3.2.76 \(\int \frac {A+B x}{x^{9/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=136 \[ \frac {2 c^{7/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}+\frac {2 c^3 (b B-A c)}{b^5 \sqrt {x}}-\frac {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {2 A}{9 b x^{9/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 136, 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 {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}+\frac {2 c^3 (b B-A c)}{b^5 \sqrt {x}}+\frac {2 c^{7/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {2 A}{9 b x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(9/2)*(b*x + c*x^2)),x]

[Out]

(-2*A)/(9*b*x^(9/2)) - (2*(b*B - A*c))/(7*b^2*x^(7/2)) + (2*c*(b*B - A*c))/(5*b^3*x^(5/2)) - (2*c^2*(b*B - A*c
))/(3*b^4*x^(3/2)) + (2*c^3*(b*B - A*c))/(b^5*Sqrt[x]) + (2*c^(7/2)*(b*B - A*c)*ArcTan[(Sqrt[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^{9/2} \left (b x+c x^2\right )} \, dx &=\int \frac {A+B x}{x^{11/2} (b+c x)} \, dx\\ &=-\frac {2 A}{9 b x^{9/2}}+\frac {\left (2 \left (\frac {9 b B}{2}-\frac {9 A c}{2}\right )\right ) \int \frac {1}{x^{9/2} (b+c x)} \, dx}{9 b}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {(c (b B-A c)) \int \frac {1}{x^{7/2} (b+c x)} \, dx}{b^2}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}+\frac {\left (c^2 (b B-A c)\right ) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{b^3}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}-\frac {\left (c^3 (b B-A c)\right ) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{b^4}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}+\frac {2 c^3 (b B-A c)}{b^5 \sqrt {x}}+\frac {\left (c^4 (b B-A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{b^5}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}+\frac {2 c^3 (b B-A c)}{b^5 \sqrt {x}}+\frac {\left (2 c^4 (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{b^5}\\ &=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{5 b^3 x^{5/2}}-\frac {2 c^2 (b B-A c)}{3 b^4 x^{3/2}}+\frac {2 c^3 (b B-A c)}{b^5 \sqrt {x}}+\frac {2 c^{7/2} (b B-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 = 0.01, size = 44, normalized size = 0.32 \begin {gather*} \frac {2 \left (\, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};-\frac {c x}{b}\right ) (9 A c x-9 b B x)-7 A b\right )}{63 b^2 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(9/2)*(b*x + c*x^2)),x]

[Out]

(2*(-7*A*b + (-9*b*B*x + 9*A*c*x)*Hypergeometric2F1[-7/2, 1, -5/2, -((c*x)/b)]))/(63*b^2*x^(9/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 139, normalized size = 1.02 \begin {gather*} \frac {2 \left (b B c^{7/2}-A c^{9/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{11/2}}-\frac {2 \left (35 A b^4-45 A b^3 c x+63 A b^2 c^2 x^2-105 A b c^3 x^3+315 A c^4 x^4+45 b^4 B x-63 b^3 B c x^2+105 b^2 B c^2 x^3-315 b B c^3 x^4\right )}{315 b^5 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(9/2)*(b*x + c*x^2)),x]

[Out]

(-2*(35*A*b^4 + 45*b^4*B*x - 45*A*b^3*c*x - 63*b^3*B*c*x^2 + 63*A*b^2*c^2*x^2 + 105*b^2*B*c^2*x^3 - 105*A*b*c^
3*x^3 - 315*b*B*c^3*x^4 + 315*A*c^4*x^4))/(315*b^5*x^(9/2)) + (2*(b*B*c^(7/2) - A*c^(9/2))*ArcTan[(Sqrt[c]*Sqr
t[x])/Sqrt[b]])/b^(11/2)

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fricas [A]  time = 0.43, size = 291, normalized size = 2.14 \begin {gather*} \left [-\frac {315 \, {\left (B b c^{3} - A c^{4}\right )} x^{5} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (35 \, A b^{4} - 315 \, {\left (B b c^{3} - A c^{4}\right )} x^{4} + 105 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} x^{3} - 63 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} x^{2} + 45 \, {\left (B b^{4} - A b^{3} c\right )} x\right )} \sqrt {x}}{315 \, b^{5} x^{5}}, -\frac {2 \, {\left (315 \, {\left (B b c^{3} - A c^{4}\right )} x^{5} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (35 \, A b^{4} - 315 \, {\left (B b c^{3} - A c^{4}\right )} x^{4} + 105 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} x^{3} - 63 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} x^{2} + 45 \, {\left (B b^{4} - A b^{3} c\right )} x\right )} \sqrt {x}\right )}}{315 \, b^{5} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-1/315*(315*(B*b*c^3 - A*c^4)*x^5*sqrt(-c/b)*log((c*x - 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(35*A*b^4
- 315*(B*b*c^3 - A*c^4)*x^4 + 105*(B*b^2*c^2 - A*b*c^3)*x^3 - 63*(B*b^3*c - A*b^2*c^2)*x^2 + 45*(B*b^4 - A*b^3
*c)*x)*sqrt(x))/(b^5*x^5), -2/315*(315*(B*b*c^3 - A*c^4)*x^5*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) + (35*A
*b^4 - 315*(B*b*c^3 - A*c^4)*x^4 + 105*(B*b^2*c^2 - A*b*c^3)*x^3 - 63*(B*b^3*c - A*b^2*c^2)*x^2 + 45*(B*b^4 -
A*b^3*c)*x)*sqrt(x))/(b^5*x^5)]

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giac [A]  time = 0.16, size = 128, normalized size = 0.94 \begin {gather*} \frac {2 \, {\left (B b c^{4} - A c^{5}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{5}} + \frac {2 \, {\left (315 \, B b c^{3} x^{4} - 315 \, A c^{4} x^{4} - 105 \, B b^{2} c^{2} x^{3} + 105 \, A b c^{3} x^{3} + 63 \, B b^{3} c x^{2} - 63 \, A b^{2} c^{2} x^{2} - 45 \, B b^{4} x + 45 \, A b^{3} c x - 35 \, A b^{4}\right )}}{315 \, b^{5} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

2*(B*b*c^4 - A*c^5)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5) + 2/315*(315*B*b*c^3*x^4 - 315*A*c^4*x^4 - 105
*B*b^2*c^2*x^3 + 105*A*b*c^3*x^3 + 63*B*b^3*c*x^2 - 63*A*b^2*c^2*x^2 - 45*B*b^4*x + 45*A*b^3*c*x - 35*A*b^4)/(
b^5*x^(9/2))

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maple [A]  time = 0.07, size = 150, normalized size = 1.10 \begin {gather*} -\frac {2 A \,c^{5} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{5}}+\frac {2 B \,c^{4} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{4}}-\frac {2 A \,c^{4}}{b^{5} \sqrt {x}}+\frac {2 B \,c^{3}}{b^{4} \sqrt {x}}+\frac {2 A \,c^{3}}{3 b^{4} x^{\frac {3}{2}}}-\frac {2 B \,c^{2}}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 A \,c^{2}}{5 b^{3} x^{\frac {5}{2}}}+\frac {2 B c}{5 b^{2} x^{\frac {5}{2}}}+\frac {2 A c}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 B}{7 b \,x^{\frac {7}{2}}}-\frac {2 A}{9 b \,x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(9/2)/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 1.16, size = 126, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (B b c^{4} - A c^{5}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{5}} - \frac {2 \, {\left (35 \, A b^{4} - 315 \, {\left (B b c^{3} - A c^{4}\right )} x^{4} + 105 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} x^{3} - 63 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} x^{2} + 45 \, {\left (B b^{4} - A b^{3} c\right )} x\right )}}{315 \, b^{5} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(9/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

2*(B*b*c^4 - A*c^5)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5) - 2/315*(35*A*b^4 - 315*(B*b*c^3 - A*c^4)*x^4
+ 105*(B*b^2*c^2 - A*b*c^3)*x^3 - 63*(B*b^3*c - A*b^2*c^2)*x^2 + 45*(B*b^4 - A*b^3*c)*x)/(b^5*x^(9/2))

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mupad [B]  time = 1.11, size = 109, normalized size = 0.80 \begin {gather*} -\frac {\frac {2\,A}{9\,b}-\frac {2\,x\,\left (A\,c-B\,b\right )}{7\,b^2}-\frac {2\,c^2\,x^3\,\left (A\,c-B\,b\right )}{3\,b^4}+\frac {2\,c^3\,x^4\,\left (A\,c-B\,b\right )}{b^5}+\frac {2\,c\,x^2\,\left (A\,c-B\,b\right )}{5\,b^3}}{x^{9/2}}-\frac {2\,c^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(9/2)*(b*x + c*x^2)),x)

[Out]

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

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sympy [A]  time = 126.31, size = 360, normalized size = 2.65 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{9 x^{\frac {9}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{9 x^{\frac {9}{2}}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{9 b x^{\frac {9}{2}}} + \frac {2 A c}{7 b^{2} x^{\frac {7}{2}}} - \frac {2 A c^{2}}{5 b^{3} x^{\frac {5}{2}}} + \frac {2 A c^{3}}{3 b^{4} x^{\frac {3}{2}}} - \frac {2 A c^{4}}{b^{5} \sqrt {x}} + \frac {i A c^{4} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {11}{2}} \sqrt {\frac {1}{c}}} - \frac {i A c^{4} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {11}{2}} \sqrt {\frac {1}{c}}} - \frac {2 B}{7 b x^{\frac {7}{2}}} + \frac {2 B c}{5 b^{2} x^{\frac {5}{2}}} - \frac {2 B c^{2}}{3 b^{3} x^{\frac {3}{2}}} + \frac {2 B c^{3}}{b^{4} \sqrt {x}} - \frac {i B c^{3} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {9}{2}} \sqrt {\frac {1}{c}}} + \frac {i B c^{3} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {9}{2}} \sqrt {\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(9/2)/(c*x**2+b*x),x)

[Out]

Piecewise((zoo*(-2*A/(11*x**(11/2)) - 2*B/(9*x**(9/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(11*x**(11/2)) - 2*B/(9*
x**(9/2)))/c, Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/b, Eq(c, 0)), (-2*A/(9*b*x**(9/2)) + 2*A*c/(7
*b**2*x**(7/2)) - 2*A*c**2/(5*b**3*x**(5/2)) + 2*A*c**3/(3*b**4*x**(3/2)) - 2*A*c**4/(b**5*sqrt(x)) + I*A*c**4
*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(11/2)*sqrt(1/c)) - I*A*c**4*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(
11/2)*sqrt(1/c)) - 2*B/(7*b*x**(7/2)) + 2*B*c/(5*b**2*x**(5/2)) - 2*B*c**2/(3*b**3*x**(3/2)) + 2*B*c**3/(b**4*
sqrt(x)) - I*B*c**3*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(9/2)*sqrt(1/c)) + I*B*c**3*log(I*sqrt(b)*sqrt(1/c
) + sqrt(x))/(b**(9/2)*sqrt(1/c)), True))

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