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

Optimal. Leaf size=85 \[ \frac{b B-3 A c}{b^2 c \sqrt{x}}+\frac{(b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2} \sqrt{c}}-\frac{b B-A c}{b c \sqrt{x} (b+c x)} \]

[Out]

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

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Rubi [A]  time = 0.0448304, antiderivative size = 85, 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} \[ \frac{b B-3 A c}{b^2 c \sqrt{x}}+\frac{(b B-3 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2} \sqrt{c}}-\frac{b B-A c}{b c \sqrt{x} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0183328, size = 59, normalized size = 0.69 \[ \frac{(b+c x) (b B-3 A c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{c x}{b}\right )+b (A c-b B)}{b^2 c \sqrt{x} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 87, normalized size = 1. \begin{align*} -2\,{\frac{A}{{b}^{2}\sqrt{x}}}-{\frac{Ac}{{b}^{2} \left ( cx+b \right ) }\sqrt{x}}+{\frac{B}{b \left ( cx+b \right ) }\sqrt{x}}-3\,{\frac{Ac}{{b}^{2}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }+{\frac{B}{b}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A/b^2/x^(1/2)-1/b^2*x^(1/2)/(c*x+b)*A*c+1/b*x^(1/2)/(c*x+b)*B-3/b^2/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2
))*A*c+1/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^(1/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64694, size = 475, normalized size = 5.59 \begin{align*} \left [\frac{{\left ({\left (B b c - 3 \, A c^{2}\right )} x^{2} +{\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt{-b c} \log \left (\frac{c x - b + 2 \, \sqrt{-b c} \sqrt{x}}{c x + b}\right ) - 2 \,{\left (2 \, A b^{2} c -{\left (B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt{x}}{2 \,{\left (b^{3} c^{2} x^{2} + b^{4} c x\right )}}, -\frac{{\left ({\left (B b c - 3 \, A c^{2}\right )} x^{2} +{\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (2 \, A b^{2} c -{\left (B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt{x}}{b^{3} c^{2} x^{2} + b^{4} c x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((B*b*c - 3*A*c^2)*x^2 + (B*b^2 - 3*A*b*c)*x)*sqrt(-b*c)*log((c*x - b + 2*sqrt(-b*c)*sqrt(x))/(c*x + b))
- 2*(2*A*b^2*c - (B*b^2*c - 3*A*b*c^2)*x)*sqrt(x))/(b^3*c^2*x^2 + b^4*c*x), -(((B*b*c - 3*A*c^2)*x^2 + (B*b^2
- 3*A*b*c)*x)*sqrt(b*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (2*A*b^2*c - (B*b^2*c - 3*A*b*c^2)*x)*sqrt(x))/(b^3*c
^2*x^2 + b^4*c*x)]

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Sympy [A]  time = 48.6808, size = 884, normalized size = 10.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x**(1/2)/(c*x**2+b*x)**2,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**2, Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**2, Eq(c, 0)), (-4*I*A*b**(3/2)*c*sqrt(1/c)/(2*I*b**(7
/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) - 6*I*A*sqrt(b)*c**2*x*sqrt(1/c)/(2*I*b**(7/2)
*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) - 3*A*b*c*sqrt(x)*log(-I*sqrt(b)*sqrt(1/c) + sqrt
(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) + 3*A*b*c*sqrt(x)*log(I*sqrt(b)
*sqrt(1/c) + sqrt(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) - 3*A*c**2*x**
(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(
1/c)) + 3*A*c**2*x**(3/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*
c**2*x**(3/2)*sqrt(1/c)) + 2*I*B*b**(3/2)*c*x*sqrt(1/c)/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*
x**(3/2)*sqrt(1/c)) + B*b**2*sqrt(x)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2
*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) - B*b**2*sqrt(x)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I*b**(7/2)*c*sqrt(
x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) + B*b*c*x**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(2*I
*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)) - B*b*c*x**(3/2)*log(I*sqrt(b)*sqrt(1/c)
+ sqrt(x))/(2*I*b**(7/2)*c*sqrt(x)*sqrt(1/c) + 2*I*b**(5/2)*c**2*x**(3/2)*sqrt(1/c)), True))

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Giac [A]  time = 1.14531, size = 81, normalized size = 0.95 \begin{align*} \frac{{\left (B b - 3 \, A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b^{2}} + \frac{B b x - 3 \, A c x - 2 \, A b}{{\left (c x^{\frac{3}{2}} + b \sqrt{x}\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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