3.1265 \(\int \frac{(A+B x) (d+e x)^{3/2}}{\sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=339 \[ -\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (5 A c e-4 b B e+3 B c d) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{15 c^{5/2} e \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (10 A c e (2 c d-b e)+B \left (8 b^2 e^2-13 b c d e+3 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{5/2} e \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (5 A c e-4 b B e+3 B c d)}{15 c^2}+\frac{2 B \sqrt{b x+c x^2} (d+e x)^{3/2}}{5 c} \]

[Out]

(2*(3*B*c*d - 4*b*B*e + 5*A*c*e)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(15*c^2) + (2*B*(d + e*x)^(3/2)*Sqrt[b*x + c
*x^2])/(5*c) + (2*Sqrt[-b]*(10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 - 13*b*c*d*e + 8*b^2*e^2))*Sqrt[x]*Sqrt[1 +
(c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*c^(5/2)*e*Sqrt[1 + (e*x
)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*d*(c*d - b*e)*(3*B*c*d - 4*b*B*e + 5*A*c*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sq
rt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*c^(5/2)*e*Sqrt[d + e*x]*Sqrt[b
*x + c*x^2])

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Rubi [A]  time = 0.477889, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {832, 843, 715, 112, 110, 117, 116} \[ \frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (10 A c e (2 c d-b e)+B \left (8 b^2 e^2-13 b c d e+3 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{5/2} e \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (5 A c e-4 b B e+3 B c d)}{15 c^2}-\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (5 A c e-4 b B e+3 B c d) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{5/2} e \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 B \sqrt{b x+c x^2} (d+e x)^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*B*c*d - 4*b*B*e + 5*A*c*e)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(15*c^2) + (2*B*(d + e*x)^(3/2)*Sqrt[b*x + c
*x^2])/(5*c) + (2*Sqrt[-b]*(10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 - 13*b*c*d*e + 8*b^2*e^2))*Sqrt[x]*Sqrt[1 +
(c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*c^(5/2)*e*Sqrt[1 + (e*x
)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*d*(c*d - b*e)*(3*B*c*d - 4*b*B*e + 5*A*c*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sq
rt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*c^(5/2)*e*Sqrt[d + e*x]*Sqrt[b
*x + c*x^2])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 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) (d+e x)^{3/2}}{\sqrt{b x+c x^2}} \, dx &=\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{2 \int \frac{\sqrt{d+e x} \left (-\frac{1}{2} (b B-5 A c) d+\frac{1}{2} (3 B c d-4 b B e+5 A c e) x\right )}{\sqrt{b x+c x^2}} \, dx}{5 c}\\ &=\frac{2 (3 B c d-4 b B e+5 A c e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{4 \int \frac{-\frac{1}{4} d \left (6 b B c d-15 A c^2 d-4 b^2 B e+5 A b c e\right )+\frac{1}{4} \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2-13 b c d e+8 b^2 e^2\right )\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c^2}\\ &=\frac{2 (3 B c d-4 b B e+5 A c e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}-\frac{(d (c d-b e) (3 B c d-4 b B e+5 A c e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c^2 e}+\frac{\left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2-13 b c d e+8 b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{15 c^2 e}\\ &=\frac{2 (3 B c d-4 b B e+5 A c e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}-\frac{\left (d (c d-b e) (3 B c d-4 b B e+5 A c e) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{15 c^2 e \sqrt{b x+c x^2}}+\frac{\left (\left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2-13 b c d e+8 b^2 e^2\right )\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{15 c^2 e \sqrt{b x+c x^2}}\\ &=\frac{2 (3 B c d-4 b B e+5 A c e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{\left (\left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2-13 b c d e+8 b^2 e^2\right )\right ) \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}{15 c^2 e \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left (d (c d-b e) (3 B c d-4 b B e+5 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}{15 c^2 e \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=\frac{2 (3 B c d-4 b B e+5 A c e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c^2}+\frac{2 B (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 c}+\frac{2 \sqrt{-b} \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2-13 b c d e+8 b^2 e^2\right )\right ) \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 )}{15 c^{5/2} e \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{2 \sqrt{-b} d (c d-b e) (3 B c d-4 b B e+5 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 )}{15 c^{5/2} e \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.86511, size = 356, normalized size = 1.05 \[ \frac{2 \sqrt{x} \left (-\frac{i x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (b e-c d) \left (-b c (10 A e+9 B d)+15 A c^2 d+8 b^2 B e\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )}{b}+\frac{(b+c x) (d+e x) \left (10 A c e (2 c d-b e)+B \left (8 b^2 e^2-13 b c d e+3 c^2 d^2\right )\right )}{c e \sqrt{x}}+i x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (10 A c e (2 c d-b e)+B \left (8 b^2 e^2-13 b c d e+3 c^2 d^2\right )\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\sqrt{x} (b+c x) (d+e x) (5 A c e+B (-4 b e+6 c d+3 c e x))\right )}{15 c^2 \sqrt{x (b+c x)} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(((10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 - 13*b*c*d*e + 8*b^2*e^2))*(b + c*x)*(d + e*x))/(c*e*Sqrt[
x]) + Sqrt[x]*(b + c*x)*(d + e*x)*(5*A*c*e + B*(6*c*d - 4*b*e + 3*c*e*x)) + I*Sqrt[b/c]*(10*A*c*e*(2*c*d - b*e
) + B*(3*c^2*d^2 - 13*b*c*d*e + 8*b^2*e^2))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x*EllipticE[I*ArcSinh[Sqrt[b/c
]/Sqrt[x]], (c*d)/(b*e)] - (I*Sqrt[b/c]*(-(c*d) + b*e)*(15*A*c^2*d + 8*b^2*B*e - b*c*(9*B*d + 10*A*e))*Sqrt[1
+ b/(c*x)]*Sqrt[1 + d/(e*x)]*x*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])/b))/(15*c^2*Sqrt[x*(b + c
*x)]*Sqrt[d + e*x])

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Maple [B]  time = 0.02, size = 1144, normalized size = 3.4 \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)*(e*x+d)^(3/2)/(c*x^2+b*x)^(1/2),x)

[Out]

2/15*(e*x+d)^(1/2)*(x*(c*x+b))^(1/2)*(5*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Ellipt
icF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d*e^2-5*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*
(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^2*e+10*A*((c*x+b)/b)^(1/2)*(-(e*x+d)
*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*e^3-30*A*((c*x+b)/
b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^
2*d*e^2+20*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b
*e-c*d))^(1/2))*b*c^3*d^2*e-4*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+
b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d*e^2+7*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/
2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e-3*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^3-8*B*((c*x+b)/b)^(1/2)*(-(
e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*e^3+21*B*((c*x
+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^
3*c*d*e^2-16*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/
(b*e-c*d))^(1/2))*b^2*c^2*d^2*e+3*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((
c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^3+3*B*x^4*c^4*e^3+5*A*x^3*c^4*e^3-B*x^3*b*c^3*e^3+9*B*x^3*c^4*d
*e^2+5*A*x^2*b*c^3*e^3+5*A*x^2*c^4*d*e^2-4*B*x^2*b^2*c^2*e^3+5*B*x^2*b*c^3*d*e^2+6*B*x^2*c^4*d^2*e+5*A*x*b*c^3
*d*e^2-4*B*x*b^2*c^2*d*e^2+6*B*x*b*c^3*d^2*e)/e/x/(c*e*x^2+b*e*x+c*d*x+b*d)/c^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{\frac{3}{2}}}{\sqrt{c x^{2} + b x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x + d)^(3/2)/sqrt(c*x^2 + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{\frac{3}{2}}}{\sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**(3/2)/sqrt(x*(b + c*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{\frac{3}{2}}}{\sqrt{c x^{2} + b x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x + d)^(3/2)/sqrt(c*x^2 + b*x), x)