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

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

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Rubi [A]  time = 0.14, 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 = {734, 843, 620, 206, 724} \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 206

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

Rule 620

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 724

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 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 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} \, 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)) \operatorname {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) \operatorname {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 = 0.40, size = 137, normalized size = 1.06 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {(b e-2 c d) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c} \sqrt {\frac {c x}{b}+1}}+\frac {2 \sqrt {d} \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}+e \sqrt {x}\right )}{e^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(e*Sqrt[x] + ((-2*c*d + b*e)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[c]*Sqrt[1 +
(c*x)/b]) + (2*Sqrt[d]*Sqrt[c*d - b*e]*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[b + c*
x]))/(e^2*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.53, size = 171, normalized size = 1.33 \begin {gather*} \frac {(2 c d-b e) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{2 \sqrt {c} e^2}+\frac {2 \sqrt {d} \sqrt {c d-b e} \tanh ^{-1}\left (-\frac {e \sqrt {b x+c x^2}}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {c d-b e}}\right )}{e^2}+\frac {\sqrt {b x+c x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 488, normalized size = 3.78 \begin {gather*} \left [\frac {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 {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right )}{2 \, c e^{2}}, \frac {2 \, \sqrt {c x^{2} + b x} c e + 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}}{{\left (c d - b e\right )} x}\right ) - {\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 \, c e^{2}}, \frac {\sqrt {c x^{2} + b x} c e + {\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 d^{2} - b d e} c \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right )}{c e^{2}}, \frac {\sqrt {c x^{2} + b x} c e + 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}}{{\left (c d - b e\right )} x}\right ) + {\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right )}{c e^{2}}\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((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)))/(c*e^2), 1/2*
(2*sqrt(c*x^2 + b*x)*c*e + 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 - b*e
)*x)) - (2*c*d - b*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)))/(c*e^2), (sqrt(c*x^2 + b*x)*c*e +
(2*c*d - b*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*d^2 - b*d*e)*c*log((b*d + (2*c*d - b*
e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)))/(c*e^2), (sqrt(c*x^2 + b*x)*c*e + 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 - b*e)*x)) + (2*c*d - b*e)*sqrt(-c)*arctan(sqrt(
c*x^2 + b*x)*sqrt(-c)/(c*x)))/(c*e^2)]

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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,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.08, size = 490, normalized size = 3.80 \begin {gather*} \frac {b d \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{2}}-\frac {c \,d^{2} \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{3}}+\frac {b \ln \left (\frac {\left (x +\frac {d}{e}\right ) c +\frac {b e -2 c d}{2 e}}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\right )}{2 \sqrt {c}\, e}-\frac {\sqrt {c}\, d \ln \left (\frac {\left (x +\frac {d}{e}\right ) c +\frac {b e -2 c d}{2 e}}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\right )}{e^{2}}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)+1/2/e*ln((1/2*(b*e-2*c*d)/e+c*(d/e+x))/c^(1/2)+(
c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)*b-1/e^2*ln((1/2*(b*e-2*c*d)/e+c*(d/e+x))/c^(
1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(1/2)*d+1/e^2*d/(-d*(b*e-c*d)/e^2)^(1/2)*ln(
(-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e
-c*d)/e^2)^(1/2))/(d/e+x))*b-1/e^3*d^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2
*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*c

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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(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d zero or nonzero?

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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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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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