\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^4} \, dx\) [1939]

Optimal result

Integrand size = 37, antiderivative size = 235 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {15 c d \left (a-\frac {c d^2}{e^2}\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e}+\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {15 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^2 \text {arctanh}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 e^{7/2}} \]

[Out]

5/2*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/e^2/(e*x+d)-2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/e/(e*x+d
)^3+15/8*(-a*e^2+c*d^2)^2*arctanh(1/2*(2*c*d*e*x+a*e^2+c*d^2)/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c
*d*e*x^2)^(1/2))*c^(1/2)*d^(1/2)/e^(7/2)+15/4*c*d*(a-c*d^2/e^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/e

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {676, 678, 635, 212} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {15 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^2 \text {arctanh}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 e^{7/2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}+\frac {15 c d \left (a-\frac {c d^2}{e^2}\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e} \]

[In]

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

[Out]

(15*c*d*(a - (c*d^2)/e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e) + (5*c*d*(a*d*e + (c*d^2 + a*e^2)
*x + c*d*e*x^2)^(3/2))/(2*e^2*(d + e*x)) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(d + e*x)^3) +
(15*Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^2*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d
*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(8*e^(7/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 676

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

Rule 678

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {(5 c d) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^2} \, dx}{e} \\ & = \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}}{2 e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}-\frac {\left (15 c d \left (c d^2-a e^2\right )\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{d+e x} \, dx}{4 e^2} \\ & = -\frac {15 c d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\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}}{2 e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {\left (15 c d \left (c d^2-a e^2\right )^2\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e^3} \\ & = -\frac {15 c d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\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}}{2 e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {\left (15 c d \left (c d^2-a e^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{4 e^3} \\ & = -\frac {15 c d \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\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}}{2 e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac {15 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 e^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {((a e+c d x) (d+e x))^{5/2} \left (\frac {\sqrt {e} \left (-8 a^2 e^4+a c d e^2 (25 d+9 e x)+c^2 d^2 \left (-15 d^2-5 d e x+2 e^2 x^2\right )\right )}{(a e+c d x)^2 (d+e x)^3}+\frac {15 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{(a e+c d x)^{5/2} (d+e x)^{5/2}}\right )}{4 e^{7/2}} \]

[In]

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

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*((Sqrt[e]*(-8*a^2*e^4 + a*c*d*e^2*(25*d + 9*e*x) + c^2*d^2*(-15*d^2 - 5*d*e*x
+ 2*e^2*x^2)))/((a*e + c*d*x)^2*(d + e*x)^3) + (15*Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^2*ArcTanh[(Sqrt[c]*Sqrt[d]
*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c*d*x])])/((a*e + c*d*x)^(5/2)*(d + e*x)^(5/2))))/(4*e^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(569\) vs. \(2(207)=414\).

Time = 3.55 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.43

method	result	size
default	\(\frac {-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{4}}+\frac {6 c d e \left (\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{3}}-\frac {8 c d e \left (\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {10 c d e \left (\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+\frac {\left (e^{2} a -c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 c d e}-\frac {3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (e^{2} a -c \,d^{2}\right )^{2} \ln \left (\frac {\frac {e^{2} a}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{16 c d e}\right )}{2}\right )}{3 \left (e^{2} a -c \,d^{2}\right )}\right )}{e^{2} a -c \,d^{2}}\right )}{e^{2} a -c \,d^{2}}}{e^{4}}\)	\(570\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

1/e^4*(-2/(a*e^2-c*d^2)/(x+d/e)^4*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(7/2)+6*c*d*e/(a*e^2-c*d^2)*(2/(a*e^
2-c*d^2)/(x+d/e)^3*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(7/2)-8*c*d*e/(a*e^2-c*d^2)*(2/3/(a*e^2-c*d^2)/(x+d
/e)^2*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(7/2)-10/3*c*d*e/(a*e^2-c*d^2)*(1/5*(c*d*e*(x+d/e)^2+(a*e^2-c*d^
2)*(x+d/e))^(5/2)+1/2*(a*e^2-c*d^2)*(1/8*(2*c*d*e*(x+d/e)+e^2*a-c*d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x
+d/e))^(3/2)-3/16*(a*e^2-c*d^2)^2/c/d/e*(1/4*(2*c*d*e*(x+d/e)+e^2*a-c*d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2
)*(x+d/e))^(1/2)-1/8*(a*e^2-c*d^2)^2/c/d/e*ln((1/2*e^2*a-1/2*c*d^2+c*d*e*(x+d/e))/(c*d*e)^(1/2)+(c*d*e*(x+d/e)
^2+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)))))))

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 554, normalized size of antiderivative = 2.36 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\left [\frac {15 \, {\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )} \sqrt {\frac {c d}{e}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, {\left (2 \, c d e^{2} x + c d^{2} e + a e^{3}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {c d}{e}} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \, {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 15 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} - {\left (5 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{16 \, {\left (e^{4} x + d e^{3}\right )}}, -\frac {15 \, {\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )} \sqrt {-\frac {c d}{e}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {c d}{e}}}{2 \, {\left (c^{2} d^{2} e x^{2} + a c d^{2} e + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 15 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} - {\left (5 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{8 \, {\left (e^{4} x + d e^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/16*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(c*d/e)*log(8*c^
2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*(2*c*d*e^2*x + c*d^2*e + a*e^3)*sqrt(c*d*e*x^2 + a*d*e +
(c*d^2 + a*e^2)*x)*sqrt(c*d/e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(2*c^2*d^2*e^2*x^2 - 15*c^2*d^4 + 25*a*c*d^
2*e^2 - 8*a^2*e^4 - (5*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^4*x + d*e^3
), -1/8*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-c*d/e)*arcta
n(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d/e)/(c^2*d^2*e*x^2 + a*
c*d^2*e + (c^2*d^3 + a*c*d*e^2)*x)) - 2*(2*c^2*d^2*e^2*x^2 - 15*c^2*d^4 + 25*a*c*d^2*e^2 - 8*a^2*e^4 - (5*c^2*
d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^4*x + d*e^3)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e*(a*e^2-c*d^2)>0)', see `assu
me?` for mor

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {1}{4} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} {\left (\frac {2 \, c^{2} d^{2} x}{e^{2}} - \frac {7 \, c^{3} d^{4} e^{5} - 9 \, a c^{2} d^{2} e^{7}}{c d e^{8}}\right )} - \frac {15 \, {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \log \left ({\left | -c d^{2} - a e^{2} - 2 \, \sqrt {c d e} {\left (\sqrt {c d e} x - \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e}\right )} \right |}\right )}{8 \, \sqrt {c d e} e^{3}} - \frac {2 \, {\left (\sqrt {c d e} c^{3} d^{6} - 3 \, \sqrt {c d e} a c^{2} d^{4} e^{2} + 3 \, \sqrt {c d e} a^{2} c d^{2} e^{4} - \sqrt {c d e} a^{3} e^{6}\right )}}{\sqrt {c d e} {\left ({\left (\sqrt {c d e} x - \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e}\right )} e + \sqrt {c d e} d\right )} e^{3}} \]

[In]

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

[Out]

1/4*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*(2*c^2*d^2*x/e^2 - (7*c^3*d^4*e^5 - 9*a*c^2*d^2*e^7)/(c*d*e^8)
) - 15/8*(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*log(abs(-c*d^2 - a*e^2 - 2*sqrt(c*d*e)*(sqrt(c*d*e)*x - sqr
t(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e))))/(sqrt(c*d*e)*e^3) - 2*(sqrt(c*d*e)*c^3*d^6 - 3*sqrt(c*d*e)*a*c^2*d
^4*e^2 + 3*sqrt(c*d*e)*a^2*c*d^2*e^4 - sqrt(c*d*e)*a^3*e^6)/(sqrt(c*d*e)*((sqrt(c*d*e)*x - sqrt(c*d*e*x^2 + c*
d^2*x + a*e^2*x + a*d*e))*e + sqrt(c*d*e)*d)*e^3)

Mupad [F(-1)]

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

[In]

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

[Out]

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