3.20.22 \(\int (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2} \, dx\) [1922]

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

[Out]

1/8*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/e+3/128*(-a*e^2+c*d^2)^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^(5/2
)-3/64*(-a*e^2+c*d^2)^2*(2*c*d*e*x+a*e^2+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^2

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Rubi [A]
time = 0.05, 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 = {626, 635, 212} \begin {gather*} \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} \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 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 626

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 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]

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 ) \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 )}{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 = 0.57, size = 240, normalized size = 1.03 \begin {gather*} \frac {\left (c d^2-a e^2\right )^4 ((a e+c d x) (d+e x))^{3/2} \left (-\frac {\sqrt {c} \sqrt {d} \sqrt {e} (a e+c d x)^2 \left (3 e^3-\frac {11 c d e^2 (d+e x)}{a e+c d x}-\frac {11 c^2 d^2 e (d+e x)^2}{(a e+c d x)^2}+\frac {3 c^3 d^3 (d+e x)^3}{(a e+c d x)^3}\right )}{\left (c d^2-a e^2\right )^4 (d+e x)}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{(a e+c d x)^{3/2} (d+e x)^{3/2}}\right )}{64 c^{5/2} d^{5/2} e^{5/2}} \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]

((c*d^2 - a*e^2)^4*((a*e + c*d*x)*(d + e*x))^(3/2)*(-((Sqrt[c]*Sqrt[d]*Sqrt[e]*(a*e + c*d*x)^2*(3*e^3 - (11*c*
d*e^2*(d + e*x))/(a*e + c*d*x) - (11*c^2*d^2*e*(d + e*x)^2)/(a*e + c*d*x)^2 + (3*c^3*d^3*(d + e*x)^3)/(a*e + c
*d*x)^3))/((c*d^2 - a*e^2)^4*(d + e*x))) + (3*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c*d*
x])])/((a*e + c*d*x)^(3/2)*(d + e*x)^(3/2))))/(64*c^(5/2)*d^(5/2)*e^(5/2))

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Maple [A]
time = 0.63, size = 247, normalized size = 1.06

method	result	size
default	\(\frac {\left (2 c d e x +e^{2} a +c \,d^{2}\right ) \left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )^{\frac {3}{2}}}{8 c d e}+\frac {3 \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \left (\frac {\left (2 c d e x +e^{2} a +c \,d^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}{4 c d e}+\frac {\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{8 c d e \sqrt {c d e}}\right )}{16 c d e}\)	\(247\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 3.57, size = 647, normalized size = 2.79 \begin {gather*} \left [\frac {{\left (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^{\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 (2 \, c^{4} d^{6} x e^{2} - 3 \, c^{4} d^{7} e + 2 \, a^{2} c^{2} d^{2} x e^{6} - 3 \, a^{3} c d e^{7} + {\left (24 \, a c^{3} d^{3} x^{2} + 11 \, a^{2} c^{2} d^{3}\right )} e^{5} + 4 \, {\left (4 \, c^{4} d^{4} x^{3} + 11 \, a c^{3} d^{4} x\right )} e^{4} + {\left (24 \, c^{4} d^{5} x^{2} + 11 \, a c^{3} d^{5}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-3\right )}}{256 \, c^{3} d^{3}}, -\frac {{\left (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^{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 (2 \, c^{4} d^{6} x e^{2} - 3 \, c^{4} d^{7} e + 2 \, a^{2} c^{2} d^{2} x e^{6} - 3 \, a^{3} c d e^{7} + {\left (24 \, a c^{3} d^{3} x^{2} + 11 \, a^{2} c^{2} d^{3}\right )} e^{5} + 4 \, {\left (4 \, c^{4} d^{4} x^{3} + 11 \, a c^{3} d^{4} x\right )} e^{4} + {\left (24 \, c^{4} d^{5} x^{2} + 11 \, a c^{3} d^{5}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-3\right )}}{128 \, c^{3} d^{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^(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*(2*c^4*d^6*x*e^2 - 3*c^4*d^7*e + 2*a^
2*c^2*d^2*x*e^6 - 3*a^3*c*d*e^7 + (24*a*c^3*d^3*x^2 + 11*a^2*c^2*d^3)*e^5 + 4*(4*c^4*d^4*x^3 + 11*a*c^3*d^4*x)
*e^4 + (24*c^4*d^5*x^2 + 11*a*c^3*d^5)*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-3)/(c^3*d^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*sqr
t(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*(2*c^4*d^6*x*e^2 - 3*c^4*d^7*e + 2*a^2*c^2*d^2*x*e^6 - 3*a^3*c*d*e^7 + (24*
a*c^3*d^3*x^2 + 11*a^2*c^2*d^3)*e^5 + 4*(4*c^4*d^4*x^3 + 11*a*c^3*d^4*x)*e^4 + (24*c^4*d^5*x^2 + 11*a*c^3*d^5)
*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-3)/(c^3*d^3)]

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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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Giac [A]
time = 0.77, 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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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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