3.319 \(\int \frac {(a+b x)^{9/2}}{x^3} \, dx\)

Optimal. Leaf size=114 \[ -\frac {63}{4} a^{5/2} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {63}{20} b^2 (a+b x)^{5/2}+\frac {21}{4} a b^2 (a+b x)^{3/2}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {9 b (a+b x)^{7/2}}{4 x} \]

[Out]

21/4*a*b^2*(b*x+a)^(3/2)+63/20*b^2*(b*x+a)^(5/2)-9/4*b*(b*x+a)^(7/2)/x-1/2*(b*x+a)^(9/2)/x^2-63/4*a^(5/2)*b^2*
arctanh((b*x+a)^(1/2)/a^(1/2))+63/4*a^2*b^2*(b*x+a)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 50, 63, 208} \[ \frac {63}{4} a^2 b^2 \sqrt {a+b x}-\frac {63}{4} a^{5/2} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {63}{20} b^2 (a+b x)^{5/2}+\frac {21}{4} a b^2 (a+b x)^{3/2}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {9 b (a+b x)^{7/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(9/2)/x^3,x]

[Out]

(63*a^2*b^2*Sqrt[a + b*x])/4 + (21*a*b^2*(a + b*x)^(3/2))/4 + (63*b^2*(a + b*x)^(5/2))/20 - (9*b*(a + b*x)^(7/
2))/(4*x) - (a + b*x)^(9/2)/(2*x^2) - (63*a^(5/2)*b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/4

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 208

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

Rubi steps

\begin {align*} \int \frac {(a+b x)^{9/2}}{x^3} \, dx &=-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{4} (9 b) \int \frac {(a+b x)^{7/2}}{x^2} \, dx\\ &=-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 b^2\right ) \int \frac {(a+b x)^{5/2}}{x} \, dx\\ &=\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a b^2\right ) \int \frac {(a+b x)^{3/2}}{x} \, dx\\ &=\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a^2 b^2\right ) \int \frac {\sqrt {a+b x}}{x} \, dx\\ &=\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{8} \left (63 a^3 b^2\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx\\ &=\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}+\frac {1}{4} \left (63 a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )\\ &=\frac {63}{4} a^2 b^2 \sqrt {a+b x}+\frac {21}{4} a b^2 (a+b x)^{3/2}+\frac {63}{20} b^2 (a+b x)^{5/2}-\frac {9 b (a+b x)^{7/2}}{4 x}-\frac {(a+b x)^{9/2}}{2 x^2}-\frac {63}{4} a^{5/2} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.31 \[ -\frac {2 b^2 (a+b x)^{11/2} \, _2F_1\left (3,\frac {11}{2};\frac {13}{2};\frac {b x}{a}+1\right )}{11 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(9/2)/x^3,x]

[Out]

(-2*b^2*(a + b*x)^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, 1 + (b*x)/a])/(11*a^3)

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fricas [A]  time = 0.48, size = 180, normalized size = 1.58 \[ \left [\frac {315 \, a^{\frac {5}{2}} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{4} x^{4} + 56 \, a b^{3} x^{3} + 288 \, a^{2} b^{2} x^{2} - 85 \, a^{3} b x - 10 \, a^{4}\right )} \sqrt {b x + a}}{40 \, x^{2}}, \frac {315 \, \sqrt {-a} a^{2} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, b^{4} x^{4} + 56 \, a b^{3} x^{3} + 288 \, a^{2} b^{2} x^{2} - 85 \, a^{3} b x - 10 \, a^{4}\right )} \sqrt {b x + a}}{20 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(9/2)/x^3,x, algorithm="fricas")

[Out]

[1/40*(315*a^(5/2)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*b^4*x^4 + 56*a*b^3*x^3 + 288*a^
2*b^2*x^2 - 85*a^3*b*x - 10*a^4)*sqrt(b*x + a))/x^2, 1/20*(315*sqrt(-a)*a^2*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(
-a)/a) + (8*b^4*x^4 + 56*a*b^3*x^3 + 288*a^2*b^2*x^2 - 85*a^3*b*x - 10*a^4)*sqrt(b*x + a))/x^2]

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giac [A]  time = 1.10, size = 112, normalized size = 0.98 \[ \frac {\frac {315 \, a^{3} b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} + 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 240 \, \sqrt {b x + a} a^{2} b^{3} - \frac {5 \, {\left (17 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{3} - 15 \, \sqrt {b x + a} a^{4} b^{3}\right )}}{b^{2} x^{2}}}{20 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(9/2)/x^3,x, algorithm="giac")

[Out]

1/20*(315*a^3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 8*(b*x + a)^(5/2)*b^3 + 40*(b*x + a)^(3/2)*a*b^3 +
240*sqrt(b*x + a)*a^2*b^3 - 5*(17*(b*x + a)^(3/2)*a^3*b^3 - 15*sqrt(b*x + a)*a^4*b^3)/(b^2*x^2))/b

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maple [A]  time = 0.01, size = 86, normalized size = 0.75 \[ 2 \left (\left (-\frac {63 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {\frac {15 \sqrt {b x +a}\, a}{8}-\frac {17 \left (b x +a \right )^{\frac {3}{2}}}{8}}{b^{2} x^{2}}\right ) a^{3}+6 \sqrt {b x +a}\, a^{2}+\left (b x +a \right )^{\frac {3}{2}} a +\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}\right ) b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(9/2)/x^3,x)

[Out]

2*b^2*(1/5*(b*x+a)^(5/2)+(b*x+a)^(3/2)*a+6*(b*x+a)^(1/2)*a^2+a^3*((-17/8*(b*x+a)^(3/2)+15/8*(b*x+a)^(1/2)*a)/x
^2/b^2-63/8*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2)))

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maxima [A]  time = 2.95, size = 131, normalized size = 1.15 \[ \frac {63}{8} \, a^{\frac {5}{2}} b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} b^{2} + 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} + 12 \, \sqrt {b x + a} a^{2} b^{2} - \frac {17 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{2} - 15 \, \sqrt {b x + a} a^{4} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(9/2)/x^3,x, algorithm="maxima")

[Out]

63/8*a^(5/2)*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/5*(b*x + a)^(5/2)*b^2 + 2*(b*x +
a)^(3/2)*a*b^2 + 12*sqrt(b*x + a)*a^2*b^2 - 1/4*(17*(b*x + a)^(3/2)*a^3*b^2 - 15*sqrt(b*x + a)*a^4*b^2)/((b*x
+ a)^2 - 2*(b*x + a)*a + a^2)

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mupad [B]  time = 0.05, size = 117, normalized size = 1.03 \[ \frac {2\,b^2\,{\left (a+b\,x\right )}^{5/2}}{5}+\frac {\frac {15\,a^4\,b^2\,\sqrt {a+b\,x}}{4}-\frac {17\,a^3\,b^2\,{\left (a+b\,x\right )}^{3/2}}{4}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+12\,a^2\,b^2\,\sqrt {a+b\,x}+2\,a\,b^2\,{\left (a+b\,x\right )}^{3/2}+\frac {a^{5/2}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,63{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(9/2)/x^3,x)

[Out]

(2*b^2*(a + b*x)^(5/2))/5 + ((15*a^4*b^2*(a + b*x)^(1/2))/4 - (17*a^3*b^2*(a + b*x)^(3/2))/4)/((a + b*x)^2 - 2
*a*(a + b*x) + a^2) + 12*a^2*b^2*(a + b*x)^(1/2) + (a^(5/2)*b^2*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*63i)/4 + 2*
a*b^2*(a + b*x)^(3/2)

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sympy [A]  time = 8.99, size = 184, normalized size = 1.61 \[ - \frac {63 a^{\frac {5}{2}} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {a^{5}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {19 a^{4} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {203 a^{3} b^{\frac {3}{2}}}{20 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {86 a^{2} b^{\frac {5}{2}} \sqrt {x}}{5 \sqrt {\frac {a}{b x} + 1}} + \frac {16 a b^{\frac {7}{2}} x^{\frac {3}{2}}}{5 \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {9}{2}} x^{\frac {5}{2}}}{5 \sqrt {\frac {a}{b x} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(9/2)/x**3,x)

[Out]

-63*a**(5/2)*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - a**5/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) - 19*a**4*s
qrt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1)) + 203*a**3*b**(3/2)/(20*sqrt(x)*sqrt(a/(b*x) + 1)) + 86*a**2*b**(5/2)*sq
rt(x)/(5*sqrt(a/(b*x) + 1)) + 16*a*b**(7/2)*x**(3/2)/(5*sqrt(a/(b*x) + 1)) + 2*b**(9/2)*x**(5/2)/(5*sqrt(a/(b*
x) + 1))

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