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

Optimal. Leaf size=66 \[ -5 a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {(a+b x)^{5/2}}{x}+\frac {5}{3} b (a+b x)^{3/2}+5 a b \sqrt {a+b x} \]

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Rubi [A]  time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 50, 63, 208} \begin {gather*} -5 a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {(a+b x)^{5/2}}{x}+\frac {5}{3} b (a+b x)^{3/2}+5 a b \sqrt {a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^2,x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.50 \begin {gather*} \frac {2 b (a+b x)^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {b x}{a}+1\right )}{7 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 64, normalized size = 0.97 \begin {gather*} \frac {\sqrt {a+b x} \left (-15 a^2+10 a (a+b x)+2 (a+b x)^2\right )}{3 x}-5 a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(5/2)/x^2,x]

[Out]

(Sqrt[a + b*x]*(-15*a^2 + 10*a*(a + b*x) + 2*(a + b*x)^2))/(3*x) - 5*a^(3/2)*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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fricas [A]  time = 0.80, size = 126, normalized size = 1.91 \begin {gather*} \left [\frac {15 \, a^{\frac {3}{2}} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{6 \, x}, \frac {15 \, \sqrt {-a} a b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{3 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a^(3/2)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*b^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x
+ a))/x, 1/3*(15*sqrt(-a)*a*b*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (2*b^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x + a
))/x]

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giac [A]  time = 1.09, size = 74, normalized size = 1.12 \begin {gather*} \frac {\frac {15 \, a^{2} b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} + 12 \, \sqrt {b x + a} a b^{2} - \frac {3 \, \sqrt {b x + a} a^{2} b}{x}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*a^2*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2*(b*x + a)^(3/2)*b^2 + 12*sqrt(b*x + a)*a*b^2 - 3*s
qrt(b*x + a)*a^2*b/x)/b

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maple [A]  time = 0.01, size = 61, normalized size = 0.92 \begin {gather*} 2 \left (\left (-\frac {5 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b x +a}}{2 b x}\right ) a^{2}+2 \sqrt {b x +a}\, a +\frac {\left (b x +a \right )^{\frac {3}{2}}}{3}\right ) b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 71, normalized size = 1.08 \begin {gather*} \frac {5}{2} \, a^{\frac {3}{2}} b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} b + 4 \, \sqrt {b x + a} a b - \frac {\sqrt {b x + a} a^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*a^(3/2)*b*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/3*(b*x + a)^(3/2)*b + 4*sqrt(b*x +
a)*a*b - sqrt(b*x + a)*a^2/x

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mupad [B]  time = 0.11, size = 58, normalized size = 0.88 \begin {gather*} \frac {2\,b\,{\left (a+b\,x\right )}^{3/2}}{3}-\frac {a^2\,\sqrt {a+b\,x}}{x}+4\,a\,b\,\sqrt {a+b\,x}+a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/x^2,x)

[Out]

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

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sympy [A]  time = 3.75, size = 99, normalized size = 1.50 \begin {gather*} - \frac {a^{\frac {5}{2}} \sqrt {1 + \frac {b x}{a}}}{x} + \frac {14 a^{\frac {3}{2}} b \sqrt {1 + \frac {b x}{a}}}{3} + \frac {5 a^{\frac {3}{2}} b \log {\left (\frac {b x}{a} \right )}}{2} - 5 a^{\frac {3}{2}} b \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {2 \sqrt {a} b^{2} x \sqrt {1 + \frac {b x}{a}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)*sqrt(1 + b*x/a)/x + 14*a**(3/2)*b*sqrt(1 + b*x/a)/3 + 5*a**(3/2)*b*log(b*x/a)/2 - 5*a**(3/2)*b*log(s
qrt(1 + b*x/a) + 1) + 2*sqrt(a)*b**2*x*sqrt(1 + b*x/a)/3

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