3.5.71 \(\int \frac {x^2}{(d+e x) \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [471]

Optimal. Leaf size=195 \[ \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d e^2}+\frac {2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 \left (c d^2-a e^2\right ) (d+e x)}-\frac {\left (3 c d^2+a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{3/2} d^{3/2} e^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1652, 806, 635, 212} \begin {gather*} -\frac {\left (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 )}{2 c^{3/2} d^{3/2} e^{5/2}}+\frac {2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^2 (d+e x) \left (c d^2-a e^2\right )}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(c*d*e^2) + (2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e
^2*(c*d^2 - a*e^2)*(d + e*x)) - ((3*c*d^2 + a*e^2)*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])])/(2*c^(3/2)*d^(3/2)*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 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 806

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

Rule 1652

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m
+ q + 2*p + 1))), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q + e*f*(m + p + q)*(d + e*x)^(q - 2)*(b*d - 2*a*e +
(2*c*d - b*e)*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] &&
NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 201, normalized size = 1.03 \begin {gather*} \frac {\sqrt {c} \sqrt {d} \sqrt {e} \left (-a^2 e^3 (d+e x)+c^2 d^3 x (3 d+e x)+a c d e \left (3 d^2-e^2 x^2\right )\right )-\left (3 c^2 d^4-2 a c d^2 e^2-a^2 e^4\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 )}{c^{3/2} d^{3/2} e^{5/2} \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^2/((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]*(-(a^2*e^3*(d + e*x)) + c^2*d^3*x*(3*d + e*x) + a*c*d*e*(3*d^2 - e^2*x^2)) - (3*c^2*d
^4 - 2*a*c*d^2*e^2 - a^2*e^4)*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqr
t[d]*Sqrt[d + e*x])])/(c^(3/2)*d^(3/2)*e^(5/2)*(c*d^2 - a*e^2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [A]
time = 0.09, size = 259, normalized size = 1.33

method	result	size
default	\(\frac {\frac {\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{c d e}-\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 c d e \sqrt {c d e}}}{e}-\frac {d \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{e^{2} \sqrt {c d e}}-\frac {2 d^{2} \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{e^{3} \left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(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/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/e^2*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/e^3*d^2/(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^2/(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 = 4.03, size = 563, normalized size = 2.89 \begin {gather*} \left [\frac {{\left (3 \, c^{2} d^{4} x e + 3 \, c^{2} d^{5} - 2 \, a c d^{2} x e^{3} - 2 \, a c d^{3} e^{2} - a^{2} x e^{5} - a^{2} d e^{4}\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 (c^{2} d^{3} x e^{2} + 3 \, c^{2} d^{4} e - a c d x e^{4} - a c d^{2} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{4 \, {\left (c^{3} d^{4} x e^{4} + c^{3} d^{5} e^{3} - a c^{2} d^{2} x e^{6} - a c^{2} d^{3} e^{5}\right )}}, \frac {{\left (3 \, c^{2} d^{4} x e + 3 \, c^{2} d^{5} - 2 \, a c d^{2} x e^{3} - 2 \, a c d^{3} e^{2} - a^{2} x e^{5} - a^{2} d e^{4}\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 (c^{2} d^{3} x e^{2} + 3 \, c^{2} d^{4} e - a c d x e^{4} - a c d^{2} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{2 \, {\left (c^{3} d^{4} x e^{4} + c^{3} d^{5} e^{3} - a c^{2} d^{2} x e^{6} - a c^{2} d^{3} e^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((3*c^2*d^4*x*e + 3*c^2*d^5 - 2*a*c*d^2*x*e^3 - 2*a*c*d^3*e^2 - a^2*x*e^5 - a^2*d*e^4)*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*(c^2*d^3*x*e^2 + 3*c^2*d^4*e -
a*c*d*x*e^4 - a*c*d^2*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))/(c^3*d^4*x*e^4 + c^3*d^5*e^3 - a*c^2*d
^2*x*e^6 - a*c^2*d^3*e^5), 1/2*((3*c^2*d^4*x*e + 3*c^2*d^5 - 2*a*c*d^2*x*e^3 - 2*a*c*d^3*e^2 - a^2*x*e^5 - a^2
*d*e^4)*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*(c^2*d^3*x*e^2 + 3*c^2*d^4*e - a*c*d*x*e
^4 - a*c*d^2*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))/(c^3*d^4*x*e^4 + c^3*d^5*e^3 - a*c^2*d^2*x*e^6
- a*c^2*d^3*e^5)]

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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 )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.73, size = 202, normalized size = 1.04 \begin {gather*} \frac {2 \, d^{2} e^{\left (-2\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 {\sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} e^{\left (-2\right )}}{c d} + \frac {{\left (3 \, \sqrt {c d} c d^{2} e^{\frac {1}{2}} + \sqrt {c d} a e^{\frac {5}{2}}\right )} e^{\left (-3\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 )}{2 \, c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(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^2*e^(-2)/(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) + s
qrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*e^(-2)/(c*d) + 1/2*(3*sqrt(c*d)*c*d^2*e^(1/2) + sqrt(c*d)*a*e^(5/2)
)*e^(-3)*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^2*d^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{\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^2/((d + e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

int(x^2/((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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