3.1956 \(\int \frac{(d+e x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0991254, antiderivative size = 180, normalized size of antiderivative = 1., 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 = {668, 640, 621, 206} \[ \frac{3 e \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2}+\frac{3 \sqrt{e} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{5/2} d^{5/2}}-\frac{2 (d+e x)^2}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

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 668

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]

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 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 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])

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 ) \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^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*}

Mathematica [C]  time = 0.0510216, size = 93, normalized size = 0.52 \[ -\frac{2 (d+e x)^2 \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{c d \left (\frac{c d (d+e x)}{c d^2-a e^2}\right )^{3/2} \sqrt{(d+e x) (a e+c d x)}} \]

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]

(-2*(d + e*x)^2*Hypergeometric2F1[-3/2, -1/2, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(c*d*((c*d*(d + e*x)
)/(c*d^2 - a*e^2))^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B]  time = 0.056, size = 1047, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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)-9/4*d/c/
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-3/2*e/c*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-9/4*d^5*c/(-a^2*e^4+
2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+3/2*e/c*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c
)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)+2*e^2/d/c^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2)*a-5/2*e^2*d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a+e^2*x^2/d/c/(a*d
*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-3/4*e^4/d^3/c^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^2-3/4*e^8/d^3/c^
3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^4+1/2*e^6/d/c^2/(-a^2*e^4+2*a*c*d
^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^3+e^4*d/c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a
*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^2-3/2*e^3/d^2/c^2*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^
2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)*a-3/2*e^7/d^2/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*
d^2)*x+c*d*e*x^2)^(1/2)*x*a^3+3/2*e^3/d^2/c^2*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a+5/2*e^5/c/(-a^2*e^4+
2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a^2-1/2*e^3*d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d
^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a-9/2*e*d^4*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(1/2)*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.82326, size = 932, normalized size = 5.18 \begin{align*} \left [\frac{3 \,{\left (a c d^{2} e - a^{2} e^{3} +{\left (c^{2} d^{3} - a c d e^{2}\right )} x\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}{\left (c d e x - 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 (a c d^{2} e - a^{2} e^{3} +{\left (c^{2} d^{3} - a c d e^{2}\right )} x\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}{\left (c d e x - 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{align*}

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*(a*c*d^2*e - a^2*e^3 + (c^2*d^3 - a*c*d*e^2)*x)*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*d*e*x - 2*c*d^2 + 3*
a*e^2))/(c^3*d^3*x + a*c^2*d^2*e), -1/2*(3*(a*c*d^2*e - a^2*e^3 + (c^2*d^3 - a*c*d*e^2)*x)*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*d*e*x - 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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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 [B]  time = 1.28745, size = 478, normalized size = 2.66 \begin{align*} \frac{{\left (\frac{{\left (c^{3} d^{5} e^{3} - 2 \, a c^{2} d^{3} e^{5} + a^{2} c d e^{7}\right )} x}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}} - \frac{c^{3} d^{6} e^{2} - 5 \, a c^{2} d^{4} e^{4} + 7 \, a^{2} c d^{2} e^{6} - 3 \, a^{3} e^{8}}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}}\right )} x - \frac{2 \, c^{3} d^{7} e - 7 \, a c^{2} d^{5} e^{3} + 8 \, a^{2} c d^{3} e^{5} - 3 \, a^{3} d e^{7}}{c^{4} d^{6} e - 2 \, a c^{3} d^{4} e^{3} + a^{2} c^{2} d^{2} e^{5}}}{\sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}} - \frac{3 \,{\left (c d^{2} e - a e^{3}\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 )}{2 \, c^{3} d^{3}} \end{align*}

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]

(((c^3*d^5*e^3 - 2*a*c^2*d^3*e^5 + a^2*c*d*e^7)*x/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5) - (c^3*d^6*e
^2 - 5*a*c^2*d^4*e^4 + 7*a^2*c*d^2*e^6 - 3*a^3*e^8)/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5))*x - (2*c^
3*d^7*e - 7*a*c^2*d^5*e^3 + 8*a^2*c*d^3*e^5 - 3*a^3*d*e^7)/(c^4*d^6*e - 2*a*c^3*d^4*e^3 + a^2*c^2*d^2*e^5))/sq
rt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x) - 3/2*(c*d^2*e - a*e^3)*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)