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

Optimal. Leaf size=82 \[ \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}+\frac {\sqrt {a+b x} (A b-4 a B)}{4 a x}-\frac {A (a+b x)^{3/2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 82, 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, 208} \[ \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}+\frac {\sqrt {a+b x} (A b-4 a B)}{4 a x}-\frac {A (a+b x)^{3/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^3,x]

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.99 \[ \frac {-b x^2 \sqrt {\frac {b x}{a}+1} (4 a B-A b) \tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )-(a+b x) (2 a (A+2 B x)+A b x)}{4 a x^2 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*x)*(A*b*x + 2*a*(A + 2*B*x))) - b*(-(A*b) + 4*a*B)*x^2*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]])
/(4*a*x^2*Sqrt[a + b*x])

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fricas [A]  time = 0.60, size = 158, normalized size = 1.93 \[ \left [-\frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {a} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x\right )} \sqrt {b x + a}}{8 \, a^{2} x^{2}}, \frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x\right )} \sqrt {b x + a}}{4 \, a^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.25, size = 110, normalized size = 1.34 \[ \frac {\frac {{\left (4 \, B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x + a} B a^{2} b^{2} + {\left (b x + a\right )}^{\frac {3}{2}} A b^{3} + \sqrt {b x + a} A a b^{3}}{a b^{2} x^{2}}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 75, normalized size = 0.91 \[ 2 \left (\frac {\left (A b -4 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}+\frac {-\frac {\left (A b +4 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a}+\left (-\frac {A b}{8}+\frac {B a}{2}\right ) \sqrt {b x +a}}{b^{2} x^{2}}\right ) b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*((-1/8*(A*b+4*B*a)/a*(b*x+a)^(3/2)+(1/2*B*a-1/8*A*b)*(b*x+a)^(1/2))/x^2/b^2+1/8*(A*b-4*B*a)/a^(3/2)*arctan
h((b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 1.96, size = 120, normalized size = 1.46 \[ -\frac {1}{8} \, b^{2} {\left (\frac {2 \, {\left ({\left (4 \, B a + A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - {\left (4 \, B a^{2} - A a b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{2} a b - 2 \, {\left (b x + a\right )} a^{2} b + a^{3} b} - \frac {{\left (4 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^2*(2*((4*B*a + A*b)*(b*x + a)^(3/2) - (4*B*a^2 - A*a*b)*sqrt(b*x + a))/((b*x + a)^2*a*b - 2*(b*x + a)*a
^2*b + a^3*b) - (4*B*a - A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(3/2)*b))

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mupad [B]  time = 0.12, size = 94, normalized size = 1.15 \[ \frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-4\,B\,a\right )}{4\,a^{3/2}}-\frac {\left (\frac {A\,b^2}{4}-B\,a\,b\right )\,\sqrt {a+b\,x}+\frac {\left (A\,b^2+4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^{3/2}}{4\,a}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 36.42, size = 372, normalized size = 4.54 \[ - \frac {10 A a^{2} b^{2} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {6 A a b^{2} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {3 A a b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 A a b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {A b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {A b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} - \frac {A b \sqrt {a + b x}}{a x} - \frac {B a b \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {B a b \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {2 B b \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {B \sqrt {a + b x}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*A*a**2*b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*A*a*b**2*(a + b*x)**(3/2)/(-8*
a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*a*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/
8 - 3*A*a*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - A*b**2*sqrt(a**(-3))*log(-a**2*sqrt(a
**(-3)) + sqrt(a + b*x))/2 + A*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 - A*b*sqrt(a + b*x
)/(a*x) - B*a*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + B*a*b*sqrt(a**(-3))*log(a**2*sqrt(a
**(-3)) + sqrt(a + b*x))/2 + 2*B*b*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - B*sqrt(a + b*x)/x

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