Integrand size = 24, antiderivative size = 81 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 A \sqrt {b x+c x^2}}{\sqrt {x}}+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}}-2 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \] Output:
2*A*(c*x^2+b*x)^(1/2)/x^(1/2)+2/3*B*(c*x^2+b*x)^(3/2)/c/x^(3/2)-2*A*b^(1/2 )*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x} (b B+3 A c+B c x)-3 A \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{3 c \sqrt {x (b+c x)}} \] Input:
Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^(3/2),x]
Output:
(2*Sqrt[x]*Sqrt[b + c*x]*(Sqrt[b + c*x]*(b*B + 3*A*c + B*c*x) - 3*A*Sqrt[b ]*c*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(3*c*Sqrt[x*(b + c*x)])
Time = 0.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1221, 1131, 1136, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle A \int \frac {\sqrt {c x^2+b x}}{x^{3/2}}dx+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}}\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle A \left (b \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle A \left (2 b \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}+\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}\right )+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle A \left (\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )\right )+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}}\) |
Input:
Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^(3/2),x]
Output:
(2*B*(b*x + c*x^2)^(3/2))/(3*c*x^(3/2)) + A*((2*Sqrt[b*x + c*x^2])/Sqrt[x] - 2*Sqrt[b]*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x ] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
Time = 1.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (3 A \sqrt {b}\, c \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-B c x \sqrt {c x +b}-3 A c \sqrt {c x +b}-B b \sqrt {c x +b}\right )}{3 \sqrt {x}\, \sqrt {c x +b}\, c}\) | \(79\) |
Input:
int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(x*(c*x+b))^(1/2)*(3*A*b^(1/2)*c*arctanh((c*x+b)^(1/2)/b^(1/2))-B*c*x *(c*x+b)^(1/2)-3*A*c*(c*x+b)^(1/2)-B*b*(c*x+b)^(1/2))/x^(1/2)/(c*x+b)^(1/2 )/c
Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.86 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\left [\frac {3 \, A \sqrt {b} c x \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (B c x + B b + 3 \, A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, c x}, \frac {2 \, {\left (3 \, A \sqrt {-b} c x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (B c x + B b + 3 \, A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{3 \, c x}\right ] \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(3/2),x, algorithm="fricas")
Output:
[1/3*(3*A*sqrt(b)*c*x*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sq rt(x))/x^2) + 2*(B*c*x + B*b + 3*A*c)*sqrt(c*x^2 + b*x)*sqrt(x))/(c*x), 2/ 3*(3*A*sqrt(-b)*c*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) + (B*c* x + B*b + 3*A*c)*sqrt(c*x^2 + b*x)*sqrt(x))/(c*x)]
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac {3}{2}}}\, dx \] Input:
integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**(3/2),x)
Output:
Integral(sqrt(x*(b + c*x))*(A + B*x)/x**(3/2), x)
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x} {\left (B x + A\right )}}{x^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(3/2),x, algorithm="maxima")
Output:
A*integrate(sqrt(c*x + b)/x, x) + 2/3*(B*c*x + B*b)*sqrt(c*x + b)/c
Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \, A b \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} B c^{2} + 3 \, \sqrt {c x + b} A c^{3}\right )}}{3 \, c^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(3/2),x, algorithm="giac")
Output:
2*A*b*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) + 2/3*((c*x + b)^(3/2)*B*c^2 + 3*sqrt(c*x + b)*A*c^3)/c^3
Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^{3/2}} \,d x \] Input:
int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(3/2),x)
Output:
int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(3/2), x)
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {6 \sqrt {c x +b}\, a c +2 \sqrt {c x +b}\, b^{2}+2 \sqrt {c x +b}\, b c x +3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a c -3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a c}{3 c} \] Input:
int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(3/2),x)
Output:
(6*sqrt(b + c*x)*a*c + 2*sqrt(b + c*x)*b**2 + 2*sqrt(b + c*x)*b*c*x + 3*sq rt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c - 3*sqrt(b)*log(sqrt(b + c*x) + sqr t(b))*a*c)/(3*c)