\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx\) [30]

Optimal result

Integrand size = 46, antiderivative size = 253 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=-\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt {d+e x} (f+g x)}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac {5 c^3 d^3 \arctan \left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d f-a e g} \sqrt {d+e x}}\right )}{8 g^{7/2} \sqrt {c d f-a e g}} \] Output:

-5/8*c^2*d^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/g^3/(e*x+d)^(1/2)/(g* 
x+f)-5/12*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/g^2/(e*x+d)^(3/2)/(g 
*x+f)^2-1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/g/(e*x+d)^(5/2)/(g*x+f 
)^3+5/8*c^3*d^3*arctan(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))/g^(7/2)/(-a*e*g+c*d*f)^(1/2)
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\frac {\sqrt {(a e+c d x) (d+e x)} \left (-\frac {\sqrt {g} \left (8 a^2 e^2 g^2+2 a c d e g (5 f+13 g x)+c^2 d^2 \left (15 f^2+40 f g x+33 g^2 x^2\right )\right )}{(f+g x)^3}+\frac {15 c^3 d^3 \arctan \left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )}{\sqrt {c d f-a e g} \sqrt {a e+c d x}}\right )}{24 g^{7/2} \sqrt {d+e x}} \] Input:

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*( 
f + g*x)^4),x]
 

Output:

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(-((Sqrt[g]*(8*a^2*e^2*g^2 + 2*a*c*d*e*g*(5 
*f + 13*g*x) + c^2*d^2*(15*f^2 + 40*f*g*x + 33*g^2*x^2)))/(f + g*x)^3) + ( 
15*c^3*d^3*ArcTan[(Sqrt[g]*Sqrt[a*e + c*d*x])/Sqrt[c*d*f - a*e*g]])/(Sqrt[ 
c*d*f - a*e*g]*Sqrt[a*e + c*d*x])))/(24*g^(7/2)*Sqrt[d + e*x])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {1249, 1249, 1249, 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.

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

\(\Big \downarrow \) 1249

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

\(\Big \downarrow \) 1249

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

\(\Big \downarrow \) 1249

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

\(\Big \downarrow \) 1255

\(\displaystyle \frac {5 c d \left (\frac {3 c d \left (\frac {c d e^2 \int \frac {1}{(c d f-a e g) e^2+\frac {g \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right ) e^2}{d+e x}}d\frac {\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {d+e x}}}{g}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x} (f+g x)}\right )}{4 g}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2} (f+g x)^2}\right )}{6 g}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}\)

\(\Big \downarrow \) 218

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

Input:

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g* 
x)^4),x]
 

Output:

-1/3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(g*(d + e*x)^(5/2)*(f + 
 g*x)^3) + (5*c*d*(-1/2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(g*( 
d + e*x)^(3/2)*(f + g*x)^2) + (3*c*d*(-(Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c 
*d*e*x^2]/(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])])/(g^ 
(3/2)*Sqrt[c*d*f - a*e*g])))/(4*g)))/(6*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 1249

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

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 = 2.89 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.70

method	result	size
default	\(-\frac {\sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) c^{3} d^{3} g^{3} x^{3}+45 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) c^{3} d^{3} f \,g^{2} x^{2}+45 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) c^{3} d^{3} f^{2} g x +15 \,\operatorname {arctanh}\left (\frac {g \sqrt {c d x +a e}}{\sqrt {\left (a e g -d f c \right ) g}}\right ) c^{3} d^{3} f^{3}+33 c^{2} d^{2} g^{2} x^{2} \sqrt {c d x +a e}\, \sqrt {\left (a e g -d f c \right ) g}+26 a c d e \,g^{2} x \sqrt {c d x +a e}\, \sqrt {\left (a e g -d f c \right ) g}+40 c^{2} d^{2} f g x \sqrt {c d x +a e}\, \sqrt {\left (a e g -d f c \right ) g}+8 \sqrt {\left (a e g -d f c \right ) g}\, \sqrt {c d x +a e}\, a^{2} e^{2} g^{2}+10 \sqrt {\left (a e g -d f c \right ) g}\, \sqrt {c d x +a e}\, a c d e f g +15 \sqrt {\left (a e g -d f c \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} f^{2}\right )}{24 \sqrt {e x +d}\, \sqrt {c d x +a e}\, g^{3} \left (g x +f \right )^{3} \sqrt {\left (a e g -d f c \right ) g}}\)	\(431\)

Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^4,x,meth 
od=_RETURNVERBOSE)
 

Output:

-1/24*((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^3*d^3*g^3*x^3+45*arctanh(g*(c*d*x+a*e)^(1/2)/((a*e*g-c* 
d*f)*g)^(1/2))*c^3*d^3*f*g^2*x^2+45*arctanh(g*(c*d*x+a*e)^(1/2)/((a*e*g-c* 
d*f)*g)^(1/2))*c^3*d^3*f^2*g*x+15*arctanh(g*(c*d*x+a*e)^(1/2)/((a*e*g-c*d* 
f)*g)^(1/2))*c^3*d^3*f^3+33*c^2*d^2*g^2*x^2*(c*d*x+a*e)^(1/2)*((a*e*g-c*d* 
f)*g)^(1/2)+26*a*c*d*e*g^2*x*(c*d*x+a*e)^(1/2)*((a*e*g-c*d*f)*g)^(1/2)+40* 
c^2*d^2*f*g*x*(c*d*x+a*e)^(1/2)*((a*e*g-c*d*f)*g)^(1/2)+8*((a*e*g-c*d*f)*g 
)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*e^2*g^2+10*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a* 
e)^(1/2)*a*c*d*e*f*g+15*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2* 
f^2)/(e*x+d)^(1/2)/(c*d*x+a*e)^(1/2)/g^3/(g*x+f)^3/((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. 550 vs. \(2 (221) = 442\).

Time = 0.13 (sec) , antiderivative size = 1141, normalized size of antiderivative = 4.51 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx =\text {Too large to display} \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^4, 
x, algorithm="fricas")
 

Output:

[-1/48*(15*(c^3*d^3*e*g^3*x^4 + c^3*d^4*f^3 + (3*c^3*d^3*e*f*g^2 + c^3*d^4 
*g^3)*x^3 + 3*(c^3*d^3*e*f^2*g + c^3*d^4*f*g^2)*x^2 + (c^3*d^3*e*f^3 + 3*c 
^3*d^4*f^2*g)*x)*sqrt(-c*d*f*g + a*e*g^2)*log(-(c*d*e*g*x^2 - c*d^2*f + 2* 
a*d*e*g - (c*d*e*f - (c*d^2 + 2*a*e^2)*g)*x - 2*sqrt(c*d*e*x^2 + a*d*e + ( 
c*d^2 + a*e^2)*x)*sqrt(-c*d*f*g + a*e*g^2)*sqrt(e*x + d))/(e*g*x^2 + d*f + 
 (e*f + d*g)*x)) + 2*(15*c^3*d^3*f^3*g - 5*a*c^2*d^2*e*f^2*g^2 - 2*a^2*c*d 
*e^2*f*g^3 - 8*a^3*e^3*g^4 + 33*(c^3*d^3*f*g^3 - a*c^2*d^2*e*g^4)*x^2 + 2* 
(20*c^3*d^3*f^2*g^2 - 7*a*c^2*d^2*e*f*g^3 - 13*a^2*c*d*e^2*g^4)*x)*sqrt(c* 
d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c*d^2*f^4*g^4 - a*d*e 
*f^3*g^5 + (c*d*e*f*g^7 - a*e^2*g^8)*x^4 + (3*c*d*e*f^2*g^6 - a*d*e*g^8 + 
(c*d^2 - 3*a*e^2)*f*g^7)*x^3 + 3*(c*d*e*f^3*g^5 - a*d*e*f*g^7 + (c*d^2 - a 
*e^2)*f^2*g^6)*x^2 + (c*d*e*f^4*g^4 - 3*a*d*e*f^2*g^6 + (3*c*d^2 - a*e^2)* 
f^3*g^5)*x), -1/24*(15*(c^3*d^3*e*g^3*x^4 + c^3*d^4*f^3 + (3*c^3*d^3*e*f*g 
^2 + c^3*d^4*g^3)*x^3 + 3*(c^3*d^3*e*f^2*g + c^3*d^4*f*g^2)*x^2 + (c^3*d^3 
*e*f^3 + 3*c^3*d^4*f^2*g)*x)*sqrt(c*d*f*g - a*e*g^2)*arctan(-sqrt(c*d*e*x^ 
2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d*f*g - a*e*g^2)*sqrt(e*x + d)/(c*d^ 
2*f - a*d*e*g + (c*d*e*f - a*e^2*g)*x)) + (15*c^3*d^3*f^3*g - 5*a*c^2*d^2* 
e*f^2*g^2 - 2*a^2*c*d*e^2*f*g^3 - 8*a^3*e^3*g^4 + 33*(c^3*d^3*f*g^3 - a*c^ 
2*d^2*e*g^4)*x^2 + 2*(20*c^3*d^3*f^2*g^2 - 7*a*c^2*d^2*e*f*g^3 - 13*a^2*c* 
d*e^2*g^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\text {Timed out} \] Input:

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(5/2)/(g*x+ 
f)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{4}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^4, 
x, algorithm="maxima")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*( 
g*x + f)^4), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\frac {5 \, c^{3} d^{3} {\left | e \right |} \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} g}{\sqrt {c d f g - a e g^{2}} e}\right )}{8 \, \sqrt {c d f g - a e g^{2}} e g^{3}} - \frac {15 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{5} d^{5} e^{4} f^{2} {\left | e \right |} - 30 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{4} d^{4} e^{5} f g {\left | e \right |} + 15 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a^{2} c^{3} d^{3} e^{6} g^{2} {\left | e \right |} + 40 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{4} d^{4} e^{2} f g {\left | e \right |} - 40 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c^{3} d^{3} e^{3} g^{2} {\left | e \right |} + 33 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} c^{3} d^{3} g^{2} {\left | e \right |}}{24 \, {\left (c d e^{2} f - a e^{3} g + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} g\right )}^{3} g^{3}} \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^4, 
x, algorithm="giac")
 

Output:

5/8*c^3*d^3*abs(e)*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))/(sqrt(c*d*f*g - a*e*g^2)*e*g^3) - 1/24*(15*sqrt((e* 
x + d)*c*d*e - c*d^2*e + a*e^3)*c^5*d^5*e^4*f^2*abs(e) - 30*sqrt((e*x + d) 
*c*d*e - c*d^2*e + a*e^3)*a*c^4*d^4*e^5*f*g*abs(e) + 15*sqrt((e*x + d)*c*d 
*e - c*d^2*e + a*e^3)*a^2*c^3*d^3*e^6*g^2*abs(e) + 40*((e*x + d)*c*d*e - c 
*d^2*e + a*e^3)^(3/2)*c^4*d^4*e^2*f*g*abs(e) - 40*((e*x + d)*c*d*e - c*d^2 
*e + a*e^3)^(3/2)*a*c^3*d^3*e^3*g^2*abs(e) + 33*((e*x + d)*c*d*e - c*d^2*e 
 + a*e^3)^(5/2)*c^3*d^3*g^2*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)^3*g^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (f+g\,x\right )}^4\,{\left (d+e\,x\right )}^{5/2}} \,d x \] Input:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^4*(d + e*x)^( 
5/2)),x)
 

Output:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^4*(d + e*x)^( 
5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.04 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx=\frac {-15 \sqrt {g}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{3} d^{3} f^{3}-45 \sqrt {g}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{3} d^{3} f^{2} g x -45 \sqrt {g}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{3} d^{3} f \,g^{2} x^{2}-15 \sqrt {g}\, \sqrt {-a e g +c d f}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {g}\, \sqrt {-a e g +c d f}}\right ) c^{3} d^{3} g^{3} x^{3}-8 \sqrt {c d x +a e}\, a^{3} e^{3} g^{4}-2 \sqrt {c d x +a e}\, a^{2} c d \,e^{2} f \,g^{3}-26 \sqrt {c d x +a e}\, a^{2} c d \,e^{2} g^{4} x -5 \sqrt {c d x +a e}\, a \,c^{2} d^{2} e \,f^{2} g^{2}-14 \sqrt {c d x +a e}\, a \,c^{2} d^{2} e f \,g^{3} x -33 \sqrt {c d x +a e}\, a \,c^{2} d^{2} e \,g^{4} x^{2}+15 \sqrt {c d x +a e}\, c^{3} d^{3} f^{3} g +40 \sqrt {c d x +a e}\, c^{3} d^{3} f^{2} g^{2} x +33 \sqrt {c d x +a e}\, c^{3} d^{3} f \,g^{3} x^{2}}{24 g^{4} \left (a e \,g^{4} x^{3}-c d f \,g^{3} x^{3}+3 a e f \,g^{3} x^{2}-3 c d \,f^{2} g^{2} x^{2}+3 a e \,f^{2} g^{2} x -3 c d \,f^{3} g x +a e \,f^{3} g -c d \,f^{4}\right )} \] Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^4,x)
 

Output:

( - 15*sqrt(g)*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**3*d**3*f**3 - 45*sqrt(g)*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**3*d**3*f 
**2*g*x - 45*sqrt(g)*sqrt( - a*e*g + c*d*f)*atan((sqrt(a*e + c*d*x)*g)/(sq 
rt(g)*sqrt( - a*e*g + c*d*f)))*c**3*d**3*f*g**2*x**2 - 15*sqrt(g)*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**3*d**3*g**3*x**3 - 8*sqrt(a*e + c*d*x)*a**3*e**3*g**4 - 2*sqrt(a*e + 
c*d*x)*a**2*c*d*e**2*f*g**3 - 26*sqrt(a*e + c*d*x)*a**2*c*d*e**2*g**4*x - 
5*sqrt(a*e + c*d*x)*a*c**2*d**2*e*f**2*g**2 - 14*sqrt(a*e + c*d*x)*a*c**2* 
d**2*e*f*g**3*x - 33*sqrt(a*e + c*d*x)*a*c**2*d**2*e*g**4*x**2 + 15*sqrt(a 
*e + c*d*x)*c**3*d**3*f**3*g + 40*sqrt(a*e + c*d*x)*c**3*d**3*f**2*g**2*x 
+ 33*sqrt(a*e + c*d*x)*c**3*d**3*f*g**3*x**2)/(24*g**4*(a*e*f**3*g + 3*a*e 
*f**2*g**2*x + 3*a*e*f*g**3*x**2 + a*e*g**4*x**3 - c*d*f**4 - 3*c*d*f**3*g 
*x - 3*c*d*f**2*g**2*x**2 - c*d*f*g**3*x**3))