3.17.30 \(\int \frac {a+b x}{(c+d x) \sqrt {e+f x}} \, dx\)

Optimal. Leaf size=74 \[ \frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} \sqrt {d e-c f}}+\frac {2 b \sqrt {e+f x}}{d f} \]

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Rubi [A]  time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {80, 63, 208} \begin {gather*} \frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} \sqrt {d e-c f}}+\frac {2 b \sqrt {e+f x}}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((c + d*x)*Sqrt[e + f*x]),x]

[Out]

(2*b*Sqrt[e + f*x])/(d*f) + (2*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(3/2)*Sqrt[d*e
- c*f])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 208

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

Rubi steps

\begin {align*} \int \frac {a+b x}{(c+d x) \sqrt {e+f x}} \, dx &=\frac {2 b \sqrt {e+f x}}{d f}+\frac {\left (2 \left (-\frac {1}{2} b c f+\frac {a d f}{2}\right )\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d f}\\ &=\frac {2 b \sqrt {e+f x}}{d f}-\frac {(2 (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{d f}\\ &=\frac {2 b \sqrt {e+f x}}{d f}+\frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} \sqrt {d e-c f}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 74, normalized size = 1.00 \begin {gather*} \frac {2 (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} \sqrt {d e-c f}}+\frac {2 b \sqrt {e+f x}}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((c + d*x)*Sqrt[e + f*x]),x]

[Out]

(2*b*Sqrt[e + f*x])/(d*f) + (2*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(3/2)*Sqrt[d*e
- c*f])

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IntegrateAlgebraic [A]  time = 0.10, size = 84, normalized size = 1.14 \begin {gather*} \frac {2 b \sqrt {e+f x}}{d f}-\frac {2 (a d-b c) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{3/2} \sqrt {c f-d e}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)/((c + d*x)*Sqrt[e + f*x]),x]

[Out]

(2*b*Sqrt[e + f*x])/(d*f) - (2*(-(b*c) + a*d)*ArcTan[(Sqrt[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/
(d^(3/2)*Sqrt[-(d*e) + c*f])

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fricas [A]  time = 1.38, size = 209, normalized size = 2.82 \begin {gather*} \left [-\frac {\sqrt {d^{2} e - c d f} {\left (b c - a d\right )} f \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {d^{2} e - c d f} \sqrt {f x + e}}{d x + c}\right ) - 2 \, {\left (b d^{2} e - b c d f\right )} \sqrt {f x + e}}{d^{3} e f - c d^{2} f^{2}}, -\frac {2 \, {\left (\sqrt {-d^{2} e + c d f} {\left (b c - a d\right )} f \arctan \left (\frac {\sqrt {-d^{2} e + c d f} \sqrt {f x + e}}{d f x + d e}\right ) - {\left (b d^{2} e - b c d f\right )} \sqrt {f x + e}\right )}}{d^{3} e f - c d^{2} f^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.20, size = 70, normalized size = 0.95 \begin {gather*} -\frac {2 \, {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{\sqrt {c d f - d^{2} e} d} + \frac {2 \, \sqrt {f x + e} b}{d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*c - a*d)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d) + 2*sqrt(f*x + e)*b/(d*f)

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maple [A]  time = 0.01, size = 96, normalized size = 1.30 \begin {gather*} \frac {2 a \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}-\frac {2 b c \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d}+\frac {2 \sqrt {f x +e}\, b}{d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(f*x+e)^(1/2)/d/f+2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a-2/d/((c*f-d*e)*d)^(1
/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b*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((b*x+a)/(d*x+c)/(f*x+e)^(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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 0.08, size = 62, normalized size = 0.84 \begin {gather*} \frac {2\,b\,\sqrt {e+f\,x}}{d\,f}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}}{\sqrt {c\,f-d\,e}}\right )\,\left (a\,d-b\,c\right )}{d^{3/2}\,\sqrt {c\,f-d\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*(e + f*x)^(1/2))/(d*f) + (2*atan((d^(1/2)*(e + f*x)^(1/2))/(c*f - d*e)^(1/2))*(a*d - b*c))/(d^(3/2)*(c*f
- d*e)^(1/2))

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sympy [A]  time = 17.34, size = 66, normalized size = 0.89 \begin {gather*} \frac {2 b \sqrt {e + f x}}{d f} - \frac {2 \left (a d - b c\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {d}{c f - d e}} \sqrt {e + f x}} \right )}}{d \sqrt {\frac {d}{c f - d e}} \left (c f - d e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*sqrt(e + f*x)/(d*f) - 2*(a*d - b*c)*atan(1/(sqrt(d/(c*f - d*e))*sqrt(e + f*x)))/(d*sqrt(d/(c*f - d*e))*(c*
f - d*e))

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