3.1258 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=346 \[ -\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (-2 A c e-3 b B e+8 B c d) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{3 \sqrt{c} e^3 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{b x+c x^2} \left (d^2 (-A c e-3 b B e+4 B c d)+e x (B d (5 c d-4 b e)-A e (2 c d-b e))\right )}{3 d e^2 (d+e x)^{3/2} (c d-b e)}+\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (B d (8 c d-7 b e)-A e (2 c d-b e)) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 d e^3 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)} \]

[Out]

(-2*(d^2*(4*B*c*d - 3*b*B*e - A*c*e) + e*(B*d*(5*c*d - 4*b*e) - A*e*(2*c*d - b*e))*x)*Sqrt[b*x + c*x^2])/(3*d*
e^2*(c*d - b*e)*(d + e*x)^(3/2)) + (2*Sqrt[-b]*Sqrt[c]*(B*d*(8*c*d - 7*b*e) - A*e*(2*c*d - b*e))*Sqrt[x]*Sqrt[
1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*d*e^3*(c*d - b*e)*Sq
rt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*(8*B*c*d - 3*b*B*e - 2*A*c*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[
1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*Sqrt[c]*e^3*Sqrt[d + e*x]*Sqrt[b*x
+ c*x^2])

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Rubi [A]  time = 0.380008, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {810, 843, 715, 112, 110, 117, 116} \[ -\frac{2 \sqrt{b x+c x^2} \left (d^2 (-A c e-3 b B e+4 B c d)+e x (B d (5 c d-4 b e)-A e (2 c d-b e))\right )}{3 d e^2 (d+e x)^{3/2} (c d-b e)}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (-2 A c e-3 b B e+8 B c d) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 \sqrt{c} e^3 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (B d (8 c d-7 b e)-A e (2 c d-b e)) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 d e^3 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/(d + e*x)^(5/2),x]

[Out]

(-2*(d^2*(4*B*c*d - 3*b*B*e - A*c*e) + e*(B*d*(5*c*d - 4*b*e) - A*e*(2*c*d - b*e))*x)*Sqrt[b*x + c*x^2])/(3*d*
e^2*(c*d - b*e)*(d + e*x)^(3/2)) + (2*Sqrt[-b]*Sqrt[c]*(B*d*(8*c*d - 7*b*e) - A*e*(2*c*d - b*e))*Sqrt[x]*Sqrt[
1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*d*e^3*(c*d - b*e)*Sq
rt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*(8*B*c*d - 3*b*B*e - 2*A*c*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[
1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*Sqrt[c]*e^3*Sqrt[d + e*x]*Sqrt[b*x
+ c*x^2])

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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 715

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(Sqrt[x]*Sqrt[b + c*x])/Sqrt[
b*x + c*x^2], Int[(d + e*x)^m/(Sqrt[x]*Sqrt[b + c*x]), x], x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] &
& NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 112

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*x]*Sqrt[
1 + (d*x)/c])/(Sqrt[c + d*x]*Sqrt[1 + (f*x)/e]), Int[Sqrt[1 + (f*x)/e]/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]), x], x] /
; FreeQ[{b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 110

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Sqrt[e]*Rt[-(b/d)
, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/b, x] /; FreeQ[{b, c, d, e, f}, x] &&
NeQ[d*e - c*f, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !LtQ[-(b/d), 0]

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{b x+c x^2}}{(d+e x)^{5/2}} \, dx &=-\frac{2 \left (d^2 (4 B c d-3 b B e-A c e)+e (B d (5 c d-4 b e)-A e (2 c d-b e)) x\right ) \sqrt{b x+c x^2}}{3 d e^2 (c d-b e) (d+e x)^{3/2}}-\frac{2 \int \frac{-\frac{1}{2} b d (4 B c d-3 b B e-A c e)-\frac{1}{2} c (B d (8 c d-7 b e)-A e (2 c d-b e)) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 d e^2 (c d-b e)}\\ &=-\frac{2 \left (d^2 (4 B c d-3 b B e-A c e)+e (B d (5 c d-4 b e)-A e (2 c d-b e)) x\right ) \sqrt{b x+c x^2}}{3 d e^2 (c d-b e) (d+e x)^{3/2}}-\frac{(8 B c d-3 b B e-2 A c e) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 e^3}+\frac{(c (B d (8 c d-7 b e)-A e (2 c d-b e))) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{3 d e^3 (c d-b e)}\\ &=-\frac{2 \left (d^2 (4 B c d-3 b B e-A c e)+e (B d (5 c d-4 b e)-A e (2 c d-b e)) x\right ) \sqrt{b x+c x^2}}{3 d e^2 (c d-b e) (d+e x)^{3/2}}-\frac{\left ((8 B c d-3 b B e-2 A c e) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{3 e^3 \sqrt{b x+c x^2}}+\frac{\left (c (B d (8 c d-7 b e)-A e (2 c d-b e)) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{3 d e^3 (c d-b e) \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (d^2 (4 B c d-3 b B e-A c e)+e (B d (5 c d-4 b e)-A e (2 c d-b e)) x\right ) \sqrt{b x+c x^2}}{3 d e^2 (c d-b e) (d+e x)^{3/2}}+\frac{\left (c (B d (8 c d-7 b e)-A e (2 c d-b e)) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x}\right ) \int \frac{\sqrt{1+\frac{e x}{d}}}{\sqrt{x} \sqrt{1+\frac{c x}{b}}} \, dx}{3 d e^3 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left ((8 B c d-3 b B e-2 A c e) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}\right ) \int \frac{1}{\sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}} \, dx}{3 e^3 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (d^2 (4 B c d-3 b B e-A c e)+e (B d (5 c d-4 b e)-A e (2 c d-b e)) x\right ) \sqrt{b x+c x^2}}{3 d e^2 (c d-b e) (d+e x)^{3/2}}+\frac{2 \sqrt{-b} \sqrt{c} (B d (8 c d-7 b e)-A e (2 c d-b e)) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x} E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 d e^3 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{2 \sqrt{-b} (8 B c d-3 b B e-2 A c e) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}} F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 \sqrt{c} e^3 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.74913, size = 346, normalized size = 1. \[ \frac{2 \left (e x \sqrt{\frac{b}{c}} (b+c x) \left (A e \left (c d (d+2 e x)-b e^2 x\right )+B d (b e (3 d+4 e x)-c d (4 d+5 e x))\right )+(d+e x) \left (-i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (4 B d-A e) (c d-b e) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (A e (2 c d-b e)+B d (7 b e-8 c d)) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\sqrt{\frac{b}{c}} (b+c x) (d+e x) (A e (b e-2 c d)+B d (8 c d-7 b e))\right )\right )}{3 d e^3 \sqrt{\frac{b}{c}} \sqrt{x (b+c x)} (d+e x)^{3/2} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/(d + e*x)^(5/2),x]

[Out]

(2*(Sqrt[b/c]*e*x*(b + c*x)*(A*e*(-(b*e^2*x) + c*d*(d + 2*e*x)) + B*d*(b*e*(3*d + 4*e*x) - c*d*(4*d + 5*e*x)))
+ (d + e*x)*(Sqrt[b/c]*(B*d*(8*c*d - 7*b*e) + A*e*(-2*c*d + b*e))*(b + c*x)*(d + e*x) - I*b*e*(A*e*(2*c*d - b
*e) + B*d*(-8*c*d + 7*b*e))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]]
, (c*d)/(b*e)] - I*b*e*(4*B*d - A*e)*(c*d - b*e)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSi
nh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])))/(3*Sqrt[b/c]*d*e^3*(c*d - b*e)*Sqrt[x*(b + c*x)]*(d + e*x)^(3/2))

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Maple [B]  time = 0.053, size = 1959, normalized size = 5.7 \begin{align*} \text{result 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)^(5/2),x)

[Out]

2/3*(-7*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*
d))^(1/2)*(-c*x/b)^(1/2)+3*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*d*e^3*((c*x+b)/b)^(1/2)*
(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-3*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^2*e
^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+A*x^3*b*c^2*e^4-2*A*x^2*b*c^2*d*e^3-4*B*x^2*b
^2*c*d*e^3+2*B*x^2*b*c^2*d^2*e^2-A*x*b*c^2*d^2*e^2-3*B*x*b^2*c*d^2*e^2+4*B*x*b*c^2*d^3*e-4*B*x^3*b*c^2*d*e^3-2
*A*x^3*c^3*d*e^3+5*B*x^3*c^3*d^2*e^2+A*x^2*b^2*c*e^4-A*x^2*c^3*d^2*e^2+4*B*x^2*c^3*d^3*e+15*B*EllipticE(((c*x+
b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/
2)-8*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^2*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)*(-c*x/b)^(1/2)-11*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d^2*e^2*((c*x+b)/b)^(1
/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+8*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^2
*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c
*d))^(1/2))*x*b^3*e^4*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+A*EllipticE(((c*x+b)/b)^(1
/2),(b*e/(b*e-c*d))^(1/2))*b^3*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-7*B*Ellipti
cE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b
)^(1/2)-8*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^4*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*
d))^(1/2)*(-c*x/b)^(1/2)+3*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*d^2*e^2*((c*x+b)/b)^(1/2)*
(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+8*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^4*(
(c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-3*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(
1/2))*x*b^2*c*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*A*EllipticE(((c*x+b)/b)^(1
/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*A*E
llipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)
*(-c*x/b)^(1/2)-2*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*
x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^3*e*((c*
x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2
))*b^2*c*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-2*A*EllipticF(((c*x+b)/b)^(1/2)
,(b*e/(b*e-c*d))^(1/2))*b*c^2*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+15*B*Ellipti
cE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b
)^(1/2)-11*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e
-c*d))^(1/2)*(-c*x/b)^(1/2))*(x*(c*x+b))^(1/2)/c/(b*e-c*d)/d/(c*x+b)/x/e^3/(e*x+d)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \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)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)/(e*x + d)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x}{\left (B x + A\right )} \sqrt{e x + d}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x)*(B*x + A)*sqrt(e*x + d)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^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 (A + B x\right )}{\left (d + e x\right )^{\frac{5}{2}}}\, 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)**(5/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/(d + e*x)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \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)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)/(e*x + d)^(5/2), x)