3.173 \(\int \frac{A+B x}{x^{3/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=69 \[ -\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{2 \sqrt{c} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2}}-\frac{2 A}{3 b x^{3/2}} \]

[Out]

(-2*A)/(3*b*x^(3/2)) - (2*(b*B - A*c))/(b^2*Sqrt[x]) - (2*Sqrt[c]*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]
])/b^(5/2)

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Rubi [A]  time = 0.0417975, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \[ -\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{2 \sqrt{c} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2}}-\frac{2 A}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(b*x + c*x^2)),x]

[Out]

(-2*A)/(3*b*x^(3/2)) - (2*(b*B - A*c))/(b^2*Sqrt[x]) - (2*Sqrt[c]*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]
])/b^(5/2)

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]

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 51

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

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 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]

Rubi steps

\begin{align*} \int \frac{A+B x}{x^{3/2} \left (b x+c x^2\right )} \, dx &=\int \frac{A+B x}{x^{5/2} (b+c x)} \, dx\\ &=-\frac{2 A}{3 b x^{3/2}}+\frac{\left (2 \left (\frac{3 b B}{2}-\frac{3 A c}{2}\right )\right ) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{3 b}\\ &=-\frac{2 A}{3 b x^{3/2}}-\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{(c (b B-A c)) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{b^2}\\ &=-\frac{2 A}{3 b x^{3/2}}-\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{(2 c (b B-A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{b^2}\\ &=-\frac{2 A}{3 b x^{3/2}}-\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{2 \sqrt{c} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0134692, size = 44, normalized size = 0.64 \[ \frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{c x}{b}\right ) (6 A c x-6 b B x)-2 A b}{3 b^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(3/2)*(b*x + c*x^2)),x]

[Out]

(-2*A*b + (-6*b*B*x + 6*A*c*x)*Hypergeometric2F1[-1/2, 1, 1/2, -((c*x)/b)])/(3*b^2*x^(3/2))

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Maple [A]  time = 0.012, size = 78, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{3\,b}{x}^{-{\frac{3}{2}}}}+2\,{\frac{Ac}{{b}^{2}\sqrt{x}}}-2\,{\frac{B}{b\sqrt{x}}}+2\,{\frac{A{c}^{2}}{{b}^{2}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-2\,{\frac{Bc}{b\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(c*x^2+b*x),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78254, size = 335, normalized size = 4.86 \begin{align*} \left [-\frac{3 \,{\left (B b - A c\right )} x^{2} \sqrt{-\frac{c}{b}} \log \left (\frac{c x + 2 \, b \sqrt{x} \sqrt{-\frac{c}{b}} - b}{c x + b}\right ) + 2 \,{\left (A b + 3 \,{\left (B b - A c\right )} x\right )} \sqrt{x}}{3 \, b^{2} x^{2}}, \frac{2 \,{\left (3 \,{\left (B b - A c\right )} x^{2} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) -{\left (A b + 3 \,{\left (B b - A c\right )} x\right )} \sqrt{x}\right )}}{3 \, b^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-1/3*(3*(B*b - A*c)*x^2*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(A*b + 3*(B*b - A*c)
*x)*sqrt(x))/(b^2*x^2), 2/3*(3*(B*b - A*c)*x^2*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (A*b + 3*(B*b - A*c
)*x)*sqrt(x))/(b^2*x^2)]

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Sympy [A]  time = 15.2149, size = 248, normalized size = 3.59 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{5 x^{\frac{5}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}\right ) & \text{for}\: b = 0 \wedge c = 0 \\\frac{- \frac{2 A}{5 x^{\frac{5}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}}{c} & \text{for}\: b = 0 \\\frac{- \frac{2 A}{3 x^{\frac{3}{2}}} - \frac{2 B}{\sqrt{x}}}{b} & \text{for}\: c = 0 \\- \frac{2 A}{3 b x^{\frac{3}{2}}} + \frac{2 A c}{b^{2} \sqrt{x}} - \frac{i A c \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{c}}} + \frac{i A c \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{c}}} - \frac{2 B}{b \sqrt{x}} + \frac{i B \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 )}}{b^{\frac{3}{2}} \sqrt{\frac{1}{c}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(3/2)/(c*x**2+b*x),x)

[Out]

Piecewise((zoo*(-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/c, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b, Eq(c, 0)), (-2*A/(3*b*x**(3/2)) + 2*A*c/(b**2*sqrt(
x)) - I*A*c*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(5/2)*sqrt(1/c)) + I*A*c*log(I*sqrt(b)*sqrt(1/c) + sqrt(x)
)/(b**(5/2)*sqrt(1/c)) - 2*B/(b*sqrt(x)) + I*B*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(3/2)*sqrt(1/c)) - I*B*
log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(3/2)*sqrt(1/c)), True))

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Giac [A]  time = 1.10225, size = 74, normalized size = 1.07 \begin{align*} -\frac{2 \,{\left (B b c - A c^{2}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b^{2}} - \frac{2 \,{\left (3 \, B b x - 3 \, A c x + A b\right )}}{3 \, b^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

-2*(B*b*c - A*c^2)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^2) - 2/3*(3*B*b*x - 3*A*c*x + A*b)/(b^2*x^(3/2))