\(\int \frac {A+B x}{x^{11/2} (a+b x)} \, dx\) [353]

Optimal result

Integrand size = 18, antiderivative size = 136 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=-\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {2 b^{7/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \]

[Out]

-2/9*A/a/x^(9/2)+2/7*(A*b-B*a)/a^2/x^(7/2)-2/5*b*(A*b-B*a)/a^3/x^(5/2)+2/3*b^2*(A*b-B*a)/a^4/x^(3/2)-2*b^(7/2)
*(A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(11/2)-2*b^3*(A*b-B*a)/a^5/x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 53, 65, 211} \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=-\frac {2 b^{7/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 A}{9 a x^{9/2}} \]

[In]

Int[(A + B*x)/(x^(11/2)*(a + b*x)),x]

[Out]

(-2*A)/(9*a*x^(9/2)) + (2*(A*b - a*B))/(7*a^2*x^(7/2)) - (2*b*(A*b - a*B))/(5*a^3*x^(5/2)) + (2*b^2*(A*b - a*B
))/(3*a^4*x^(3/2)) - (2*b^3*(A*b - a*B))/(a^5*Sqrt[x]) - (2*b^(7/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[
a]])/a^(11/2)

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 211

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A}{9 a x^{9/2}}+\frac {\left (2 \left (-\frac {9 A b}{2}+\frac {9 a B}{2}\right )\right ) \int \frac {1}{x^{9/2} (a+b x)} \, dx}{9 a} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}+\frac {(b (A b-a B)) \int \frac {1}{x^{7/2} (a+b x)} \, dx}{a^2} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}-\frac {\left (b^2 (A b-a B)\right ) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{a^3} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}+\frac {\left (b^3 (A b-a B)\right ) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{a^4} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {\left (b^4 (A b-a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{a^5} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {\left (2 b^4 (A b-a B)\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{a^5} \\ & = -\frac {2 A}{9 a x^{9/2}}+\frac {2 (A b-a B)}{7 a^2 x^{7/2}}-\frac {2 b (A b-a B)}{5 a^3 x^{5/2}}+\frac {2 b^2 (A b-a B)}{3 a^4 x^{3/2}}-\frac {2 b^3 (A b-a B)}{a^5 \sqrt {x}}-\frac {2 b^{7/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=-\frac {2 \left (315 A b^4 x^4-105 a b^3 x^3 (A+3 B x)+21 a^2 b^2 x^2 (3 A+5 B x)-9 a^3 b x (5 A+7 B x)+5 a^4 (7 A+9 B x)\right )}{315 a^5 x^{9/2}}+\frac {2 b^{7/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \]

[In]

Integrate[(A + B*x)/(x^(11/2)*(a + b*x)),x]

[Out]

(-2*(315*A*b^4*x^4 - 105*a*b^3*x^3*(A + 3*B*x) + 21*a^2*b^2*x^2*(3*A + 5*B*x) - 9*a^3*b*x*(5*A + 7*B*x) + 5*a^
4*(7*A + 9*B*x)))/(315*a^5*x^(9/2)) + (2*b^(7/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(11/2)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {2 b^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{5} \sqrt {a b}}-\frac {2 A}{9 a \,x^{\frac {9}{2}}}-\frac {2 \left (-A b +B a \right )}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 b^{3} \left (A b -B a \right )}{a^{5} \sqrt {x}}-\frac {2 b \left (A b -B a \right )}{5 a^{3} x^{\frac {5}{2}}}+\frac {2 b^{2} \left (A b -B a \right )}{3 a^{4} x^{\frac {3}{2}}}\)	\(114\)
default	\(-\frac {2 b^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{5} \sqrt {a b}}-\frac {2 A}{9 a \,x^{\frac {9}{2}}}-\frac {2 \left (-A b +B a \right )}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 b^{3} \left (A b -B a \right )}{a^{5} \sqrt {x}}-\frac {2 b \left (A b -B a \right )}{5 a^{3} x^{\frac {5}{2}}}+\frac {2 b^{2} \left (A b -B a \right )}{3 a^{4} x^{\frac {3}{2}}}\)	\(114\)
risch	\(-\frac {2 \left (315 A \,b^{4} x^{4}-315 B a \,b^{3} x^{4}-105 A a \,b^{3} x^{3}+105 B \,a^{2} b^{2} x^{3}+63 A \,a^{2} b^{2} x^{2}-63 B \,a^{3} b \,x^{2}-45 A \,a^{3} b x +45 B \,a^{4} x +35 A \,a^{4}\right )}{315 a^{5} x^{\frac {9}{2}}}-\frac {2 b^{4} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{5} \sqrt {a b}}\)	\(127\)

[In]

int((B*x+A)/x^(11/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-2*b^4*(A*b-B*a)/a^5/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))-2/9*A/a/x^(9/2)-2/7*(-A*b+B*a)/a^2/x^(7/2)-2*b^
3*(A*b-B*a)/a^5/x^(1/2)-2/5*b*(A*b-B*a)/a^3/x^(5/2)+2/3*b^2*(A*b-B*a)/a^4/x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.14 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=\left [-\frac {315 \, {\left (B a b^{3} - A b^{4}\right )} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{315 \, a^{5} x^{5}}, -\frac {2 \, {\left (315 \, {\left (B a b^{3} - A b^{4}\right )} x^{5} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}\right )}}{315 \, a^{5} x^{5}}\right ] \]

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="fricas")

[Out]

[-1/315*(315*(B*a*b^3 - A*b^4)*x^5*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(35*A*a^4
- 315*(B*a*b^3 - A*b^4)*x^4 + 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 - A*a^3
*b)*x)*sqrt(x))/(a^5*x^5), -2/315*(315*(B*a*b^3 - A*b^4)*x^5*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) + (35*A
*a^4 - 315*(B*a*b^3 - A*b^4)*x^4 + 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 -
A*a^3*b)*x)*sqrt(x))/(a^5*x^5)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (131) = 262\).

Time = 44.23 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.45 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{9 x^{\frac {9}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{9 x^{\frac {9}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{9 a x^{\frac {9}{2}}} + \frac {2 A b}{7 a^{2} x^{\frac {7}{2}}} - \frac {2 A b^{2}}{5 a^{3} x^{\frac {5}{2}}} + \frac {2 A b^{3}}{3 a^{4} x^{\frac {3}{2}}} - \frac {A b^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{5} \sqrt {- \frac {a}{b}}} + \frac {A b^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{5} \sqrt {- \frac {a}{b}}} - \frac {2 A b^{4}}{a^{5} \sqrt {x}} - \frac {2 B}{7 a x^{\frac {7}{2}}} + \frac {2 B b}{5 a^{2} x^{\frac {5}{2}}} - \frac {2 B b^{2}}{3 a^{3} x^{\frac {3}{2}}} + \frac {B b^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{4} \sqrt {- \frac {a}{b}}} - \frac {B b^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{4} \sqrt {- \frac {a}{b}}} + \frac {2 B b^{3}}{a^{4} \sqrt {x}} & \text {otherwise} \end {cases} \]

[In]

integrate((B*x+A)/x**(11/2)/(b*x+a),x)

[Out]

Piecewise((zoo*(-2*A/(11*x**(11/2)) - 2*B/(9*x**(9/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(11*x**(11/2)) - 2*B/(9*
x**(9/2)))/b, Eq(a, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/a, Eq(b, 0)), (-2*A/(9*a*x**(9/2)) + 2*A*b/(7
*a**2*x**(7/2)) - 2*A*b**2/(5*a**3*x**(5/2)) + 2*A*b**3/(3*a**4*x**(3/2)) - A*b**4*log(sqrt(x) - sqrt(-a/b))/(
a**5*sqrt(-a/b)) + A*b**4*log(sqrt(x) + sqrt(-a/b))/(a**5*sqrt(-a/b)) - 2*A*b**4/(a**5*sqrt(x)) - 2*B/(7*a*x**
(7/2)) + 2*B*b/(5*a**2*x**(5/2)) - 2*B*b**2/(3*a**3*x**(3/2)) + B*b**3*log(sqrt(x) - sqrt(-a/b))/(a**4*sqrt(-a
/b)) - B*b**3*log(sqrt(x) + sqrt(-a/b))/(a**4*sqrt(-a/b)) + 2*B*b**3/(a**4*sqrt(x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{4} - A b^{5}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} - \frac {2 \, {\left (35 \, A a^{4} - 315 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} + 105 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 63 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 45 \, {\left (B a^{4} - A a^{3} b\right )} x\right )}}{315 \, a^{5} x^{\frac {9}{2}}} \]

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="maxima")

[Out]

2*(B*a*b^4 - A*b^5)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 2/315*(35*A*a^4 - 315*(B*a*b^3 - A*b^4)*x^4
+ 105*(B*a^2*b^2 - A*a*b^3)*x^3 - 63*(B*a^3*b - A*a^2*b^2)*x^2 + 45*(B*a^4 - A*a^3*b)*x)/(a^5*x^(9/2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{4} - A b^{5}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} + \frac {2 \, {\left (315 \, B a b^{3} x^{4} - 315 \, A b^{4} x^{4} - 105 \, B a^{2} b^{2} x^{3} + 105 \, A a b^{3} x^{3} + 63 \, B a^{3} b x^{2} - 63 \, A a^{2} b^{2} x^{2} - 45 \, B a^{4} x + 45 \, A a^{3} b x - 35 \, A a^{4}\right )}}{315 \, a^{5} x^{\frac {9}{2}}} \]

[In]

integrate((B*x+A)/x^(11/2)/(b*x+a),x, algorithm="giac")

[Out]

2*(B*a*b^4 - A*b^5)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) + 2/315*(315*B*a*b^3*x^4 - 315*A*b^4*x^4 - 105
*B*a^2*b^2*x^3 + 105*A*a*b^3*x^3 + 63*B*a^3*b*x^2 - 63*A*a^2*b^2*x^2 - 45*B*a^4*x + 45*A*a^3*b*x - 35*A*a^4)/(
a^5*x^(9/2))

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^{11/2} (a+b x)} \, dx=-\frac {\frac {2\,A}{9\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{7\,a^2}-\frac {2\,b^2\,x^3\,\left (A\,b-B\,a\right )}{3\,a^4}+\frac {2\,b^3\,x^4\,\left (A\,b-B\,a\right )}{a^5}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{5\,a^3}}{x^{9/2}}-\frac {2\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{11/2}} \]

[In]

int((A + B*x)/(x^(11/2)*(a + b*x)),x)

[Out]

- ((2*A)/(9*a) - (2*x*(A*b - B*a))/(7*a^2) - (2*b^2*x^3*(A*b - B*a))/(3*a^4) + (2*b^3*x^4*(A*b - B*a))/a^5 + (
2*b*x^2*(A*b - B*a))/(5*a^3))/x^(9/2) - (2*b^(7/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b - B*a))/a^(11/2)