3.526 \(\int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 63, 217, 206} \[ 2 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 b \sqrt {a+b x}}{\sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^(5/2),x]

[Out]

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

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 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.75 \[ -\frac {2 a \sqrt {a+b x} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{a}\right )}{3 x^{3/2} \sqrt {\frac {b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^(5/2),x]

[Out]

(-2*a*Sqrt[a + b*x]*Hypergeometric2F1[-3/2, -3/2, -1/2, -((b*x)/a)])/(3*x^(3/2)*Sqrt[1 + (b*x)/a])

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fricas [A]  time = 0.47, size = 109, normalized size = 1.70 \[ \left [\frac {3 \, b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (4 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (4 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*b^(3/2)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(4*b*x + a)*sqrt(b*x + a)*sqrt(x))/x^
2, -2/3*(3*sqrt(-b)*b*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (4*b*x + a)*sqrt(b*x + a)*sqrt(x))/x^2]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 67, normalized size = 1.05 \[ \frac {\sqrt {\left (b x +a \right ) x}\, b^{\frac {3}{2}} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b x +a}\, \sqrt {x}}-\frac {2 \sqrt {b x +a}\, \left (4 b x +a \right )}{3 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^(5/2),x)

[Out]

-2/3*(b*x+a)^(1/2)*(4*b*x+a)/x^(3/2)+b^(3/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))*((b*x+a)*x)^(1/2)/(b*x+
a)^(1/2)/x^(1/2)

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maxima [A]  time = 2.93, size = 67, normalized size = 1.05 \[ -b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right ) - \frac {2 \, \sqrt {b x + a} b}{\sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}}{x^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/x^(5/2),x)

[Out]

int((a + b*x)^(3/2)/x^(5/2), x)

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sympy [A]  time = 3.04, size = 71, normalized size = 1.11 \[ - \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x**(5/2),x)

[Out]

-2*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 8*b**(3/2)*sqrt(a/(b*x) + 1)/3 - b**(3/2)*log(a/(b*x)) + 2*b**(3/2)*log
(sqrt(a/(b*x) + 1) + 1)

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