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

Optimal. Leaf size=207 \[ \frac {\left (c d^2-a e^2\right ) \left (a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{3/2} d^{3/2} e^{5/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 c d e (d+e x)}-\frac {1}{4} \left (\frac {a}{c d}+\frac {3 d}{e^2}\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {794, 664, 621, 206} \begin {gather*} \frac {\left (c d^2-a e^2\right ) \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 )}{8 c^{3/2} d^{3/2} e^{5/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 c d e (d+e x)}-\frac {1}{4} \left (\frac {a}{c d}+\frac {3 d}{e^2}\right ) \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*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(d + e*x),x]

[Out]

-((a/(c*d) + (3*d)/e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/4 + (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^
2)^(3/2)/(2*c*d*e*(d + e*x)) + ((c*d^2 - a*e^2)*(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])])/(8*c^(3/2)*d^(3/2)*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 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 664

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

Rule 794

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.66, size = 197, normalized size = 0.95 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\frac {\sqrt {c d} \sqrt {c d^2-a e^2} \left (a e^2+3 c d^2\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}}}+\sqrt {c} \sqrt {d} \sqrt {e} \left (a e^2+c d (2 e x-3 d)\right )\right )}{4 c^{3/2} d^{3/2} e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.22, size = 330, normalized size = 1.59 \begin {gather*} -\frac {\sqrt {c d e} \left (-a^2 e^4-2 a c d^2 e^2+3 c^2 d^4\right ) \log \left (a^2 e^4+8 c d e x \sqrt {c d e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 a c d^2 e^2-4 a c d e^3 x+c^2 d^4-4 c^2 d^3 e x-8 c^2 d^2 e^2 x^2\right )}{16 c^2 d^2 e^3}+\frac {\left (-a^2 e^4-2 a c d^2 e^2+3 c^2 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \left (2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 x \sqrt {c d e}\right )}{a e^2+c d^2}\right )}{8 c^{3/2} d^{3/2} e^{5/2}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (a e^2-3 c d^2+2 c d e x\right )}{4 c d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*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])/(4*c*d*e^2) + ((3*c^2*d^4 - 2*a*c
*d^2*e^2 - a^2*e^4)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*(-2*Sqrt[c*d*e]*x + 2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*
d*e*x^2]))/(c*d^2 + a*e^2)])/(8*c^(3/2)*d^(3/2)*e^(5/2)) - (Sqrt[c*d*e]*(3*c^2*d^4 - 2*a*c*d^2*e^2 - a^2*e^4)*
Log[c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4 - 4*c^2*d^3*e*x - 4*a*c*d*e^3*x - 8*c^2*d^2*e^2*x^2 + 8*c*d*e*Sqrt[c*d*e
]*x*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]])/(16*c^2*d^2*e^3)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 418, normalized size = 2.02 \begin {gather*} \left [-\frac {{\left (3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4}\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 (2 \, c^{2} d^{2} e^{2} x - 3 \, c^{2} d^{3} e + a c d e^{3}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{16 \, c^{2} d^{2} e^{3}}, -\frac {{\left (3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4}\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 (2 \, c^{2} d^{2} e^{2} x - 3 \, c^{2} d^{3} e + a c d e^{3}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{8 \, c^{2} d^{2} e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((3*c^2*d^4 - 2*a*c*d^2*e^2 - a^2*e^4)*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*(2*c^2*d^2*e^2*x - 3*c^2*d^3*e + a*c*d*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c
^2*d^2*e^3), -1/8*((3*c^2*d^4 - 2*a*c*d^2*e^2 - a^2*e^4)*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*(2*c^2*d^2*e^2*x - 3*c^2*d^3*e + a*c*d*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e
^3)]

________________________________________________________________________________________

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*(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.Evaluation time: 1.9Error: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 0.01, size = 516, normalized size = 2.49 \begin {gather*} -\frac {a^{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 )}{8 \sqrt {c d e}\, c d}-\frac {a d \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}}+\frac {a d \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 )}{4 \sqrt {c d e}}+\frac {c \,d^{3} \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^{2}}-\frac {c \,d^{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 )}{8 \sqrt {c d e}\, e^{2}}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, x}{2 e}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a}{4 c d}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d}{4 e^{2}}-\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}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*(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?

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2)/(e*x+d),x)

[Out]

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

________________________________________________________________________________________