3.390 \(\int \frac{\sqrt{b x+c x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=301 \[ \frac{4 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{3 e^2 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 d e^2 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)}+\frac{2 \sqrt{b x+c x^2} (2 c d-b e)}{3 d e \sqrt{d+e x} (c d-b e)}-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.280329, antiderivative size = 301, 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, 834, 843, 715, 112, 110, 117, 116} \[ \frac{4 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 e^2 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{3 d e^2 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1} (c d-b e)}+\frac{2 \sqrt{b x+c x^2} (2 c d-b e)}{3 d e \sqrt{d+e x} (c d-b e)}-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x)^(5/2),x]

[Out]

(-2*Sqrt[b*x + c*x^2])/(3*e*(d + e*x)^(3/2)) + (2*(2*c*d - b*e)*Sqrt[b*x + c*x^2])/(3*d*e*(c*d - b*e)*Sqrt[d +
 e*x]) - (2*Sqrt[-b]*Sqrt[c]*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*S
qrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*d*e^2*(c*d - b*e)*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) + (4*Sqrt[-b]*Sqrt[
c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(3*
e^2*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 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
 + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
 + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*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] && LtQ[m, -1] && (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{\sqrt{b x+c x^2}}{(d+e x)^{5/2}} \, dx &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{\int \frac{b+2 c x}{(d+e x)^{3/2} \sqrt{b x+c x^2}} \, dx}{3 e}\\ &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{2 (2 c d-b e) \sqrt{b x+c x^2}}{3 d e (c d-b e) \sqrt{d+e x}}-\frac{2 \int \frac{\frac{b c d}{2}+\frac{1}{2} c (2 c d-b e) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 d e (c d-b e)}\\ &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{2 (2 c d-b e) \sqrt{b x+c x^2}}{3 d e (c d-b e) \sqrt{d+e x}}+\frac{(2 c) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{3 e^2}-\frac{(c (2 c d-b e)) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{3 d e^2 (c d-b e)}\\ &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{2 (2 c d-b e) \sqrt{b x+c x^2}}{3 d e (c d-b e) \sqrt{d+e x}}+\frac{\left (2 c \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{3 e^2 \sqrt{b x+c x^2}}-\frac{\left (c (2 c d-b e) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{3 d e^2 (c d-b e) \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{2 (2 c d-b e) \sqrt{b x+c x^2}}{3 d e (c d-b e) \sqrt{d+e x}}-\frac{\left (c (2 c d-b e) \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}{3 d e^2 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (2 c \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}{3 e^2 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \sqrt{b x+c x^2}}{3 e (d+e x)^{3/2}}+\frac{2 (2 c d-b e) \sqrt{b x+c x^2}}{3 d e (c d-b e) \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{c} (2 c d-b e) \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 )}{3 d e^2 (c d-b e) \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{4 \sqrt{-b} \sqrt{c} \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 )}{3 e^2 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.14323, size = 265, normalized size = 0.88 \[ -\frac{2 \left (e x (b+c x) \left (b e^2 x-c d (d+2 e x)\right )+(d+e x) \left (i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (b e-c d) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-i c e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (b e-2 c d) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+(b+c x) (d+e x) (2 c d-b e)\right )\right )}{3 d e^2 \sqrt{x (b+c x)} (d+e x)^{3/2} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/(d + e*x)^(5/2),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.293, size = 887, normalized size = 3. \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)^(1/2)/(e*x+d)^(5/2),x)

[Out]

2/3*(2*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)*(-c*x/b)^(1/2)-2*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^2*d^2*e*((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))*x*b^3*e^3*((c*x+
b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-3*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x
*b^2*c*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*EllipticE(((c*x+b)/b)^(1/2),(b*e/
(b*e-c*d))^(1/2))*x*b*c^2*d^2*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*EllipticF(((c*
x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^2*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)
-2*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^3*((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*d*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*
e-c*d))^(1/2)*(-c*x/b)^(1/2)-3*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c*d^2*e*((c*x+b)/b)^(1/2
)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+2*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^2*d^3*(
(c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+x^3*b*c^2*e^3-2*x^3*c^3*d*e^2+x^2*b^2*c*e^3-2*x^2
*b*c^2*d*e^2-x^2*c^3*d^2*e-x*b*c^2*d^2*e)*(x*(c*x+b))^(1/2)/c/(b*e-c*d)/d/(c*x+b)/x/e^2/(e*x+d)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)/(e*x + d)^(5/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )}}{\left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(1/2)/(e*x+d)**(5/2),x)

[Out]

Integral(sqrt(x*(b + c*x))/(d + e*x)**(5/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x)/(e*x + d)^(5/2), x)