3.1256 \(\int \frac{(A+B x) \sqrt{b x+c x^2}}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=318 \[ -\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (-10 A c e+b B e+8 B c d) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{15 c^{3/2} e^3 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (5 A c e (2 c d-b e)-B \left (-2 b^2 e^2-3 b c d e+8 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^3 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (-5 A c e-b B e+4 B c d-3 B c e x)}{15 c e^2} \]

[Out]

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

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Rubi [A]  time = 0.329367, antiderivative size = 318, 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 = {814, 843, 715, 112, 110, 117, 116} \[ -\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (5 A c e (2 c d-b e)-B \left (-2 b^2 e^2-3 b c d e+8 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^3 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}-\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) (-10 A c e+b B e+8 B c d) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 c^{3/2} e^3 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} (-5 A c e-b B e+4 B c d-3 B c e x)}{15 c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{(A+B x) \sqrt{b x+c x^2}}{\sqrt{d+e x}} \, dx &=-\frac{2 \sqrt{d+e x} (4 B c d-b B e-5 A c e-3 B c e x) \sqrt{b x+c x^2}}{15 c e^2}-\frac{2 \int \frac{-\frac{1}{2} b d (4 B c d-b B e-5 A c e)+\frac{1}{2} \left (5 A c e (2 c d-b e)-B \left (8 c^2 d^2-3 b c d e-2 b^2 e^2\right )\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c e^2}\\ &=-\frac{2 \sqrt{d+e x} (4 B c d-b B e-5 A c e-3 B c e x) \sqrt{b x+c x^2}}{15 c e^2}-\frac{(d (c d-b e) (8 B c d+b B e-10 A c e)) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 c e^3}-\frac{\left (5 A c e (2 c d-b e)-B \left (8 c^2 d^2-3 b c d e-2 b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{15 c e^3}\\ &=-\frac{2 \sqrt{d+e x} (4 B c d-b B e-5 A c e-3 B c e x) \sqrt{b x+c x^2}}{15 c e^2}-\frac{\left (d (c d-b e) (8 B c d+b B e-10 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 e^3 \sqrt{b x+c x^2}}-\frac{\left (\left (5 A c e (2 c d-b e)-B \left (8 c^2 d^2-3 b c d e-2 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 e^3 \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{d+e x} (4 B c d-b B e-5 A c e-3 B c e x) \sqrt{b x+c x^2}}{15 c e^2}-\frac{\left (\left (5 A c e (2 c d-b e)-B \left (8 c^2 d^2-3 b c d e-2 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 e^3 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left (d (c d-b e) (8 B c d+b B e-10 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 e^3 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{d+e x} (4 B c d-b B e-5 A c e-3 B c e x) \sqrt{b x+c x^2}}{15 c e^2}-\frac{2 \sqrt{-b} \left (5 A c e (2 c d-b e)-B \left (8 c^2 d^2-3 b c d e-2 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^{3/2} e^3 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{2 \sqrt{-b} d (c d-b e) (8 B c d+b B e-10 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^{3/2} e^3 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.56536, size = 344, normalized size = 1.08 \[ -\frac{2 \left (-b e x (b+c x) (d+e x) (5 A c e+B (b e-4 c d+3 c e x))-\sqrt{\frac{b}{c}} \left (i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (c d-b e) (5 A c e-2 B (b e+2 c d)) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-i b e x^{3/2} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (5 A c e (2 c d-b e)+B \left (2 b^2 e^2+3 b c d e-8 c^2 d^2\right )\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\sqrt{\frac{b}{c}} (b+c x) (d+e x) \left (5 A c e (b e-2 c d)+B \left (-2 b^2 e^2-3 b c d e+8 c^2 d^2\right )\right )\right )\right )}{15 b c e^3 \sqrt{x (b+c x)} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 1129, normalized size = 3.6 \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)^(1/2),x)

[Out]

-2/15*(x*(c*x+b))^(1/2)*(e*x+d)^(1/2)*(10*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Elli
pticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d*e^2-10*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+5*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^3-15*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+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*c^3*d^2*e-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+8*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^3-2*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-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+11*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-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*c^3*d^3-3*B*x^4*c^4*e^3-5*A*x^3*c^4*e^3-4*B*x^3*b*c^3*e^3+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-B*x^2*b^2*c^2*e^3+4*B*x^2*c^4*d^2*e-5*A*x*b*c^3*d*e^2-B*x*b^2*c^2*d*e^2+
4*B*x*b*c^3*d^2*e)/x/(c*e*x^2+b*e*x+c*d*x+b*d)/e^3/c^3

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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