3.3.90 \(\int \frac {\sqrt {b x+c x^2}}{(d+e x)^2} \, dx\) [290]

Optimal. Leaf size=140 \[ -\frac {\sqrt {b x+c x^2}}{e (d+e x)}+\frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{e^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 \sqrt {d} e^2 \sqrt {c d-b e}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 140, 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 = {746, 857, 634, 212, 738} \begin {gather*} -\frac {(2 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 )}{2 \sqrt {d} e^2 \sqrt {c d-b e}}-\frac {\sqrt {b x+c x^2}}{e (d+e x)}+\frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x)^2,x]

[Out]

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

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 746

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

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Mathematica [A]
time = 0.47, size = 159, normalized size = 1.14 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (-\frac {e}{d+e x}+\frac {(2 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 )}{\sqrt {d} \sqrt {-c d+b e} \sqrt {x} \sqrt {b+c x}}-\frac {2 \sqrt {c} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(588\) vs. \(2(118)=236\).
time = 0.47, size = 589, normalized size = 4.21

method	result	size
default	\(\frac {\frac {e^{2} \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}\right )^{\frac {3}{2}}}{d \left (b e -c d \right ) \left (x +\frac {d}{e}\right )}-\frac {e \left (b e -2 c d \right ) \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}\right )}{2 e \sqrt {c}}+\frac {d \left (b e -c d \right ) \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}\right )}{2 d \left (b e -c d \right )}-\frac {2 c \,e^{2} \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{4 c}+\frac {\left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{d \left (b e -c d \right )}}{e^{2}}\)	\(589\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^2*(1/d/(b*e-c*d)*e^2/(x+d/e)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(3/2)-1/2*e*(b*e-2*c*d)
/d/(b*e-c*d)*((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)))-2*c/d/(b*e-c*d)*e^2*(1/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d
))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)+1/8*(-4*c*d*(b*e-c*d)/e^2-1/e^2*(b*e-2*c*d)^2
)/c^(3/2)*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)
)))

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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)^2,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 = 3.42, size = 836, normalized size = 5.97 \begin {gather*} \left [\frac {2 \, {\left (c d^{3} - b d x e^{2} + {\left (c d^{2} x - b d^{2}\right )} e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - {\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {c d^{2} - b d e} \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 ) - 2 \, {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c d^{3} e^{2} - b d x e^{4} + {\left (c d^{2} x - b d^{2}\right )} e^{3}\right )}}, -\frac {{\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x - b x e}\right ) - {\left (c d^{3} - b d x e^{2} + {\left (c d^{2} x - b d^{2}\right )} e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{c d^{3} e^{2} - b d x e^{4} + {\left (c d^{2} x - b d^{2}\right )} e^{3}}, -\frac {4 \, {\left (c d^{3} - b d x e^{2} + {\left (c d^{2} x - b d^{2}\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {c d^{2} - b d e} \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 ) + 2 \, {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c d^{3} e^{2} - b d x e^{4} + {\left (c d^{2} x - b d^{2}\right )} e^{3}\right )}}, -\frac {{\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{c d x - b x e}\right ) + 2 \, {\left (c d^{3} - b d x e^{2} + {\left (c d^{2} x - b d^{2}\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (c d^{2} e - b d e^{2}\right )} \sqrt {c x^{2} + b x}}{c d^{3} e^{2} - b d x e^{4} + {\left (c d^{2} x - b d^{2}\right )} e^{3}}\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)^2,x, algorithm="fricas")

[Out]

[1/2*(2*(c*d^3 - b*d*x*e^2 + (c*d^2*x - b*d^2)*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - (2*c*
d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(c*d^2 - b*d*e)*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)) - 2*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d*x*e^4 + (c*d^2*x - b*d^2
)*e^3), -((2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2
+ b*x)/(c*d*x - b*x*e)) - (c*d^3 - b*d*x*e^2 + (c*d^2*x - b*d^2)*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*
x)*sqrt(c)) + (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d*x*e^4 + (c*d^2*x - b*d^2)*e^3), -1/2*(4*
(c*d^3 - b*d*x*e^2 + (c*d^2*x - b*d^2)*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*c*d^2 - b*x*e
^2 + (2*c*d*x - b*d)*e)*sqrt(c*d^2 - b*d*e)*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)) + 2*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d*x*e^4 + (c*d^2*x - b*d^2)*e^3), -((
2*c*d^2 - b*x*e^2 + (2*c*d*x - b*d)*e)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/(c*
d*x - b*x*e)) + 2*(c*d^3 - b*d*x*e^2 + (c*d^2*x - b*d^2)*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x))
+ (c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d*x*e^4 + (c*d^2*x - b*d^2)*e^3)]

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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 )}}{\left (d + e x\right )^{2}}\, 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)**2,x)

[Out]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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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}}{{\left (d+e\,x\right )}^2} \,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)^2,x)

[Out]

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

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