\(\int \frac {(d+e x)^{3/2}}{(f+g x)^3 (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [48]

Optimal result

Integrand size = 46, antiderivative size = 273 \[ \int \frac {(d+e x)^{3/2}}{(f+g x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {15 c^2 d^2 \sqrt {d+e x}}{4 (c d f-a e g)^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\sqrt {d+e x}}{2 (c d f-a e g) (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 c d \sqrt {d+e x}}{4 (c d f-a e g)^2 (f+g x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {15 c^2 d^2 \sqrt {g} \arctan \left (\frac {\sqrt {c d f-a e g} \sqrt {d+e x}}{\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{4 (c d f-a e g)^{7/2}} \] Output:

-15/4*c^2*d^2*(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)+1/2*(e*x+d)^(1/2)/(-a*e*g+c*d*f)/(g*x+f)^2/(a*d*e+(a*e^2+c*d^2) 
*x+c*d*e*x^2)^(1/2)+5/4*c*d*(e*x+d)^(1/2)/(-a*e*g+c*d*f)^2/(g*x+f)/(a*d*e+ 
(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+15/4*c^2*d^2*g^(1/2)*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 = 0.86 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^{3/2}}{(f+g x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d+e x} \left (\sqrt {c d f-a e g} \left (-2 a^2 e^2 g^2+a c d e g (9 f+5 g x)+c^2 d^2 \left (8 f^2+25 f g x+15 g^2 x^2\right )\right )+15 c^2 d^2 \sqrt {g} \sqrt {a e+c d x} (f+g x)^2 \arctan \left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{4 (c d f-a e g)^{7/2} \sqrt {(a e+c d x) (d+e x)} (f+g x)^2} \] Input:

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

Output:

-1/4*(Sqrt[d + e*x]*(Sqrt[c*d*f - a*e*g]*(-2*a^2*e^2*g^2 + a*c*d*e*g*(9*f 
+ 5*g*x) + c^2*d^2*(8*f^2 + 25*f*g*x + 15*g^2*x^2)) + 15*c^2*d^2*Sqrt[g]*S 
qrt[a*e + c*d*x]*(f + g*x)^2*ArcTan[(Sqrt[g]*Sqrt[a*e + c*d*x])/Sqrt[c*d*f 
 - a*e*g]]))/((c*d*f - a*e*g)^(7/2)*Sqrt[(a*e + c*d*x)*(d + e*x)]*(f + g*x 
)^2)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {1252, 1254, 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.

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

\(\Big \downarrow \) 1252

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

\(\Big \downarrow \) 1254

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

\(\Big \downarrow \) 1254

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

\(\Big \downarrow \) 1255

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-2*Sqrt[d + e*x])/((c*d*f - a*e*g)*(f + g*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^ 
2)*x + c*d*e*x^2]) - (5*g*(Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(2* 
(c*d*f - a*e*g)*Sqrt[d + e*x]*(f + g*x)^2) + (3*c*d*(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*Ar 
cTan[(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))))/(4*(c*d*f - a*e*g 
))))/(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 = 2.91 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {\sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) \sqrt {c d x +a e}\, c^{2} d^{2} g^{3} x^{2}+30 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) \sqrt {c d x +a e}\, c^{2} d^{2} f \,g^{2} x -15 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} g^{2} x^{2}+15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) \sqrt {c d x +a e}\, c^{2} d^{2} f^{2} g -5 \sqrt {\left (a e g -d f c \right ) g}\, a c d e \,g^{2} x -25 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} f g x +2 \sqrt {\left (a e g -d f c \right ) g}\, a^{2} e^{2} g^{2}-9 \sqrt {\left (a e g -d f c \right ) g}\, a c d e f g -8 \sqrt {\left (a e g -d f c \right ) g}\, c^{2} d^{2} f^{2}\right )}{4 \sqrt {e x +d}\, \left (c d x +a e \right ) \left (a e g -d f c \right )^{3} \left (g x +f \right )^{2} \sqrt {\left (a e g -d f c \right ) g}}\)	\(369\)

Input:

int((e*x+d)^(3/2)/(g*x+f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2),x,meth 
od=_RETURNVERBOSE)
 

Output:

-1/4/(e*x+d)^(1/2)*((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+30*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*((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*g^2*x^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^2*g-5*((a*e*g-c 
*d*f)*g)^(1/2)*a*c*d*e*g^2*x-25*((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*f*g*x+2*(( 
a*e*g-c*d*f)*g)^(1/2)*a^2*e^2*g^2-9*((a*e*g-c*d*f)*g)^(1/2)*a*c*d*e*f*g-8* 
((a*e*g-c*d*f)*g)^(1/2)*c^2*d^2*f^2)/(c*d*x+a*e)/(a*e*g-c*d*f)^3/(g*x+f)^2 
/((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. 910 vs. \(2 (241) = 482\).

Time = 0.25 (sec) , antiderivative size = 1863, normalized size of antiderivative = 6.82 \[ \int \frac {(d+e x)^{3/2}}{(f+g x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/8*(15*(c^3*d^3*e*g^2*x^4 + a*c^2*d^3*e*f^2 + (2*c^3*d^3*e*f*g + (c^3*d 
^4 + a*c^2*d^2*e^2)*g^2)*x^3 + (c^3*d^3*e*f^2 + a*c^2*d^3*e*g^2 + 2*(c^3*d 
^4 + a*c^2*d^2*e^2)*f*g)*x^2 + (2*a*c^2*d^3*e*f*g + (c^3*d^4 + a*c^2*d^2*e 
^2)*f^2)*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 + 8*c^2*d^2*f^2 + 9*a* 
c*d*e*f*g - 2*a^2*e^2*g^2 + 5*(5*c^2*d^2*f*g + a*c*d*e*g^2)*x)*sqrt(c*d*e* 
x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a*c^3*d^4*e*f^5 - 3*a^2*c 
^2*d^3*e^2*f^4*g + 3*a^3*c*d^2*e^3*f^3*g^2 - a^4*d*e^4*f^2*g^3 + (c^4*d^4* 
e*f^3*g^2 - 3*a*c^3*d^3*e^2*f^2*g^3 + 3*a^2*c^2*d^2*e^3*f*g^4 - a^3*c*d*e^ 
4*g^5)*x^4 + (2*c^4*d^4*e*f^4*g + (c^4*d^5 - 5*a*c^3*d^3*e^2)*f^3*g^2 - 3* 
(a*c^3*d^4*e - a^2*c^2*d^2*e^3)*f^2*g^3 + (3*a^2*c^2*d^3*e^2 + a^3*c*d*e^4 
)*f*g^4 - (a^3*c*d^2*e^3 + a^4*e^5)*g^5)*x^3 + (c^4*d^4*e*f^5 - a^4*d*e^4* 
g^5 + (2*c^4*d^5 - a*c^3*d^3*e^2)*f^4*g - (5*a*c^3*d^4*e + 3*a^2*c^2*d^2*e 
^3)*f^3*g^2 + (3*a^2*c^2*d^3*e^2 + 5*a^3*c*d*e^4)*f^2*g^3 + (a^3*c*d^2*e^3 
 - 2*a^4*e^5)*f*g^4)*x^2 - (2*a^4*d*e^4*f*g^4 - (c^4*d^5 + a*c^3*d^3*e^2)* 
f^5 + (a*c^3*d^4*e + 3*a^2*c^2*d^2*e^3)*f^4*g + 3*(a^2*c^2*d^3*e^2 - a^3*c 
*d*e^4)*f^3*g^2 - (5*a^3*c*d^2*e^3 - a^4*e^5)*f^2*g^3)*x), -1/4*(15*(c^3*d 
^3*e*g^2*x^4 + a*c^2*d^3*e*f^2 + (2*c^3*d^3*e*f*g + (c^3*d^4 + a*c^2*d^...
 

Sympy [F(-1)]

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

integrate((e*x+d)**(3/2)/(g*x+f)**3/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)** 
(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^{3/2}}{(f+g x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {15 \, c^{2} d^{2} g \arctan \left (\frac {\sqrt {c d x + a e} g}{\sqrt {c d f g - a e g^{2}}}\right )}{4 \, {\left (c^{3} d^{3} f^{3} - 3 \, a c^{2} d^{2} e f^{2} g + 3 \, a^{2} c d e^{2} f g^{2} - a^{3} e^{3} g^{3}\right )} \sqrt {c d f g - a e g^{2}}} - \frac {2 \, c^{2} d^{2}}{{\left (c^{3} d^{3} f^{3} - 3 \, a c^{2} d^{2} e f^{2} g + 3 \, a^{2} c d e^{2} f g^{2} - a^{3} e^{3} g^{3}\right )} \sqrt {c d x + a e}} - \frac {9 \, \sqrt {c d x + a e} c^{3} d^{3} f g - 9 \, \sqrt {c d x + a e} a c^{2} d^{2} e g^{2} + 7 \, {\left (c d x + a e\right )}^{\frac {3}{2}} c^{2} d^{2} g^{2}}{4 \, {\left (c^{3} d^{3} f^{3} - 3 \, a c^{2} d^{2} e f^{2} g + 3 \, a^{2} c d e^{2} f g^{2} - a^{3} e^{3} g^{3}\right )} {\left (c d f - a e g + {\left (c d x + a e\right )} g\right )}^{2}} \] Input:

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

Output:

-15/4*c^2*d^2*g*arctan(sqrt(c*d*x + a*e)*g/sqrt(c*d*f*g - a*e*g^2))/((c^3* 
d^3*f^3 - 3*a*c^2*d^2*e*f^2*g + 3*a^2*c*d*e^2*f*g^2 - a^3*e^3*g^3)*sqrt(c* 
d*f*g - a*e*g^2)) - 2*c^2*d^2/((c^3*d^3*f^3 - 3*a*c^2*d^2*e*f^2*g + 3*a^2* 
c*d*e^2*f*g^2 - a^3*e^3*g^3)*sqrt(c*d*x + a*e)) - 1/4*(9*sqrt(c*d*x + a*e) 
*c^3*d^3*f*g - 9*sqrt(c*d*x + a*e)*a*c^2*d^2*e*g^2 + 7*(c*d*x + a*e)^(3/2) 
*c^2*d^2*g^2)/((c^3*d^3*f^3 - 3*a*c^2*d^2*e*f^2*g + 3*a^2*c*d*e^2*f*g^2 - 
a^3*e^3*g^3)*(c*d*f - a*e*g + (c*d*x + a*e)*g)^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.09 \[ \int \frac {(d+e x)^{3/2}}{(f+g x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {-15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{2} d^{2} f^{2}-30 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{2} d^{2} f g x -15 \sqrt {g}\, \sqrt {c d x +a e}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{2} d^{2} g^{2} x^{2}-2 a^{3} e^{3} g^{3}+11 a^{2} c d \,e^{2} f \,g^{2}+5 a^{2} c d \,e^{2} g^{3} x -a \,c^{2} d^{2} e \,f^{2} g +20 a \,c^{2} d^{2} e f \,g^{2} x +15 a \,c^{2} d^{2} e \,g^{3} x^{2}-8 c^{3} d^{3} f^{3}-25 c^{3} d^{3} f^{2} g x -15 c^{3} d^{3} f \,g^{2} x^{2}}{4 \sqrt {c d x +a e}\, \left (a^{4} e^{4} g^{6} x^{2}-4 a^{3} c d \,e^{3} f \,g^{5} x^{2}+6 a^{2} c^{2} d^{2} e^{2} f^{2} g^{4} x^{2}-4 a \,c^{3} d^{3} e \,f^{3} g^{3} x^{2}+c^{4} d^{4} f^{4} g^{2} x^{2}+2 a^{4} e^{4} f \,g^{5} x -8 a^{3} c d \,e^{3} f^{2} g^{4} x +12 a^{2} c^{2} d^{2} e^{2} f^{3} g^{3} x -8 a \,c^{3} d^{3} e \,f^{4} g^{2} x +2 c^{4} d^{4} f^{5} g x +a^{4} e^{4} f^{2} g^{4}-4 a^{3} c d \,e^{3} f^{3} g^{3}+6 a^{2} c^{2} d^{2} e^{2} f^{4} g^{2}-4 a \,c^{3} d^{3} e \,f^{5} g +c^{4} d^{4} f^{6}\right )} \] Input:

int((e*x+d)^(3/2)/(g*x+f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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)))*c**2*d**2*f**2 - 30*sqrt(g)*sqr 
t(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)*sq 
rt( - 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*g**2*x**2 - 2*a**3*e**3*g**3 + 11*a**2*c*d*e**2*f*g**2 + 
 5*a**2*c*d*e**2*g**3*x - a*c**2*d**2*e*f**2*g + 20*a*c**2*d**2*e*f*g**2*x 
 + 15*a*c**2*d**2*e*g**3*x**2 - 8*c**3*d**3*f**3 - 25*c**3*d**3*f**2*g*x - 
 15*c**3*d**3*f*g**2*x**2)/(4*sqrt(a*e + c*d*x)*(a**4*e**4*f**2*g**4 + 2*a 
**4*e**4*f*g**5*x + a**4*e**4*g**6*x**2 - 4*a**3*c*d*e**3*f**3*g**3 - 8*a* 
*3*c*d*e**3*f**2*g**4*x - 4*a**3*c*d*e**3*f*g**5*x**2 + 6*a**2*c**2*d**2*e 
**2*f**4*g**2 + 12*a**2*c**2*d**2*e**2*f**3*g**3*x + 6*a**2*c**2*d**2*e**2 
*f**2*g**4*x**2 - 4*a*c**3*d**3*e*f**5*g - 8*a*c**3*d**3*e*f**4*g**2*x - 4 
*a*c**3*d**3*e*f**3*g**3*x**2 + c**4*d**4*f**6 + 2*c**4*d**4*f**5*g*x + c* 
*4*d**4*f**4*g**2*x**2))