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

Optimal. Leaf size=232 \[ \frac {3 \left (c d^2-a e^2\right )^4 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{5/2} d^{5/2} e^{5/2}}-\frac {3 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c^2 d^2 e^2}+\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{8 c d e} \]

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Rubi [A]  time = 0.09, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {612, 621, 206} \begin {gather*} -\frac {3 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c^2 d^2 e^2}+\frac {3 \left (c d^2-a e^2\right )^4 \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 )}{128 c^{5/2} d^{5/2} e^{5/2}}+\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{8 c d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(c*d^2 - a*e^2)^2*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(64*c^2*d^2*e^2
) + ((c*d^2 + a*e^2 + 2*c*d*e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(8*c*d*e) + (3*(c*d^2 - a*e^2)
^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])
])/(128*c^(5/2)*d^(5/2)*e^(5/2))

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 612

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

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]

Rubi steps

\begin {align*} \int \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, dx &=\frac {\left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{8 c d e}-\frac {\left (3 \left (c d^2-a e^2\right )^2\right ) \int \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{16 c d e}\\ &=-\frac {3 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c^2 d^2 e^2}+\frac {\left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{8 c d e}+\frac {\left (3 \left (c d^2-a e^2\right )^4\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{128 c^2 d^2 e^2}\\ &=-\frac {3 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c^2 d^2 e^2}+\frac {\left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{8 c d e}+\frac {\left (3 \left (c d^2-a e^2\right )^4\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 )}{64 c^2 d^2 e^2}\\ &=-\frac {3 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c^2 d^2 e^2}+\frac {\left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{8 c d e}+\frac {3 \left (c d^2-a e^2\right )^4 \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 )}{128 c^{5/2} d^{5/2} e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 317, normalized size = 1.37 \begin {gather*} \frac {\sqrt {c d} \left (3 \left (c d^2-a e^2\right )^{9/2} \sqrt {a e+c d x} \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 )-\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d} (d+e x) \left (3 a^4 e^7+a^3 c d e^5 (e x-11 d)-a^2 c^2 d^2 e^3 \left (11 d^2+55 d e x+26 e^2 x^2\right )+a c^3 d^3 e \left (3 d^3-13 d^2 e x-68 d e^2 x^2-40 e^3 x^3\right )+c^4 d^4 x \left (3 d^3-2 d^2 e x-24 d e^2 x^2-16 e^3 x^3\right )\right )\right )}{64 c^{7/2} d^{7/2} e^{5/2} \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*d]*(-(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(3*a^4*e^7 + a^3*c*d*e^5*(-11*d + e*x) - a^2*c^2*d^2
*e^3*(11*d^2 + 55*d*e*x + 26*e^2*x^2) + a*c^3*d^3*e*(3*d^3 - 13*d^2*e*x - 68*d*e^2*x^2 - 40*e^3*x^3) + c^4*d^4
*x*(3*d^3 - 2*d^2*e*x - 24*d*e^2*x^2 - 16*e^3*x^3))) + 3*(c*d^2 - a*e^2)^(9/2)*Sqrt[a*e + c*d*x]*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])])
)/(64*c^(7/2)*d^(7/2)*e^(5/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [B]  time = 20.13, size = 24152, normalized size = 104.10 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.46, size = 666, normalized size = 2.87 \begin {gather*} \left [\frac {3 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\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 (16 \, c^{4} d^{4} e^{4} x^{3} - 3 \, c^{4} d^{7} e + 11 \, a c^{3} d^{5} e^{3} + 11 \, a^{2} c^{2} d^{3} e^{5} - 3 \, a^{3} c d e^{7} + 24 \, {\left (c^{4} d^{5} e^{3} + a c^{3} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (c^{4} d^{6} e^{2} + 22 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{256 \, c^{3} d^{3} e^{3}}, -\frac {3 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\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 (16 \, c^{4} d^{4} e^{4} x^{3} - 3 \, c^{4} d^{7} e + 11 \, a c^{3} d^{5} e^{3} + 11 \, a^{2} c^{2} d^{3} e^{5} - 3 \, a^{3} c d e^{7} + 24 \, {\left (c^{4} d^{5} e^{3} + a c^{3} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (c^{4} d^{6} e^{2} + 22 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{128 \, c^{3} d^{3} e^{3}}\right ] \end {gather*}

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, algorithm="fricas")

[Out]

[1/256*(3*(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*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*(16*c^4*d^4*e^4*x^3 - 3*c^4*d^7*e + 11*a*c^3*d^5*e^
3 + 11*a^2*c^2*d^3*e^5 - 3*a^3*c*d*e^7 + 24*(c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^2 + 2*(c^4*d^6*e^2 + 22*a*c^3*d^4*
e^4 + a^2*c^2*d^2*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^3), -1/128*(3*(c^4*d^8 - 4*a
*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*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*(16*c^4*d^4*e^4*x^3 - 3*c^4*d^7*e + 11*a*c^3*d^5*e^3 + 11*a^2*c^2*d^3*e^5 - 3*a^3*c*d*e^7 + 24
*(c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^2 + 2*(c^4*d^6*e^2 + 22*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x)*sqrt(c*d*e*x^2 +
a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^3)]

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giac [A]  time = 0.53, size = 298, normalized size = 1.28 \begin {gather*} \frac {1}{64} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (2 \, {\left (4 \, {\left (2 \, c d x e + \frac {3 \, {\left (c^{4} d^{5} e^{3} + a c^{3} d^{3} e^{5}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x + \frac {{\left (c^{4} d^{6} e^{2} + 22 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x - \frac {{\left (3 \, c^{4} d^{7} e - 11 \, a c^{3} d^{5} e^{3} - 11 \, a^{2} c^{2} d^{3} e^{5} + 3 \, a^{3} c d e^{7}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} - \frac {3 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -c d^{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 )} \sqrt {c d} e^{\frac {1}{2}} - a e^{2} \right |}\right )}{128 \, \sqrt {c d} c^{2} d^{2}} \end {gather*}

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, algorithm="giac")

[Out]

1/64*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*(2*c*d*x*e + 3*(c^4*d^5*e^3 + a*c^3*d^3*e^5)*e^(-3)/(c^
3*d^3))*x + (c^4*d^6*e^2 + 22*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*e^(-3)/(c^3*d^3))*x - (3*c^4*d^7*e - 11*a*c^3*d
^5*e^3 - 11*a^2*c^2*d^3*e^5 + 3*a^3*c*d*e^7)*e^(-3)/(c^3*d^3)) - 3/128*(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*
d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*e^(-5/2)*log(abs(-c*d^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))*sqrt(c*d)*e^(1/2) - a*e^2))/(sqrt(c*d)*c^2*d^2)

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maple [B]  time = 0.05, size = 671, normalized size = 2.89 \begin {gather*} \frac {3 a^{4} e^{6} \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 )}{128 \sqrt {c d e}\, c^{2} d^{2}}-\frac {3 a^{3} e^{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 )}{32 \sqrt {c d e}\, c}+\frac {9 a^{2} d^{2} e^{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 )}{64 \sqrt {c d e}}-\frac {3 a 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 )}{32 \sqrt {c d e}}+\frac {3 c^{2} d^{6} \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 )}{128 \sqrt {c d e}\, e^{2}}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e^{3} x}{32 c d}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a d e x}{16}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c \,d^{3} x}{32 e}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{3} e^{4}}{64 c^{2} d^{2}}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e^{2}}{64 c}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,d^{2}}{64}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c \,d^{4}}{64 e^{2}}+\frac {\left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) \left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}}}{8 c d e} \end {gather*}

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)

[Out]

1/8*(2*c*d*e*x+a*e^2+c*d^2)*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)/c/d/e-3/32/c/d*e^3*(c*d*e*x^2+a*d*e+(a*e^2
+c*d^2)*x)^(1/2)*x*a^2+3/16*d*e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a-3/32*c*d^3/e*(c*d*e*x^2+a*d*e+(a*e
^2+c*d^2)*x)^(1/2)*x-3/64/c^2/d^2*e^4*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^3+3/64/c*e^2*(c*d*e*x^2+a*d*e+
(a*e^2+c*d^2)*x)^(1/2)*a^2+3/64*d^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-3/64*c*d^4/e^2*(c*d*e*x^2+a*d*e+
(a*e^2+c*d^2)*x)^(1/2)+3/128/c^2/d^2*e^6*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^4-3/32/c*e^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)*a^3+9/64*d^2*e^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^2-3/32*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)*a+3/128*c^2*d^6/e^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)

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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(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 [B]  time = 0.81, size = 225, normalized size = 0.97 \begin {gather*} \frac {\left (\frac {c\,d^2}{2}+c\,x\,d\,e+\frac {a\,e^2}{2}\right )\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{4\,c\,d\,e}-\frac {\left (\frac {3\,{\left (c\,d^2+a\,e^2\right )}^2}{4}-3\,a\,c\,d^2\,e^2\right )\,\left (\left (\frac {x}{2}+\frac {c\,d^2+a\,e^2}{4\,c\,d\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}-\frac {\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{4}-a\,c\,d^2\,e^2\right )}{2\,{\left (c\,d\,e\right )}^{3/2}}\right )}{4\,c\,d\,e} \end {gather*}

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)

[Out]

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

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

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)

[Out]

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

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