3.3.89 \(\int \frac {\sqrt {b x+c x^2}}{d+e x} \, dx\) [289]

Optimal. Leaf size=129 \[ \frac {\sqrt {b x+c x^2}}{e}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c} e^2}+\frac {\sqrt {d} \sqrt {c d-b e} \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{e^2} \]

[Out]

-(-b*e+2*c*d)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/e^2/c^(1/2)+arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+
c*d)^(1/2)/(c*x^2+b*x)^(1/2))*d^(1/2)*(-b*e+c*d)^(1/2)/e^2+(c*x^2+b*x)^(1/2)/e

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Rubi [A]
time = 0.09, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {748, 857, 634, 212, 738} \begin {gather*} -\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c} e^2}+\frac {\sqrt {d} \sqrt {c d-b e} \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{e^2}+\frac {\sqrt {b x+c x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

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 857

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

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Mathematica [A]
time = 0.28, size = 161, normalized size = 1.25 \begin {gather*} \frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} e \sqrt {x} \sqrt {b+c x}+2 \sqrt {c} \sqrt {d} \sqrt {-c d+b e} \tan ^{-1}\left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )+(2 c d-b e) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{\sqrt {c} e^2 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*e*Sqrt[x]*Sqrt[b + c*x] + 2*Sqrt[c]*Sqrt[d]*Sqrt[-(c*d) + b*e]*ArcTan[(-(e*Sqr
t[x]*Sqrt[b + c*x]) + Sqrt[c]*(d + e*x))/(Sqrt[d]*Sqrt[-(c*d) + b*e])] + (2*c*d - b*e)*Log[-(Sqrt[c]*Sqrt[x])
+ Sqrt[b + c*x]]))/(Sqrt[c]*e^2*Sqrt[x*(b + c*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(283\) vs. \(2(109)=218\).
time = 0.50, size = 284, normalized size = 2.20

method	result	size
default	\(\frac {\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}\right )}{2 e \sqrt {c}}+\frac {d \left (b e -c d \right ) \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}}{e}\)	\(284\)
risch	\(\frac {x \left (c x +b \right )}{e \sqrt {x \left (c x +b \right )}}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) b}{2 e \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) \sqrt {c}\, d}{e^{2}}+\frac {d \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right ) b}{e^{2} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}-\frac {d^{2} \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right ) c}{e^{3} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*((c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*ln((1/2/e*(b*e-2*c*d)+c*(x+
d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)+d*(b*e-c*d)/e^2/(-d*(b*e-c*
d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+1/e*(b*e-2*c*d)*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(x+d/e)^2+1/e*(b*e-
2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*d-%e*b>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.29, size = 502, normalized size = 3.89 \begin {gather*} \left [\frac {{\left (2 \, \sqrt {c x^{2} + b x} c e - {\left (2 \, c d - b e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, \sqrt {c d^{2} - b d e} c \log \left (\frac {2 \, c d x - b x e + b d + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{x e + d}\right )\right )} e^{\left (-2\right )}}{2 \, c}, \frac {{\left (4 \, \sqrt {-c d^{2} + b d e} c \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x - b x e}\right ) + 2 \, \sqrt {c x^{2} + b x} c e - {\left (2 \, c d - b e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )\right )} e^{\left (-2\right )}}{2 \, c}, \frac {{\left ({\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c e + \sqrt {c d^{2} - b d e} c \log \left (\frac {2 \, c d x - b x e + b d + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{x e + d}\right )\right )} e^{\left (-2\right )}}{c}, \frac {{\left (2 \, \sqrt {-c d^{2} + b d e} c \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x - b x e}\right ) + {\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c e\right )} e^{\left (-2\right )}}{c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*sqrt(c*x^2 + b*x)*c*e - (2*c*d - b*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*sqrt(c*
d^2 - b*d*e)*c*log((2*c*d*x - b*x*e + b*d + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(x*e + d)))*e^(-2)/c, 1/2
*(4*sqrt(-c*d^2 + b*d*e)*c*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/(c*d*x - b*x*e)) + 2*sqrt(c*x^2 + b*
x)*c*e - (2*c*d - b*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)))*e^(-2)/c, ((2*c*d - b*e)*sqrt(-c)
*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*c*e + sqrt(c*d^2 - b*d*e)*c*log((2*c*d*x - b*x*e
+ b*d + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(x*e + d)))*e^(-2)/c, (2*sqrt(-c*d^2 + b*d*e)*c*arctan(-sqrt
(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/(c*d*x - b*x*e)) + (2*c*d - b*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)
/(c*x)) + sqrt(c*x^2 + b*x)*c*e)*e^(-2)/c]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (b + c x\right )}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/(d + e*x),x)

[Out]

int((b*x + c*x^2)^(1/2)/(d + e*x), x)

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