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

Optimal. Leaf size=86 \[ 5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )+5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}} \]

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Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, 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*} 5 b^2 \sqrt {x} \sqrt {a+b x}+5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^(5/2),x]

[Out]

5*b^2*Sqrt[x]*Sqrt[a + b*x] - (10*b*(a + b*x)^(3/2))/(3*Sqrt[x]) - (2*(a + b*x)^(5/2))/(3*x^(3/2)) + 5*a*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 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)^{5/2}}{x^{5/2}} \, dx &=-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+\frac {1}{3} (5 b) \int \frac {(a+b x)^{3/2}}{x^{3/2}} \, dx\\ &=-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+\frac {1}{2} \left (5 a b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx\\ &=5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+\left (5 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )\\ &=5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+\left (5 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )\\ &=5 b^2 \sqrt {x} \sqrt {a+b x}-\frac {10 b (a+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (a+b x)^{5/2}}{3 x^{3/2}}+5 a 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 = 50, normalized size = 0.58 \begin {gather*} -\frac {2 a^2 \sqrt {a+b x} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{a}\right )}{3 x^{3/2} \sqrt {\frac {b x}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^(5/2),x]

[Out]

(-2*a^2*Sqrt[a + b*x]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((b*x)/a)])/(3*x^(3/2)*Sqrt[1 + (b*x)/a])

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IntegrateAlgebraic [A]  time = 0.16, size = 69, normalized size = 0.80 \begin {gather*} \frac {\sqrt {a+b x} \left (-2 a^2-14 a b x+3 b^2 x^2\right )}{3 x^{3/2}}-5 a b^{3/2} \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-2*a^2 - 14*a*b*x + 3*b^2*x^2))/(3*x^(3/2)) - 5*a*b^(3/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*
x]]

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fricas [A]  time = 1.53, size = 138, normalized size = 1.60 \begin {gather*} \left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(5/2)/x^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 82, normalized size = 0.95 \begin {gather*} \frac {5 \sqrt {\left (b x +a \right ) x}\, a \,b^{\frac {3}{2}} \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 (-3 b^{2} x^{2}+14 a b x +2 a^{2}\right )}{3 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b*x+a)^(1/2)*(-3*b^2*x^2+14*a*b*x+2*a^2)/x^(3/2)+5/2*a*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 = 2.95, size = 100, normalized size = 1.16 \begin {gather*} -\frac {5}{2} \, a 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 {4 \, \sqrt {b x + a} a b}{\sqrt {x}} - \frac {\sqrt {b x + a} a b^{2}}{{\left (b - \frac {b x + a}{x}\right )} \sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} a}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/x^(5/2),x)

[Out]

int((a + b*x)^(5/2)/x^(5/2), x)

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sympy [A]  time = 5.62, size = 99, normalized size = 1.15 \begin {gather*} - \frac {2 a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {14 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )}}{2} + 5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} + b^{\frac {5}{2}} x \sqrt {\frac {a}{b x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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