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

Optimal. Leaf size=178 \[ \frac {35 \sqrt {b} (3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}}+\frac {35 (3 A b-a B)}{8 a^5 \sqrt {x}}-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \begin {gather*} \frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {35 (3 A b-a B)}{8 a^5 \sqrt {x}}+\frac {35 \sqrt {b} (3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*(3*A*b - a*B))/(24*a^4*b*x^(3/2)) + (35*(3*A*b - a*B))/(8*a^5*Sqrt[x]) + (A*b - a*B)/(3*a*b*x^(3/2)*(a +
b*x)^3) + (3*A*b - a*B)/(4*a^2*b*x^(3/2)*(a + b*x)^2) + (7*(3*A*b - a*B))/(8*a^3*b*x^(3/2)*(a + b*x)) + (35*Sq
rt[b]*(3*A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(11/2))

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^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {A+B x}{x^{5/2} (a+b x)^4} \, dx\\ &=\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}-\frac {\left (-\frac {9 A b}{2}+\frac {3 a B}{2}\right ) \int \frac {1}{x^{5/2} (a+b x)^3} \, dx}{3 a b}\\ &=\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {(7 (3 A b-a B)) \int \frac {1}{x^{5/2} (a+b x)^2} \, dx}{8 a^2 b}\\ &=\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}+\frac {(35 (3 A b-a B)) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{16 a^3 b}\\ &=-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}-\frac {(35 (3 A b-a B)) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{16 a^4}\\ &=-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {35 (3 A b-a B)}{8 a^5 \sqrt {x}}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}+\frac {(35 b (3 A b-a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{16 a^5}\\ &=-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {35 (3 A b-a B)}{8 a^5 \sqrt {x}}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}+\frac {(35 b (3 A b-a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{8 a^5}\\ &=-\frac {35 (3 A b-a B)}{24 a^4 b x^{3/2}}+\frac {35 (3 A b-a B)}{8 a^5 \sqrt {x}}+\frac {A b-a B}{3 a b x^{3/2} (a+b x)^3}+\frac {3 A b-a B}{4 a^2 b x^{3/2} (a+b x)^2}+\frac {7 (3 A b-a B)}{8 a^3 b x^{3/2} (a+b x)}+\frac {35 \sqrt {b} (3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.34 \begin {gather*} \frac {\frac {3 a^3 (A b-a B)}{(a+b x)^3}+(3 a B-9 A b) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {b x}{a}\right )}{9 a^4 b x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^3*(A*b - a*B))/(a + b*x)^3 + (-9*A*b + 3*a*B)*Hypergeometric2F1[-3/2, 3, -1/2, -((b*x)/a)])/(9*a^4*b*x^(
3/2))

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IntegrateAlgebraic [A]  time = 0.26, size = 148, normalized size = 0.83 \begin {gather*} \frac {-16 a^4 A-48 a^4 B x+144 a^3 A b x-231 a^3 b B x^2+693 a^2 A b^2 x^2-280 a^2 b^2 B x^3+840 a A b^3 x^3-105 a b^3 B x^4+315 A b^4 x^4}{24 a^5 x^{3/2} (a+b x)^3}-\frac {35 \left (a \sqrt {b} B-3 A b^{3/2}\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 a^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a^4*A + 144*a^3*A*b*x - 48*a^4*B*x + 693*a^2*A*b^2*x^2 - 231*a^3*b*B*x^2 + 840*a*A*b^3*x^3 - 280*a^2*b^2*
B*x^3 + 315*A*b^4*x^4 - 105*a*b^3*B*x^4)/(24*a^5*x^(3/2)*(a + b*x)^3) - (35*(-3*A*b^(3/2) + a*Sqrt[b]*B)*ArcTa
n[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*a^(11/2))

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fricas [A]  time = 0.45, size = 482, normalized size = 2.71 \begin {gather*} \left [-\frac {105 \, {\left ({\left (B a b^{3} - 3 \, A b^{4}\right )} x^{5} + 3 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3} + {\left (B a^{4} - 3 \, A a^{3} b\right )} x^{2}\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 (16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x\right )} \sqrt {x}}{48 \, {\left (a^{5} b^{3} x^{5} + 3 \, a^{6} b^{2} x^{4} + 3 \, a^{7} b x^{3} + a^{8} x^{2}\right )}}, \frac {105 \, {\left ({\left (B a b^{3} - 3 \, A b^{4}\right )} x^{5} + 3 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3} + {\left (B a^{4} - 3 \, A a^{3} b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{5} b^{3} x^{5} + 3 \, a^{6} b^{2} x^{4} + 3 \, a^{7} b x^{3} + a^{8} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(105*((B*a*b^3 - 3*A*b^4)*x^5 + 3*(B*a^2*b^2 - 3*A*a*b^3)*x^4 + 3*(B*a^3*b - 3*A*a^2*b^2)*x^3 + (B*a^4
- 3*A*a^3*b)*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(16*A*a^4 + 105*(B*a*b^3 -
3*A*b^4)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3)*x^3 + 231*(B*a^3*b - 3*A*a^2*b^2)*x^2 + 48*(B*a^4 - 3*A*a^3*b)*x)*s
qrt(x))/(a^5*b^3*x^5 + 3*a^6*b^2*x^4 + 3*a^7*b*x^3 + a^8*x^2), 1/24*(105*((B*a*b^3 - 3*A*b^4)*x^5 + 3*(B*a^2*b
^2 - 3*A*a*b^3)*x^4 + 3*(B*a^3*b - 3*A*a^2*b^2)*x^3 + (B*a^4 - 3*A*a^3*b)*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b
*sqrt(x))) - (16*A*a^4 + 105*(B*a*b^3 - 3*A*b^4)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3)*x^3 + 231*(B*a^3*b - 3*A*a^
2*b^2)*x^2 + 48*(B*a^4 - 3*A*a^3*b)*x)*sqrt(x))/(a^5*b^3*x^5 + 3*a^6*b^2*x^4 + 3*a^7*b*x^3 + a^8*x^2)]

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giac [A]  time = 0.17, size = 136, normalized size = 0.76 \begin {gather*} -\frac {35 \, {\left (B a b - 3 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} - \frac {105 \, B a b^{3} x^{4} - 315 \, A b^{4} x^{4} + 280 \, B a^{2} b^{2} x^{3} - 840 \, A a b^{3} x^{3} + 231 \, B a^{3} b x^{2} - 693 \, A a^{2} b^{2} x^{2} + 48 \, B a^{4} x - 144 \, A a^{3} b x + 16 \, A a^{4}}{24 \, {\left (b x^{\frac {3}{2}} + a \sqrt {x}\right )}^{3} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35/8*(B*a*b - 3*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/24*(105*B*a*b^3*x^4 - 315*A*b^4*x^4 +
280*B*a^2*b^2*x^3 - 840*A*a*b^3*x^3 + 231*B*a^3*b*x^2 - 693*A*a^2*b^2*x^2 + 48*B*a^4*x - 144*A*a^3*b*x + 16*A*
a^4)/((b*x^(3/2) + a*sqrt(x))^3*a^5)

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maple [A]  time = 0.08, size = 190, normalized size = 1.07 \begin {gather*} \frac {41 A \,b^{4} x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} a^{5}}-\frac {19 B \,b^{3} x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} a^{4}}+\frac {35 A \,b^{3} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} a^{4}}-\frac {17 B \,b^{2} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} a^{3}}+\frac {55 A \,b^{2} \sqrt {x}}{8 \left (b x +a \right )^{3} a^{3}}-\frac {29 B b \sqrt {x}}{8 \left (b x +a \right )^{3} a^{2}}+\frac {105 A \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{5}}-\frac {35 B b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{4}}+\frac {8 A b}{a^{5} \sqrt {x}}-\frac {2 B}{a^{4} \sqrt {x}}-\frac {2 A}{3 a^{4} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

41/8/a^5*b^4/(b*x+a)^3*x^(5/2)*A-19/8/a^4*b^3/(b*x+a)^3*x^(5/2)*B+35/3/a^4*b^3/(b*x+a)^3*A*x^(3/2)-17/3/a^3*b^
2/(b*x+a)^3*B*x^(3/2)+55/8/a^3*b^2/(b*x+a)^3*x^(1/2)*A-29/8/a^2*b/(b*x+a)^3*x^(1/2)*B+105/8/a^5*b^2/(a*b)^(1/2
)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-35/8/a^4*b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B-2/3*A/a^4/x^(3/2)
+8/a^5/x^(1/2)*A*b-2/a^4/x^(1/2)*B

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maxima [A]  time = 1.40, size = 158, normalized size = 0.89 \begin {gather*} -\frac {16 \, A a^{4} + 105 \, {\left (B a b^{3} - 3 \, A b^{4}\right )} x^{4} + 280 \, {\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 231 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 48 \, {\left (B a^{4} - 3 \, A a^{3} b\right )} x}{24 \, {\left (a^{5} b^{3} x^{\frac {9}{2}} + 3 \, a^{6} b^{2} x^{\frac {7}{2}} + 3 \, a^{7} b x^{\frac {5}{2}} + a^{8} x^{\frac {3}{2}}\right )}} - \frac {35 \, {\left (B a b - 3 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(16*A*a^4 + 105*(B*a*b^3 - 3*A*b^4)*x^4 + 280*(B*a^2*b^2 - 3*A*a*b^3)*x^3 + 231*(B*a^3*b - 3*A*a^2*b^2)*
x^2 + 48*(B*a^4 - 3*A*a^3*b)*x)/(a^5*b^3*x^(9/2) + 3*a^6*b^2*x^(7/2) + 3*a^7*b*x^(5/2) + a^8*x^(3/2)) - 35/8*(
B*a*b - 3*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5)

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mupad [B]  time = 1.25, size = 145, normalized size = 0.81 \begin {gather*} \frac {\frac {2\,x\,\left (3\,A\,b-B\,a\right )}{a^2}-\frac {2\,A}{3\,a}+\frac {35\,b^2\,x^3\,\left (3\,A\,b-B\,a\right )}{3\,a^4}+\frac {35\,b^3\,x^4\,\left (3\,A\,b-B\,a\right )}{8\,a^5}+\frac {77\,b\,x^2\,\left (3\,A\,b-B\,a\right )}{8\,a^3}}{a^3\,x^{3/2}+b^3\,x^{9/2}+3\,a^2\,b\,x^{5/2}+3\,a\,b^2\,x^{7/2}}+\frac {35\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (3\,A\,b-B\,a\right )}{8\,a^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x*(3*A*b - B*a))/a^2 - (2*A)/(3*a) + (35*b^2*x^3*(3*A*b - B*a))/(3*a^4) + (35*b^3*x^4*(3*A*b - B*a))/(8*a^
5) + (77*b*x^2*(3*A*b - B*a))/(8*a^3))/(a^3*x^(3/2) + b^3*x^(9/2) + 3*a^2*b*x^(5/2) + 3*a*b^2*x^(7/2)) + (35*b
^(1/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(3*A*b - B*a))/(8*a^(11/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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