\(\int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx\) [548]

Optimal result

Integrand size = 15, antiderivative size = 92 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=\frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}} \]

[Out]

5/8*a^3*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(1/2)+5/12*a*(b*x+a)^(3/2)*x^(1/2)+1/3*(b*x+a)^(5/2)*x^(1/2)+
5/8*a^2*x^(1/2)*(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {52, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}}+\frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2} \]

[In]

Int[(a + b*x)^(5/2)/Sqrt[x],x]

[Out]

(5*a^2*Sqrt[x]*Sqrt[a + b*x])/8 + (5*a*Sqrt[x]*(a + b*x)^(3/2))/12 + (Sqrt[x]*(a + b*x)^(5/2))/3 + (5*a^3*ArcT
anh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*Sqrt[b])

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 {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {1}{6} (5 a) \int \frac {(a+b x)^{3/2}}{\sqrt {x}} \, dx \\ & = \frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {1}{8} \left (5 a^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx \\ & = \frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {1}{16} \left (5 a^3\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = \frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {1}{8} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {1}{8} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = \frac {5}{8} a^2 \sqrt {x} \sqrt {a+b x}+\frac {5}{12} a \sqrt {x} (a+b x)^{3/2}+\frac {1}{3} \sqrt {x} (a+b x)^{5/2}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=\frac {1}{24} \sqrt {x} \sqrt {a+b x} \left (33 a^2+26 a b x+8 b^2 x^2\right )-\frac {5 a^3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{8 \sqrt {b}} \]

[In]

Integrate[(a + b*x)^(5/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(33*a^2 + 26*a*b*x + 8*b^2*x^2))/24 - (5*a^3*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(
8*Sqrt[b])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {\left (8 b^{2} x^{2}+26 a b x +33 a^{2}\right ) \sqrt {x}\, \sqrt {b x +a}}{24}+\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 \sqrt {b}\, \sqrt {x}\, \sqrt {b x +a}}\)	\(84\)
default	\(\frac {\left (b x +a \right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 a \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 a \left (\sqrt {x}\, \sqrt {b x +a}+\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4}\right )}{6}\)	\(94\)

[In]

int((b*x+a)^(5/2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(8*b^2*x^2+26*a*b*x+33*a^2)*x^(1/2)*(b*x+a)^(1/2)+5/16*a^3*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1
/2)*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b}\right ] \]

[In]

integrate((b*x+a)^(5/2)/x^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)
*sqrt(b*x + a)*sqrt(x))/b, -1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 26
*a*b^2*x + 33*a^2*b)*sqrt(b*x + a)*sqrt(x))/b]

Sympy [A] (verification not implemented)

Time = 5.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=\frac {11 a^{\frac {5}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}}}{8} + \frac {13 a^{\frac {3}{2}} b x^{\frac {3}{2}} \sqrt {1 + \frac {b x}{a}}}{12} + \frac {\sqrt {a} b^{2} x^{\frac {5}{2}} \sqrt {1 + \frac {b x}{a}}}{3} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}} \]

[In]

integrate((b*x+a)**(5/2)/x**(1/2),x)

[Out]

11*a**(5/2)*sqrt(x)*sqrt(1 + b*x/a)/8 + 13*a**(3/2)*b*x**(3/2)*sqrt(1 + b*x/a)/12 + sqrt(a)*b**2*x**(5/2)*sqrt
(1 + b*x/a)/3 + 5*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (64) = 128\).

Time = 0.32 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=-\frac {5 \, a^{3} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{16 \, \sqrt {b}} - \frac {\frac {15 \, \sqrt {b x + a} a^{3} b^{2}}{\sqrt {x}} - \frac {40 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} + \frac {33 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{3} - \frac {3 \, {\left (b x + a\right )} b^{2}}{x} + \frac {3 \, {\left (b x + a\right )}^{2} b}{x^{2}} - \frac {{\left (b x + a\right )}^{3}}{x^{3}}\right )}} \]

[In]

integrate((b*x+a)^(5/2)/x^(1/2),x, algorithm="maxima")

[Out]

-5/16*a^3*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/sqrt(b) - 1/24*(15*sqrt(b*
x + a)*a^3*b^2/sqrt(x) - 40*(b*x + a)^(3/2)*a^3*b/x^(3/2) + 33*(b*x + a)^(5/2)*a^3/x^(5/2))/(b^3 - 3*(b*x + a)
*b^2/x + 3*(b*x + a)^2*b/x^2 - (b*x + a)^3/x^3)

Giac [A] (verification not implemented)

none

Time = 77.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx=-\frac {{\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b} + \frac {5 \, a}{b}\right )} + \frac {15 \, a^{2}}{b}\right )}\right )} b}{24 \, {\left | b \right |}} \]

[In]

integrate((b*x+a)^(5/2)/x^(1/2),x, algorithm="giac")

[Out]

-1/24*(15*a^3*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/sqrt(b) - sqrt((b*x + a)*b - a*b)*sqr
t(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b + 5*a/b) + 15*a^2/b))*b/abs(b)

Mupad [F(-1)]

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

[In]

int((a + b*x)^(5/2)/x^(1/2),x)

[Out]

int((a + b*x)^(5/2)/x^(1/2), x)