3.20.56 \(\int \frac {(d+e x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [1956]

Optimal. Leaf size=180 \[ -\frac {2 (d+e x)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {3 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2}+\frac {3 \sqrt {e} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{5/2} d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 180, 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 = {682, 654, 635, 212} \begin {gather*} \frac {3 \sqrt {e} \left (c d^2-a e^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^{5/2} d^{5/2}}+\frac {3 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2}-\frac {2 (d+e x)^2}{c d \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[(d + e*x)^3/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^2)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3*e*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d
*e*x^2])/(c^2*d^2) + (3*Sqrt[e]*(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^(5/2)*d^(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 654

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 682

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*(p + 1))), x] - Dist[e^2*((m + p)/(c*(p + 1))), Int[(d + e*x)^(m - 2)*(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] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.49, size = 134, normalized size = 0.74 \begin {gather*} \frac {-c d (d+e x) \left (-3 a e^2+c d (2 d-e x)\right )+3 \sqrt {\frac {c d}{e}} e \left (-c d^2+a e^2\right ) \sqrt {a e+c d x} \sqrt {d+e x} \log \left (\sqrt {a e+c d x}-\sqrt {\frac {c d}{e}} \sqrt {d+e x}\right )}{c^3 d^3 \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(959\) vs. \(2(158)=316\).
time = 0.76, size = 960, normalized size = 5.33

method	result	size
default	\(e^{3} \left (\frac {x^{2}}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {3 \left (e^{2} a +c \,d^{2}\right ) \left (-\frac {x}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 c d e x +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )}{2 c d e}+\frac {\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 )}{c d e \sqrt {c d e}}\right )}{2 c d e}-\frac {2 a \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 c d e x +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )}{c}\right )+3 d \,e^{2} \left (-\frac {x}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 c d e x +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )}{2 c d e}+\frac {\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 )}{c d e \sqrt {c d e}}\right )+3 d^{2} e \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 c d e x +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )+\frac {2 d^{3} \left (2 c d e x +e^{2} a +c \,d^{2}\right )}{\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\)	\(960\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^3*(x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-3/2*(a*e^2+c*d^2)/c/d/e*(-x/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*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d
/e*(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)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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))-2*a/c*(-1
/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(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)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))+3*d*e^2*(-x/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*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(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)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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))+3*d^2*e*(-1/c/d/e/(a*d*e+
(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(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)/(
a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+2*d^3*(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)/(a*d*e+(
a*e^2+c*d^2)*x+c*d*e*x^2)^(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)^3/(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 = 2.78, size = 444, normalized size = 2.47 \begin {gather*} \left [\frac {3 \, {\left (c^{2} d^{3} x - a c d x e^{2} + a c d^{2} e - a^{2} e^{3}\right )} \sqrt {\frac {1}{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 \, {\left (2 \, c^{2} d^{2} x e + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {\frac {1}{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 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (c d x e - 2 \, c d^{2} + 3 \, a e^{2}\right )}}{4 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}, -\frac {3 \, {\left (c^{2} d^{3} x - a c d x e^{2} + a c d^{2} e - a^{2} e^{3}\right )} \sqrt {-\frac {e}{c d}} \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 {-\frac {e}{c d}}}{2 \, {\left (c d^{2} x e + a x e^{3} + {\left (c d x^{2} + a d\right )} e^{2}\right )}}\right ) - 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (c d x e - 2 \, c d^{2} + 3 \, a e^{2}\right )}}{2 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(c^2*d^3*x - a*c*d*x*e^2 + a*c*d^2*e - a^2*e^3)*sqrt(1/(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*(2*c^2*d^2*x*e + c^2*d^3 + a*c*d*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sq
rt(1/(c*d))*e^(1/2) + 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)*e^2) + 4*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(c*d*
x*e - 2*c*d^2 + 3*a*e^2))/(c^3*d^3*x + a*c^2*d^2*e), -1/2*(3*(c^2*d^3*x - a*c*d*x*e^2 + a*c*d^2*e - a^2*e^3)*s
qrt(-e/(c*d))*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(-e/(c*d)
)/(c*d^2*x*e + a*x*e^3 + (c*d*x^2 + a*d)*e^2)) - 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(c*d*x*e - 2*c*
d^2 + 3*a*e^2))/(c^3*d^3*x + a*c^2*d^2*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.60, size = 226, normalized size = 1.26 \begin {gather*} \frac {\sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} e}{c^{2} d^{2}} - \frac {3 \, {\left (\sqrt {c d} c d^{2} e^{\frac {3}{2}} - \sqrt {c d} a e^{\frac {7}{2}}\right )} e^{\left (-1\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^{3} d^{3}} + \frac {2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}}{{\left ({\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 + \sqrt {c d} a e^{\frac {3}{2}}\right )} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*e/(c^2*d^2) - 3/2*(sqrt(c*d)*c*d^2*e^(3/2) - sqrt(c*d)*a*e^(7/2))*
e^(-1)*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^3*d^3) + 2*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)/(((sqrt(c*d)*x*e^(1/2) - sqr
t(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*c*d + sqrt(c*d)*a*e^(3/2))*c^2*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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