3.5.70 \(\int \frac {x^3}{(d+e x) \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [470]

Optimal. Leaf size=271 \[ -\frac {3 \left (3 c d^2+a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 e^3}-\frac {2 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 \left (c d^2-a e^2\right ) (d+e x)}+\frac {(d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d e^3}+\frac {3 \left (5 c^2 d^4+2 a c d^2 e^2+a^2 e^4\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 c^{5/2} d^{5/2} e^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 298, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {863, 832, 793, 635, 212} \begin {gather*} \frac {3 \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\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 )}{8 c^{5/2} d^{5/2} e^{7/2}}-\frac {\left (\left (5 c d^2-3 a e^2\right ) \left (a e^2+3 c d^2\right )-2 c d e x \left (5 c d^2-a e^2\right )\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^2 d^2 e^3 \left (c d^2-a e^2\right )}-\frac {2 d x^2 \left (c d x \left (c d^2-a e^2\right )+a e \left (c d^2-a e^2\right )\right )}{e \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x^2*(a*e*(c*d^2 - a*e^2) + c*d*(c*d^2 - a*e^2)*x))/(e*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x +
c*d*e*x^2]) - (((5*c*d^2 - 3*a*e^2)*(3*c*d^2 + a*e^2) - 2*c*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*c^2*d^2*e^3*(c*d^2 - a*e^2)) + (3*(5*c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*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*c^(5/2)*d^(
5/2)*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 793

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

Rule 832

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

Rule 863

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

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Mathematica [A]
time = 0.48, size = 277, normalized size = 1.02 \begin {gather*} \frac {\sqrt {c} \sqrt {d} \sqrt {e} \left (3 a^3 e^5 (d+e x)+a^2 c d e^3 \left (4 d^2+5 d e x+e^2 x^2\right )+c^3 d^4 x \left (-15 d^2-5 d e x+2 e^2 x^2\right )-a c^2 d^2 e \left (15 d^3+d^2 e x-4 d e^2 x^2+2 e^3 x^3\right )\right )+3 \left (5 c^3 d^6-3 a c^2 d^4 e^2-a^2 c d^2 e^4-a^3 e^6\right ) \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{4 c^{5/2} d^{5/2} e^{7/2} \left (c d^2-a e^2\right ) \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(243)=486\).
time = 0.09, size = 514, normalized size = 1.90

method	result	size
default	\(\frac {\frac {x \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{2 c d e}-\frac {3 \left (a \,e^{2}+c \,d^{2}\right ) \left (\frac {\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{c d e}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 c d e \sqrt {c d e}}\right )}{4 c d e}-\frac {a \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 c \sqrt {c d e}}}{e}-\frac {d \left (\frac {\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{c d e}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 c d e \sqrt {c d e}}\right )}{e^{2}}+\frac {d^{2} \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{e^{3} \sqrt {c d e}}+\frac {2 d^{3} \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{e^{4} \left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\)	\(514\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/e*(1/2*x/c/d/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-3/4*(a*e^2+c*d^2)/c/d/e*(1/c/d/e*(a*d*e+(a*e^2+c*d^2)
*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2
)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))-1/2*a/c*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^
2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))-d/e^2*(1/c/d/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)
/c/d/e*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))+
d^2/e^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)+
2*d^3/e^4/(a*e^2-c*d^2)/(x+d/e)*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 3.70, size = 729, normalized size = 2.69 \begin {gather*} \left [\frac {3 \, {\left (5 \, c^{3} d^{6} x e + 5 \, c^{3} d^{7} - 3 \, a c^{2} d^{4} x e^{3} - 3 \, a c^{2} d^{5} e^{2} - a^{2} c d^{2} x e^{5} - a^{2} c d^{3} e^{4} - a^{3} x e^{7} - a^{3} d e^{6}\right )} \sqrt {c d} e^{\frac {1}{2}} \log \left (8 \, c^{2} d^{3} x e + c^{2} d^{4} + 8 \, a c d x e^{3} + a^{2} e^{4} + 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {c d} e^{\frac {1}{2}} + 2 \, {\left (4 \, c^{2} d^{2} x^{2} + 3 \, a c d^{2}\right )} e^{2}\right ) - 4 \, {\left (5 \, c^{3} d^{5} x e^{2} + 15 \, c^{3} d^{6} e - 2 \, a c^{2} d^{3} x e^{4} - 3 \, a^{2} c d x e^{6} + {\left (2 \, a c^{2} d^{2} x^{2} - 3 \, a^{2} c d^{2}\right )} e^{5} - 2 \, {\left (c^{3} d^{4} x^{2} + 2 \, a c^{2} d^{4}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{16 \, {\left (c^{4} d^{5} x e^{5} + c^{4} d^{6} e^{4} - a c^{3} d^{3} x e^{7} - a c^{3} d^{4} e^{6}\right )}}, -\frac {3 \, {\left (5 \, c^{3} d^{6} x e + 5 \, c^{3} d^{7} - 3 \, a c^{2} d^{4} x e^{3} - 3 \, a c^{2} d^{5} e^{2} - a^{2} c d^{2} x e^{5} - a^{2} c d^{3} e^{4} - a^{3} x e^{7} - a^{3} d e^{6}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{3} x e + a c d x e^{3} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e^{2}\right )}}\right ) + 2 \, {\left (5 \, c^{3} d^{5} x e^{2} + 15 \, c^{3} d^{6} e - 2 \, a c^{2} d^{3} x e^{4} - 3 \, a^{2} c d x e^{6} + {\left (2 \, a c^{2} d^{2} x^{2} - 3 \, a^{2} c d^{2}\right )} e^{5} - 2 \, {\left (c^{3} d^{4} x^{2} + 2 \, a c^{2} d^{4}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{8 \, {\left (c^{4} d^{5} x e^{5} + c^{4} d^{6} e^{4} - a c^{3} d^{3} x e^{7} - a c^{3} d^{4} e^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(5*c^3*d^6*x*e + 5*c^3*d^7 - 3*a*c^2*d^4*x*e^3 - 3*a*c^2*d^5*e^2 - a^2*c*d^2*x*e^5 - a^2*c*d^3*e^4 -
a^3*x*e^7 - a^3*d*e^6)*sqrt(c*d)*e^(1/2)*log(8*c^2*d^3*x*e + c^2*d^4 + 8*a*c*d*x*e^3 + a^2*e^4 + 4*sqrt(c*d^2*
x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(2*c*d*x*e + c*d^2 + a*e^2)*sqrt(c*d)*e^(1/2) + 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)
*e^2) - 4*(5*c^3*d^5*x*e^2 + 15*c^3*d^6*e - 2*a*c^2*d^3*x*e^4 - 3*a^2*c*d*x*e^6 + (2*a*c^2*d^2*x^2 - 3*a^2*c*d
^2)*e^5 - 2*(c^3*d^4*x^2 + 2*a*c^2*d^4)*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))/(c^4*d^5*x*e^5 + c^4
*d^6*e^4 - a*c^3*d^3*x*e^7 - a*c^3*d^4*e^6), -1/8*(3*(5*c^3*d^6*x*e + 5*c^3*d^7 - 3*a*c^2*d^4*x*e^3 - 3*a*c^2*
d^5*e^2 - a^2*c*d^2*x*e^5 - a^2*c*d^3*e^4 - a^3*x*e^7 - a^3*d*e^6)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d^2*x + a*x*
e^2 + (c*d*x^2 + a*d)*e)*(2*c*d*x*e + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^3*x*e + a*c*d*x*e^3 + (c^2*d^2*x^2 +
a*c*d^2)*e^2)) + 2*(5*c^3*d^5*x*e^2 + 15*c^3*d^6*e - 2*a*c^2*d^3*x*e^4 - 3*a^2*c*d*x*e^6 + (2*a*c^2*d^2*x^2 -
3*a^2*c*d^2)*e^5 - 2*(c^3*d^4*x^2 + 2*a*c^2*d^4)*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))/(c^4*d^5*x*
e^5 + c^4*d^6*e^4 - a*c^3*d^3*x*e^7 - a*c^3*d^4*e^6)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(e*x+d)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 1.17, size = 249, normalized size = 0.92 \begin {gather*} -\frac {2 \, d^{3} e^{\left (-3\right )}}{\sqrt {c d} d e^{\frac {1}{2}} + {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} e} + \frac {1}{4} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (\frac {2 \, x e^{\left (-2\right )}}{c d} - \frac {{\left (7 \, c d^{2} e^{5} + 3 \, a e^{7}\right )} e^{\left (-8\right )}}{c^{2} d^{2}}\right )} - \frac {3 \, {\left (5 \, \sqrt {c d} c^{2} d^{4} e^{\frac {1}{2}} + 2 \, \sqrt {c d} a c d^{2} e^{\frac {5}{2}} + \sqrt {c d} a^{2} e^{\frac {9}{2}}\right )} e^{\left (-4\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{8 \, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3}{\left (d+e\,x\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((d + e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

int(x^3/((d + e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)), x)

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