\(\int \frac {A+B x}{x^{7/2} (a+b x)} \, dx\) [351]

Optimal result

Integrand size = 18, antiderivative size = 90 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}-\frac {2 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, 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, 211} \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {2 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 A}{5 a x^{5/2}} \]

[In]

Int[(A + B*x)/(x^(7/2)*(a + b*x)),x]

[Out]

(-2*A)/(5*a*x^(5/2)) + (2*(A*b - a*B))/(3*a^2*x^(3/2)) - (2*b*(A*b - a*B))/(a^3*Sqrt[x]) - (2*b^(3/2)*(A*b - a
*B)*ArcTan[(Sqrt[b]*Sqrt[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 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}{5 a x^{5/2}}+\frac {\left (2 \left (-\frac {5 A b}{2}+\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{5 a} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}+\frac {(b (A b-a B)) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{a^2} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}-\frac {\left (b^2 (A b-a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{a^3} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}-\frac {\left (2 b^2 (A b-a B)\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{a^3} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}-\frac {2 b^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {2 \left (15 A b^2 x^2-5 a b x (A+3 B x)+a^2 (3 A+5 B x)\right )}{15 a^3 x^{5/2}}+\frac {2 b^{3/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[In]

Integrate[(A + B*x)/(x^(7/2)*(a + b*x)),x]

[Out]

(-2*(15*A*b^2*x^2 - 5*a*b*x*(A + 3*B*x) + a^2*(3*A + 5*B*x)))/(15*a^3*x^(5/2)) + (2*b^(3/2)*(-(A*b) + a*B)*Arc
Tan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b \left (A b -B a \right )}{a^{3} \sqrt {x}}\)	\(76\)
default	\(-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b \left (A b -B a \right )}{a^{3} \sqrt {x}}\)	\(76\)
risch	\(-\frac {2 \left (15 A \,b^{2} x^{2}-15 B a b \,x^{2}-5 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{3} x^{\frac {5}{2}}}-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\)	\(79\)

[In]

int((B*x+A)/x^(7/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\left [-\frac {15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, -\frac {2 \, {\left (15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \]

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a),x, algorithm="fricas")

[Out]

[-1/15*(15*(B*a*b - A*b^2)*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(3*A*a^2 - 15*
(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*x^3), -2/15*(15*(B*a*b - A*b^2)*x^3*sqrt(b/a)*arctan(
a*sqrt(b/a)/(b*sqrt(x))) + (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (87) = 174\).

Time = 5.40 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.91 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{5 a x^{\frac {5}{2}}} + \frac {2 A b}{3 a^{2} x^{\frac {3}{2}}} - \frac {A b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} + \frac {A b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} - \frac {2 A b^{2}}{a^{3} \sqrt {x}} - \frac {2 B}{3 a x^{\frac {3}{2}}} + \frac {B b \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} - \frac {B b \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} + \frac {2 B b}{a^{2} \sqrt {x}} & \text {otherwise} \end {cases} \]

[In]

integrate((B*x+A)/x**(7/2)/(b*x+a),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a),x, algorithm="maxima")

[Out]

2*(B*a*b^2 - A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(
B*a^2 - A*a*b)*x)/(a^3*x^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {2 \, {\left (15 \, B a b x^{2} - 15 \, A b^{2} x^{2} - 5 \, B a^{2} x + 5 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]

[In]

integrate((B*x+A)/x^(7/2)/(b*x+a),x, algorithm="giac")

[Out]

2*(B*a*b^2 - A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2/15*(15*B*a*b*x^2 - 15*A*b^2*x^2 - 5*B*a^2*
x + 5*A*a*b*x - 3*A*a^2)/(a^3*x^(5/2))

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{3\,a^2}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{a^3}}{x^{5/2}}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{7/2}} \]

[In]

int((A + B*x)/(x^(7/2)*(a + b*x)),x)

[Out]

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