3.452 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{x^3 (d+e x)} \, dx\)

Optimal. Leaf size=256 \[ -\frac {\left (-a^2 e^4+6 a c d^2 e^2+3 c^2 d^4\right ) \tanh ^{-1}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 \sqrt {a} d^{3/2} \sqrt {e}}+c^{3/2} d^{3/2} \sqrt {e} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac {\left (x \left (a e^2+5 c d^2\right )+2 a d e\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 d x^2} \]

[Out]

-1/8*(-a^2*e^4+6*a*c*d^2*e^2+3*c^2*d^4)*arctanh(1/2*(2*a*d*e+(a*e^2+c*d^2)*x)/a^(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))/d^(3/2)/a^(1/2)/e^(1/2)+c^(3/2)*d^(3/2)*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))*e^(1/2)-1/4*(2*a*d*e+(a*e^2+5*c*d^2)*x)*(a*d*e
+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/d/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {849, 810, 843, 621, 206, 724} \[ -\frac {\left (-a^2 e^4+6 a c d^2 e^2+3 c^2 d^4\right ) \tanh ^{-1}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 \sqrt {a} d^{3/2} \sqrt {e}}+c^{3/2} d^{3/2} \sqrt {e} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac {\left (x \left (a e^2+5 c d^2\right )+2 a d e\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*d*e + (5*c*d^2 + a*e^2)*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*d*x^2) + c^(3/2)*d^(3/2)*Sqr
t[e]*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
])] - ((3*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d]*Sqrt[e]*
Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(8*Sqrt[a]*d^(3/2)*Sqrt[e])

Rule 206

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

Rule 621

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 724

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

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
 - b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
 - c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 849

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

Rubi steps

\begin {align*} \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x^3 (d+e x)} \, dx &=\int \frac {(a e+c d x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{x^3} \, dx\\ &=-\frac {\left (2 a d e+\left (5 c d^2+a e^2\right ) x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 d x^2}-\frac {\int \frac {-\frac {1}{2} a e \left (3 c^2 d^4+6 a c d^2 e^2-a^2 e^4\right )-4 a c^2 d^3 e^2 x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 a d e}\\ &=-\frac {\left (2 a d e+\left (5 c d^2+a e^2\right ) x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 d x^2}+\left (c^2 d^2 e\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx+\frac {\left (3 c^2 d^4+6 a c d^2 e^2-a^2 e^4\right ) \int \frac {1}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 d}\\ &=-\frac {\left (2 a d e+\left (5 c d^2+a e^2\right ) x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 d x^2}+\left (2 c^2 d^2 e\right ) \operatorname {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 )-\frac {\left (3 c^2 d^4+6 a c d^2 e^2-a^2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{4 a d e-x^2} \, dx,x,\frac {2 a d e-\left (-c d^2-a e^2\right ) x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{4 d}\\ &=-\frac {\left (2 a d e+\left (5 c d^2+a e^2\right ) x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 d x^2}+c^{3/2} d^{3/2} \sqrt {e} \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 )-\frac {\left (3 c^2 d^4+6 a c d^2 e^2-a^2 e^4\right ) \tanh ^{-1}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 \sqrt {a} d^{3/2} \sqrt {e}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.45, size = 285, normalized size = 1.11 \[ \frac {\sqrt {a e+c d x} \left (-\frac {\sqrt {d+e x} \left (-a^2 e^4+6 a c d^2 e^2+3 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a e+c d x}}{\sqrt {a} \sqrt {e} \sqrt {d+e x}}\right )}{\sqrt {a}}+\frac {8 e (c d)^{5/2} \sqrt {c d^2-a e^2} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )}{c^{3/2}}-\frac {\sqrt {d} \sqrt {e} (d+e x) \sqrt {a e+c d x} \left (a e (2 d+e x)+5 c d^2 x\right )}{x^2}\right )}{4 d^{3/2} \sqrt {e} \sqrt {(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*e + c*d*x]*(-((Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x]*(d + e*x)*(5*c*d^2*x + a*e*(2*d + e*x)))/x^2) + (8*(c
*d)^(5/2)*e*Sqrt[c*d^2 - a*e^2]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*
e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/c^(3/2) - ((3*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*Sqrt[d + e*x]*A
rcTanh[(Sqrt[d]*Sqrt[a*e + c*d*x])/(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])])/Sqrt[a]))/(4*d^(3/2)*Sqrt[e]*Sqrt[(a*e +
c*d*x)*(d + e*x)])

________________________________________________________________________________________

fricas [A]  time = 5.63, size = 1375, normalized size = 5.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(8*sqrt(c*d*e)*a*c*d^3*e*x^2*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*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) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - (3*c^2
*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*sqrt(a*d*e)*x^2*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 +
 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*
d*e^3)*x)/x^2) - 4*(2*a^2*d^2*e^2 + (5*a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/
(a*d^2*e*x^2), -1/16*(16*sqrt(-c*d*e)*a*c*d^3*e*x^2*arctan(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^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) + (3*c^2*d^
4 + 6*a*c*d^2*e^2 - a^2*e^4)*sqrt(a*d*e)*x^2*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 + 4*
sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*d*e
^3)*x)/x^2) + 4*(2*a^2*d^2*e^2 + (5*a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*
d^2*e*x^2), 1/8*(4*sqrt(c*d*e)*a*c*d^3*e*x^2*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*sqr
t(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) + 8*(c^2*d^3*e + a*c*d*e^3)*x
) + (3*c^2*d^4 + 6*a*c*d^2*e^2 - a^2*e^4)*sqrt(-a*d*e)*x^2*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)
*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*d^2*e^2 + (a*c*d^3*e + a^2*d*e^3)*x)) -
2*(2*a^2*d^2*e^2 + (5*a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^2), -1
/8*(8*sqrt(-c*d*e)*a*c*d^3*e*x^2*arctan(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^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) - (3*c^2*d^4 + 6*a*c*d^2*e^2 -
 a^2*e^4)*sqrt(-a*d*e)*x^2*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x
)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*d^2*e^2 + (a*c*d^3*e + a^2*d*e^3)*x)) + 2*(2*a^2*d^2*e^2 + (5*a*c*d^3*e
+ a^2*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^2)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.42Error: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 0.02, size = 1604, normalized size = 6.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d^4*e^7*a^3/c*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e)
)^(1/2))/(c*d*e)^(1/2)-1/16/d^4*e^7*a^3/c*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e
^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)-1/4/d/a^2/e^2/x*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)*c-3/4*e/d^2*c/a*(c*d
*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x-3/4*e^2*d*a/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(c
*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/x)*c-3/8*d^3/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/x)*c^2+3/4/d^3/a/x*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)-1/4*e^3/d^
2*a*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+1/4/a^2/e*c^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x+3/4*d^2/a/
e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*c^2-1/3/d^3*e^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+1/8*e*
c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+7/8*e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*c-5/12*e^2/d^3*(
c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)+1/4*d/a^2/e^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*c^2+3/4*d/a*(c*d*
e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*c^2-1/2/d^2/a/e/x^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)-1/4/d^3*e^4*a
*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-1/8/d^4*e^5*a^2/c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/
2)-3/16/d^2*e^5*a^2*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e
))^(1/2))/(c*d*e)^(1/2)+3/16*e^3*a*c*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*
e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)+1/4/d*e^2*c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-1/16*d^2*
e*c^2*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d
*e)^(1/2)+1/4/d^3*e^4*a*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x+1/8/d^4*e^5*a^2/c*(c*d*e*x^2+a*d*e+(a*e^2+c*
d^2)*x)^(1/2)+3/16*e^5/d^2*a^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x
)^(1/2))/(c*d*e)^(1/2)+17/16*e*d^2*c^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+
c*d^2)*x)^(1/2))/(c*d*e)^(1/2)-3/16*e^3*a*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e
^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*c-1/2*e^2/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*c+1/8*e^4/d*a^2/(a*d*e
)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}}}{{\left (e x + d\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{x^3\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}{x^{3} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**(3/2)/(x**3*(d + e*x)), x)

________________________________________________________________________________________