3.1169 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=200 \[ \frac{b^2 (A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{16 d^{5/2} (c d-b e)^{5/2}}-\frac{\sqrt{b x+c x^2} (x (2 c d-b e)+b d) (A b e-2 A c d+b B d)}{8 d^2 (d+e x)^2 (c d-b e)^2}+\frac{\left (b x+c x^2\right )^{3/2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)} \]

[Out]

-((b*B*d - 2*A*c*d + A*b*e)*(b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(8*d^2*(c*d - b*e)^2*(d + e*x)^2) + ((B
*d - A*e)*(b*x + c*x^2)^(3/2))/(3*d*(c*d - b*e)*(d + e*x)^3) + (b^2*(b*B*d - 2*A*c*d + A*b*e)*ArcTanh[(b*d + (
2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(16*d^(5/2)*(c*d - b*e)^(5/2))

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Rubi [A]  time = 0.180779, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {806, 720, 724, 206} \[ \frac{b^2 (A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{16 d^{5/2} (c d-b e)^{5/2}}-\frac{\sqrt{b x+c x^2} (x (2 c d-b e)+b d) (A b e-2 A c d+b B d)}{8 d^2 (d+e x)^2 (c d-b e)^2}+\frac{\left (b x+c x^2\right )^{3/2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/(d + e*x)^4,x]

[Out]

-((b*B*d - 2*A*c*d + A*b*e)*(b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(8*d^2*(c*d - b*e)^2*(d + e*x)^2) + ((B
*d - A*e)*(b*x + c*x^2)^(3/2))/(3*d*(c*d - b*e)*(d + e*x)^3) + (b^2*(b*B*d - 2*A*c*d + A*b*e)*ArcTanh[(b*d + (
2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(16*d^(5/2)*(c*d - b*e)^(5/2))

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 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]

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{(A+B x) \sqrt{b x+c x^2}}{(d+e x)^4} \, dx &=\frac{(B d-A e) \left (b x+c x^2\right )^{3/2}}{3 d (c d-b e) (d+e x)^3}-\frac{(b B d-2 A c d+A b e) \int \frac{\sqrt{b x+c x^2}}{(d+e x)^3} \, dx}{2 d (c d-b e)}\\ &=-\frac{(b B d-2 A c d+A b e) (b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{8 d^2 (c d-b e)^2 (d+e x)^2}+\frac{(B d-A e) \left (b x+c x^2\right )^{3/2}}{3 d (c d-b e) (d+e x)^3}+\frac{\left (b^2 (b B d-2 A c d+A b e)\right ) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{16 d^2 (c d-b e)^2}\\ &=-\frac{(b B d-2 A c d+A b e) (b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{8 d^2 (c d-b e)^2 (d+e x)^2}+\frac{(B d-A e) \left (b x+c x^2\right )^{3/2}}{3 d (c d-b e) (d+e x)^3}-\frac{\left (b^2 (b B d-2 A c d+A b e)\right ) \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 )}{8 d^2 (c d-b e)^2}\\ &=-\frac{(b B d-2 A c d+A b e) (b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{8 d^2 (c d-b e)^2 (d+e x)^2}+\frac{(B d-A e) \left (b x+c x^2\right )^{3/2}}{3 d (c d-b e) (d+e x)^3}+\frac{b^2 (b B d-2 A c d+A 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 )}{16 d^{5/2} (c d-b e)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.489272, size = 199, normalized size = 1. \[ \frac{\sqrt{x (b+c x)} \left (\frac{3 (d+e x) (A b e-2 A c d+b B d) \left (b^2 (d+e x)^2 \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )+\sqrt{d} \sqrt{x} \sqrt{b+c x} \sqrt{b e-c d} (-b d+b e x-2 c d x)\right )}{d^{3/2} \sqrt{b+c x} (b e-c d)^{3/2}}+8 x^{3/2} (b+c x) (A e-B d)\right )}{24 d \sqrt{x} (d+e x)^3 (b e-c d)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/(d + e*x)^4,x]

[Out]

(Sqrt[x*(b + c*x)]*(8*(-(B*d) + A*e)*x^(3/2)*(b + c*x) + (3*(b*B*d - 2*A*c*d + A*b*e)*(d + e*x)*(Sqrt[d]*Sqrt[
-(c*d) + b*e]*Sqrt[x]*Sqrt[b + c*x]*(-(b*d) - 2*c*d*x + b*e*x) + b^2*(d + e*x)^2*ArcTan[(Sqrt[-(c*d) + b*e]*Sq
rt[x])/(Sqrt[d]*Sqrt[b + c*x])]))/(d^(3/2)*(-(c*d) + b*e)^(3/2)*Sqrt[b + c*x])))/(24*d*(-(c*d) + b*e)*Sqrt[x]*
(d + e*x)^3)

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Maple [B]  time = 0.014, size = 6956, normalized size = 34.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/(e*x+d)^4,x)

[Out]

result too large to display

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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((B*x+A)*(c*x^2+b*x)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.15934, size = 2462, normalized size = 12.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

[1/48*(3*(A*b^3*d^3*e + (B*b^3 - 2*A*b^2*c)*d^4 + (A*b^3*e^4 + (B*b^3 - 2*A*b^2*c)*d*e^3)*x^3 + 3*(A*b^3*d*e^3
+ (B*b^3 - 2*A*b^2*c)*d^2*e^2)*x^2 + 3*(A*b^3*d^2*e^2 + (B*b^3 - 2*A*b^2*c)*d^3*e)*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*(3*A*b^3*d^3*e^2 - 3*(B*b^2*
c - 2*A*b*c^2)*d^5 + 3*(B*b^3 - 3*A*b^2*c)*d^4*e + (8*B*c^3*d^5 - 3*A*b^3*d*e^4 - 2*(11*B*b*c^2 - 2*A*c^3)*d^4
*e + (17*B*b^2*c - 8*A*b*c^2)*d^3*e^2 - (3*B*b^3 - 7*A*b^2*c)*d^2*e^3)*x^2 - 2*(4*A*b^3*d^2*e^3 - (B*b*c^2 + 6
*A*c^3)*d^5 + (5*B*b^2*c + 13*A*b*c^2)*d^4*e - (4*B*b^3 + 11*A*b^2*c)*d^3*e^2)*x)*sqrt(c*x^2 + b*x))/(c^3*d^9
- 3*b*c^2*d^8*e + 3*b^2*c*d^7*e^2 - b^3*d^6*e^3 + (c^3*d^6*e^3 - 3*b*c^2*d^5*e^4 + 3*b^2*c*d^4*e^5 - b^3*d^3*e
^6)*x^3 + 3*(c^3*d^7*e^2 - 3*b*c^2*d^6*e^3 + 3*b^2*c*d^5*e^4 - b^3*d^4*e^5)*x^2 + 3*(c^3*d^8*e - 3*b*c^2*d^7*e
^2 + 3*b^2*c*d^6*e^3 - b^3*d^5*e^4)*x), 1/24*(3*(A*b^3*d^3*e + (B*b^3 - 2*A*b^2*c)*d^4 + (A*b^3*e^4 + (B*b^3 -
2*A*b^2*c)*d*e^3)*x^3 + 3*(A*b^3*d*e^3 + (B*b^3 - 2*A*b^2*c)*d^2*e^2)*x^2 + 3*(A*b^3*d^2*e^2 + (B*b^3 - 2*A*b
^2*c)*d^3*e)*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)) + (3*A*b^
3*d^3*e^2 - 3*(B*b^2*c - 2*A*b*c^2)*d^5 + 3*(B*b^3 - 3*A*b^2*c)*d^4*e + (8*B*c^3*d^5 - 3*A*b^3*d*e^4 - 2*(11*B
*b*c^2 - 2*A*c^3)*d^4*e + (17*B*b^2*c - 8*A*b*c^2)*d^3*e^2 - (3*B*b^3 - 7*A*b^2*c)*d^2*e^3)*x^2 - 2*(4*A*b^3*d
^2*e^3 - (B*b*c^2 + 6*A*c^3)*d^5 + (5*B*b^2*c + 13*A*b*c^2)*d^4*e - (4*B*b^3 + 11*A*b^2*c)*d^3*e^2)*x)*sqrt(c*
x^2 + b*x))/(c^3*d^9 - 3*b*c^2*d^8*e + 3*b^2*c*d^7*e^2 - b^3*d^6*e^3 + (c^3*d^6*e^3 - 3*b*c^2*d^5*e^4 + 3*b^2*
c*d^4*e^5 - b^3*d^3*e^6)*x^3 + 3*(c^3*d^7*e^2 - 3*b*c^2*d^6*e^3 + 3*b^2*c*d^5*e^4 - b^3*d^4*e^5)*x^2 + 3*(c^3*
d^8*e - 3*b*c^2*d^7*e^2 + 3*b^2*c*d^6*e^3 - b^3*d^5*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 (A + B x\right )}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/(d + e*x)**4, x)

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Giac [B]  time = 1.43199, size = 2099, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

1/8*(B*b^3*d - 2*A*b^2*c*d + A*b^3*e)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*
d*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*sqrt(-c*d^2 + b*d*e)) + 1/24*(96*(sqrt(c)*x - sqrt(c*x^2 + b*x))^
4*B*c^(7/2)*d^5*e + 64*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*c^4*d^6 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*c^
3*d^4*e^2 - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b*c^3*d^5*e + 32*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^4*d^
5*e + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b*c^(7/2)*d^6 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b*c^(5/2)
*d^4*e^2 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^(7/2)*d^4*e^2 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^
2*c^(5/2)*d^5*e + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c^(7/2)*d^5*e + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))*
B*b^2*c^3*d^6 - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b*c^2*d^3*e^3 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B
*b^2*c^2*d^4*e^2 + 16*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c^3*d^4*e^2 - 84*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B
*b^3*c^2*d^5*e + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*c^3*d^5*e + 8*B*b^3*c^(5/2)*d^6 - 96*(sqrt(c)*x - sq
rt(c*x^2 + b*x))^4*A*b*c^(5/2)*d^3*e^3 - 6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*c^(3/2)*d^4*e^2 - 36*(sqrt(
c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(5/2)*d^4*e^2 - 14*B*b^4*c^(3/2)*d^5*e + 4*A*b^3*c^(5/2)*d^5*e + 48*(sqrt(
c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c*d^2*e^4 + 58*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*c*d^3*e^3 - 84*(sqrt(
c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*c^2*d^3*e^3 + 18*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b^4*c*d^4*e^2 - 24*(sqrt(
c)*x - sqrt(c*x^2 + b*x))*A*b^3*c^2*d^4*e^2 + 33*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^3*sqrt(c)*d^2*e^4 + 78*
(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^(3/2)*d^2*e^4 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^4*sqrt(c)*d
^3*e^3 - 6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^3*c^(3/2)*d^3*e^3 + 3*B*b^5*sqrt(c)*d^4*e^2 - 4*A*b^4*c^(3/2)
*d^4*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^3*d*e^5 + 6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^2*c*d*e^5
+ 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*d^2*e^4 + 74*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^3*c*d^2*e^4 +
3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b^5*d^3*e^3 + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^4*c*d^3*e^3 - 15*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^4*A*b^3*sqrt(c)*d*e^5 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^4*sqrt(c)*d^2*e^4
+ 3*A*b^5*sqrt(c)*d^3*e^3 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^3*e^6 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))
^3*A*b^4*d*e^5 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^5*d^2*e^4)/((c^2*d^4*e^3 - 2*b*c*d^3*e^4 + b^2*d^2*e^5)
*((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)^3)