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

Optimal. Leaf size=231 \[ -\frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (2 c d-b e) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{\sqrt {c} e^2 \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {4 \sqrt {-b} \sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{e^2 \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}-\frac {2 \sqrt {b x+c x^2}}{e \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {732, 843, 715, 112, 110, 117, 116} \[ -\frac {2 \sqrt {-b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (2 c d-b e) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{\sqrt {c} e^2 \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {4 \sqrt {-b} \sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{e^2 \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}}-\frac {2 \sqrt {b x+c x^2}}{e \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*x + c*x^2])/(e*Sqrt[d + e*x]) + (4*Sqrt[-b]*Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*Ellipti
cE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(e^2*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*(
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)])/(Sqrt[c]*e^2*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

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 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 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)])

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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.52, size = 195, normalized size = 0.84 \[ \frac {2 \left (-i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+2 i b e x^{3/2} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )+\sqrt {\frac {b}{c}} (b+c x) (2 d+e x)\right )}{e^2 \sqrt {\frac {b}{c}} \sqrt {x (b+c x)} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Sqrt[b/c]*(b + c*x)*(2*d + e*x) + (2*I)*b*e*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticE[I*ArcSin
h[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] - I*b*e*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSinh[Sqr
t[b/c]/Sqrt[x]], (c*d)/(b*e)]))/(Sqrt[b/c]*e^2*Sqrt[x*(b + c*x)]*Sqrt[d + e*x])

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + b x}}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 353, normalized size = 1.53 \[ \frac {2 \sqrt {\left (c x +b \right ) x}\, \sqrt {e x +d}\, \left (-c^{2} e \,x^{2}-2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b^{2} e \EllipticE \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )+\sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b^{2} e \EllipticF \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )+2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b c d \EllipticE \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )-2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b c d \EllipticF \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )-b c e x \right )}{\left (c e \,x^{2}+b e x +c d x +b d \right ) c \,e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(1/2)/(e*x+d)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + b x}}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c\,x^2+b\,x}}{{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/(d + e*x)^(3/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/(d + e*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (b + c x\right )}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(1/2)/(e*x+d)**(3/2),x)

[Out]

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

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