### 3.291 $$\int \frac{\sqrt{b x+c x^2}}{(d+e x)^3} \, dx$$

Optimal. Leaf size=127 $\frac{\sqrt{b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac{b^2 \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0876072, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {720, 724, 206} $\frac{\sqrt{b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac{b^2 \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x)^3,x]

[Out]

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

Rule 720

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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{\sqrt{b x+c x^2}}{(d+e x)^3} \, dx &=\frac{(b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{4 d (c d-b e) (d+e x)^2}-\frac{b^2 \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{8 d (c d-b e)}\\ &=\frac{(b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{4 d (c d-b e) (d+e x)^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac{-b d-(2 c d-b e) x}{\sqrt{b x+c x^2}}\right )}{4 d (c d-b e)}\\ &=\frac{(b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{4 d (c d-b e) (d+e x)^2}-\frac{b^2 \tanh ^{-1}\left (\frac{b d+(2 c d-b e) x}{2 \sqrt{d} \sqrt{c d-b e} \sqrt{b x+c x^2}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.216279, size = 121, normalized size = 0.95 $\frac{\sqrt{x (b+c x)} \left (\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{x} \sqrt{b+c x} (b e-c d)^{3/2}}+\frac{\sqrt{d} (b (d-e x)+2 c d x)}{(d+e x)^2 (c d-b e)}\right )}{4 d^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.23, size = 1963, normalized size = 15.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(1/2)/(e*x+d)^3,x)

[Out]

1/2/e/d/(b*e-c*d)/(d/e+x)^2*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(3/2)+1/4*e/d^2/(b*e-c*d)^2/(d
/e+x)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(3/2)*b-1/2/d/(b*e-c*d)^2/(d/e+x)*(c*(d/e+x)^2+(b*e-
2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(3/2)*c-1/4*e/d^2/(b*e-c*d)^2*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)
/e^2)^(1/2)*b^2+3/4/d/(b*e-c*d)^2*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*b*c-1/2/e/(b*e-c*d
)^2*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*c^2+1/4/d/(b*e-c*d)^2*ln((1/2*(b*e-2*c*d)/e+(d/e
+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(1/2)*b^2-3/4/e/(b*e-c*d)^2*ln((1/
2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(3/2)*b+1/2/e^
2*d/(b*e-c*d)^2*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(
1/2))*c^(5/2)-1/8/d/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b
*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*b^3+5/8/e/(b*e-c*d)^2/(
-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(
b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*b^2*c-1/e^2*d/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-
2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c
*d)/e^2)^(1/2))/(d/e+x))*b*c^2+1/2/e^3*d^2/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*
c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*
c^3-1/4*e/d^2/(b*e-c*d)^2*c*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*b+1/2/d/(b*e-c*d)^2*c^
2*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x-1/2/e*c/d/(b*e-c*d)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(
d/e+x)-d*(b*e-c*d)/e^2)^(1/2)-1/4/e*c^(1/2)/d/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(c*(d/e+x)^2+
(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*b+1/2/e^2*c^(3/2)/(b*e-c*d)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(
1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))-1/2/e^2*c/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*l
n((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b
*e-c*d)/e^2)^(1/2))/(d/e+x))*b+1/2/e^3*c^2*d/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*
c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+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((c*x^2+b*x)^(1/2)/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.27907, size = 977, normalized size = 7.69 \begin{align*} \left [-\frac{{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \sqrt{c d^{2} - b d e} \log \left (\frac{b d +{\left (2 \, c d - b e\right )} x + 2 \, \sqrt{c d^{2} - b d e} \sqrt{c x^{2} + b x}}{e x + d}\right ) - 2 \,{\left (b c d^{3} - b^{2} d^{2} e +{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{8 \,{\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} +{\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}}, -\frac{{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \sqrt{-c d^{2} + b d e} \arctan \left (-\frac{\sqrt{-c d^{2} + b d e} \sqrt{c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) -{\left (b c d^{3} - b^{2} d^{2} e +{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} x\right )} \sqrt{c x^{2} + b x}}{4 \,{\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} +{\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

[-1/8*((b^2*e^2*x^2 + 2*b^2*d*e*x + b^2*d^2)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b
*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(b*c*d^3 - b^2*d^2*e + (2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2)*x)*sqrt(c
*x^2 + b*x))/(c^2*d^6 - 2*b*c*d^5*e + b^2*d^4*e^2 + (c^2*d^4*e^2 - 2*b*c*d^3*e^3 + b^2*d^2*e^4)*x^2 + 2*(c^2*d
^5*e - 2*b*c*d^4*e^2 + b^2*d^3*e^3)*x), -1/4*((b^2*e^2*x^2 + 2*b^2*d*e*x + b^2*d^2)*sqrt(-c*d^2 + b*d*e)*arcta
n(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) - (b*c*d^3 - b^2*d^2*e + (2*c^2*d^3 - 3*b*c*d^2*e +
b^2*d*e^2)*x)*sqrt(c*x^2 + b*x))/(c^2*d^6 - 2*b*c*d^5*e + b^2*d^4*e^2 + (c^2*d^4*e^2 - 2*b*c*d^3*e^3 + b^2*d^
2*e^4)*x^2 + 2*(c^2*d^5*e - 2*b*c*d^4*e^2 + b^2*d^3*e^3)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )}}{\left (d + e x\right )^{3}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(1/2)/(e*x+d)**3,x)

[Out]

Integral(sqrt(x*(b + c*x))/(d + e*x)**3, x)

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Giac [B]  time = 1.40329, size = 552, normalized size = 4.35 \begin{align*} -\frac{b^{2} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} + b d e}}\right )}{4 \,{\left (c d^{2} - b d e\right )} \sqrt{-c d^{2} + b d e}} + \frac{8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} c^{2} d^{2} e + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} c^{\frac{5}{2}} d^{3} + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b c^{2} d^{3} - 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} b c d e^{2} - 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{2} c d^{2} e + 2 \, b^{2} c^{\frac{3}{2}} d^{3} - 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b^{2} \sqrt{c} d e^{2} - b^{3} \sqrt{c} d^{2} e +{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} b^{2} e^{3} -{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{3} d e^{2}}{4 \,{\left (c d^{2} e^{2} - b d e^{3}\right )}{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} d + b d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

-1/4*b^2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c*d^2 - b*d*e)*sqrt(-
c*d^2 + b*d*e)) + 1/4*(8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^2*d^2*e + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*c^(
5/2)*d^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b*c^2*d^3 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c*d*e^2 - 4*(sq
rt(c)*x - sqrt(c*x^2 + b*x))*b^2*c*d^2*e + 2*b^2*c^(3/2)*d^3 - 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^2*sqrt(c)
*d*e^2 - b^3*sqrt(c)*d^2*e + (sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^2*e^3 - (sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*d
*e^2)/((c*d^2*e^2 - b*d*e^3)*((sqrt(c)*x - sqrt(c*x^2 + b*x))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c)*
d + b*d)^2)