3.17.7 \(\int \frac {A+B x}{(a+b x) \sqrt {d+e x}} \, dx\)

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

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Rubi [A]  time = 0.04, 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 \sqrt {d+e x}}{b e}-\frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.17, size = 74, normalized size = 1.00 \begin {gather*} \frac {2 (a B-A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}+\frac {2 B \sqrt {d+e x}}{b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.57, size = 209, normalized size = 2.82 \begin {gather*} \left [-\frac {\sqrt {b^{2} d - a b e} {\left (B a - A b\right )} e \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (B b^{2} d - B a b e\right )} \sqrt {e x + d}}{b^{3} d e - a b^{2} e^{2}}, -\frac {2 \, {\left (\sqrt {-b^{2} d + a b e} {\left (B a - A b\right )} e \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (B b^{2} d - B a b e\right )} \sqrt {e x + d}\right )}}{b^{3} d e - a b^{2} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.27, size = 69, normalized size = 0.93 \begin {gather*} \frac {2 \, \sqrt {x e + d} B e^{\left (-1\right )}}{b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*B*e^(-1)/b - 2*(B*a - A*b)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*
b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*(e*x+d)^(1/2)/b/e+2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A-2/b/((a*e-b*d)*b)^(1
/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 17.56, size = 66, normalized size = 0.89 \begin {gather*} \frac {2 B \sqrt {d + e x}}{b e} + \frac {2 \left (- A b + B a\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {b}{a e - b d}} \sqrt {d + e x}} \right )}}{b \sqrt {\frac {b}{a e - b d}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*sqrt(d + e*x)/(b*e) + 2*(-A*b + B*a)*atan(1/(sqrt(b/(a*e - b*d))*sqrt(d + e*x)))/(b*sqrt(b/(a*e - b*d))*(a
*e - b*d))

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