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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0350025, antiderivative size = 114, normalized size of antiderivative = 1., 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*}

Mathematica [C]  time = 0.139489, 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)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 86, normalized size = 0.8 \begin{align*} 2\,{b}^{2} \left ( 1/5\, \left ( bx+a \right ) ^{5/2}+a \left ( bx+a \right ) ^{3/2}+6\,{a}^{2}\sqrt{bx+a}+{a}^{3} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( -{\frac{17\, \left ( bx+a \right ) ^{3/2}}{8}}+{\frac{15\,a\sqrt{bx+a}}{8}} \right ) }-{\frac{63}{8\,\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \right ) \end{align*}

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.56112, size = 431, normalized size = 3.78 \begin{align*} \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 ] \end{align*}

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]

________________________________________________________________________________________

Sympy [A]  time = 12.614, size = 184, normalized size = 1.61 \begin{align*} - \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}} \end{align*}

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))

________________________________________________________________________________________

Giac [A]  time = 1.26392, size = 151, normalized size = 1.32 \begin{align*} \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} \end{align*}

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