3.367 \(\int \frac{A+B x}{\sqrt{x} (a+b x)^3} \, dx\)

Optimal. Leaf size=100 \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{5/2} b^{3/2}}+\frac{\sqrt{x} (a B+3 A b)}{4 a^2 b (a+b x)}+\frac{\sqrt{x} (A b-a B)}{2 a b (a+b x)^2} \]

[Out]

((A*b - a*B)*Sqrt[x])/(2*a*b*(a + b*x)^2) + ((3*A*b + a*B)*Sqrt[x])/(4*a^2*b*(a + b*x)) + ((3*A*b + a*B)*ArcTa
n[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(5/2)*b^(3/2))

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Rubi [A]  time = 0.036757, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{5/2} b^{3/2}}+\frac{\sqrt{x} (a B+3 A b)}{4 a^2 b (a+b x)}+\frac{\sqrt{x} (A b-a B)}{2 a b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[x]*(a + b*x)^3),x]

[Out]

((A*b - a*B)*Sqrt[x])/(2*a*b*(a + b*x)^2) + ((3*A*b + a*B)*Sqrt[x])/(4*a^2*b*(a + b*x)) + ((3*A*b + a*B)*ArcTa
n[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(5/2)*b^(3/2))

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} (a+b x)^3} \, dx &=\frac{(A b-a B) \sqrt{x}}{2 a b (a+b x)^2}+\frac{(3 A b+a B) \int \frac{1}{\sqrt{x} (a+b x)^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b (a+b x)^2}+\frac{(3 A b+a B) \sqrt{x}}{4 a^2 b (a+b x)}+\frac{(3 A b+a B) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 a^2 b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b (a+b x)^2}+\frac{(3 A b+a B) \sqrt{x}}{4 a^2 b (a+b x)}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 a^2 b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b (a+b x)^2}+\frac{(3 A b+a B) \sqrt{x}}{4 a^2 b (a+b x)}+\frac{(3 A b+a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{5/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.104543, size = 91, normalized size = 0.91 \[ \frac{\sqrt{x} \left (\frac{a^2 (A b-a B)}{(a+b x)^2}-\frac{1}{2} (-a B-3 A b) \left (\frac{a}{a+b x}+\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{x}}\right )\right )}{2 a^3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[x]*(a + b*x)^3),x]

[Out]

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

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Maple [A]  time = 0.01, size = 95, normalized size = 1. \begin{align*} 2\,{\frac{1}{ \left ( bx+a \right ) ^{2}} \left ( 1/8\,{\frac{ \left ( 3\,Ab+Ba \right ){x}^{3/2}}{{a}^{2}}}+1/8\,{\frac{ \left ( 5\,Ab-Ba \right ) \sqrt{x}}{ab}} \right ) }+{\frac{3\,A}{4\,{a}^{2}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{4\,ab}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^3/x^(1/2),x)

[Out]

2*(1/8*(3*A*b+B*a)/a^2*x^(3/2)+1/8*(5*A*b-B*a)/a/b*x^(1/2))/(b*x+a)^2+3/4/a^2/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*
b)^(1/2))*A+1/4/a/b/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(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)/(b*x+a)^3/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.22719, size = 633, normalized size = 6.33 \begin{align*} \left [-\frac{{\left (B a^{3} + 3 \, A a^{2} b +{\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) + 2 \,{\left (B a^{3} b - 5 \, A a^{2} b^{2} -{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}, -\frac{{\left (B a^{3} + 3 \, A a^{2} b +{\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (B a^{3} b - 5 \, A a^{2} b^{2} -{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/x^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((B*a^3 + 3*A*a^2*b + (B*a*b^2 + 3*A*b^3)*x^2 + 2*(B*a^2*b + 3*A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*s
qrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(B*a^3*b - 5*A*a^2*b^2 - (B*a^2*b^2 + 3*A*a*b^3)*x)*sqrt(x))/(a^3*b^4*x^2 +
2*a^4*b^3*x + a^5*b^2), -1/4*((B*a^3 + 3*A*a^2*b + (B*a*b^2 + 3*A*b^3)*x^2 + 2*(B*a^2*b + 3*A*a*b^2)*x)*sqrt(a
*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (B*a^3*b - 5*A*a^2*b^2 - (B*a^2*b^2 + 3*A*a*b^3)*x)*sqrt(x))/(a^3*b^4*x^2
+ 2*a^4*b^3*x + a^5*b^2)]

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Sympy [A]  time = 66.9465, size = 1501, normalized size = 15.01 \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)/(b*x+a)**3/x**(1/2),x)

[Out]

Piecewise((zoo*(-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2))), Eq(a, 0) & Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a
**3, Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2)))/b**3, Eq(a, 0)), (10*I*A*a**(3/2)*b**2*sqrt(x)*sqrt(1/
b)/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + 6*I*A*s
qrt(a)*b**3*x**(3/2)*sqrt(1/b)/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b*
*4*x**2*sqrt(1/b)) + 3*A*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/
2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) - 3*A*a**2*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a
**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + 6*A*a*b**2*x*log
(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*
b**4*x**2*sqrt(1/b)) - 6*A*a*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**
(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + 3*A*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))
/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) - 3*A*b**3*
x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a*
*(5/2)*b**4*x**2*sqrt(1/b)) - 2*I*B*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*
b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + 2*I*B*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*I*a**(9/2)*b*
*2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + B*a**3*log(-I*sqrt(a)*sqrt
(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1
/b)) - B*a**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b)
+ 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) + 2*B*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqr
t(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) - 2*B*a**2*b*x*log(I*sqrt(a)*sqrt(
1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/
b)) + B*a*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sq
rt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)) - B*a*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(9/2)*b
**2*sqrt(1/b) + 16*I*a**(7/2)*b**3*x*sqrt(1/b) + 8*I*a**(5/2)*b**4*x**2*sqrt(1/b)), True))

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Giac [A]  time = 1.15222, size = 111, normalized size = 1.11 \begin{align*} \frac{{\left (B a + 3 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{2} b} + \frac{B a b x^{\frac{3}{2}} + 3 \, A b^{2} x^{\frac{3}{2}} - B a^{2} \sqrt{x} + 5 \, A a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^3/x^(1/2),x, algorithm="giac")

[Out]

1/4*(B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/4*(B*a*b*x^(3/2) + 3*A*b^2*x^(3/2) - B*a^2
*sqrt(x) + 5*A*a*b*sqrt(x))/((b*x + a)^2*a^2*b)