### 3.387 $$\int \sqrt{d+e x} \sqrt{b x+c x^2} \, dx$$

Optimal. Leaf size=308 $\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{15 c^{3/2} e^2 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{4 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^2 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{b x+c x^2} (d+e x)^{3/2}}{5 e}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (2 c d-b e)}{15 c e}$

[Out]

(-2*(2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(15*c*e) + (2*(d + e*x)^(3/2)*Sqrt[b*x + c*x^2])/(5*e) - (4
*Sqrt[-b]*(c^2*d^2 - 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)])/(15*c^(3/2)*e^2*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*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])/Sqrt[-b]], (b*e)
/(c*d)])/(15*c^(3/2)*e^2*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.341274, antiderivative size = 308, 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, 832, 843, 715, 112, 110, 117, 116} $-\frac{4 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^2 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (2 c d-b e) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^2 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{b x+c x^2} (d+e x)^{3/2}}{5 e}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (2 c d-b e)}{15 c e}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[d + e*x]*Sqrt[b*x + c*x^2],x]

[Out]

(-2*(2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(15*c*e) + (2*(d + e*x)^(3/2)*Sqrt[b*x + c*x^2])/(5*e) - (4
*Sqrt[-b]*(c^2*d^2 - 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)])/(15*c^(3/2)*e^2*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (2*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])/Sqrt[-b]], (b*e)
/(c*d)])/(15*c^(3/2)*e^2*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 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 \sqrt{d+e x} \sqrt{b x+c x^2} \, dx &=\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}-\frac{\int \frac{\sqrt{d+e x} (b d+(2 c d-b e) x)}{\sqrt{b x+c x^2}} \, dx}{5 e}\\ &=-\frac{2 (2 c d-b e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c e}+\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}-\frac{2 \int \frac{\frac{1}{2} b d (c d+b e)+\left (c^2 d^2-b c d e+b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c e}\\ &=-\frac{2 (2 c d-b e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c e}+\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}+\frac{(d (c d-b e) (2 c d-b e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c e^2}-\frac{\left (2 \left (c^2 d^2-b c d e+b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{15 c e^2}\\ &=-\frac{2 (2 c d-b e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c e}+\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}+\frac{\left (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}{15 c e^2 \sqrt{b x+c x^2}}-\frac{\left (2 \left (c^2 d^2-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}{15 c e^2 \sqrt{b x+c x^2}}\\ &=-\frac{2 (2 c d-b e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c e}+\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}-\frac{\left (2 \left (c^2 d^2-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}{15 c e^2 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (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}{15 c e^2 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 (2 c d-b e) \sqrt{d+e x} \sqrt{b x+c x^2}}{15 c e}+\frac{2 (d+e x)^{3/2} \sqrt{b x+c x^2}}{5 e}-\frac{4 \sqrt{-b} \left (c^2 d^2-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 )}{15 c^{3/2} e^2 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{2 \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 )}{15 c^{3/2} e^2 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.1264, size = 294, normalized size = 0.95 $\frac{2 \left (b e x (b+c x) (d+e x) (b e+c (d+3 e x))+\sqrt{\frac{b}{c}} \left (i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (2 b^2 e^2-3 b c d e+c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-2 i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (b^2 e^2-b c d e+c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )-2 \sqrt{\frac{b}{c}} (b+c x) (d+e x) \left (b^2 e^2-b c d e+c^2 d^2\right )\right )\right )}{15 b c e^2 \sqrt{x (b+c x)} \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[d + e*x]*Sqrt[b*x + c*x^2],x]

[Out]

(2*(b*e*x*(b + c*x)*(d + e*x)*(b*e + c*(d + 3*e*x)) + Sqrt[b/c]*(-2*Sqrt[b/c]*(c^2*d^2 - b*c*d*e + b^2*e^2)*(b
+ c*x)*(d + e*x) - (2*I)*b*e*(c^2*d^2 - b*c*d*e + b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*Ellipt
icE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] + I*b*e*(c^2*d^2 - 3*b*c*d*e + 2*b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqr
t[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])))/(15*b*c*e^2*Sqrt[x*(b + c*x)]*S
qrt[d + e*x])

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Maple [B]  time = 0.289, size = 681, normalized size = 2.2 \begin{align*}{\frac{2}{15\,x \left ( ce{x}^{2}+bxe+cdx+bd \right ){c}^{3}{e}^{2}}\sqrt{ex+d}\sqrt{x \left ( cx+b \right ) } \left ( \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}-3\,\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+2\,\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}+2\,\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}-4\,\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}+4\,\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-2\,\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}+3\,{x}^{4}{c}^{4}{e}^{3}+4\,{x}^{3}b{c}^{3}{e}^{3}+4\,{x}^{3}{c}^{4}d{e}^{2}+{x}^{2}{b}^{2}{c}^{2}{e}^{3}+5\,{x}^{2}b{c}^{3}d{e}^{2}+{x}^{2}{c}^{4}{d}^{2}e+x{b}^{2}{c}^{2}d{e}^{2}+xb{c}^{3}{d}^{2}e \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(e*x+d)^(1/2)*(x*(c*x+b))^(1/2)*((-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^2-3*(-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+2*(-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+2*(-c*x/b)^(1/2)*Ellipt
icE(((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^3-4*(-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+4*(-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-2*(-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+3*x^4*c^4*e^3+4*x^3*b*c^3*e^3+4*x^3*c^4*d*e^2+x^2*b^2*c^2*e^3+5*x^2*b
*c^3*d*e^2+x^2*c^4*d^2*e+x*b^2*c^2*d*e^2+x*b*c^3*d^2*e)/x/(c*e*x^2+b*e*x+c*d*x+b*d)/c^3/e^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + b x} \sqrt{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*sqrt(d + e*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + b x} \sqrt{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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