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

Optimal. Leaf size=113 \[ \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{\sqrt {d} e \sqrt {c d-b e}} \]

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Rubi [A]  time = 0.08, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {843, 620, 206, 724} \begin {gather*} \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A 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 )}{\sqrt {d} e \sqrt {c d-b e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(Sqrt[c]*e) - ((B*d - A*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqr
t[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(Sqrt[d]*e*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 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 {A+B x}{(d+e x) \sqrt {b x+c x^2}} \, dx &=\frac {B \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{e}+\frac {(-B d+A e) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{e}\\ &=\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{e}-\frac {(2 (-B d+A 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}\\ &=\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A 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 )}{\sqrt {d} e \sqrt {c d-b e}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 131, normalized size = 1.16 \begin {gather*} \frac {2 \sqrt {x} \left (\frac {\sqrt {b+c x} (A e-B d) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {b} B \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c}}\right )}{e \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.48, size = 124, normalized size = 1.10 \begin {gather*} -\frac {2 (B d-A e) \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 \sqrt {c d-b e}}-\frac {B \log \left (-2 \sqrt {c} e \sqrt {b x+c x^2}+b e+2 c e x\right )}{\sqrt {c} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 534, normalized size = 4.73 \begin {gather*} \left [\frac {{\left (B c d^{2} - B b d e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - {\left (B c d - A c e\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 )}{c^{2} d^{2} e - b c d e^{2}}, -\frac {2 \, {\left (B c d - A c 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}}{{\left (c d - b e\right )} x}\right ) - {\left (B c d^{2} - B b d e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{2} d^{2} e - b c d e^{2}}, -\frac {2 \, {\left (B c d^{2} - B b d e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (B c d - A c e\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 )}{c^{2} d^{2} e - b c d e^{2}}, -\frac {2 \, {\left ({\left (B c d - A c 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}}{{\left (c d - b e\right )} x}\right ) + {\left (B c d^{2} - B b d e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right )\right )}}{c^{2} d^{2} e - b c d e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((B*c*d^2 - B*b*d*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - (B*c*d - A*c*e)*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)))/(c^2*d^2*e - b*c*d*e^2),
-(2*(B*c*d - A*c*e)*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)) - (B*
c*d^2 - B*b*d*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)))/(c^2*d^2*e - b*c*d*e^2), -(2*(B*c*d^2 -
B*b*d*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (B*c*d - A*c*e)*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)))/(c^2*d^2*e - b*c*d*e^2), -2*((B*c*d - A*
c*e)*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)) + (B*c*d^2 - B*b*d*e
)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)))/(c^2*d^2*e - b*c*d*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((B*x+A)/(e*x+d)/(c*x^2+b*x)^(1/2),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 = 0.05, size = 298, normalized size = 2.64 \begin {gather*} -\frac {A \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}+\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 {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((B*x+A)/(e*x+d)/(c*x^2+b*x)^(1/2),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-(2*c*d)/e^2)^2>0)', see `
assume?` for more details)Is (b/e-(2*c*d)/e^2)^2    -(4*c       *((c*d^2)/e^2        -(b*d)/e))     /e^2 zero
or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{\sqrt {c\,x^2+b\,x}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*x)/((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 {A + B x}{\sqrt {x \left (b + c x\right )} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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