∫ (−a+bx)5∕2 _________________

3.4.33 \(\int \frac {(-a+b x)^{5/2}}{x^3} \, dx\) [333]

Optimal. Leaf size=86 \[ \frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right ) \]

[Out]

-5/4*b*(b*x-a)^(3/2)/x-1/2*(b*x-a)^(5/2)/x^2-15/4*b^2*arctan((b*x-a)^(1/2)/a^(1/2))*a^(1/2)+15/4*b^2*(b*x-a)^(

1/2)

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Rubi [A]
time = 0.02, antiderivative size = 86, 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 = {43, 52, 65, 211} \begin {gather*} \frac {15}{4} b^2 \sqrt {b x-a}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(15*b^2*Sqrt[-a + b*x])/4 - (5*b*(-a + b*x)^(3/2))/(4*x) - (-a + b*x)^(5/2)/(2*x^2) - (15*Sqrt[a]*b^2*ArcTan[S

qrt[-a + b*x]/Sqrt[a]])/4

Rule 43

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, n

}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]

&& PosQ[a/b]

Rubi steps

\begin {align*} \int \frac {(-a+b x)^{5/2}}{x^3} \, dx &=-\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{4} (5 b) \int \frac {(-a+b x)^{3/2}}{x^2} \, dx\\ &=-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{8} \left (15 b^2\right ) \int \frac {\sqrt {-a+b x}}{x} \, dx\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{8} \left (15 a b^2\right ) \int \frac {1}{x \sqrt {-a+b x}} \, dx\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{4} (15 a b) \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 67, normalized size = 0.78 \begin {gather*} \frac {1}{4} \left (\frac {\sqrt {-a+b x} \left (-2 a^2+9 a b x+8 b^2 x^2\right )}{x^2}-15 \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((Sqrt[-a + b*x]*(-2*a^2 + 9*a*b*x + 8*b^2*x^2))/x^2 - 15*Sqrt[a]*b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/4

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 11.81, size = 266, normalized size = 3.09 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I b^{\frac {3}{2}} \left (-2 a^3 x^2 \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}-15 \sqrt {a} \sqrt {b} x^{\frac {5}{2}} \text {ArcCosh}\left [\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}}\right ] \left (a-b x\right )^2+11 a^2 b x^3 \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}-a b^2 x^4 \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}-8 b^3 x^5 \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}\right )}{4 x^{\frac {5}{2}} \left (a-b x\right )^2},\text {Abs}\left [\frac {a}{b x}\right ]>1\right \}\right \},\frac {a^3}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {1-\frac {a}{b x}}}-\frac {11 a^2 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {1-\frac {a}{b x}}}+\frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {1-\frac {a}{b x}}}+\frac {15 \sqrt {a} b^2 \text {ArcSin}\left [\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}}\right ]}{4}+\frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {1-\frac {a}{b x}}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(-a + b*x)^(5/2)/x^3,x]')

[Out]

Piecewise[{{I / 4 b ^ (3 / 2) (-2 a ^ 3 x ^ 2 ((a - b x) / (b x)) ^ (3 / 2) - 15 Sqrt[a] Sqrt[b] x ^ (5 / 2) A

rcCosh[Sqrt[a] / (Sqrt[b] Sqrt[x])] (a - b x) ^ 2 + 11 a ^ 2 b x ^ 3 ((a - b x) / (b x)) ^ (3 / 2) - a b ^ 2 x

^ 4 ((a - b x) / (b x)) ^ (3 / 2) - 8 b ^ 3 x ^ 5 ((a - b x) / (b x)) ^ (3 / 2)) / (x ^ (5 / 2) (a - b x) ^ 2

), Abs[a / (b x)] > 1}}, a ^ 3 / (2 Sqrt[b] x ^ (5 / 2) Sqrt[1 - a / (b x)]) - 11 a ^ 2 Sqrt[b] / (4 x ^ (3 /

2) Sqrt[1 - a / (b x)]) + a b ^ (3 / 2) / (4 Sqrt[x] Sqrt[1 - a / (b x)]) + 15 Sqrt[a] b ^ 2 ArcSin[Sqrt[a] /

(Sqrt[b] Sqrt[x])] / 4 + 2 b ^ (5 / 2) Sqrt[x] / Sqrt[1 - a / (b x)]]

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Maple [A]
time = 0.10, size = 70, normalized size = 0.81


method	result	size

risch	\(\frac {a \left (-b x +a \right ) \left (-9 b x +2 a \right )}{4 x^{2} \sqrt {b x -a}}-\frac {15 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right ) \sqrt {a}}{4}+2 b^{2} \sqrt {b x -a}\)	\(67\)
derivativedivides	\(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(70\)
default	\(2 b^{2} \left (\sqrt {b x -a}-a \left (\frac {-\frac {9 \left (b x -a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {15 \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(70\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*x-a)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

2*b^2*((b*x-a)^(1/2)-a*((-9/8*(b*x-a)^(3/2)-7/8*a*(b*x-a)^(1/2))/b^2/x^2+15/8*arctan((b*x-a)^(1/2)/a^(1/2))/a^

(1/2)))

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Maxima [A]
time = 0.35, size = 97, normalized size = 1.13 \begin {gather*} -\frac {15}{4} \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b^{2} + \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{2} + 7 \, \sqrt {b x - a} a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*sqrt(a)*b^2*arctan(sqrt(b*x - a)/sqrt(a)) + 2*sqrt(b*x - a)*b^2 + 1/4*(9*(b*x - a)^(3/2)*a*b^2 + 7*sqrt(

b*x - a)*a^2*b^2)/((b*x - a)^2 + 2*(b*x - a)*a + a^2)

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Fricas [A]
time = 0.31, size = 139, normalized size = 1.62 \begin {gather*} \left [\frac {15 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, x^{2}}, -\frac {15 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, x^{2}}\right ] \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*sqrt(-a)*b^2*x^2*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) + 2*(8*b^2*x^2 + 9*a*b*x - 2*a^2)*sqrt

(b*x - a))/x^2, -1/4*(15*sqrt(a)*b^2*x^2*arctan(sqrt(b*x - a)/sqrt(a)) - (8*b^2*x^2 + 9*a*b*x - 2*a^2)*sqrt(b*

x - a))/x^2]

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Sympy [A]
time = 2.41, size = 267, normalized size = 3.10 \begin {gather*} \begin {cases} - \frac {15 i \sqrt {a} b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {i a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {11 i a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {15 \sqrt {a} b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} + \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {- \frac {a}{b x} + 1}} & \text {otherwise} \end {cases} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-15*I*sqrt(a)*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - I*a**3/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1

)) + 11*I*a**2*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) - 1)) - I*a*b**(3/2)/(4*sqrt(x)*sqrt(a/(b*x) - 1)) - 2*I*b**(5

/2)*sqrt(x)/sqrt(a/(b*x) - 1), Abs(a/(b*x)) > 1), (15*sqrt(a)*b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/4 + a**3/(2

*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) - 11*a**2*sqrt(b)/(4*x**(3/2)*sqrt(-a/(b*x) + 1)) + a*b**(3/2)/(4*sqrt(x

)*sqrt(-a/(b*x) + 1)) + 2*b**(5/2)*sqrt(x)/sqrt(-a/(b*x) + 1), True))

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Giac [A]
time = 0.01, size = 106, normalized size = 1.23 \begin {gather*} \frac {2 \sqrt {-a+b x} b^{3}+\frac {9 \sqrt {-a+b x} \left (-a+b x\right ) b^{3} a+7 \sqrt {-a+b x} b^{3} a^{2}}{4 \left (-a+b x+a\right )^{2}}-\frac {15 b^{3} a \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}}}{b} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*sqrt(a)*b^3*arctan(sqrt(b*x - a)/sqrt(a)) - 8*sqrt(b*x - a)*b^3 - (9*(b*x - a)^(3/2)*a*b^3 + 7*sqrt(b

*x - a)*a^2*b^3)/(b^2*x^2))/b

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Mupad [B]
time = 0.09, size = 69, normalized size = 0.80 \begin {gather*} 2\,b^2\,\sqrt {b\,x-a}-\frac {15\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4}+\frac {9\,a\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}+\frac {7\,a^2\,\sqrt {b\,x-a}}{4\,x^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*(b*x - a)^(1/2) - (15*a^(1/2)*b^2*atan((b*x - a)^(1/2)/a^(1/2)))/4 + (9*a*(b*x - a)^(3/2))/(4*x^2) + (7*

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

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