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

Optimal. Leaf size=190 \[ \frac {2 (a+b x) (A b-a B)}{3 a^2 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A (a+b x)}{5 a x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b (a+b x) (A b-a B)}{a^3 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \[ -\frac {2 b (a+b x) (A b-a B)}{a^3 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) (A b-a B)}{3 a^2 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A (a+b x)}{5 a x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.32 \[ -\frac {2 (a+b x) \left (\, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {b x}{a}\right ) (5 a B x-5 A b x)+3 a A\right )}{15 a^2 x^{5/2} \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x)*(3*a*A + (-5*A*b*x + 5*a*B*x)*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x)/a)]))/(15*a^2*x^(5/2)*Sqr
t[(a + b*x)^2])

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fricas [A]  time = 1.13, size = 195, normalized size = 1.03 \[ \left [-\frac {15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, -\frac {2 \, {\left (15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),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)]

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giac [A]  time = 0.16, size = 122, normalized size = 0.64 \[ \frac {2 \, {\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {2 \, {\left (15 \, B a b x^{2} \mathrm {sgn}\left (b x + a\right ) - 15 \, A b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, B a^{2} x \mathrm {sgn}\left (b x + a\right ) + 5 \, A a b x \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 131, normalized size = 0.69 \[ -\frac {2 \left (b x +a \right ) \left (15 A \,b^{3} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-15 B a \,b^{2} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 \sqrt {a b}\, A \,b^{2} x^{2}-15 \sqrt {a b}\, B a b \,x^{2}-5 \sqrt {a b}\, A a b x +5 \sqrt {a b}\, B \,a^{2} x +3 \sqrt {a b}\, A \,a^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {a b}\, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

-2/15*(b*x+a)*(15*A*x^(5/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*b^3-15*B*x^(5/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*a
*b^2+15*A*x^2*(a*b)^(1/2)*b^2-15*B*x^2*(a*b)^(1/2)*a*b-5*A*x*(a*b)^(1/2)*a*b+5*B*x*(a*b)^(1/2)*a^2+3*A*a^2*(a*
b)^(1/2))/((b*x+a)^2)^(1/2)/a^3/(a*b)^(1/2)/x^(5/2)

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maxima [B]  time = 1.65, size = 305, normalized size = 1.61 \[ \frac {5 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{2} + 3 \, {\left (5 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x\right )} \sqrt {x} - \frac {10 \, {\left ({\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{2} - 3 \, {\left (5 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x\right )}}{\sqrt {x}} - \frac {10 \, {\left (3 \, {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{2} - {\left (5 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )}}{x^{\frac {3}{2}}} - \frac {2 \, {\left (5 \, {\left (3 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (5 \, B a^{5} - 7 \, A a^{4} b\right )} x\right )}}{x^{\frac {5}{2}}} - \frac {2 \, {\left (5 \, A a^{4} b x^{2} + 3 \, A a^{5} x\right )}}{x^{\frac {7}{2}}}}{15 \, {\left (a^{5} b x + a^{6}\right )}} + \frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {{\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} x^{\frac {3}{2}} + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} \sqrt {x}}{3 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x}{x^{7/2}\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*((a + b*x)^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^(7/2)*((a + b*x)^2)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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