\(\int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx\) [395]

Optimal result

Integrand size = 18, antiderivative size = 112 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=\frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}-\frac {b^2 (A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \]

[Out]

-1/3*A*(b*x+a)^(3/2)/a/x^3-1/8*b^2*(A*b-2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(5/2)+1/4*(A*b-2*B*a)*(b*x+a)^
(1/2)/a/x^2+1/8*b*(A*b-2*B*a)*(b*x+a)^(1/2)/a^2/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 43, 44, 65, 214} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=-\frac {b^2 (A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {b \sqrt {a+b x} (A b-2 a B)}{8 a^2 x}+\frac {\sqrt {a+b x} (A b-2 a B)}{4 a x^2}-\frac {A (a+b x)^{3/2}}{3 a x^3} \]

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^4,x]

[Out]

((A*b - 2*a*B)*Sqrt[a + b*x])/(4*a*x^2) + (b*(A*b - 2*a*B)*Sqrt[a + b*x])/(8*a^2*x) - (A*(a + b*x)^(3/2))/(3*a
*x^3) - (b^2*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 214

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

Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {\left (-\frac {3 A b}{2}+3 a B\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{3 a} \\ & = \frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}-\frac {A (a+b x)^{3/2}}{3 a x^3}-\frac {(b (A b-2 a B)) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{8 a} \\ & = \frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {\left (b^2 (A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^2} \\ & = \frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}+\frac {(b (A b-2 a B)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^2} \\ & = \frac {(A b-2 a B) \sqrt {a+b x}}{4 a x^2}+\frac {b (A b-2 a B) \sqrt {a+b x}}{8 a^2 x}-\frac {A (a+b x)^{3/2}}{3 a x^3}-\frac {b^2 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=\frac {\sqrt {a+b x} \left (3 A b^2 x^2-2 a b x (A+3 B x)-4 a^2 (2 A+3 B x)\right )}{24 a^2 x^3}+\frac {b^2 (-A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \]

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x))/x^4,x]

[Out]

(Sqrt[a + b*x]*(3*A*b^2*x^2 - 2*a*b*x*(A + 3*B*x) - 4*a^2*(2*A + 3*B*x)))/(24*a^2*x^3) + (b^2*(-(A*b) + 2*a*B)
*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2))

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-3 A \,b^{2} x^{2}+6 B a b \,x^{2}+2 a A b x +12 a^{2} B x +8 a^{2} A \right )}{24 x^{3} a^{2}}-\frac {b^{2} \left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\)	\(82\)
pseudoelliptic	\(-\frac {\frac {3 b^{2} x^{3} \left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8}+\sqrt {b x +a}\, \left (\frac {b x \left (3 B x +A \right ) a^{\frac {3}{2}}}{4}+\left (\frac {3 B x}{2}+A \right ) a^{\frac {5}{2}}-\frac {3 A \sqrt {a}\, b^{2} x^{2}}{8}\right )}{3 a^{\frac {5}{2}} x^{3}}\)	\(82\)
derivativedivides	\(2 b^{2} \left (-\frac {-\frac {\left (A b -2 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {A b \left (b x +a \right )^{\frac {3}{2}}}{6 a}+\left (\frac {A b}{16}-\frac {B a}{8}\right ) \sqrt {b x +a}}{b^{3} x^{3}}-\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\)	\(92\)
default	\(2 b^{2} \left (-\frac {-\frac {\left (A b -2 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {A b \left (b x +a \right )^{\frac {3}{2}}}{6 a}+\left (\frac {A b}{16}-\frac {B a}{8}\right ) \sqrt {b x +a}}{b^{3} x^{3}}-\frac {\left (A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\)	\(92\)

[In]

int((B*x+A)*(b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/24*(b*x+a)^(1/2)*(-3*A*b^2*x^2+6*B*a*b*x^2+2*A*a*b*x+12*B*a^2*x+8*A*a^2)/x^3/a^2-1/8*b^2*(A*b-2*B*a)*arctan
h((b*x+a)^(1/2)/a^(1/2))/a^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=\left [-\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{3} + 3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, -\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, A a^{3} + 3 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[-1/48*(3*(2*B*a*b^2 - A*b^3)*sqrt(a)*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^3 + 3*(2*B*a
^2*b - A*a*b^2)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt(b*x + a))/(a^3*x^3), -1/24*(3*(2*B*a*b^2 - A*b^3)*sqrt(-a)
*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*A*a^3 + 3*(2*B*a^2*b - A*a*b^2)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt
(b*x + a))/(a^3*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (99) = 198\).

Time = 41.63 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.11 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=- \frac {A a}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A b^{\frac {3}{2}}}{24 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A b^{\frac {5}{2}}}{8 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {5}{2}}} - \frac {B a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 B \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {B b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {B b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**4,x)

[Out]

-A*a/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 5*A*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) + A*b**(3/2)/(24*a*x
**(3/2)*sqrt(a/(b*x) + 1)) + A*b**(5/2)/(8*a**2*sqrt(x)*sqrt(a/(b*x) + 1)) - A*b**3*asinh(sqrt(a)/(sqrt(b)*sqr
t(x)))/(8*a**(5/2)) - B*a/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) - 3*B*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1))
- B*b**(3/2)/(4*a*sqrt(x)*sqrt(a/(b*x) + 1)) + B*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=-\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (8 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b + 3 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 3 \, {\left (2 \, B a^{3} - A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{3} a^{2} b - 3 \, {\left (b x + a\right )}^{2} a^{3} b + 3 \, {\left (b x + a\right )} a^{4} b - a^{5} b} + \frac {3 \, {\left (2 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

-1/48*b^3*(2*(8*(b*x + a)^(3/2)*A*a*b + 3*(2*B*a - A*b)*(b*x + a)^(5/2) - 3*(2*B*a^3 - A*a^2*b)*sqrt(b*x + a))
/((b*x + a)^3*a^2*b - 3*(b*x + a)^2*a^3*b + 3*(b*x + a)*a^4*b - a^5*b) + 3*(2*B*a - A*b)*log((sqrt(b*x + a) -
sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(5/2)*b))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=-\frac {\frac {3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {6 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{3} - 6 \, \sqrt {b x + a} B a^{3} b^{3} - 3 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{4} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{4} + 3 \, \sqrt {b x + a} A a^{2} b^{4}}{a^{2} b^{3} x^{3}}}{24 \, b} \]

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(3*(2*B*a*b^3 - A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (6*(b*x + a)^(5/2)*B*a*b^3 - 6*sq
rt(b*x + a)*B*a^3*b^3 - 3*(b*x + a)^(5/2)*A*b^4 + 8*(b*x + a)^(3/2)*A*a*b^4 + 3*sqrt(b*x + a)*A*a^2*b^4)/(a^2*
b^3*x^3))/b

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^4} \, dx=\frac {\left (\frac {A\,b^3}{8}-\frac {B\,a\,b^2}{4}\right )\,\sqrt {a+b\,x}-\frac {\left (A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^{5/2}}{8\,a^2}+\frac {A\,b^3\,{\left (a+b\,x\right )}^{3/2}}{3\,a}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-2\,B\,a\right )}{8\,a^{5/2}} \]

[In]

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

[Out]

(((A*b^3)/8 - (B*a*b^2)/4)*(a + b*x)^(1/2) - ((A*b^3 - 2*B*a*b^2)*(a + b*x)^(5/2))/(8*a^2) + (A*b^3*(a + b*x)^
(3/2))/(3*a))/(3*a*(a + b*x)^2 - 3*a^2*(a + b*x) - (a + b*x)^3 + a^3) - (b^2*atanh((a + b*x)^(1/2)/a^(1/2))*(A
*b - 2*B*a))/(8*a^(5/2))