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

Optimal. Leaf size=49 \[ \frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {2 A}{b \sqrt {x}} \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {781, 78, 63, 205} \begin {gather*} \frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

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

Rule 205

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

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )} \, dx &=\int \frac {A+B x}{x^{3/2} (b+c x)} \, dx\\ &=-\frac {2 A}{b \sqrt {x}}+\frac {\left (2 \left (\frac {b B}{2}-\frac {A c}{2}\right )\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{b}\\ &=-\frac {2 A}{b \sqrt {x}}+\frac {\left (4 \left (\frac {b B}{2}-\frac {A c}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{b}\\ &=-\frac {2 A}{b \sqrt {x}}+\frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 1.00 \begin {gather*} \frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 49, normalized size = 1.00 \begin {gather*} \frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 112, normalized size = 2.29 \begin {gather*} \left [-\frac {2 \, A b c \sqrt {x} - {\left (B b - A c\right )} \sqrt {-b c} x \log \left (\frac {c x - b + 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right )}{b^{2} c x}, -\frac {2 \, {\left (A b c \sqrt {x} + {\left (B b - A c\right )} \sqrt {b c} x \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right )\right )}}{b^{2} c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*A*b*c*sqrt(x) - (B*b - A*c)*sqrt(-b*c)*x*log((c*x - b + 2*sqrt(-b*c)*sqrt(x))/(c*x + b)))/(b^2*c*x), -2*(
A*b*c*sqrt(x) + (B*b - A*c)*sqrt(b*c)*x*arctan(sqrt(b*c)/(c*sqrt(x))))/(b^2*c*x)]

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giac [A]  time = 0.15, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {2 \, A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b - A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b) - 2*A/(b*sqrt(x))

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maple [A]  time = 0.06, size = 53, normalized size = 1.08 \begin {gather*} -\frac {2 A c \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b}+\frac {2 B \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/b/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A*c+2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*B-2*A/b/x^(
1/2)

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maxima [A]  time = 1.19, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {2 \, A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b - A*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b) - 2*A/(b*sqrt(x))

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mupad [B]  time = 0.07, size = 50, normalized size = 1.02 \begin {gather*} \frac {2\,B\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b}\,\sqrt {c}}-\frac {2\,A\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2\,A}{b\,\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*atan((c^(1/2)*x^(1/2))/b^(1/2)))/(b^(1/2)*c^(1/2)) - (2*A*c^(1/2)*atan((c^(1/2)*x^(1/2))/b^(1/2)))/b^(3/2
) - (2*A)/(b*x^(1/2))

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sympy [A]  time = 3.40, size = 216, normalized size = 4.41 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{b \sqrt {x}} + \frac {i A \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {3}{2}} \sqrt {\frac {1}{c}}} - \frac {i A \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {3}{2}} \sqrt {\frac {1}{c}}} - \frac {i B \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{\sqrt {b} c \sqrt {\frac {1}{c}}} + \frac {i B \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{\sqrt {b} c \sqrt {\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(b, 0) & Eq(c, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/c,
Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b, Eq(c, 0)), (-2*A/(b*sqrt(x)) + I*A*log(-I*sqrt(b)*sqrt(1/c) + sqrt
(x))/(b**(3/2)*sqrt(1/c)) - I*A*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(3/2)*sqrt(1/c)) - I*B*log(-I*sqrt(b)*s
qrt(1/c) + sqrt(x))/(sqrt(b)*c*sqrt(1/c)) + I*B*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(sqrt(b)*c*sqrt(1/c)), True
))

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