\(\int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx\) [549]

Optimal result

Integrand size = 15, antiderivative size = 89 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=\frac {15}{4} a b \sqrt {x} \sqrt {a+b x}+\frac {5}{2} b \sqrt {x} (a+b x)^{3/2}-\frac {2 (a+b x)^{5/2}}{\sqrt {x}}+\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {49, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=\frac {15}{4} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{5/2}}{\sqrt {x}}+\frac {5}{2} b \sqrt {x} (a+b x)^{3/2}+\frac {15}{4} a b \sqrt {x} \sqrt {a+b x} \]

[In]

Int[(a + b*x)^(5/2)/x^(3/2),x]

[Out]

(15*a*b*Sqrt[x]*Sqrt[a + b*x])/4 + (5*b*Sqrt[x]*(a + b*x)^(3/2))/2 - (2*(a + b*x)^(5/2))/Sqrt[x] + (15*a^2*Sqr
t[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/4

Rule 49

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 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 212

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

Rule 223

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-8 a^2+9 a b x+2 b^2 x^2\right )}{4 \sqrt {x}}+\frac {15}{2} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]

[In]

Integrate[(a + b*x)^(5/2)/x^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-2 b^{2} x^{2}-9 a b x +8 a^{2}\right )}{4 \sqrt {x}}+\frac {15 a^{2} \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 \sqrt {x}\, \sqrt {b x +a}}\)	\(84\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{4 \, x}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=- \frac {2 a^{\frac {5}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + \frac {a^{\frac {3}{2}} b \sqrt {x}}{4 \sqrt {1 + \frac {b x}{a}}} + \frac {11 \sqrt {a} b^{2} x^{\frac {3}{2}}}{4 \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4} + \frac {b^{3} x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-2*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + a**(3/2)*b*sqrt(x)/(4*sqrt(1 + b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(4*s
qrt(1 + b*x/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/4 + b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=-\frac {15}{8} \, a^{2} \sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right ) - \frac {2 \, \sqrt {b x + a} a^{2}}{\sqrt {x}} - \frac {\frac {7 \, \sqrt {b x + a} a^{2} b^{2}}{\sqrt {x}} - \frac {9 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b}{x^{\frac {3}{2}}}}{4 \, {\left (b^{2} - \frac {2 \, {\left (b x + a\right )} b}{x} + \frac {{\left (b x + a\right )}^{2}}{x^{2}}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 75.94 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=-\frac {{\left (\frac {15 \, a^{2} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left ({\left (2 \, b x + 7 \, a\right )} {\left (b x + a\right )} - 15 \, a^{2}\right )} \sqrt {b x + a}}{\sqrt {{\left (b x + a\right )} b - a b}}\right )} b^{2}}{4 \, {\left | b \right |}} \]

[In]

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

[Out]

-1/4*(15*a^2*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/sqrt(b) - ((2*b*x + 7*a)*(b*x + a) - 1
5*a^2)*sqrt(b*x + a)/sqrt((b*x + a)*b - a*b))*b^2/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2}}{x^{3/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{x^{3/2}} \,d x \]

[In]

int((a + b*x)^(5/2)/x^(3/2),x)

[Out]

int((a + b*x)^(5/2)/x^(3/2), x)