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

Optimal. Leaf size=98 \[ 3 a^2 b (a+b x)^{3/2}+9 a^3 b \sqrt{a+b x}-9 a^{7/2} b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{(a+b x)^{9/2}}{x}+\frac{9}{7} b (a+b x)^{7/2}+\frac{9}{5} a b (a+b x)^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0340381, antiderivative size = 98, 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} \[ 3 a^2 b (a+b x)^{3/2}+9 a^3 b \sqrt{a+b x}-9 a^{7/2} b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{(a+b x)^{9/2}}{x}+\frac{9}{7} b (a+b x)^{7/2}+\frac{9}{5} a b (a+b x)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(9/2)/x^2,x]

[Out]

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

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

Mathematica [C]  time = 0.0257217, size = 33, normalized size = 0.34 \[ \frac{2 b (a+b x)^{11/2} \, _2F_1\left (2,\frac{11}{2};\frac{13}{2};\frac{b x}{a}+1\right )}{11 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(9/2)/x^2,x]

[Out]

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

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Maple [A]  time = 0.007, size = 84, normalized size = 0.9 \begin{align*} 2\,b \left ( 1/7\, \left ( bx+a \right ) ^{7/2}+2/5\,a \left ( bx+a \right ) ^{5/2}+{a}^{2} \left ( bx+a \right ) ^{3/2}+4\,\sqrt{bx+a}{a}^{3}+{a}^{4} \left ( -1/2\,{\frac{\sqrt{bx+a}}{bx}}-9/2\,{\frac{1}{\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^2,x)

[Out]

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

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66573, size = 420, normalized size = 4.29 \begin{align*} \left [\frac{315 \, a^{\frac{7}{2}} b x \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (10 \, b^{4} x^{4} + 58 \, a b^{3} x^{3} + 156 \, a^{2} b^{2} x^{2} + 388 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt{b x + a}}{70 \, x}, \frac{315 \, \sqrt{-a} a^{3} b x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (10 \, b^{4} x^{4} + 58 \, a b^{3} x^{3} + 156 \, a^{2} b^{2} x^{2} + 388 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt{b x + a}}{35 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/70*(315*a^(7/2)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(10*b^4*x^4 + 58*a*b^3*x^3 + 156*a^2*b
^2*x^2 + 388*a^3*b*x - 35*a^4)*sqrt(b*x + a))/x, 1/35*(315*sqrt(-a)*a^3*b*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(10*b^4*x^4 + 58*a*b^3*x^3 + 156*a^2*b^2*x^2 + 388*a^3*b*x - 35*a^4)*sqrt(b*x + a))/x]

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Sympy [A]  time = 12.7312, size = 150, normalized size = 1.53 \begin{align*} - \frac{a^{\frac{9}{2}} \sqrt{1 + \frac{b x}{a}}}{x} + \frac{388 a^{\frac{7}{2}} b \sqrt{1 + \frac{b x}{a}}}{35} + \frac{9 a^{\frac{7}{2}} b \log{\left (\frac{b x}{a} \right )}}{2} - 9 a^{\frac{7}{2}} b \log{\left (\sqrt{1 + \frac{b x}{a}} + 1 \right )} + \frac{156 a^{\frac{5}{2}} b^{2} x \sqrt{1 + \frac{b x}{a}}}{35} + \frac{58 a^{\frac{3}{2}} b^{3} x^{2} \sqrt{1 + \frac{b x}{a}}}{35} + \frac{2 \sqrt{a} b^{4} x^{3} \sqrt{1 + \frac{b x}{a}}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(9/2)*sqrt(1 + b*x/a)/x + 388*a**(7/2)*b*sqrt(1 + b*x/a)/35 + 9*a**(7/2)*b*log(b*x/a)/2 - 9*a**(7/2)*b*log
(sqrt(1 + b*x/a) + 1) + 156*a**(5/2)*b**2*x*sqrt(1 + b*x/a)/35 + 58*a**(3/2)*b**3*x**2*sqrt(1 + b*x/a)/35 + 2*
sqrt(a)*b**4*x**3*sqrt(1 + b*x/a)/7

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Giac [A]  time = 1.24112, size = 140, normalized size = 1.43 \begin{align*} \frac{\frac{315 \, a^{4} b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 10 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{2} + 28 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{2} + 70 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{2} + 280 \, \sqrt{b x + a} a^{3} b^{2} - \frac{35 \, \sqrt{b x + a} a^{4} b}{x}}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(315*a^4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 10*(b*x + a)^(7/2)*b^2 + 28*(b*x + a)^(5/2)*a*b^2
+ 70*(b*x + a)^(3/2)*a^2*b^2 + 280*sqrt(b*x + a)*a^3*b^2 - 35*sqrt(b*x + a)*a^4*b/x)/b