\(\int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx\) [450]

Optimal result

Integrand size = 18, antiderivative size = 98 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\frac {-5 A b+2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}}-\frac {5 A b-2 a B}{a^3 \sqrt {a+b x}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

1/3*(-5*A*b+2*B*a)/a^2/(b*x+a)^(3/2)-A/a/x/(b*x+a)^(3/2)+(5*A*b-2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(7/2)+
(-5*A*b+2*B*a)/a^3/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 53, 65, 214} \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {5 A b-2 a B}{a^3 \sqrt {a+b x}}-\frac {5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}} \]

[In]

Int[(A + B*x)/(x^2*(a + b*x)^(5/2)),x]

[Out]

-1/3*(5*A*b - 2*a*B)/(a^2*(a + b*x)^(3/2)) - A/(a*x*(a + b*x)^(3/2)) - (5*A*b - 2*a*B)/(a^3*Sqrt[a + b*x]) + (
(5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(7/2)

Rule 53

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 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 x (a+b x)^{3/2}}+\frac {\left (-\frac {5 A b}{2}+a B\right ) \int \frac {1}{x (a+b x)^{5/2}} \, dx}{a} \\ & = -\frac {5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}}-\frac {(5 A b-2 a B) \int \frac {1}{x (a+b x)^{3/2}} \, dx}{2 a^2} \\ & = -\frac {5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}}-\frac {5 A b-2 a B}{a^3 \sqrt {a+b x}}-\frac {(5 A b-2 a B) \int \frac {1}{x \sqrt {a+b x}} \, dx}{2 a^3} \\ & = -\frac {5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}}-\frac {5 A b-2 a B}{a^3 \sqrt {a+b x}}-\frac {(5 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 )}{a^3 b} \\ & = -\frac {5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac {A}{a x (a+b x)^{3/2}}-\frac {5 A b-2 a B}{a^3 \sqrt {a+b x}}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\frac {-15 A b^2 x^2+2 a b x (-10 A+3 B x)+a^2 (-3 A+8 B x)}{3 a^3 x (a+b x)^{3/2}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[In]

Integrate[(A + B*x)/(x^2*(a + b*x)^(5/2)),x]

[Out]

(-15*A*b^2*x^2 + 2*a*b*x*(-10*A + 3*B*x) + a^2*(-3*A + 8*B*x))/(3*a^3*x*(a + b*x)^(3/2)) + ((5*A*b - 2*a*B)*Ar
cTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(7/2)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(-\frac {-5 x \left (b x +a \right )^{\frac {3}{2}} \left (A b -\frac {2 B a}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {20 x \left (-\frac {3 B x}{10}+A \right ) b \,a^{\frac {3}{2}}}{3}+\left (-\frac {8 B x}{3}+A \right ) a^{\frac {5}{2}}+5 A \sqrt {a}\, b^{2} x^{2}}{\left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}} x}\)	\(82\)
risch	\(-\frac {A \sqrt {b x +a}}{a^{3} x}-\frac {-\frac {2 \left (-4 A b +2 B a \right )}{\sqrt {b x +a}}+\frac {4 a \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}-\frac {2 \left (5 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 a^{3}}\)	\(86\)
derivativedivides	\(-\frac {2 \left (2 A b -B a \right )}{a^{3} \sqrt {b x +a}}-\frac {2 \left (A b -B a \right )}{3 a^{2} \left (b x +a \right )^{\frac {3}{2}}}+\frac {-\frac {A \sqrt {b x +a}}{x}+\frac {\left (5 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}}{a^{3}}\)	\(88\)
default	\(-\frac {2 \left (2 A b -B a \right )}{a^{3} \sqrt {b x +a}}-\frac {2 \left (A b -B a \right )}{3 a^{2} \left (b x +a \right )^{\frac {3}{2}}}+\frac {-\frac {A \sqrt {b x +a}}{x}+\frac {\left (5 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}}{a^{3}}\)	\(88\)

[In]

int((B*x+A)/x^2/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-(-5*x*(b*x+a)^(3/2)*(A*b-2/5*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))+20/3*x*(-3/10*B*x+A)*b*a^(3/2)+(-8/3*B*x+A)*
a^(5/2)+5*A*a^(1/2)*b^2*x^2)/(b*x+a)^(3/2)/a^(7/2)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.37 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, A a^{3} - 3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{6 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, \frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (3 \, A a^{3} - 3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{3 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \]

[In]

integrate((B*x+A)/x^2/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*((2*B*a*b^2 - 5*A*b^3)*x^3 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^2 + (2*B*a^3 - 5*A*a^2*b)*x)*sqrt(a)*log((b*
x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*A*a^3 - 3*(2*B*a^2*b - 5*A*a*b^2)*x^2 - 4*(2*B*a^3 - 5*A*a^2*b)*x
)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), 1/3*(3*((2*B*a*b^2 - 5*A*b^3)*x^3 + 2*(2*B*a^2*b - 5*A*a
*b^2)*x^2 + (2*B*a^3 - 5*A*a^2*b)*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (3*A*a^3 - 3*(2*B*a^2*b - 5*A
*a*b^2)*x^2 - 4*(2*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1520 vs. \(2 (90) = 180\).

Time = 23.35 (sec) , antiderivative size = 1520, normalized size of antiderivative = 15.51 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((B*x+A)/x**2/(b*x+a)**(5/2),x)

[Out]

A*(-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x
**4) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2
)*b**3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(3
3/2)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b*
*2*x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b**2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1
8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**15*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(6*a**(3
9/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b
*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**14*b**3*x
**3*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**
14*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33
/2)*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 +
6*a**(33/2)*b**3*x**4) + 30*a**13*b**4*x**4*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1
8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)) + B*(8*a**7*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9
*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 3*a**7*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)
*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 6*a**7*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(1
5/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 14*a**6*b*x*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(1
5/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a**6*b*x*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b
**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*a**6*b*x*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a*
*(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 6*a**5*b**2*x**2*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x +
9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a**5*b**2*x**2*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x +
9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*a**5*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*
a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 3*a**4*b**3*x**3*log(b*x/a)/(3*a**(19/2) + 9*
a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 6*a**4*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(3*
a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=-\frac {1}{6} \, b {\left (\frac {2 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b - 3 \, {\left (2 \, B a - 5 \, A b\right )} {\left (b x + a\right )}^{2} + 2 \, {\left (2 \, B a^{2} - 5 \, A a b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {5}{2}} a^{3} b - {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b} - \frac {3 \, {\left (2 \, B a - 5 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]

[In]

integrate((B*x+A)/x^2/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*(2*(2*B*a^3 - 2*A*a^2*b - 3*(2*B*a - 5*A*b)*(b*x + a)^2 + 2*(2*B*a^2 - 5*A*a*b)*(b*x + a))/((b*x + a)^(
5/2)*a^3*b - (b*x + a)^(3/2)*a^4*b) - 3*(2*B*a - 5*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)
))/(a^(7/2)*b))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {\sqrt {b x + a} A}{a^{3} x} + \frac {2 \, {\left (3 \, {\left (b x + a\right )} B a + B a^{2} - 6 \, {\left (b x + a\right )} A b - A a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3}} \]

[In]

integrate((B*x+A)/x^2/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

(2*B*a - 5*A*b)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - sqrt(b*x + a)*A/(a^3*x) + 2/3*(3*(b*x + a)*B*a
+ B*a^2 - 6*(b*x + a)*A*b - A*a*b)/((b*x + a)^(3/2)*a^3)

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (5\,A\,b-2\,B\,a\right )}{a^{7/2}}-\frac {\frac {2\,\left (A\,b-B\,a\right )}{3\,a}+\frac {2\,\left (5\,A\,b-2\,B\,a\right )\,\left (a+b\,x\right )}{3\,a^2}-\frac {\left (5\,A\,b-2\,B\,a\right )\,{\left (a+b\,x\right )}^2}{a^3}}{a\,{\left (a+b\,x\right )}^{3/2}-{\left (a+b\,x\right )}^{5/2}} \]

[In]

int((A + B*x)/(x^2*(a + b*x)^(5/2)),x)

[Out]

(atanh((a + b*x)^(1/2)/a^(1/2))*(5*A*b - 2*B*a))/a^(7/2) - ((2*(A*b - B*a))/(3*a) + (2*(5*A*b - 2*B*a)*(a + b*
x))/(3*a^2) - ((5*A*b - 2*B*a)*(a + b*x)^2)/a^3)/(a*(a + b*x)^(3/2) - (a + b*x)^(5/2))