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

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

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Rubi [A]  time = 0.04, antiderivative size = 85, 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, 50, 63, 205} \begin {gather*} -\frac {\sqrt {x} (A b-3 a B)}{a b^2}+\frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}+\frac {x^{3/2} (A b-a B)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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)^2} \, dx &=\frac {(A b-a B) x^{3/2}}{a b (a+b x)}-\frac {\left (\frac {A b}{2}-\frac {3 a B}{2}\right ) \int \frac {\sqrt {x}}{a+b x} \, dx}{a b}\\ &=-\frac {(A b-3 a B) \sqrt {x}}{a b^2}+\frac {(A b-a B) x^{3/2}}{a b (a+b x)}+\frac {(A b-3 a B) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 b^2}\\ &=-\frac {(A b-3 a B) \sqrt {x}}{a b^2}+\frac {(A b-a B) x^{3/2}}{a b (a+b x)}+\frac {(A b-3 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=-\frac {(A b-3 a B) \sqrt {x}}{a b^2}+\frac {(A b-a B) x^{3/2}}{a b (a+b x)}+\frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 67, normalized size = 0.79 \begin {gather*} \frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}+\frac {\sqrt {x} (3 a B-A b+2 b B x)}{b^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 67, normalized size = 0.79 \begin {gather*} \frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}+\frac {\sqrt {x} (3 a B-A b+2 b B x)}{b^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[x]*(A + B*x))/(a + b*x)^2,x]

[Out]

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

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fricas [A]  time = 1.10, size = 198, normalized size = 2.33 \begin {gather*} \left [\frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - 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 (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{2 \, {\left (a b^{4} x + a^{2} b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{a b^{4} x + a^{2} b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.26, size = 65, normalized size = 0.76 \begin {gather*} \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {B a \sqrt {x} - A b \sqrt {x}}{{\left (b x + a\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 87, normalized size = 1.02 \begin {gather*} \frac {A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}-\frac {3 B a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {A \sqrt {x}}{\left (b x +a \right ) b}+\frac {B a \sqrt {x}}{\left (b x +a \right ) b^{2}}+\frac {2 B \sqrt {x}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.00, size = 65, normalized size = 0.76 \begin {gather*} \frac {{\left (B a - A b\right )} \sqrt {x}}{b^{3} x + a b^{2}} + \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 62, normalized size = 0.73 \begin {gather*} \frac {2\,B\,\sqrt {x}}{b^2}-\frac {\sqrt {x}\,\left (A\,b-B\,a\right )}{x\,b^3+a\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-3\,B\,a\right )}{\sqrt {a}\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.28, size = 782, normalized size = 9.20 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{a^{2}} & \text {for}\: b = 0 \\- \frac {2 i A \sqrt {a} b^{2} \sqrt {x} \sqrt {\frac {1}{b}}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {A a b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {A a b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {A b^{2} x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {A b^{2} x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {6 i B a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {4 i B \sqrt {a} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {3 B a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {3 B a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} - \frac {3 B a b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} + \frac {3 B a b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{3} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{4} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x**(1/2)/(b*x+a)**2,x)

[Out]

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

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