Integrand size = 20, antiderivative size = 85 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=-2 (a c+b d) \sqrt {c+\frac {d}{x}}-\frac {2}{3} a \left (c+\frac {d}{x}\right )^{3/2}+b c \sqrt {c+\frac {d}{x}} x+\sqrt {c} (2 a c+3 b d) \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right ) \] Output:
-2*(a*c+b*d)*(c+d/x)^(1/2)-2/3*a*(c+d/x)^(3/2)+b*c*(c+d/x)^(1/2)*x+c^(1/2) *(2*a*c+3*b*d)*arctanh((c+d/x)^(1/2)/c^(1/2))
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=\frac {\sqrt {c+\frac {d}{x}} (3 b x (-2 d+c x)-2 a (d+4 c x))}{3 x}+\sqrt {c} (2 a c+3 b d) \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right ) \] Input:
Integrate[((c + d/x)^(3/2)*(a + b*x))/x,x]
Output:
(Sqrt[c + d/x]*(3*b*x*(-2*d + c*x) - 2*a*(d + 4*c*x)))/(3*x) + Sqrt[c]*(2* a*c + 3*b*d)*ArcTanh[Sqrt[c + d/x]/Sqrt[c]]
Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1016, 899, 87, 60, 60, 73, 221}
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.
\(\displaystyle \int \frac {(a+b x) \left (c+\frac {d}{x}\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 1016 |
\(\displaystyle \int \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2}dx\) |
\(\Big \downarrow \) 899 |
\(\displaystyle -\int \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2} x^2d\frac {1}{x}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {b x \left (c+\frac {d}{x}\right )^{5/2}}{c}-\frac {(2 a c+3 b d) \int \left (c+\frac {d}{x}\right )^{3/2} xd\frac {1}{x}}{2 c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {b x \left (c+\frac {d}{x}\right )^{5/2}}{c}-\frac {(2 a c+3 b d) \left (c \int \sqrt {c+\frac {d}{x}} xd\frac {1}{x}+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )}{2 c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {b x \left (c+\frac {d}{x}\right )^{5/2}}{c}-\frac {(2 a c+3 b d) \left (c \left (c \int \frac {x}{\sqrt {c+\frac {d}{x}}}d\frac {1}{x}+2 \sqrt {c+\frac {d}{x}}\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )}{2 c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {b x \left (c+\frac {d}{x}\right )^{5/2}}{c}-\frac {(2 a c+3 b d) \left (c \left (\frac {2 c \int \frac {1}{\frac {1}{d x^2}-\frac {c}{d}}d\sqrt {c+\frac {d}{x}}}{d}+2 \sqrt {c+\frac {d}{x}}\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )}{2 c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {b x \left (c+\frac {d}{x}\right )^{5/2}}{c}-\frac {(2 a c+3 b d) \left (c \left (2 \sqrt {c+\frac {d}{x}}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )}{2 c}\) |
Input:
Int[((c + d/x)^(3/2)*(a + b*x))/x,x]
Output:
(b*(c + d/x)^(5/2)*x)/c - ((2*a*c + 3*b*d)*((2*(c + d/x)^(3/2))/3 + c*(2*S qrt[c + d/x] - 2*Sqrt[c]*ArcTanh[Sqrt[c + d/x]/Sqrt[c]])))/(2*c)
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^( p_.), x_Symbol] :> Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ [{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] || !I ntegerQ[p])
Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24
method | result | size |
risch | \(-\frac {\left (-3 b c \,x^{2}+8 a c x +6 b d x +2 a d \right ) \sqrt {\frac {c x +d}{x}}}{3 x}+\frac {\left (2 a c +3 b d \right ) \sqrt {c}\, \ln \left (\frac {\frac {d}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+d x}\right ) \sqrt {\frac {c x +d}{x}}\, \sqrt {\left (c x +d \right ) x}}{2 c x +2 d}\) | \(105\) |
default | \(-\frac {\sqrt {\frac {c x +d}{x}}\, \left (-12 \sqrt {c \,x^{2}+d x}\, c^{\frac {5}{2}} a \,x^{3}-18 \sqrt {c \,x^{2}+d x}\, c^{\frac {3}{2}} b d \,x^{3}-6 \ln \left (\frac {2 \sqrt {c \,x^{2}+d x}\, \sqrt {c}+2 c x +d}{2 \sqrt {c}}\right ) a \,c^{2} d \,x^{3}-9 \ln \left (\frac {2 \sqrt {c \,x^{2}+d x}\, \sqrt {c}+2 c x +d}{2 \sqrt {c}}\right ) b c \,d^{2} x^{3}+12 \left (c \,x^{2}+d x \right )^{\frac {3}{2}} c^{\frac {3}{2}} a x +12 \left (c \,x^{2}+d x \right )^{\frac {3}{2}} \sqrt {c}\, b d x +4 a \left (c \,x^{2}+d x \right )^{\frac {3}{2}} d \sqrt {c}\right )}{6 x^{2} \sqrt {\left (c x +d \right ) x}\, d \sqrt {c}}\) | \(205\) |
Input:
int((c+d/x)^(3/2)*(b*x+a)/x,x,method=_RETURNVERBOSE)
Output:
-1/3*(-3*b*c*x^2+8*a*c*x+6*b*d*x+2*a*d)/x*((c*x+d)/x)^(1/2)+1/2*(2*a*c+3*b *d)*c^(1/2)*ln((1/2*d+c*x)/c^(1/2)+(c*x^2+d*x)^(1/2))/(c*x+d)*((c*x+d)/x)^ (1/2)*((c*x+d)*x)^(1/2)
Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.99 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=\left [\frac {3 \, {\left (2 \, a c + 3 \, b d\right )} \sqrt {c} x \log \left (2 \, c x + 2 \, \sqrt {c} x \sqrt {\frac {c x + d}{x}} + d\right ) + 2 \, {\left (3 \, b c x^{2} - 2 \, a d - 2 \, {\left (4 \, a c + 3 \, b d\right )} x\right )} \sqrt {\frac {c x + d}{x}}}{6 \, x}, -\frac {3 \, {\left (2 \, a c + 3 \, b d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {-c} x \sqrt {\frac {c x + d}{x}}}{c x + d}\right ) - {\left (3 \, b c x^{2} - 2 \, a d - 2 \, {\left (4 \, a c + 3 \, b d\right )} x\right )} \sqrt {\frac {c x + d}{x}}}{3 \, x}\right ] \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x,x, algorithm="fricas")
Output:
[1/6*(3*(2*a*c + 3*b*d)*sqrt(c)*x*log(2*c*x + 2*sqrt(c)*x*sqrt((c*x + d)/x ) + d) + 2*(3*b*c*x^2 - 2*a*d - 2*(4*a*c + 3*b*d)*x)*sqrt((c*x + d)/x))/x, -1/3*(3*(2*a*c + 3*b*d)*sqrt(-c)*x*arctan(sqrt(-c)*x*sqrt((c*x + d)/x)/(c *x + d)) - (3*b*c*x^2 - 2*a*d - 2*(4*a*c + 3*b*d)*x)*sqrt((c*x + d)/x))/x]
Time = 16.01 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.08 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=- a c \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + \frac {d}{x}} & \text {for}\: d \neq 0 \\- \sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + a d \left (\begin {cases} - \frac {\sqrt {c}}{x} & \text {for}\: d = 0 \\- \frac {2 \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) + b \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {d}} \right )} + b c \sqrt {d} \sqrt {x} \sqrt {\frac {c x}{d} + 1} - b d \left (\begin {cases} \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 \sqrt {c + \frac {d}{x}} & \text {for}\: d \neq 0 \\- \sqrt {c} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((c+d/x)**(3/2)*(b*x+a)/x,x)
Output:
-a*c*Piecewise((2*c*atan(sqrt(c + d/x)/sqrt(-c))/sqrt(-c) + 2*sqrt(c + d/x ), Ne(d, 0)), (-sqrt(c)*log(x), True)) + a*d*Piecewise((-sqrt(c)/x, Eq(d, 0)), (-2*(c + d/x)**(3/2)/(3*d), True)) + b*sqrt(c)*d*asinh(sqrt(c)*sqrt(x )/sqrt(d)) + b*c*sqrt(d)*sqrt(x)*sqrt(c*x/d + 1) - b*d*Piecewise((2*c*atan (sqrt(c + d/x)/sqrt(-c))/sqrt(-c) + 2*sqrt(c + d/x), Ne(d, 0)), (-sqrt(c)* log(x), True))
Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.55 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=-\frac {1}{3} \, {\left (3 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x}} - \sqrt {c}}{\sqrt {c + \frac {d}{x}} + \sqrt {c}}\right ) + 2 \, {\left (c + \frac {d}{x}\right )}^{\frac {3}{2}} + 6 \, \sqrt {c + \frac {d}{x}} c\right )} a + \frac {1}{2} \, {\left (2 \, \sqrt {c + \frac {d}{x}} c x - 3 \, \sqrt {c} d \log \left (\frac {\sqrt {c + \frac {d}{x}} - \sqrt {c}}{\sqrt {c + \frac {d}{x}} + \sqrt {c}}\right ) - 4 \, \sqrt {c + \frac {d}{x}} d\right )} b \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x,x, algorithm="maxima")
Output:
-1/3*(3*c^(3/2)*log((sqrt(c + d/x) - sqrt(c))/(sqrt(c + d/x) + sqrt(c))) + 2*(c + d/x)^(3/2) + 6*sqrt(c + d/x)*c)*a + 1/2*(2*sqrt(c + d/x)*c*x - 3*s qrt(c)*d*log((sqrt(c + d/x) - sqrt(c))/(sqrt(c + d/x) + sqrt(c))) - 4*sqrt (c + d/x)*d)*b
Exception generated. \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 7.81 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=2\,a\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right )-\frac {2\,a\,{\left (c+\frac {d}{x}\right )}^{3/2}}{3}-2\,a\,c\,\sqrt {c+\frac {d}{x}}-\frac {2\,b\,x\,{\left (c+\frac {d}{x}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {c\,x}{d}\right )}{{\left (\frac {c\,x}{d}+1\right )}^{3/2}} \] Input:
int(((c + d/x)^(3/2)*(a + b*x))/x,x)
Output:
2*a*c^(3/2)*atanh((c + d/x)^(1/2)/c^(1/2)) - (2*a*(c + d/x)^(3/2))/3 - 2*a *c*(c + d/x)^(1/2) - (2*b*x*(c + d/x)^(3/2)*hypergeom([-3/2, -1/2], 1/2, - (c*x)/d))/((c*x)/d + 1)^(3/2)
Time = 0.16 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.44 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x} \, dx=\frac {-16 \sqrt {x}\, \sqrt {c x +d}\, a c x -4 \sqrt {x}\, \sqrt {c x +d}\, a d +6 \sqrt {x}\, \sqrt {c x +d}\, b c \,x^{2}-12 \sqrt {x}\, \sqrt {c x +d}\, b d x +12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +d}+\sqrt {x}\, \sqrt {c}}{\sqrt {d}}\right ) a c \,x^{2}+18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +d}+\sqrt {x}\, \sqrt {c}}{\sqrt {d}}\right ) b d \,x^{2}+5 \sqrt {c}\, b d \,x^{2}}{6 x^{2}} \] Input:
int((c+d/x)^(3/2)*(b*x+a)/x,x)
Output:
( - 16*sqrt(x)*sqrt(c*x + d)*a*c*x - 4*sqrt(x)*sqrt(c*x + d)*a*d + 6*sqrt( x)*sqrt(c*x + d)*b*c*x**2 - 12*sqrt(x)*sqrt(c*x + d)*b*d*x + 12*sqrt(c)*lo g((sqrt(c*x + d) + sqrt(x)*sqrt(c))/sqrt(d))*a*c*x**2 + 18*sqrt(c)*log((sq rt(c*x + d) + sqrt(x)*sqrt(c))/sqrt(d))*b*d*x**2 + 5*sqrt(c)*b*d*x**2)/(6* x**2)