### 3.394 $$\int \frac{(b x+c x^2)^{3/2}}{\sqrt{d+e x}} \, dx$$

Optimal. Leaf size=360 $\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (-b^2 e^2-16 b c d e+16 c^2 d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{35 c^{3/2} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (b^2 e^2-3 c e x (2 c d-b e)-11 b c d e+8 c^2 d^2\right )}{35 c e^3}-\frac{4 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{35 c^{3/2} e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} \sqrt{d+e x}}{7 e}$

[Out]

(2*Sqrt[d + e*x]*(8*c^2*d^2 - 11*b*c*d*e + b^2*e^2 - 3*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(35*c*e^3) + (2
*Sqrt[d + e*x]*(b*x + c*x^2)^(3/2))/(7*e) - (4*Sqrt[-b]*(2*c*d - b*e)*(4*c^2*d^2 - 4*b*c*d*e - 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)])/(35*c^(3/2)*e^4*
Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*Sqrt[-b]*d*(c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e - b^2*e^2)*Sqrt[x]*S
qrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(35*c^(3/2)*e^4
*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.352649, antiderivative size = 360, 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 = {734, 814, 843, 715, 112, 110, 117, 116} $\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (b^2 e^2-3 c e x (2 c d-b e)-11 b c d e+8 c^2 d^2\right )}{35 c e^3}+\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (-b^2 e^2-16 b c d e+16 c^2 d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{35 c^{3/2} e^4 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{4 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{35 c^{3/2} e^4 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} \sqrt{d+e x}}{7 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(8*c^2*d^2 - 11*b*c*d*e + b^2*e^2 - 3*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(35*c*e^3) + (2
*Sqrt[d + e*x]*(b*x + c*x^2)^(3/2))/(7*e) - (4*Sqrt[-b]*(2*c*d - b*e)*(4*c^2*d^2 - 4*b*c*d*e - 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)])/(35*c^(3/2)*e^4*
Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*Sqrt[-b]*d*(c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e - b^2*e^2)*Sqrt[x]*S
qrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(35*c^(3/2)*e^4
*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 734

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 + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, 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] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[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}}{\sqrt{d+e x}} \, dx &=\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}-\frac{3 \int \frac{(b d+(2 c d-b e) x) \sqrt{b x+c x^2}}{\sqrt{d+e x}} \, dx}{7 e}\\ &=\frac{2 \sqrt{d+e x} \left (8 c^2 d^2-11 b c d e+b^2 e^2-3 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{35 c e^3}+\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}+\frac{2 \int \frac{-\frac{1}{2} b d \left (8 c^2 d^2-11 b c d e+b^2 e^2\right )-(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{35 c e^3}\\ &=\frac{2 \sqrt{d+e x} \left (8 c^2 d^2-11 b c d e+b^2 e^2-3 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{35 c e^3}+\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}+\frac{\left (d (c d-b e) \left (16 c^2 d^2-16 b c d e-b^2 e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{35 c e^4}-\frac{\left (2 (2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{35 c e^4}\\ &=\frac{2 \sqrt{d+e x} \left (8 c^2 d^2-11 b c d e+b^2 e^2-3 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{35 c e^3}+\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}+\frac{\left (d (c d-b e) \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{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{35 c e^4 \sqrt{b x+c x^2}}-\frac{\left (2 (2 c d-b e) \left (4 c^2 d^2-4 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}{35 c e^4 \sqrt{b x+c x^2}}\\ &=\frac{2 \sqrt{d+e x} \left (8 c^2 d^2-11 b c d e+b^2 e^2-3 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{35 c e^3}+\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}-\frac{\left (2 (2 c d-b e) \left (4 c^2 d^2-4 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}{35 c e^4 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (d (c d-b e) \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{1+\frac{e x}{d}}\right ) \int \frac{1}{\sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}} \, dx}{35 c e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=\frac{2 \sqrt{d+e x} \left (8 c^2 d^2-11 b c d e+b^2 e^2-3 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{35 c e^3}+\frac{2 \sqrt{d+e x} \left (b x+c x^2\right )^{3/2}}{7 e}-\frac{4 \sqrt{-b} (2 c d-b e) \left (4 c^2 d^2-4 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 )}{35 c^{3/2} e^4 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{2 \sqrt{-b} d (c d-b e) \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{1+\frac{e x}{d}} F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{35 c^{3/2} e^4 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.02002, size = 380, normalized size = 1.06 $\frac{2 (x (b+c x))^{3/2} \left (b e x (b+c x) (d+e x) \left (b^2 e^2+b c e (8 e x-11 d)+c^2 \left (8 d^2-6 d e x+5 e^2 x^2\right )\right )+\sqrt{\frac{b}{c}} \left (i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (3 b^2 c d e^2+2 b^3 e^3-13 b c^2 d^2 e+8 c^3 d^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-2 i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (2 b^2 c d e^2+b^3 e^3-12 b c^2 d^2 e+8 c^3 d^3\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )-2 \sqrt{\frac{b}{c}} (b+c x) (d+e x) \left (2 b^2 c d e^2+b^3 e^3-12 b c^2 d^2 e+8 c^3 d^3\right )\right )\right )}{35 b c e^4 x^2 (b+c x)^2 \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(b*e*x*(b + c*x)*(d + e*x)*(b^2*e^2 + b*c*e*(-11*d + 8*e*x) + c^2*(8*d^2 - 6*d*e*x + 5*
e^2*x^2)) + Sqrt[b/c]*(-2*Sqrt[b/c]*(8*c^3*d^3 - 12*b*c^2*d^2*e + 2*b^2*c*d*e^2 + b^3*e^3)*(b + c*x)*(d + e*x)
- (2*I)*b*e*(8*c^3*d^3 - 12*b*c^2*d^2*e + 2*b^2*c*d*e^2 + b^3*e^3)*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*(8*c^3*d^3 - 13*b*c^2*d^2*e + 3*b^2*c*d*e^2 + 2
*b^3*e^3)*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)])))/
(35*b*c*e^4*x^2*(b + c*x)^2*Sqrt[d + e*x])

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Maple [B]  time = 0.271, size = 918, normalized size = 2.6 \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)^(3/2)/(e*x+d)^(1/2),x)

[Out]

2/35*(x*(c*x+b))^(1/2)*(e*x+d)^(1/2)*(5*x^5*c^5*e^4+2*((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^5*e^4+2*((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*c*d*e^3-28*((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^2*d^2*e^2+40*((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^3*d^3*e-16*((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^4*d^4+((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^4*c*d*e^3+15*((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^2*d^2*e^2-32*((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^3*d^3*e+16*((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^4*d^4+13*x^4*
b*c^4*e^4-x^4*c^5*d*e^3+9*x^3*b^2*c^3*e^4-4*x^3*b*c^4*d*e^3+2*x^3*c^5*d^2*e^2+x^2*b^3*c^2*e^4-2*x^2*b^2*c^3*d*
e^3-9*x^2*b*c^4*d^2*e^2+8*x^2*c^5*d^3*e+x*b^3*c^2*d*e^3-11*x*b^2*c^3*d^2*e^2+8*x*b*c^4*d^3*e)/c^3/e^4/x/(c*e*x
^2+b*e*x+c*d*x+b*d)

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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}}}{\sqrt{e x + d}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((c*x^2 + b*x)^(3/2)/sqrt(e*x + d), 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}}}{\sqrt{d + e x}}\, 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)**(1/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)/sqrt(d + e*x), 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}}}{\sqrt{e x + d}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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