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

Optimal. Leaf size=63 \[ -\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+3 b \sqrt {x} \sqrt {a+b x}+3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {47, 50, 63, 217, 206} \begin {gather*} -\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+3 b \sqrt {x} \sqrt {a+b x}+3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^(3/2),x]

[Out]

3*b*Sqrt[x]*Sqrt[a + b*x] - (2*(a + b*x)^(3/2))/Sqrt[x] + 3*a*Sqrt[b]*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 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 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^{3/2}} \, dx &=-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+(3 b) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx\\ &=3 b \sqrt {x} \sqrt {a+b x}-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+\frac {1}{2} (3 a b) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=3 b \sqrt {x} \sqrt {a+b x}-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+(3 a b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=3 b \sqrt {x} \sqrt {a+b x}-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+(3 a b) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )\\ &=3 b \sqrt {x} \sqrt {a+b x}-\frac {2 (a+b x)^{3/2}}{\sqrt {x}}+3 a \sqrt {b} \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 = 46, normalized size = 0.73 \begin {gather*} -\frac {2 a \sqrt {a+b x} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x}{a}\right )}{\sqrt {x} \sqrt {\frac {b x}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 54, normalized size = 0.86 \begin {gather*} \frac {(b x-2 a) \sqrt {a+b x}}{\sqrt {x}}-3 a \sqrt {b} \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)/x^(3/2),x]

[Out]

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

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fricas [A]  time = 1.33, size = 109, normalized size = 1.73 \begin {gather*} \left [\frac {3 \, a \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, \sqrt {b x + a} {\left (b x - 2 \, a\right )} \sqrt {x}}{2 \, x}, -\frac {3 \, a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - \sqrt {b x + a} {\left (b x - 2 \, a\right )} \sqrt {x}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.02, size = 71, normalized size = 1.13 \begin {gather*} \frac {3 \sqrt {\left (b x +a \right ) x}\, a \sqrt {b}\, \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}}-\frac {\sqrt {b x +a}\, \left (-b x +2 a \right )}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x+a)^(1/2)*(-b*x+2*a)/x^(1/2)+3/2*a*b^(1/2)*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 = 2.99, size = 84, normalized size = 1.33 \begin {gather*} -\frac {3}{2} \, a \sqrt {b} \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} a}{\sqrt {x}} - \frac {\sqrt {b x + a} a b}{{\left (b - \frac {b x + a}{x}\right )} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/x^(3/2),x)

[Out]

int((a + b*x)^(3/2)/x^(3/2), x)

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sympy [A]  time = 2.72, size = 92, normalized size = 1.46 \begin {gather*} - \frac {2 a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {\sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + \frac {b^{2} x^{\frac {3}{2}}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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