### 3.290 $$\int \frac{\sqrt{b x+c x^2}}{(d+e x)^2} \, dx$$

Optimal. Leaf size=140 $-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 \sqrt{d} e^2 \sqrt{c d-b e}}-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{e^2}$

[Out]

-(Sqrt[b*x + c*x^2]/(e*(d + e*x))) + (2*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/e^2 - ((2*c*d - b*e)*A
rcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*Sqrt[d]*e^2*Sqrt[c*d - b*e])

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Rubi [A]  time = 0.110819, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {732, 843, 620, 206, 724} $-\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 \sqrt{d} e^2 \sqrt{c d-b e}}-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{e^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x)^2,x]

[Out]

-(Sqrt[b*x + c*x^2]/(e*(d + e*x))) + (2*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/e^2 - ((2*c*d - b*e)*A
rcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*Sqrt[d]*e^2*Sqrt[c*d - b*e])

Rule 732

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 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)^2} \, dx &=-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{\int \frac{b+2 c x}{(d+e x) \sqrt{b x+c x^2}} \, dx}{2 e}\\ &=-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{c \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{e^2}-\frac{(2 c d-b e) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{2 e^2}\\ &=-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{e^2}+\frac{(2 c d-b e) \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 )}{e^2}\\ &=-\frac{\sqrt{b x+c x^2}}{e (d+e x)}+\frac{2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{e^2}-\frac{(2 c d-b e) \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 )}{2 \sqrt{d} e^2 \sqrt{c d-b e}}\\ \end{align*}

Mathematica [A]  time = 0.601086, size = 147, normalized size = 1.05 $\frac{\sqrt{x (b+c x)} \left (-\frac{(2 c d-b e) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{d} \sqrt{b+c x} \sqrt{b e-c d}}+\frac{2 \sqrt{c} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{\frac{c x}{b}+1}}-\frac{e \sqrt{x}}{d+e x}\right )}{e^2 \sqrt{x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(-((e*Sqrt[x])/(d + e*x)) + (2*Sqrt[c]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[1
+ (c*x)/b]) - ((2*c*d - b*e)*ArcTan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(Sqrt[d]*Sqrt[-(c*d
) + b*e]*Sqrt[b + c*x])))/(e^2*Sqrt[x])

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Maple [B]  time = 0.213, size = 885, normalized size = 6.3 \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)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59493, size = 1775, normalized size = 12.68 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - (2*c*
d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*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*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d
*e^4)*x), -((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x
^2 + b*x)/((c*d - b*e)*x)) - (c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 +
b*x)*sqrt(c)) + (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d*e^4)*x), -1/2
*(4*(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*c*d^2 - b
*d*e + (2*c*d*e - b*e^2)*x)*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*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d*e^4)*x
), -((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*
x)/((c*d - b*e)*x)) + 2*(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(
c*x)) + (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d*e^4)*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 )^{2}}\, 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)**2,x)

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError