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

Optimal. Leaf size=131 \[ -\frac {b^{3/2} (7 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}}-\frac {b (7 A b-5 a B)}{a^4 \sqrt {x}}+\frac {7 A b-5 a B}{3 a^3 x^{3/2}}-\frac {7 A b-5 a B}{5 a^2 b x^{5/2}}+\frac {A b-a B}{a b x^{5/2} (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {27, 78, 51, 63, 205} \[ -\frac {b^{3/2} (7 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2}}+\frac {7 A b-5 a B}{3 a^3 x^{3/2}}-\frac {7 A b-5 a B}{5 a^2 b x^{5/2}}-\frac {b (7 A b-5 a B)}{a^4 \sqrt {x}}+\frac {A b-a B}{a b x^{5/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

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 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 205

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 64, normalized size = 0.49 \[ \frac {(a+b x) (5 a B-7 A b) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {b x}{a}\right )+5 a (A b-a B)}{5 a^2 b x^{5/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*(A*b - a*B) + (-7*A*b + 5*a*B)*(a + b*x)*Hypergeometric2F1[-5/2, 1, -3/2, -((b*x)/a)])/(5*a^2*b*x^(5/2)*(
a + b*x))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 319, normalized size = 2.44 \[ \left [-\frac {15 \, {\left ({\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{30 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, -\frac {15 \, {\left ({\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(15*((5*B*a*b^2 - 7*A*b^3)*x^4 + (5*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b
/a) - a)/(b*x + a)) + 2*(6*A*a^3 - 15*(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*A*a*b^2)*x^2 + 2*(5*B*a^3
- 7*A*a^2*b)*x)*sqrt(x))/(a^4*b*x^4 + a^5*x^3), -1/15*(15*((5*B*a*b^2 - 7*A*b^3)*x^4 + (5*B*a^2*b - 7*A*a*b^2)
*x^3)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) + (6*A*a^3 - 15*(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*
A*a*b^2)*x^2 + 2*(5*B*a^3 - 7*A*a^2*b)*x)*sqrt(x))/(a^4*b*x^4 + a^5*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.16, size = 110, normalized size = 0.84 \[ \frac {{\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {B a b^{2} \sqrt {x} - A b^{3} \sqrt {x}}{{\left (b x + a\right )} a^{4}} + \frac {2 \, {\left (30 \, B a b x^{2} - 45 \, A b^{2} x^{2} - 5 \, B a^{2} x + 10 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{4} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*B*a*b^2 - 7*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + (B*a*b^2*sqrt(x) - A*b^3*sqrt(x))/((b*x +
a)*a^4) + 2/15*(30*B*a*b*x^2 - 45*A*b^2*x^2 - 5*B*a^2*x + 10*A*a*b*x - 3*A*a^2)/(a^4*x^(5/2))

________________________________________________________________________________________

maple [A]  time = 0.07, size = 139, normalized size = 1.06 \[ -\frac {7 A \,b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{4}}+\frac {5 B \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}-\frac {A \,b^{3} \sqrt {x}}{\left (b x +a \right ) a^{4}}+\frac {B \,b^{2} \sqrt {x}}{\left (b x +a \right ) a^{3}}-\frac {6 A \,b^{2}}{a^{4} \sqrt {x}}+\frac {4 B b}{a^{3} \sqrt {x}}+\frac {4 A b}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 B}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 A}{5 a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.32, size = 118, normalized size = 0.90 \[ -\frac {6 \, A a^{3} - 15 \, {\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} - 10 \, {\left (5 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 2 \, {\left (5 \, B a^{3} - 7 \, A a^{2} b\right )} x}{15 \, {\left (a^{4} b x^{\frac {7}{2}} + a^{5} x^{\frac {5}{2}}\right )}} + \frac {{\left (5 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(6*A*a^3 - 15*(5*B*a*b^2 - 7*A*b^3)*x^3 - 10*(5*B*a^2*b - 7*A*a*b^2)*x^2 + 2*(5*B*a^3 - 7*A*a^2*b)*x)/(a
^4*b*x^(7/2) + a^5*x^(5/2)) + (5*B*a*b^2 - 7*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4)

________________________________________________________________________________________

mupad [B]  time = 1.19, size = 103, normalized size = 0.79 \[ -\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (7\,A\,b-5\,B\,a\right )}{15\,a^2}+\frac {b^2\,x^3\,\left (7\,A\,b-5\,B\,a\right )}{a^4}+\frac {2\,b\,x^2\,\left (7\,A\,b-5\,B\,a\right )}{3\,a^3}}{a\,x^{5/2}+b\,x^{7/2}}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (7\,A\,b-5\,B\,a\right )}{a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*A)/(5*a) - (2*x*(7*A*b - 5*B*a))/(15*a^2) + (b^2*x^3*(7*A*b - 5*B*a))/a^4 + (2*b*x^2*(7*A*b - 5*B*a))/(3
*a^3))/(a*x^(5/2) + b*x^(7/2)) - (b^(3/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(7*A*b - 5*B*a))/a^(9/2)

________________________________________________________________________________________

sympy [A]  time = 155.21, size = 1127, normalized size = 8.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/a**2, Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/b**2, Eq(a, 0)), (-12*I*A*a**(7/2)*sqrt(1/b)/(
30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 28*I*A*a**(5/2)*b*x*sqrt(1/b)/(30*I*
a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 140*I*A*a**(3/2)*b**2*x**2*sqrt(1/b)/(30*
I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 210*I*A*sqrt(a)*b**3*x**3*sqrt(1/b)/(30
*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 105*A*a*b**2*x**(5/2)*log(-I*sqrt(a)*s
qrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 105*A*a*b**2*x*
*(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1
/b)) - 105*A*b**3*x**(7/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9
/2)*b*x**(7/2)*sqrt(1/b)) + 105*A*b**3*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sq
rt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 20*I*B*a**(7/2)*x*sqrt(1/b)/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b)
+ 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 100*I*B*a**(5/2)*b*x**2*sqrt(1/b)/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b)
+ 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 150*I*B*a**(3/2)*b**2*x**3*sqrt(1/b)/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b
) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 75*B*a**2*b*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(1
1/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 75*B*a**2*b*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) +
sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) + 75*B*a*b**2*x**(7/2)*log(
-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(7/2)*sqrt(1/b)) - 75*
B*a*b**2*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(11/2)*x**(5/2)*sqrt(1/b) + 30*I*a**(9/2)*b*x**(
7/2)*sqrt(1/b)), True))

________________________________________________________________________________________