3.1957 \(\int \frac {(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac {\sqrt {e} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{3/2} d^{3/2}}-\frac {2 (d+e x)}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {652, 621, 206} \[ \frac {\sqrt {e} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{3/2} d^{3/2}}-\frac {2 (d+e x)}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 652

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

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 164, normalized size = 1.31 \[ \frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d^2-a e^2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )-2 (c d)^{3/2} (d+e x)}{(c d)^{5/2} \sqrt {(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 349, normalized size = 2.79 \[ \left [\frac {{\left (c d x + a e\right )} \sqrt {\frac {e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \, {\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {e}{c d}}\right ) - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{2 \, {\left (c^{2} d^{2} x + a c d e\right )}}, -\frac {{\left (c d x + a e\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{c^{2} d^{2} x + a c d e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.15, size = 229, normalized size = 1.83 \[ -\frac {2 \, {\left (\frac {{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}} + \frac {c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4}}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}}\right )}}{\sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}} - \frac {\sqrt {c d} e^{\frac {1}{2}} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 717, normalized size = 5.74 \[ \frac {a^{2} e^{5} x}{\left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c d}-\frac {2 a d \,e^{3} x}{\left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {3 c \,d^{3} e x}{\left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}+\frac {a^{3} e^{6}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d^{2}}-\frac {a^{2} e^{4}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c}-\frac {5 a \,d^{2} e^{2}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {3 c \,d^{4}}{2 \left (-a^{2} e^{4}+2 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}+\frac {2 \left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) d^{2}}{\left (4 a c \,d^{2} e^{2}-\left (a \,e^{2}+c \,d^{2}\right )^{2}\right ) \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}-\frac {e x}{\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c d}+\frac {e \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{\sqrt {c d e}\, c d}+\frac {a \,e^{2}}{2 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c^{2} d^{2}}-\frac {3}{2 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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