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

Optimal. Leaf size=195 \[ \frac {3 \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{5/2} d^{5/2} \sqrt {e}}+\frac {3 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^2 d^2}+\frac {(d+e x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c d} \]

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Rubi [A]  time = 0.12, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {670, 640, 621, 206} \begin {gather*} \frac {3 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^2 d^2}+\frac {3 \left (c d^2-a e^2\right )^2 \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^{5/2} d^{5/2} \sqrt {e}}+\frac {(d+e x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 640

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

Rule 670

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

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 182, normalized size = 0.93 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\frac {3 \sqrt {c d} \left (c d^2-a e^2\right )^{3/2} \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 {e} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}+\sqrt {c} \sqrt {d} \left (c d (5 d+2 e x)-3 a e^2\right )\right )}{4 c^{5/2} d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.04, size = 323, normalized size = 1.66 \begin {gather*} \frac {3 \left (a^2 e^4-2 a c d^2 e^2+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^{5/2} d^{5/2} \sqrt {e}}-\frac {3 \sqrt {c d e} \left (a^2 e^4-2 a c d^2 e^2+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^3 d^3 e}+\frac {\sqrt {a d e+a e^2 x+c d^2 x+c d e x^2} \left (-3 a e^2+5 c d^2+2 c d e x\right )}{4 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 418, normalized size = 2.14 \begin {gather*} \left [\frac {3 \, {\left (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 + 5 \, c^{2} d^{3} e - 3 \, 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^{3} d^{3} e}, -\frac {3 \, {\left (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 + 5 \, c^{2} d^{3} e - 3 \, 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^{3} d^{3} e}\right ] \end {gather*}

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

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giac [A]  time = 0.80, size = 176, normalized size = 0.90 \begin {gather*} \frac {1}{4} \, \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (\frac {2 \, x e}{c d} + \frac {{\left (5 \, c d^{2} e - 3 \, a e^{3}\right )} e^{\left (-1\right )}}{c^{2} d^{2}}\right )} - \frac {3 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\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 + 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 )}{8 \, c^{3} d^{3}} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)*(2*x*e/(c*d) + (5*c*d^2*e - 3*a*e^3)*e^(-1)/(c^2*d^2)) - 3/8*(
c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*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^3*d^3)

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maple [A]  time = 0.05, size = 318, normalized size = 1.63 \begin {gather*} \frac {3 a^{2} 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 )}{8 \sqrt {c d e}\, c^{2} d^{2}}-\frac {3 a \,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 )}{4 \sqrt {c d e}\, c}+\frac {3 d^{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}}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, e x}{2 c d}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,e^{2}}{4 c^{2} d^{2}}+\frac {5 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}}{4 c} \end {gather*}

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)^(1/2),x)

[Out]

1/2*e*x/c/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-3/4*e^2/c^2/d^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a+
5/4/c*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+3/8*e^4/c^2/d^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^2-3/4*e^2/c*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+3/8*d^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((e*x+d)^2/(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(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 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^2}{\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((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2),x)

[Out]

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

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

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)**(1/2),x)

[Out]

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

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