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

Optimal. Leaf size=156 \[ -\frac {c^{5/2} (7 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}-\frac {c^2 (7 b B-9 A c)}{b^5 \sqrt {x}}+\frac {c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac {7 b B-9 A c}{5 b^3 x^{5/2}}+\frac {7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac {b B-A c}{b c x^{7/2} (b+c x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \begin {gather*} -\frac {c^2 (7 b B-9 A c)}{b^5 \sqrt {x}}-\frac {c^{5/2} (7 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}+\frac {c (7 b B-9 A c)}{3 b^4 x^{3/2}}-\frac {7 b B-9 A c}{5 b^3 x^{5/2}}+\frac {7 b B-9 A c}{7 b^2 c x^{7/2}}-\frac {b B-A c}{b c x^{7/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 64, normalized size = 0.41 \begin {gather*} \frac {(b+c x) (7 b B-9 A c) \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};-\frac {c x}{b}\right )+7 b (A c-b B)}{7 b^2 c x^{7/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b*(-(b*B) + A*c) + (7*b*B - 9*A*c)*(b + c*x)*Hypergeometric2F1[-7/2, 1, -5/2, -((c*x)/b)])/(7*b^2*c*x^(7/2)
*(b + c*x))

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IntegrateAlgebraic [A]  time = 0.15, size = 146, normalized size = 0.94 \begin {gather*} \frac {\left (9 A c^{7/2}-7 b B c^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}+\frac {-30 A b^4+54 A b^3 c x-126 A b^2 c^2 x^2+630 A b c^3 x^3+945 A c^4 x^4-42 b^4 B x+98 b^3 B c x^2-490 b^2 B c^2 x^3-735 b B c^3 x^4}{105 b^5 x^{7/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*A*b^4 - 42*b^4*B*x + 54*A*b^3*c*x + 98*b^3*B*c*x^2 - 126*A*b^2*c^2*x^2 - 490*b^2*B*c^2*x^3 + 630*A*b*c^3*
x^3 - 735*b*B*c^3*x^4 + 945*A*c^4*x^4)/(105*b^5*x^(7/2)*(b + c*x)) + ((-7*b*B*c^(5/2) + 9*A*c^(7/2))*ArcTan[(S
qrt[c]*Sqrt[x])/Sqrt[b]])/b^(11/2)

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fricas [A]  time = 0.43, size = 372, normalized size = 2.38 \begin {gather*} \left [-\frac {105 \, {\left ({\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + {\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (30 \, A b^{4} + 105 \, {\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} + 70 \, {\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 14 \, {\left (7 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 6 \, {\left (7 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{210 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}, \frac {105 \, {\left ({\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + {\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (30 \, A b^{4} + 105 \, {\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} + 70 \, {\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 14 \, {\left (7 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 6 \, {\left (7 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{105 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/210*(105*((7*B*b*c^3 - 9*A*c^4)*x^5 + (7*B*b^2*c^2 - 9*A*b*c^3)*x^4)*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqr
t(-c/b) - b)/(c*x + b)) + 2*(30*A*b^4 + 105*(7*B*b*c^3 - 9*A*c^4)*x^4 + 70*(7*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 14*
(7*B*b^3*c - 9*A*b^2*c^2)*x^2 + 6*(7*B*b^4 - 9*A*b^3*c)*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4), 1/105*(105*((7*B*b*
c^3 - 9*A*c^4)*x^5 + (7*B*b^2*c^2 - 9*A*b*c^3)*x^4)*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (30*A*b^4 + 10
5*(7*B*b*c^3 - 9*A*c^4)*x^4 + 70*(7*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 14*(7*B*b^3*c - 9*A*b^2*c^2)*x^2 + 6*(7*B*b^4
- 9*A*b^3*c)*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4)]

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giac [A]  time = 0.16, size = 136, normalized size = 0.87 \begin {gather*} -\frac {{\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{5}} - \frac {B b c^{3} \sqrt {x} - A c^{4} \sqrt {x}}{{\left (c x + b\right )} b^{5}} - \frac {2 \, {\left (315 \, B b c^{2} x^{3} - 420 \, A c^{3} x^{3} - 70 \, B b^{2} c x^{2} + 105 \, A b c^{2} x^{2} + 21 \, B b^{3} x - 42 \, A b^{2} c x + 15 \, A b^{3}\right )}}{105 \, b^{5} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(7*B*b*c^3 - 9*A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5) - (B*b*c^3*sqrt(x) - A*c^4*sqrt(x))/((c*x +
b)*b^5) - 2/105*(315*B*b*c^2*x^3 - 420*A*c^3*x^3 - 70*B*b^2*c*x^2 + 105*A*b*c^2*x^2 + 21*B*b^3*x - 42*A*b^2*c
*x + 15*A*b^3)/(b^5*x^(7/2))

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maple [A]  time = 0.07, size = 163, normalized size = 1.04 \begin {gather*} \frac {9 A \,c^{4} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{5}}-\frac {7 B \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{4}}+\frac {A \,c^{4} \sqrt {x}}{\left (c x +b \right ) b^{5}}-\frac {B \,c^{3} \sqrt {x}}{\left (c x +b \right ) b^{4}}+\frac {8 A \,c^{3}}{b^{5} \sqrt {x}}-\frac {6 B \,c^{2}}{b^{4} \sqrt {x}}-\frac {2 A \,c^{2}}{b^{4} x^{\frac {3}{2}}}+\frac {4 B c}{3 b^{3} x^{\frac {3}{2}}}+\frac {4 A c}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 B}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 A}{7 b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^5*c^4*x^(1/2)/(c*x+b)*A-1/b^4*c^3*x^(1/2)/(c*x+b)*B+9/b^5*c^4/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*
A-7/b^4*c^3/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*B-2/7/b^2*A/x^(7/2)+4/5/b^3/x^(5/2)*A*c-2/5/b^2/x^(5/2
)*B-2*c^2/b^4/x^(3/2)*A+4/3*c/b^3/x^(3/2)*B+8*c^3/b^5/x^(1/2)*A-6*c^2/b^4/x^(1/2)*B

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maxima [A]  time = 1.19, size = 143, normalized size = 0.92 \begin {gather*} -\frac {30 \, A b^{4} + 105 \, {\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} + 70 \, {\left (7 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 14 \, {\left (7 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 6 \, {\left (7 \, B b^{4} - 9 \, A b^{3} c\right )} x}{105 \, {\left (b^{5} c x^{\frac {9}{2}} + b^{6} x^{\frac {7}{2}}\right )}} - \frac {{\left (7 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(30*A*b^4 + 105*(7*B*b*c^3 - 9*A*c^4)*x^4 + 70*(7*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 14*(7*B*b^3*c - 9*A*b^2*
c^2)*x^2 + 6*(7*B*b^4 - 9*A*b^3*c)*x)/(b^5*c*x^(9/2) + b^6*x^(7/2)) - (7*B*b*c^3 - 9*A*c^4)*arctan(c*sqrt(x)/s
qrt(b*c))/(sqrt(b*c)*b^5)

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mupad [B]  time = 1.12, size = 121, normalized size = 0.78 \begin {gather*} \frac {\frac {2\,x\,\left (9\,A\,c-7\,B\,b\right )}{35\,b^2}-\frac {2\,A}{7\,b}+\frac {2\,c^2\,x^3\,\left (9\,A\,c-7\,B\,b\right )}{3\,b^4}+\frac {c^3\,x^4\,\left (9\,A\,c-7\,B\,b\right )}{b^5}-\frac {2\,c\,x^2\,\left (9\,A\,c-7\,B\,b\right )}{15\,b^3}}{b\,x^{7/2}+c\,x^{9/2}}+\frac {c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (9\,A\,c-7\,B\,b\right )}{b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x*(9*A*c - 7*B*b))/(35*b^2) - (2*A)/(7*b) + (2*c^2*x^3*(9*A*c - 7*B*b))/(3*b^4) + (c^3*x^4*(9*A*c - 7*B*b)
)/b^5 - (2*c*x^2*(9*A*c - 7*B*b))/(15*b^3))/(b*x^(7/2) + c*x^(9/2)) + (c^(5/2)*atan((c^(1/2)*x^(1/2))/b^(1/2))
*(9*A*c - 7*B*b))/b^(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)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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