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

Optimal. Leaf size=110 \[ \frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{3 b B-5 A c}{b^3 \sqrt{x}}-\frac{\sqrt{c} (3 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}-\frac{b B-A c}{b c x^{3/2} (b+c x)} \]

[Out]

(3*b*B - 5*A*c)/(3*b^2*c*x^(3/2)) - (3*b*B - 5*A*c)/(b^3*Sqrt[x]) - (b*B - A*c)/(b*c*x^(3/2)*(b + c*x)) - (Sqr
t[c]*(3*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/b^(7/2)

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Rubi [A]  time = 0.0554794, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{3 b B-5 A c}{b^3 \sqrt{x}}-\frac{\sqrt{c} (3 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}-\frac{b B-A c}{b c x^{3/2} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*B - 5*A*c)/(3*b^2*c*x^(3/2)) - (3*b*B - 5*A*c)/(b^3*Sqrt[x]) - (b*B - A*c)/(b*c*x^(3/2)*(b + c*x)) - (Sqr
t[c]*(3*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/b^(7/2)

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}{\sqrt{x} \left (b x+c x^2\right )^2} \, dx &=\int \frac{A+B x}{x^{5/2} (b+c x)^2} \, dx\\ &=-\frac{b B-A c}{b c x^{3/2} (b+c x)}-\frac{\left (\frac{3 b B}{2}-\frac{5 A c}{2}\right ) \int \frac{1}{x^{5/2} (b+c x)} \, dx}{b c}\\ &=\frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{b B-A c}{b c x^{3/2} (b+c x)}+\frac{(3 b B-5 A c) \int \frac{1}{x^{3/2} (b+c x)} \, dx}{2 b^2}\\ &=\frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{3 b B-5 A c}{b^3 \sqrt{x}}-\frac{b B-A c}{b c x^{3/2} (b+c x)}-\frac{(c (3 b B-5 A c)) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{2 b^3}\\ &=\frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{3 b B-5 A c}{b^3 \sqrt{x}}-\frac{b B-A c}{b c x^{3/2} (b+c x)}-\frac{(c (3 b B-5 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{b^3}\\ &=\frac{3 b B-5 A c}{3 b^2 c x^{3/2}}-\frac{3 b B-5 A c}{b^3 \sqrt{x}}-\frac{b B-A c}{b c x^{3/2} (b+c x)}-\frac{\sqrt{c} (3 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0177257, size = 64, normalized size = 0.58 \[ \frac{(b+c x) (3 b B-5 A c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{c x}{b}\right )+3 b (A c-b B)}{3 b^2 c x^{3/2} (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(-(b*B) + A*c) + (3*b*B - 5*A*c)*(b + c*x)*Hypergeometric2F1[-3/2, 1, -1/2, -((c*x)/b)])/(3*b^2*c*x^(3/2)
*(b + c*x))

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Maple [A]  time = 0.017, size = 113, normalized size = 1. \begin{align*} -{\frac{2\,A}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}}+4\,{\frac{Ac}{{b}^{3}\sqrt{x}}}-2\,{\frac{B}{{b}^{2}\sqrt{x}}}+{\frac{{c}^{2}A}{{b}^{3} \left ( cx+b \right ) }\sqrt{x}}-{\frac{Bc}{{b}^{2} \left ( cx+b \right ) }\sqrt{x}}+5\,{\frac{{c}^{2}A}{{b}^{3}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-3\,{\frac{Bc}{{b}^{2}\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)/(c*x^2+b*x)^2/x^(1/2),x)

[Out]

-2/3*A/b^2/x^(3/2)+4/b^3/x^(1/2)*A*c-2/b^2/x^(1/2)*B+1/b^3*c^2*x^(1/2)/(c*x+b)*A-1/b^2*c*x^(1/2)/(c*x+b)*B+5/b
^3*c^2/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2))*A-3/b^2*c/(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)/(c*x^2+b*x)^2/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59729, size = 572, normalized size = 5.2 \begin{align*} \left [-\frac{3 \,{\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{3} +{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x + 2 \, b \sqrt{x} \sqrt{-\frac{c}{b}} - b}{c x + b}\right ) + 2 \,{\left (2 \, A b^{2} + 3 \,{\left (3 \, B b c - 5 \, A c^{2}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} - 5 \, A b c\right )} x\right )} \sqrt{x}}{6 \,{\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}, \frac{3 \,{\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{3} +{\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2}\right )} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c \sqrt{x}}\right ) -{\left (2 \, A b^{2} + 3 \,{\left (3 \, B b c - 5 \, A c^{2}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} - 5 \, A b c\right )} x\right )} \sqrt{x}}{3 \,{\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*((3*B*b*c - 5*A*c^2)*x^3 + (3*B*b^2 - 5*A*b*c)*x^2)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)
/(c*x + b)) + 2*(2*A*b^2 + 3*(3*B*b*c - 5*A*c^2)*x^2 + 2*(3*B*b^2 - 5*A*b*c)*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2)
, 1/3*(3*((3*B*b*c - 5*A*c^2)*x^3 + (3*B*b^2 - 5*A*b*c)*x^2)*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (2*A*
b^2 + 3*(3*B*b*c - 5*A*c^2)*x^2 + 2*(3*B*b^2 - 5*A*b*c)*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2)]

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Sympy [A]  time = 62.5702, size = 983, normalized size = 8.94 \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)/(c*x**2+b*x)**2/x**(1/2),x)

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/c**2, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b**2, Eq(c, 0)), (-4*I*A*b**(5/2)*sqrt(1/c)/(6*I*b*
*(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) + 20*I*A*b**(3/2)*c*x*sqrt(1/c)/(6*I*b**(9/2)*x
**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) + 30*I*A*sqrt(b)*c**2*x**2*sqrt(1/c)/(6*I*b**(9/2)*x**(
3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) + 15*A*b*c*x**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6
*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) - 15*A*b*c*x**(3/2)*log(I*sqrt(b)*sqrt(1/c
) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) + 15*A*c**2*x**(5/2)*log(-I
*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) - 15*A*c**
2*x**(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(
1/c)) - 12*I*B*b**(5/2)*x*sqrt(1/c)/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) - 18
*I*B*b**(3/2)*c*x**2*sqrt(1/c)/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) - 9*B*b**
2*x**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt
(1/c)) + 9*B*b**2*x**(3/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c) + 6*I*b**(7/2)*
c*x**(5/2)*sqrt(1/c)) - 9*B*b*c*x**(5/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3/2)*sqrt(1/c)
+ 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)) + 9*B*b*c*x**(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(6*I*b**(9/2)*x**(3
/2)*sqrt(1/c) + 6*I*b**(7/2)*c*x**(5/2)*sqrt(1/c)), True))

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Giac [A]  time = 1.15342, size = 115, normalized size = 1.05 \begin{align*} -\frac{{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} b^{3}} - \frac{B b c \sqrt{x} - A c^{2} \sqrt{x}}{{\left (c x + b\right )} b^{3}} - \frac{2 \,{\left (3 \, B b x - 6 \, A c x + A b\right )}}{3 \, b^{3} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*B*b*c - 5*A*c^2)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^3) - (B*b*c*sqrt(x) - A*c^2*sqrt(x))/((c*x + b)*
b^3) - 2/3*(3*B*b*x - 6*A*c*x + A*b)/(b^3*x^(3/2))