3.320 \(\int \frac{(a+b x)^{9/2}}{x^4} \, dx\)

Optimal. Leaf size=114 \[ -\frac{105}{8} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{21 b^2 (a+b x)^{5/2}}{8 x}+\frac{35}{8} b^3 (a+b x)^{3/2}+\frac{105}{8} a b^3 \sqrt{a+b x}-\frac{(a+b x)^{9/2}}{3 x^3}-\frac{3 b (a+b x)^{7/2}}{4 x^2} \]

[Out]

(105*a*b^3*Sqrt[a + b*x])/8 + (35*b^3*(a + b*x)^(3/2))/8 - (21*b^2*(a + b*x)^(5/2))/(8*x) - (3*b*(a + b*x)^(7/
2))/(4*x^2) - (a + b*x)^(9/2)/(3*x^3) - (105*a^(3/2)*b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/8

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Rubi [A]  time = 0.0357739, 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} \[ -\frac{105}{8} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{21 b^2 (a+b x)^{5/2}}{8 x}+\frac{35}{8} b^3 (a+b x)^{3/2}+\frac{105}{8} a b^3 \sqrt{a+b x}-\frac{(a+b x)^{9/2}}{3 x^3}-\frac{3 b (a+b x)^{7/2}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(9/2)/x^4,x]

[Out]

(105*a*b^3*Sqrt[a + b*x])/8 + (35*b^3*(a + b*x)^(3/2))/8 - (21*b^2*(a + b*x)^(5/2))/(8*x) - (3*b*(a + b*x)^(7/
2))/(4*x^2) - (a + b*x)^(9/2)/(3*x^3) - (105*a^(3/2)*b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/8

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^4} \, dx &=-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{2} (3 b) \int \frac{(a+b x)^{7/2}}{x^3} \, dx\\ &=-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{8} \left (21 b^2\right ) \int \frac{(a+b x)^{5/2}}{x^2} \, dx\\ &=-\frac{21 b^2 (a+b x)^{5/2}}{8 x}-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{16} \left (105 b^3\right ) \int \frac{(a+b x)^{3/2}}{x} \, dx\\ &=\frac{35}{8} b^3 (a+b x)^{3/2}-\frac{21 b^2 (a+b x)^{5/2}}{8 x}-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{16} \left (105 a b^3\right ) \int \frac{\sqrt{a+b x}}{x} \, dx\\ &=\frac{105}{8} a b^3 \sqrt{a+b x}+\frac{35}{8} b^3 (a+b x)^{3/2}-\frac{21 b^2 (a+b x)^{5/2}}{8 x}-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{16} \left (105 a^2 b^3\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=\frac{105}{8} a b^3 \sqrt{a+b x}+\frac{35}{8} b^3 (a+b x)^{3/2}-\frac{21 b^2 (a+b x)^{5/2}}{8 x}-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}+\frac{1}{8} \left (105 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )\\ &=\frac{105}{8} a b^3 \sqrt{a+b x}+\frac{35}{8} b^3 (a+b x)^{3/2}-\frac{21 b^2 (a+b x)^{5/2}}{8 x}-\frac{3 b (a+b x)^{7/2}}{4 x^2}-\frac{(a+b x)^{9/2}}{3 x^3}-\frac{105}{8} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0257829, size = 35, normalized size = 0.31 \[ \frac{2 b^3 (a+b x)^{11/2} \, _2F_1\left (4,\frac{11}{2};\frac{13}{2};\frac{b x}{a}+1\right )}{11 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(9/2)/x^4,x]

[Out]

(2*b^3*(a + b*x)^(11/2)*Hypergeometric2F1[4, 11/2, 13/2, 1 + (b*x)/a])/(11*a^4)

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Maple [A]  time = 0.01, size = 87, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ( 1/3\, \left ( bx+a \right ) ^{3/2}+4\,a\sqrt{bx+a}+{a}^{2} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( -{\frac{55\, \left ( bx+a \right ) ^{5/2}}{16}}+{\frac{35\,a \left ( bx+a \right ) ^{3/2}}{6}}-{\frac{41\,{a}^{2}\sqrt{bx+a}}{16}} \right ) }-{\frac{105}{16\,\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^4,x)

[Out]

2*b^3*(1/3*(b*x+a)^(3/2)+4*a*(b*x+a)^(1/2)+a^2*((-55/16*(b*x+a)^(5/2)+35/6*a*(b*x+a)^(3/2)-41/16*a^2*(b*x+a)^(
1/2))/b^3/x^3-105/16*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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61518, size = 431, normalized size = 3.78 \begin{align*} \left [\frac{315 \, a^{\frac{3}{2}} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt{b x + a}}{48 \, x^{3}}, \frac{315 \, \sqrt{-a} a b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (16 \, b^{4} x^{4} + 208 \, a b^{3} x^{3} - 165 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x - 8 \, a^{4}\right )} \sqrt{b x + a}}{24 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(9/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(315*a^(3/2)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(16*b^4*x^4 + 208*a*b^3*x^3 - 165*
a^2*b^2*x^2 - 50*a^3*b*x - 8*a^4)*sqrt(b*x + a))/x^3, 1/24*(315*sqrt(-a)*a*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-
a)/a) + (16*b^4*x^4 + 208*a*b^3*x^3 - 165*a^2*b^2*x^2 - 50*a^3*b*x - 8*a^4)*sqrt(b*x + a))/x^3]

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Sympy [A]  time = 11.0889, size = 184, normalized size = 1.61 \begin{align*} - \frac{105 a^{\frac{3}{2}} b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{8} - \frac{a^{5}}{3 \sqrt{b} x^{\frac{7}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{29 a^{4} \sqrt{b}}{12 x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{215 a^{3} b^{\frac{3}{2}}}{24 x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{43 a^{2} b^{\frac{5}{2}}}{24 \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{28 a b^{\frac{7}{2}} \sqrt{x}}{3 \sqrt{\frac{a}{b x} + 1}} + \frac{2 b^{\frac{9}{2}} x^{\frac{3}{2}}}{3 \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**4,x)

[Out]

-105*a**(3/2)*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/8 - a**5/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 29*a**4*
sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) - 215*a**3*b**(3/2)/(24*x**(3/2)*sqrt(a/(b*x) + 1)) + 43*a**2*b**(5/2)
/(24*sqrt(x)*sqrt(a/(b*x) + 1)) + 28*a*b**(7/2)*sqrt(x)/(3*sqrt(a/(b*x) + 1)) + 2*b**(9/2)*x**(3/2)/(3*sqrt(a/
(b*x) + 1))

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Giac [A]  time = 1.26577, size = 151, normalized size = 1.32 \begin{align*} \frac{\frac{315 \, a^{2} b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 16 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{4} + 192 \, \sqrt{b x + a} a b^{4} - \frac{165 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} b^{4} - 280 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3} b^{4} + 123 \, \sqrt{b x + a} a^{4} b^{4}}{b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(9/2)/x^4,x, algorithm="giac")

[Out]

1/24*(315*a^2*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 16*(b*x + a)^(3/2)*b^4 + 192*sqrt(b*x + a)*a*b^4 -
(165*(b*x + a)^(5/2)*a^2*b^4 - 280*(b*x + a)^(3/2)*a^3*b^4 + 123*sqrt(b*x + a)*a^4*b^4)/(b^3*x^3))/b