### 3.395 $$\int \frac{(b x+c x^2)^{3/2}}{(d+e x)^{3/2}} \, dx$$

Optimal. Leaf size=309 $-\frac{16 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{5 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (-7 b e+8 c d-6 c e x)}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}$

[Out]

(-2*Sqrt[d + e*x]*(8*c*d - 7*b*e - 6*c*e*x)*Sqrt[b*x + c*x^2])/(5*e^3) - (2*(b*x + c*x^2)^(3/2))/(e*Sqrt[d + e
*x]) + (2*Sqrt[-b]*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSi
n[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(5*Sqrt[c]*e^4*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (16*Sqrt[-b
]*d*(c*d - b*e)*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/S
qrt[-b]], (b*e)/(c*d)])/(5*Sqrt[c]*e^4*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.338442, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.348, Rules used = {732, 814, 843, 715, 112, 110, 117, 116} $\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (-7 b e+8 c d-6 c e x)}{5 e^3}-\frac{16 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{5 \sqrt{c} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(8*c*d - 7*b*e - 6*c*e*x)*Sqrt[b*x + c*x^2])/(5*e^3) - (2*(b*x + c*x^2)^(3/2))/(e*Sqrt[d + e
*x]) + (2*Sqrt[-b]*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSi
n[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(5*Sqrt[c]*e^4*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (16*Sqrt[-b
]*d*(c*d - b*e)*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/S
qrt[-b]], (b*e)/(c*d)])/(5*Sqrt[c]*e^4*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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 814

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

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{\left (b x+c x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}+\frac{3 \int \frac{(b+2 c x) \sqrt{b x+c x^2}}{\sqrt{d+e x}} \, dx}{e}\\ &=-\frac{2 \sqrt{d+e x} (8 c d-7 b e-6 c e x) \sqrt{b x+c x^2}}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}-\frac{2 \int \frac{-\frac{1}{2} b c d (8 c d-7 b e)-\frac{1}{2} c \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{5 c e^3}\\ &=-\frac{2 \sqrt{d+e x} (8 c d-7 b e-6 c e x) \sqrt{b x+c x^2}}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}-\frac{(8 d (c d-b e) (2 c d-b e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{5 e^4}+\frac{\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{5 e^4}\\ &=-\frac{2 \sqrt{d+e x} (8 c d-7 b e-6 c e x) \sqrt{b x+c x^2}}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}-\frac{\left (8 d (c d-b e) (2 c d-b e) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{5 e^4 \sqrt{b x+c x^2}}+\frac{\left (\left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{5 e^4 \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{d+e x} (8 c d-7 b e-6 c e x) \sqrt{b x+c x^2}}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}+\frac{\left (\left (16 c^2 d^2-16 b c d e+b^2 e^2\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}{5 e^4 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left (8 d (c d-b e) (2 c d-b 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}{5 e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{d+e x} (8 c d-7 b e-6 c e x) \sqrt{b x+c x^2}}{5 e^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{e \sqrt{d+e x}}+\frac{2 \sqrt{-b} \left (16 c^2 d^2-16 b c d e+b^2 e^2\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 )}{5 \sqrt{c} e^4 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{16 \sqrt{-b} d (c d-b e) (2 c d-b 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 )}{5 \sqrt{c} e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.68754, size = 340, normalized size = 1.1 $\frac{-2 i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-9 b c d e+8 c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+2 \left (b^2 c e \left (-16 d^2-8 d e x+3 e^2 x^2\right )+b^3 e^2 (d+e x)+b c^2 \left (-8 d^2 e x+16 d^3-11 d e^2 x^2+3 e^3 x^3\right )+c^3 x \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )\right )+2 i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )}{5 c e^4 \sqrt{x (b+c x)} \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(b^3*e^2*(d + e*x) + b^2*c*e*(-16*d^2 - 8*d*e*x + 3*e^2*x^2) + c^3*x*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^
3*x^3) + b*c^2*(16*d^3 - 8*d^2*e*x - 11*d*e^2*x^2 + 3*e^3*x^3)) + (2*I)*Sqrt[b/c]*c*e*(16*c^2*d^2 - 16*b*c*d*e
+ b^2*e^2)*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)] -
(2*I)*Sqrt[b/c]*c*e*(8*c^2*d^2 - 9*b*c*d*e + b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I
*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])/(5*c*e^4*Sqrt[x*(b + c*x)]*Sqrt[d + e*x])

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Maple [B]  time = 0.283, size = 685, normalized size = 2.2 \begin{align*} -{\frac{2}{5\,{c}^{2}x \left ( ce{x}^{2}+bxe+cdx+bd \right ){e}^{4}}\sqrt{x \left ( cx+b \right ) }\sqrt{ex+d} \left ( 8\,\sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{3}cd{e}^{2}-24\,\sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{2}{c}^{2}{d}^{2}e+16\,\sqrt{-{\frac{cx}{b}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}b{c}^{3}{d}^{3}+\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{4}{e}^{3}-17\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{3}cd{e}^{2}+32\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}{b}^{2}{c}^{2}{d}^{2}e-16\,\sqrt{-{\frac{cx}{b}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+b}{b}}},\sqrt{{\frac{be}{be-cd}}} \right ) \sqrt{{\frac{cx+b}{b}}}\sqrt{-{\frac{c \left ( ex+d \right ) }{be-cd}}}b{c}^{3}{d}^{3}-{x}^{4}{c}^{4}{e}^{3}-3\,{x}^{3}b{c}^{3}{e}^{3}+2\,{x}^{3}{c}^{4}d{e}^{2}-2\,{x}^{2}{b}^{2}{c}^{2}{e}^{3}-5\,{x}^{2}b{c}^{3}d{e}^{2}+8\,{x}^{2}{c}^{4}{d}^{2}e-7\,x{b}^{2}{c}^{2}d{e}^{2}+8\,xb{c}^{3}{d}^{2}e \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(x*(c*x+b))^(1/2)*(e*x+d)^(1/2)*(8*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*
x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^3*c*d*e^2-24*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e
-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^2*c^2*d^2*e+16*(-c*x/b)^(1/2)*EllipticF(((c*x+b
)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b*c^3*d^3+(-c*x/b)^(1/2)*Elli
pticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^4*e^3-17*(-c*x
/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b^
3*c*d*e^2+32*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(
b*e-c*d))^(1/2)*b^2*c^2*d^2*e-16*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*((c*x+b)/b)
^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*b*c^3*d^3-x^4*c^4*e^3-3*x^3*b*c^3*e^3+2*x^3*c^4*d*e^2-2*x^2*b^2*c^2*e^3-5*
x^2*b*c^3*d*e^2+8*x^2*c^4*d^2*e-7*x*b^2*c^2*d*e^2+8*x*b*c^3*d^2*e)/c^2/x/(c*e*x^2+b*e*x+c*d*x+b*d)/e^4

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x)^(3/2)*sqrt(e*x + d)/(e^2*x^2 + 2*d*e*x + d^2), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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