3.366 \(\int \frac {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx\)
Optimal. Leaf size=100 \[ \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}}-\frac {\sqrt {x} (3 a B+A b)}{4 a b^2 (a+b x)}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2} \]
[Out]
1/2*(A*b-B*a)*x^(3/2)/a/b/(b*x+a)^2+1/4*(A*b+3*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(3/2)/b^(5/2)-1/4*(A*b+3
*B*a)*x^(1/2)/a/b^2/(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, 47, 63, 205} \[ \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}}-\frac {\sqrt {x} (3 a B+A b)}{4 a b^2 (a+b x)}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[x]*(A + B*x))/(a + b*x)^3,x]
[Out]
((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)^2) - ((A*b + 3*a*B)*Sqrt[x])/(4*a*b^2*(a + b*x)) + ((A*b + 3*a*B)*ArcTa
n[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2))
Rule 47
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {\sqrt {x} (A+B x)}{(a+b x)^3} \, dx &=\frac {(A b-a B) x^{3/2}}{2 a b (a+b x)^2}+\frac {(A b+3 a B) \int \frac {\sqrt {x}}{(a+b x)^2} \, dx}{4 a b}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b (a+b x)^2}-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 (a+b x)}+\frac {(A b+3 a B) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 a b^2}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b (a+b x)^2}-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 (a+b x)}+\frac {(A b+3 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a b^2}\\ &=\frac {(A b-a B) x^{3/2}}{2 a b (a+b x)^2}-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 (a+b x)}+\frac {(A b+3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 85, normalized size = 0.85 \[ \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}}+\frac {\sqrt {x} \left (-3 a^2 B-a b (A+5 B x)+A b^2 x\right )}{4 a b^2 (a+b x)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[x]*(A + B*x))/(a + b*x)^3,x]
[Out]
(Sqrt[x]*(-3*a^2*B + A*b^2*x - a*b*(A + 5*B*x)))/(4*a*b^2*(a + b*x)^2) + ((A*b + 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x
])/Sqrt[a]])/(4*a^(3/2)*b^(5/2))
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fricas [A] time = 0.82, size = 291, normalized size = 2.91 \[ \left [-\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}, -\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*x^(1/2)/(b*x+a)^3,x, algorithm="fricas")
[Out]
[-1/8*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*s
qrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B*a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 +
2*a^3*b^4*x + a^4*b^3), -1/4*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(a
*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B*a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2
+ 2*a^3*b^4*x + a^4*b^3)]
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giac [A] time = 1.24, size = 82, normalized size = 0.82 \[ \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} - \frac {5 \, B a b x^{\frac {3}{2}} - A b^{2} x^{\frac {3}{2}} + 3 \, B a^{2} \sqrt {x} + A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*x^(1/2)/(b*x+a)^3,x, algorithm="giac")
[Out]
1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2) - 1/4*(5*B*a*b*x^(3/2) - A*b^2*x^(3/2) + 3*B*a
^2*sqrt(x) + A*a*b*sqrt(x))/((b*x + a)^2*a*b^2)
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maple [A] time = 0.02, size = 94, normalized size = 0.94 \[ \frac {A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a b}+\frac {3 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{2}}+\frac {\frac {\left (A b -5 B a \right ) x^{\frac {3}{2}}}{4 a b}-\frac {\left (A b +3 B a \right ) \sqrt {x}}{4 b^{2}}}{\left (b x +a \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*x^(1/2)/(b*x+a)^3,x)
[Out]
2*(1/8*(A*b-5*B*a)/a/b*x^(3/2)-1/8*(A*b+3*B*a)/b^2*x^(1/2))/(b*x+a)^2+1/4/b/a/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)
*b*x^(1/2))*A+3/4/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B
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maxima [A] time = 1.99, size = 94, normalized size = 0.94 \[ -\frac {{\left (5 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + {\left (3 \, B a^{2} + A a b\right )} \sqrt {x}}{4 \, {\left (a b^{4} x^{2} + 2 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*x^(1/2)/(b*x+a)^3,x, algorithm="maxima")
[Out]
-1/4*((5*B*a*b - A*b^2)*x^(3/2) + (3*B*a^2 + A*a*b)*sqrt(x))/(a*b^4*x^2 + 2*a^2*b^3*x + a^3*b^2) + 1/4*(3*B*a
+ A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2)
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mupad [B] time = 0.45, size = 84, normalized size = 0.84 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+3\,B\,a\right )}{4\,a^{3/2}\,b^{5/2}}-\frac {\frac {\sqrt {x}\,\left (A\,b+3\,B\,a\right )}{4\,b^2}-\frac {x^{3/2}\,\left (A\,b-5\,B\,a\right )}{4\,a\,b}}{a^2+2\,a\,b\,x+b^2\,x^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^(1/2)*(A + B*x))/(a + b*x)^3,x)
[Out]
(atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b + 3*B*a))/(4*a^(3/2)*b^(5/2)) - ((x^(1/2)*(A*b + 3*B*a))/(4*b^2) - (x^(3
/2)*(A*b - 5*B*a))/(4*a*b))/(a^2 + b^2*x^2 + 2*a*b*x)
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sympy [A] time = 20.73, size = 1499, normalized size = 14.99 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*x**(1/2)/(b*x+a)**3,x)
[Out]
Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/a**
3, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b**3, Eq(a, 0)), (-2*I*A*a**(3/2)*b**2*sqrt(x)*sqrt(1/b)/(8*I
*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) + 2*I*A*sqrt(a)*
b**3*x**(3/2)*sqrt(1/b)/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2
*sqrt(1/b)) + A*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x
*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - A*a**2*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**
3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) + 2*A*a*b**2*x*log(-I*sqrt(a)
*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*s
qrt(1/b)) - 2*A*a*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*
x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) + A*b**3*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2
)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - A*b**3*x**2*log(I*sqrt
(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**
2*sqrt(1/b)) - 6*I*B*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b
) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - 10*I*B*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*I*a**(7/2)*b**3*sqrt(1/b) +
16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) + 3*B*a**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt
(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - 3*B*a
**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**
(3/2)*b**5*x**2*sqrt(1/b)) + 6*B*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 1
6*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - 6*B*a**2*b*x*log(I*sqrt(a)*sqrt(1/b) + sqr
t(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) + 3*B*
a*b**2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sqrt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b)
+ 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)) - 3*B*a*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(7/2)*b**3*sq
rt(1/b) + 16*I*a**(5/2)*b**4*x*sqrt(1/b) + 8*I*a**(3/2)*b**5*x**2*sqrt(1/b)), True))
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