\(\int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx\) [526]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=-\frac {2 b \sqrt {a+b x}}{\sqrt {x}}-\frac {2 (a+b x)^{3/2}}{3 x^{3/2}}+2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 65, 223, 212} \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 b \sqrt {a+b x}}{\sqrt {x}} \]

[In]

Int[(a + b*x)^(3/2)/x^(5/2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=-\frac {2 \sqrt {a+b x} (a+4 b x)}{3 x^{3/2}}-2 b^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right ) \]

[In]

Integrate[(a + b*x)^(3/2)/x^(5/2),x]

[Out]

(-2*Sqrt[a + b*x]*(a + 4*b*x))/(3*x^(3/2)) - 2*b^(3/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {2 \sqrt {b x +a}\, \left (4 b x +a \right )}{3 x^{\frac {3}{2}}}+\frac {b^{\frac {3}{2}} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\)	\(67\)

[In]

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

[Out]

-2/3*(b*x+a)^(1/2)*(4*b*x+a)/x^(3/2)+b^(3/2)*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 = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.70 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=\left [\frac {3 \, b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (4 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (4 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=- \frac {2 a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} - b^{\frac {3}{2}} \log {\left (\frac {a}{b x} \right )} + 2 b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x} + 1} + 1 \right )} \]

[In]

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

[Out]

-2*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 8*b**(3/2)*sqrt(a/(b*x) + 1)/3 - b**(3/2)*log(a/(b*x)) + 2*b**(3/2)*log
(sqrt(a/(b*x) + 1) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=-b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right ) - \frac {2 \, \sqrt {b x + a} b}{\sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 76.88 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^{3/2}}{x^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b^{\frac {3}{2}} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) + \frac {{\left (4 \, {\left (b x + a\right )} b^{3} - 3 \, a b^{3}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*x)^(3/2)/x^(5/2),x)

[Out]

int((a + b*x)^(3/2)/x^(5/2), x)