3.308 \(\int \frac{(a+b x)^{5/2}}{x^5} \, dx\)

Optimal. Leaf size=103 \[ \frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{3/2}}-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b^3 \sqrt{a+b x}}{64 a x}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4} \]

[Out]

(-5*b^2*Sqrt[a + b*x])/(32*x^2) - (5*b^3*Sqrt[a + b*x])/(64*a*x) - (5*b*(a + b*x)^(3/2))/(24*x^3) - (a + b*x)^
(5/2)/(4*x^4) + (5*b^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0309203, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 51, 63, 208} \[ \frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{3/2}}-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b^3 \sqrt{a+b x}}{64 a x}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^5,x]

[Out]

(-5*b^2*Sqrt[a + b*x])/(32*x^2) - (5*b^3*Sqrt[a + b*x])/(64*a*x) - (5*b*(a + b*x)^(3/2))/(24*x^3) - (a + b*x)^
(5/2)/(4*x^4) + (5*b^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))

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 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 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)^{5/2}}{x^5} \, dx &=-\frac{(a+b x)^{5/2}}{4 x^4}+\frac{1}{8} (5 b) \int \frac{(a+b x)^{3/2}}{x^4} \, dx\\ &=-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4}+\frac{1}{16} \left (5 b^2\right ) \int \frac{\sqrt{a+b x}}{x^3} \, dx\\ &=-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4}+\frac{1}{64} \left (5 b^3\right ) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx\\ &=-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b^3 \sqrt{a+b x}}{64 a x}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4}-\frac{\left (5 b^4\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{128 a}\\ &=-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b^3 \sqrt{a+b x}}{64 a x}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{64 a}\\ &=-\frac{5 b^2 \sqrt{a+b x}}{32 x^2}-\frac{5 b^3 \sqrt{a+b x}}{64 a x}-\frac{5 b (a+b x)^{3/2}}{24 x^3}-\frac{(a+b x)^{5/2}}{4 x^4}+\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0112828, size = 35, normalized size = 0.34 \[ -\frac{2 b^4 (a+b x)^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{b x}{a}+1\right )}{7 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^5,x]

[Out]

(-2*b^4*(a + b*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, 1 + (b*x)/a])/(7*a^5)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 75, normalized size = 0.7 \begin{align*} 2\,{b}^{4} \left ({\frac{1}{{b}^{4}{x}^{4}} \left ( -{\frac{5\, \left ( bx+a \right ) ^{7/2}}{128\,a}}-{\frac{73\, \left ( bx+a \right ) ^{5/2}}{384}}+{\frac{55\,a \left ( bx+a \right ) ^{3/2}}{384}}-{\frac{5\,{a}^{2}\sqrt{bx+a}}{128}} \right ) }+{\frac{5}{128\,{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^5,x)

[Out]

2*b^4*((-5/128/a*(b*x+a)^(7/2)-73/384*(b*x+a)^(5/2)+55/384*a*(b*x+a)^(3/2)-5/128*a^2*(b*x+a)^(1/2))/b^4/x^4+5/
128*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2))

________________________________________________________________________________________

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)^(5/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.56401, size = 413, normalized size = 4.01 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{4} x^{4} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt{b x + a}}{384 \, a^{2} x^{4}}, -\frac{15 \, \sqrt{-a} b^{4} x^{4} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt{b x + a}}{192 \, a^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^5,x, algorithm="fricas")

[Out]

[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 +
136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4)]

________________________________________________________________________________________

Sympy [A]  time = 10.2826, size = 155, normalized size = 1.5 \begin{align*} - \frac{a^{3}}{4 \sqrt{b} x^{\frac{9}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{23 a^{2} \sqrt{b}}{24 x^{\frac{7}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{127 a b^{\frac{3}{2}}}{96 x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{133 b^{\frac{5}{2}}}{192 x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{5 b^{\frac{7}{2}}}{64 a \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{5 b^{4} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{64 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/x**5,x)

[Out]

-a**3/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 23*a**2*sqrt(b)/(24*x**(7/2)*sqrt(a/(b*x) + 1)) - 127*a*b**(3/2
)/(96*x**(5/2)*sqrt(a/(b*x) + 1)) - 133*b**(5/2)/(192*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*b**(7/2)/(64*a*sqrt(x)*s
qrt(a/(b*x) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(3/2))

________________________________________________________________________________________

Giac [A]  time = 1.2294, size = 134, normalized size = 1.3 \begin{align*} -\frac{\frac{15 \, b^{5} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{15 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{5} + 73 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{5} - 55 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{5} + 15 \, \sqrt{b x + a} a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^5,x, algorithm="giac")

[Out]

-1/192*(15*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + (15*(b*x + a)^(7/2)*b^5 + 73*(b*x + a)^(5/2)*a*b^
5 - 55*(b*x + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b