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

Optimal. Leaf size=371 \[ -\frac{\left (-a^2 e^4-2 c d e x \left (a e^2+7 c d^2\right )+12 a c d^2 e^2+5 c^2 d^4\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 d x}-\frac{\left (15 a^2 c d^2 e^4-a^3 e^6+45 a c^2 d^4 e^2+5 c^3 d^6\right ) \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 \sqrt{a} d^{3/2} \sqrt{e}}+\frac{1}{2} c^{3/2} d^{3/2} \sqrt{e} \left (5 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac{\left (3 x \left (a e^2+3 c d^2\right )+4 a d e\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 d x^3} \]

[Out]

-((5*c^2*d^4 + 12*a*c*d^2*e^2 - a^2*e^4 - 2*c*d*e*(7*c*d^2 + a*e^2)*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*
x^2])/(8*d*x) - ((4*a*d*e + 3*(3*c*d^2 + a*e^2)*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(12*d*x^3) +
(c^(3/2)*d^(3/2)*Sqrt[e]*(3*c*d^2 + 5*a*e^2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/2 - ((5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)
*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])
])/(16*Sqrt[a]*d^(3/2)*Sqrt[e])

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Rubi [A]  time = 0.470245, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {849, 810, 812, 843, 621, 206, 724} \[ -\frac{\left (-a^2 e^4-2 c d e x \left (a e^2+7 c d^2\right )+12 a c d^2 e^2+5 c^2 d^4\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 d x}-\frac{\left (15 a^2 c d^2 e^4-a^3 e^6+45 a c^2 d^4 e^2+5 c^3 d^6\right ) \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 \sqrt{a} d^{3/2} \sqrt{e}}+\frac{1}{2} c^{3/2} d^{3/2} \sqrt{e} \left (5 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac{\left (3 x \left (a e^2+3 c d^2\right )+4 a d e\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*c^2*d^4 + 12*a*c*d^2*e^2 - a^2*e^4 - 2*c*d*e*(7*c*d^2 + a*e^2)*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*
x^2])/(8*d*x) - ((4*a*d*e + 3*(3*c*d^2 + a*e^2)*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(12*d*x^3) +
(c^(3/2)*d^(3/2)*Sqrt[e]*(3*c*d^2 + 5*a*e^2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/2 - ((5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)
*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])
])/(16*Sqrt[a]*d^(3/2)*Sqrt[e])

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
]))

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 812

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

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 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 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 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]

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 )^{5/2}}{x^4 (d+e x)} \, dx &=\int \frac{(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac{\left (4 a d e+3 \left (3 c d^2+a e^2\right ) x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 d x^3}-\frac{\int \frac{\left (-\frac{1}{2} a e \left (5 c^2 d^4+12 a c d^2 e^2-a^2 e^4\right )-a c d e^2 \left (7 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}}{x^2} \, dx}{4 a d e}\\ &=-\frac{\left (5 c^2 d^4+12 a c d^2 e^2-a^2 e^4-2 c d e \left (7 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}}{8 d x}-\frac{\left (4 a d e+3 \left (3 c d^2+a e^2\right ) x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 d x^3}+\frac{\int \frac{\frac{1}{2} a e \left (5 c^3 d^6+45 a c^2 d^4 e^2+15 a^2 c d^2 e^4-a^3 e^6\right )+4 a c^2 d^3 e^2 \left (3 c d^2+5 a e^2\right ) x}{x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 a d e}\\ &=-\frac{\left (5 c^2 d^4+12 a c d^2 e^2-a^2 e^4-2 c d e \left (7 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}}{8 d x}-\frac{\left (4 a d e+3 \left (3 c d^2+a e^2\right ) x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 d x^3}+\frac{1}{2} \left (c^2 d^2 e \left (3 c d^2+5 a e^2\right )\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 (5 c^3 d^6+45 a c^2 d^4 e^2+15 a^2 c d^2 e^4-a^3 e^6\right ) \int \frac{1}{x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 d}\\ &=-\frac{\left (5 c^2 d^4+12 a c d^2 e^2-a^2 e^4-2 c d e \left (7 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}}{8 d x}-\frac{\left (4 a d e+3 \left (3 c d^2+a e^2\right ) x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 d x^3}+\left (c^2 d^2 e \left (3 c d^2+5 a e^2\right )\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 (5 c^3 d^6+45 a c^2 d^4 e^2+15 a^2 c d^2 e^4-a^3 e^6\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 )}{8 d}\\ &=-\frac{\left (5 c^2 d^4+12 a c d^2 e^2-a^2 e^4-2 c d e \left (7 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}}{8 d x}-\frac{\left (4 a d e+3 \left (3 c d^2+a e^2\right ) x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 d x^3}+\frac{1}{2} c^{3/2} d^{3/2} \sqrt{e} \left (3 c d^2+5 a e^2\right ) \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 (5 c^3 d^6+45 a c^2 d^4 e^2+15 a^2 c d^2 e^4-a^3 e^6\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 )}{16 \sqrt{a} d^{3/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 3.32594, size = 357, normalized size = 0.96 \[ \frac{\sqrt{a e+c d x} \left (-\frac{\sqrt{d} \sqrt{e} (d+e x) \sqrt{a e+c d x} \left (a^2 e^2 \left (8 d^2+14 d e x+3 e^2 x^2\right )+2 a c d^2 e x (13 d+34 e x)+3 c^2 d^3 x^2 (11 d-8 e x)\right )}{x^3}-\frac{3 \sqrt{d+e x} \left (15 a^2 c d^2 e^4-a^3 e^6+45 a c^2 d^4 e^2+5 c^3 d^6\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a e+c d x}}{\sqrt{a} \sqrt{e} \sqrt{d+e x}}\right )}{\sqrt{a}}+\frac{24 e (c d)^{5/2} \sqrt{c d^2-a e^2} \left (5 a e^2+3 c d^2\right ) \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{c^{3/2}}\right )}{24 d^{3/2} \sqrt{e} \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*e + c*d*x]*(-((Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x]*(d + e*x)*(3*c^2*d^3*x^2*(11*d - 8*e*x) + 2*a*c*d^2*e
*x*(13*d + 34*e*x) + a^2*e^2*(8*d^2 + 14*d*e*x + 3*e^2*x^2)))/x^3) + (24*(c*d)^(5/2)*e*Sqrt[c*d^2 - a*e^2]*(3*
c*d^2 + 5*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])/(Sq
rt[c*d]*Sqrt[c*d^2 - a*e^2])])/c^(3/2) - (3*(5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)*Sqrt[d
+ e*x]*ArcTanh[(Sqrt[d]*Sqrt[a*e + c*d*x])/(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])])/Sqrt[a]))/(24*d^(3/2)*Sqrt[e]*Sq
rt[(a*e + c*d*x)*(d + e*x)])

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Maple [B]  time = 0.082, size = 3144, normalized size = 8.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*e^9/d^6*a^4/c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+93/64*e^2*d*c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2)*x+387/256*e*d^4*c^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2))/(d*e*c)^(1/2)-3/64*e^7/d^4*a^3/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+15/128*e^5*a^2*c*ln((1/2*a*e^2+
1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)-1/3/d^2/a/e/x^3*(a*d*e
+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)+19/24/d^2/a*e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c+1/8*d^2/a^3/e^3*(a*d
*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c^3-3/8/d^4/a*e/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)+5/24*d^3/a^2/e^2
*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*c^3+5/8*d^4/a/e*c^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5/8*d^3/a
*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*c^3+5/6/d/a^2*c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*x-1/8/a^3
/e^3/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)*c^2+5/6/a*e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x*c^2+1/16*
e^6/d^5*a^2/c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+9/64*e^6/d^3*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d
/e+x))^(1/2)*x-3/128*e^9/d^6*a^4/c^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+15/128*e^7/d^2*a^3*ln((1/2*
a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+3/64
*e^2*d*c^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-3/256*e*d^4*c^3*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d
*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+1/8*e^5/d^4*a*(c*d*e*(d/e+x)^2+
(a*e^2-c*d^2)*(d/e+x))^(3/2)*x+3/64*e^7/d^4*a^3/c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-1/8*e^3/d^2*c*
(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-15/128*e^5*a^2*c*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c
)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+1/5*e^3/d^4*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2
)*(d/e+x))^(5/2)+7/40*e^3/d^4*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)+1/12*e^3/d^2*c*(a*d*e+(a*e^2+c*d^2)*x+c*
d*e*x^2)^(3/2)*x-15/128*e^7/d^2*a^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*
e*x^2)^(1/2))/(d*e*c)^(1/2)-1/16*e^6/d^5*a^2/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-9/64*e^6/d^3*a^2*(a*d*e
+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x-1/8*e^5/d^4*a*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x+1/16*e^6/d*a^3/(a*
d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/x)-3/64*e^3*a*
c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-1/16*e^2/d*c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+3/1
28*e*d^2*c^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-1/24*e^4/d^3*a*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3
/2)-1/8*e^5/d^2*a^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+37/48*e^2/d*c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3
/2)+493/128*e*d^2*c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+107/64*e^3*a*c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)
^(1/2)+35/24*d/a*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*c^2+5/12/d^3/a/x^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^
(7/2)-5/16*d^5/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/
2))/x)*c^3+25/24/a^2/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c^2-15/256*e^9/d^4*a^4/c*ln((1/2*a*e^2-1/2*c*d^
2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-3/64*e^8/d^5*a^3/c
*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-9/64*e^4/d*a*c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*
x+3/256*e^11/d^6*a^5/c^2*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(
d/e+x))^(1/2))/(d*e*c)^(1/2)+15/256*e^3*d^2*a*c^2*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*
(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+15/256*e^9/d^4*a^4/c*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d
*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)+625/256*e^3*d^2*a*c^2*ln((1/2*a*e^2+1/2*c*d
^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)+3/64*e^8/d^5*a^3/c*(a*d*e+(a*
e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x+1/64*e^4/d*a*c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x-3/256*e^11/d^6*a^5/c^
2*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)+1/8*d/
a^3/e^2*c^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*x-45/16*d^3*a*e^2/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*
x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/x)*c^2-1/12/d/a^2/e^2/x^2*(a*d*e+(a*e^2+c*d^2)*x+c*
d*e*x^2)^(7/2)*c-15/16*d*a^2*e^4/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2
)*x+c*d*e*x^2)^(1/2))/x)*c-5/6/d^2/a^2/e/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)*c+5/24*d^2/a^2/e*(a*d*e+(a*
e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x*c^3+3/8/d^3/a*e^2*c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )} x^{4}}\,{d x} \end{align*}

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

[Out]

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

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Fricas [A]  time = 59.4654, size = 3684, normalized size = 9.93 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

[1/96*(24*(3*a*c^2*d^5*e + 5*a^2*c*d^3*e^3)*sqrt(c*d*e)*x^3*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*(5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(a*d*e)*x^3*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*(24*a*c^2*d^4*e^2*x^3 - 8*a^3*d^3*e^3 - (3
3*a*c^2*d^5*e + 68*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2 - 2*(13*a^2*c*d^4*e^2 + 7*a^3*d^2*e^4)*x)*sqrt(c*d*e*x^2 +
a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^3), -1/96*(48*(3*a*c^2*d^5*e + 5*a^2*c*d^3*e^3)*sqrt(-c*d*e)*x^3*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^2*x^2 +
a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) + 3*(5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)*sqrt
(a*d*e)*x^3*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*(24*a*c^2*d^4*e^2
*x^3 - 8*a^3*d^3*e^3 - (33*a*c^2*d^5*e + 68*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2 - 2*(13*a^2*c*d^4*e^2 + 7*a^3*d^2
*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^3), 1/48*(3*(5*c^3*d^6 + 45*a*c^2*d^4*e^2 + 1
5*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-a*d*e)*x^3*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)) + 12*(3*a*c^2*d^5
*e + 5*a^2*c*d^3*e^3)*sqrt(c*d*e)*x^3*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) + 2*(
24*a*c^2*d^4*e^2*x^3 - 8*a^3*d^3*e^3 - (33*a*c^2*d^5*e + 68*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2 - 2*(13*a^2*c*d^4
*e^2 + 7*a^3*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^3), 1/48*(3*(5*c^3*d^6 + 45*a
*c^2*d^4*e^2 + 15*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-a*d*e)*x^3*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)) -
24*(3*a*c^2*d^5*e + 5*a^2*c*d^3*e^3)*sqrt(-c*d*e)*x^3*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)) + 2*(24*
a*c^2*d^4*e^2*x^3 - 8*a^3*d^3*e^3 - (33*a*c^2*d^5*e + 68*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2 - 2*(13*a^2*c*d^4*e^
2 + 7*a^3*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a*d^2*e*x^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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

[Out]

Exception raised: TypeError