - c*d^2*e + 
a*e^3)^(3/2)))*e^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 650, normalized size of antiderivative = 2.58 \[ \int \frac {(d+e x)^{5/2}}{(f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) a c d e f g +15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) a c d e \,g^{2} x +15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{2} d^{2} f g x +15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{2} d^{2} g^{2} x^{2}-3 a^{3} e^{3} g^{3}-11 a^{2} c d \,e^{2} f \,g^{2}-20 a^{2} c d \,e^{2} g^{3} x +16 a \,c^{2} d^{2} e \,f^{2} g +10 a \,c^{2} d^{2} e f \,g^{2} x -15 a \,c^{2} d^{2} e \,g^{3} x^{2}-2 c^{3} d^{3} f^{3}+10 c^{3} d^{3} f^{2} g x +15 c^{3} d^{3} f \,g^{2} x^{2}}{3 \sqrt {c d x +a e}\, \left (a^{4} c d \,e^{4} g^{5} x^{2}-4 a^{3} c^{2} d^{2} e^{3} f \,g^{4} x^{2}+6 a^{2} c^{3} d^{3} e^{2} f^{2} g^{3} x^{2}-4 a \,c^{4} d^{4} e \,f^{3} g^{2} x^{2}+c^{5} d^{5} f^{4} g \,x^{2}+a^{5} e^{5} g^{5} x -3 a^{4} c d \,e^{4} f \,g^{4} x +2 a^{3} c^{2} d^{2} e^{3} f^{2} g^{3} x +2 a^{2} c^{3} d^{3} e^{2} f^{3} g^{2} x -3 a \,c^{4} d^{4} e \,f^{4} g x +c^{5} d^{5} f^{5} x +a^{5} e^{5} f \,g^{4}-4 a^{4} c d \,e^{4} f^{2} g^{3}+6 a^{3} c^{2} d^{2} e^{3} f^{3} g^{2}-4 a^{2} c^{3} d^{3} e^{2} f^{4} g +a \,c^{4} d^{4} e \,f^{5}\right )} \] Input:

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

Output:

(15*sqrt(g)*sqrt(a*e + c*d*x)*sqrt( - a*e*g + c*d*f)*atan((sqrt(a*e + c*d* 
x)*g)/(sqrt(g)*sqrt( - a*e*g + c*d*f)))*a*c*d*e*f*g + 15*sqrt(g)*sqrt(a*e 
+ c*d*x)*sqrt( - a*e*g + c*d*f)*atan((sqrt(a*e + c*d*x)*g)/(sqrt(g)*sqrt( 
- a*e*g + c*d*f)))*a*c*d*e*g**2*x + 15*sqrt(g)*sqrt(a*e + c*d*x)*sqrt( - a 
*e*g + c*d*f)*atan((sqrt(a*e + c*d*x)*g)/(sqrt(g)*sqrt( - a*e*g + c*d*f))) 
*c**2*d**2*f*g*x + 15*sqrt(g)*sqrt(a*e + c*d*x)*sqrt( - a*e*g + c*d*f)*ata 
n((sqrt(a*e + c*d*x)*g)/(sqrt(g)*sqrt( - a*e*g + c*d*f)))*c**2*d**2*g**2*x 
**2 - 3*a**3*e**3*g**3 - 11*a**2*c*d*e**2*f*g**2 - 20*a**2*c*d*e**2*g**3*x 
 + 16*a*c**2*d**2*e*f**2*g + 10*a*c**2*d**2*e*f*g**2*x - 15*a*c**2*d**2*e* 
g**3*x**2 - 2*c**3*d**3*f**3 + 10*c**3*d**3*f**2*g*x + 15*c**3*d**3*f*g**2 
*x**2)/(3*sqrt(a*e + c*d*x)*(a**5*e**5*f*g**4 + a**5*e**5*g**5*x - 4*a**4* 
c*d*e**4*f**2*g**3 - 3*a**4*c*d*e**4*f*g**4*x + a**4*c*d*e**4*g**5*x**2 + 
6*a**3*c**2*d**2*e**3*f**3*g**2 + 2*a**3*c**2*d**2*e**3*f**2*g**3*x - 4*a* 
*3*c**2*d**2*e**3*f*g**4*x**2 - 4*a**2*c**3*d**3*e**2*f**4*g + 2*a**2*c**3 
*d**3*e**2*f**3*g**2*x + 6*a**2*c**3*d**3*e**2*f**2*g**3*x**2 + a*c**4*d** 
4*e*f**5 - 3*a*c**4*d**4*e*f**4*g*x - 4*a*c**4*d**4*e*f**3*g**2*x**2 + c** 
5*d**5*f**5*x + c**5*d**5*f**4*g*x**2))