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]

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

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Rubi [A]  time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.14, 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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fricas [A]  time = 0.72, size = 291, normalized size = 2.91 \[ \left [-\frac {{\left (B a^{3} + 3 \, A a^{2} b + {\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2} - {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}, -\frac {{\left (B a^{3} + 3 \, A a^{2} b + {\left (B a b^{2} + 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (B a^{3} b - 5 \, A a^{2} b^{2} - {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}}\right ] \]

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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giac [A]  time = 1.29, size = 82, normalized size = 0.82 \[ \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2} b} + \frac {B a b x^{\frac {3}{2}} + 3 \, A b^{2} x^{\frac {3}{2}} - B a^{2} \sqrt {x} + 5 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{2} b} \]

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)

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maple [A]  time = 0.01, size = 95, normalized size = 0.95 \[ \frac {3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{2}}+\frac {B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a b}+\frac {\frac {\left (3 A b +B a \right ) x^{\frac {3}{2}}}{4 a^{2}}+\frac {\left (5 A b -B a \right ) \sqrt {x}}{4 a b}}{\left (b x +a \right )^{2}} \]

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(1/(a*b)^(1/2)
*b*x^(1/2))*A+1/4/a/b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B

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maxima [A]  time = 2.03, size = 94, normalized size = 0.94 \[ \frac {{\left (B a b + 3 \, A b^{2}\right )} x^{\frac {3}{2}} - {\left (B a^{2} - 5 \, A a b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{3} x^{2} + 2 \, a^{3} b^{2} x + a^{4} b\right )}} + \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2} b} \]

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]

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

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mupad [B]  time = 0.13, size = 84, normalized size = 0.84 \[ \frac {\frac {x^{3/2}\,\left (3\,A\,b+B\,a\right )}{4\,a^2}+\frac {\sqrt {x}\,\left (5\,A\,b-B\,a\right )}{4\,a\,b}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (3\,A\,b+B\,a\right )}{4\,a^{5/2}\,b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 28.98, size = 1501, normalized size = 15.01 \[ \text {result too large to display} \]

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/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/b**3, Eq(a, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a**3, Eq(b, 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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