3.3.72 \(\int \frac {x^2 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{d+e x} \, dx\)

Optimal. Leaf size=251 \[ -\frac {\left (c d^2-a e^2\right ) \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{5/2} d^{5/2} e^{7/2}}+\frac {\left (\left (5 c d^2-3 a e^2\right ) \left (a e^2+3 c d^2\right )-2 c d e x \left (5 c d^2-a e^2\right )\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{24 c^2 d^2 e^3}+\frac {x^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 e} \]

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Rubi [A]  time = 0.26, antiderivative size = 251, normalized size of antiderivative = 1.00, 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 = {851, 832, 779, 621, 206} \begin {gather*} -\frac {\left (c d^2-a e^2\right ) \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 )}{16 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}}{24 c^2 d^2 e^3}+\frac {x^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*e) + (((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])/(24*c^2*d^2*e^3) - ((c*d^2 - a*e^2)*(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])])/(16*c^(5/2)*d^(5/2)*e^(7/2))

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 779

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[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 851

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

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{d+e x} \, dx &=\int \frac {x^2 (a e+c d x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx\\ &=\frac {x^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 e}+\frac {\int \frac {x \left (-2 a c d^2 e-\frac {1}{2} c d \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}{3 c d e}\\ &=\frac {x^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 e}+\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}}{24 c^2 d^2 e^3}-\frac {\left (\left (c d^2-a e^2\right ) \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}{16 c^2 d^2 e^3}\\ &=\frac {x^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 e}+\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}}{24 c^2 d^2 e^3}-\frac {\left (\left (c d^2-a e^2\right ) \left (5 c^2 d^4+2 a c d^2 e^2+a^2 e^4\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 )}{8 c^2 d^2 e^3}\\ &=\frac {x^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 e}+\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}}{24 c^2 d^2 e^3}-\frac {\left (c d^2-a e^2\right ) \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 )}{16 c^{5/2} d^{5/2} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 245, normalized size = 0.98 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\sqrt {c} \sqrt {d} \sqrt {e} \left (-3 a^2 e^4+2 a c d e^2 (e x-2 d)+c^2 d^2 \left (15 d^2-10 d e x+8 e^2 x^2\right )\right )-\frac {3 \sqrt {c d} \sqrt {c d^2-a e^2} \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \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 )}{\sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}\right )}{24 c^{5/2} d^{5/2} e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[c]*Sqrt[d]*Sqrt[e]*(-3*a^2*e^4 + 2*a*c*d*e^2*(-2*d + e*x) + c^2*d^2*(15*d
^2 - 10*d*e*x + 8*e^2*x^2)) - (3*Sqrt[c*d]*Sqrt[c*d^2 - a*e^2]*(5*c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*ArcSinh[(
Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d +
e*x))/(c*d^2 - a*e^2)])))/(24*c^(5/2)*d^(5/2)*e^(7/2))

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IntegrateAlgebraic [B]  time = 14.05, size = 13727, normalized size = 54.69 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.47, size = 536, normalized size = 2.14 \begin {gather*} \left [-\frac {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 {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 \, \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 {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 4 \, a c^{2} d^{3} e^{3} - 3 \, a^{2} c d e^{5} - 2 \, {\left (5 \, c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{96 \, c^{3} d^{3} e^{4}}, \frac {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 {-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 {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) + 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 4 \, a c^{2} d^{3} e^{3} - 3 \, a^{2} c d e^{5} - 2 \, {\left (5 \, c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{48 \, c^{3} d^{3} e^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(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(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*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) - 4*(8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 4*a*c^2*d^3*e^3 - 3*a^2*c*d*e^5 - 2
*(5*c^3*d^4*e^2 - a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^4), 1/48*(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(-c*d*e)*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*(8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 4*a*c^2*d^3*e^3 - 3*a^2*c*d*e^5 - 2*(5*c^3*d^4*e^2 - a*c^2*d^2*e^4)*x)
*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a sub
stitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perha
ps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing
0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution vari
able should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.W
arning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a
substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should pe
rhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Evaluation time
: 1.92Error: Bad Argument Type

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maple [B]  time = 0.01, size = 713, normalized size = 2.84 \begin {gather*} \frac {a^{3} e^{3} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c^{2} d^{2}}+\frac {a^{2} e \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c}+\frac {a \,d^{2} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}\, e}-\frac {5 a \,d^{2} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, e}-\frac {c \,d^{4} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}\, e^{3}}+\frac {3 c \,d^{4} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, e^{3}}-\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a x}{4 c d}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d x}{4 e^{2}}-\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e}{8 c^{2} d^{2}}-\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a}{2 c e}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d^{2}}{8 e^{3}}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, d^{2}}{e^{3}}+\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}}}{3 c d \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d),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(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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