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]

-1/8*b^2*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/(c*x^2+b*x)^(1/2))/d^(3/2)/(-b*e+c*d)^(3/2)
+1/4*(b*d+(-b*e+2*c*d)*x)*(c*x^2+b*x)^(1/2)/d/(-b*e+c*d)/(e*x+d)^2

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Rubi [A]  time = 0.09, antiderivative size = 127, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.21, size = 122, normalized size = 0.96 \[ \frac {\sqrt {x (b+c x)} \left (\frac {\sqrt {d} (b (d-e x)+2 c d x)}{(d+e x)^2 (c d-b e)}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x} (c d-b e)^{3/2}}\right )}{4 d^{3/2}} \]

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*ArcTanh[(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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fricas [B]  time = 1.07, size = 479, normalized size = 3.77 \[ \left [-\frac {{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - 2 \, {\left (b c d^{3} - b^{2} d^{2} e + {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} + {\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}}, -\frac {{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) - {\left (b c d^{3} - b^{2} d^{2} e + {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} + {\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}}\right ] \]

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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giac [B]  time = 0.27, size = 409, normalized size = 3.22 \[ -\frac {b^{2} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{4 \, {\left (c d^{2} - b d e\right )} \sqrt {-c d^{2} + b d e}} + \frac {8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{2} d^{2} e + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} c^{\frac {5}{2}} d^{3} + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b c^{2} d^{3} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b c d e^{2} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} c d^{2} e + 2 \, b^{2} c^{\frac {3}{2}} d^{3} - 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{2} \sqrt {c} d e^{2} - b^{3} \sqrt {c} d^{2} e + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{2} e^{3} - {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{3} d e^{2}}{4 \, {\left (c d^{2} e^{2} - b d e^{3}\right )} {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} d + b d\right )}^{2}} \]

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)

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maple [B]  time = 0.07, size = 1963, normalized size = 15.46 \[ \text {result too large to display} \]

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

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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((c*x^2+b*x)^(1/2)/(e*x+d)^3,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(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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