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

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

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Rubi [A]  time = 0.11, 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 = {732, 843, 620, 206, 724} \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 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 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)^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) \operatorname {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) \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 {\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 = 0.65, size = 171, normalized size = 1.22 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {e \sqrt {x} (c d-b e)}{d+e x}+\frac {2 \sqrt {c} (b e-c d) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}+\frac {(2 c d-b e) \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {d} \sqrt {b+c x}}\right )}{e^2 \sqrt {x} (b e-c d)} \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*(c*d - b*e)*Sqrt[x])/(d + e*x) + (2*Sqrt[c]*(-(c*d) + b*e)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sq
rt[b]])/(Sqrt[b]*Sqrt[1 + (c*x)/b]) + (Sqrt[c*d - b*e]*(2*c*d - b*e)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d
]*Sqrt[b + c*x])])/(Sqrt[d]*Sqrt[b + c*x])))/(e^2*(-(c*d) + b*e)*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.74, size = 143, normalized size = 1.02 \begin {gather*} \frac {(b e-2 c d) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{\sqrt {d} e^2 \sqrt {c d-b e}}-\frac {\sqrt {b x+c x^2}}{e (d+e x)}-\frac {\sqrt {c} \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 846, normalized size = 6.04 \begin {gather*} \left [\frac {2 \, {\left (c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x\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 d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \sqrt {c d^{2} - b d e} \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 \, {\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^{2} e^{3} + {\left (c d^{2} e^{3} - b d e^{4}\right )} x\right )}}, -\frac {{\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \sqrt {-c d^{2} + b d e} \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 (c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x\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^{2} e^{3} + {\left (c d^{2} e^{3} - b d e^{4}\right )} x}, -\frac {4 \, {\left (c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \sqrt {c d^{2} - b d e} \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 \, {\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^{2} e^{3} + {\left (c d^{2} e^{3} - b d e^{4}\right )} x\right )}}, -\frac {{\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + 2 \, {\left (c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x\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^{2} e^{3} + {\left (c d^{2} e^{3} - b d e^{4}\right )} x}\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^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - (2*c*
d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(c*d^2 - b*d*e)*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)) - 2*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d
*e^4)*x), -((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x
^2 + b*x)/((c*d - b*e)*x)) - (c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*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^2*e^3 + (c*d^2*e^3 - b*d*e^4)*x), -1/2
*(4*(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*c*d^2 - b
*d*e + (2*c*d*e - b*e^2)*x)*sqrt(c*d^2 - b*d*e)*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)) + 2*(c*d^2*e - b*d*e^2)*sqrt(c*x^2 + b*x))/(c*d^3*e^2 - b*d^2*e^3 + (c*d^2*e^3 - b*d*e^4)*x
), -((2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*
x)/((c*d - b*e)*x)) + 2*(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x)*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^2*e^3 + (c*d^2*e^3 - b*d*e^4)*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,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.53Error: Bad Argument Type

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maple [B]  time = 0.06, size = 885, normalized size = 6.32 \begin {gather*} -\frac {b^{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 )}{2 \left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e}+\frac {3 b c 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 )}{2 \left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{2}}-\frac {c^{2} 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 )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{3}}+\frac {b \sqrt {c}\, \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 )}{\left (b e -c d \right ) e}-\frac {c^{\frac {3}{2}} 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 )}{\left (b e -c d \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}}\, c x}{\left (b e -c d \right ) d}-\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}}\, b}{\left (b e -c d \right ) d}+\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}}\, c}{\left (b e -c d \right ) e}+\frac {\left (\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 )^{\frac {3}{2}}}{\left (b e -c d \right ) \left (x +\frac {d}{e}\right ) d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(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}}{{\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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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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