Integrand size = 20, antiderivative size = 77 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=-2 b c \sqrt {c+\frac {d}{x}}-\frac {2}{3} b \left (c+\frac {d}{x}\right )^{3/2}-\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}+2 b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right ) \] Output:
-2*b*c*(c+d/x)^(1/2)-2/3*b*(c+d/x)^(3/2)-2/5*a*(c+d/x)^(5/2)/d+2*b*c^(3/2) *arctanh((c+d/x)^(1/2)/c^(1/2))
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=-\frac {2 \sqrt {c+\frac {d}{x}} \left (3 a (d+c x)^2+5 b d x (d+4 c x)\right )}{15 d x^2}+2 b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right ) \] Input:
Integrate[((c + d/x)^(3/2)*(a + b*x))/x^2,x]
Output:
(-2*Sqrt[c + d/x]*(3*a*(d + c*x)^2 + 5*b*d*x*(d + 4*c*x)))/(15*d*x^2) + 2* b*c^(3/2)*ArcTanh[Sqrt[c + d/x]/Sqrt[c]]
Time = 0.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1016, 948, 90, 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.
\(\displaystyle \int \frac {(a+b x) \left (c+\frac {d}{x}\right )^{3/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1016 |
\(\displaystyle \int \frac {\left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2}}{x}dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\int \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x}\right )^{3/2} xd\frac {1}{x}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -b \int \left (c+\frac {d}{x}\right )^{3/2} xd\frac {1}{x}-\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -b \left (c \int \sqrt {c+\frac {d}{x}} xd\frac {1}{x}+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )-\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -b \left (c \left (c \int \frac {x}{\sqrt {c+\frac {d}{x}}}d\frac {1}{x}+2 \sqrt {c+\frac {d}{x}}\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )-\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -b \left (c \left (\frac {2 c \int \frac {1}{\frac {1}{d x^2}-\frac {c}{d}}d\sqrt {c+\frac {d}{x}}}{d}+2 \sqrt {c+\frac {d}{x}}\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )-\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a \left (c+\frac {d}{x}\right )^{5/2}}{5 d}-b \left (c \left (2 \sqrt {c+\frac {d}{x}}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+\frac {d}{x}\right )^{3/2}\right )\) |
Input:
Int[((c + d/x)^(3/2)*(a + b*x))/x^2,x]
Output:
(-2*a*(c + d/x)^(5/2))/(5*d) - b*((2*(c + d/x)^(3/2))/3 + c*(2*Sqrt[c + d/ x] - 2*Sqrt[c]*ArcTanh[Sqrt[c + d/x]/Sqrt[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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48
method | result | size |
risch | \(-\frac {2 \left (3 a \,c^{2} x^{2}+20 b c d \,x^{2}+6 a d x c +5 b x \,d^{2}+3 a \,d^{2}\right ) \sqrt {\frac {c x +d}{x}}}{15 x^{2} d}+\frac {b \,c^{\frac {3}{2}} \ln \left (\frac {\frac {d}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+d x}\right ) \sqrt {\frac {c x +d}{x}}\, \sqrt {\left (c x +d \right ) x}}{c x +d}\) | \(114\) |
default | \(-\frac {\sqrt {\frac {c x +d}{x}}\, \left (-30 \sqrt {c \,x^{2}+d x}\, c^{\frac {5}{2}} b \,x^{4}-15 \ln \left (\frac {2 \sqrt {c \,x^{2}+d x}\, \sqrt {c}+2 c x +d}{2 \sqrt {c}}\right ) b \,c^{2} d \,x^{4}+30 \left (c \,x^{2}+d x \right )^{\frac {3}{2}} c^{\frac {3}{2}} b \,x^{2}+6 \left (c \,x^{2}+d x \right )^{\frac {3}{2}} c^{\frac {3}{2}} a x +10 \left (c \,x^{2}+d x \right )^{\frac {3}{2}} \sqrt {c}\, b d x +6 a \left (c \,x^{2}+d x \right )^{\frac {3}{2}} d \sqrt {c}\right )}{15 x^{3} d \sqrt {\left (c x +d \right ) x}\, \sqrt {c}}\) | \(166\) |
Input:
int((c+d/x)^(3/2)*(b*x+a)/x^2,x,method=_RETURNVERBOSE)
Output:
-2/15*(3*a*c^2*x^2+20*b*c*d*x^2+6*a*c*d*x+5*b*d^2*x+3*a*d^2)/x^2/d*((c*x+d )/x)^(1/2)+b*c^(3/2)*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 = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.48 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=\left [\frac {15 \, b c^{\frac {3}{2}} d x^{2} \log \left (2 \, c x + 2 \, \sqrt {c} x \sqrt {\frac {c x + d}{x}} + d\right ) - 2 \, {\left (3 \, a d^{2} + {\left (3 \, a c^{2} + 20 \, b c d\right )} x^{2} + {\left (6 \, a c d + 5 \, b d^{2}\right )} x\right )} \sqrt {\frac {c x + d}{x}}}{15 \, d x^{2}}, -\frac {2 \, {\left (15 \, b \sqrt {-c} c d x^{2} \arctan \left (\frac {\sqrt {-c} x \sqrt {\frac {c x + d}{x}}}{c x + d}\right ) + {\left (3 \, a d^{2} + {\left (3 \, a c^{2} + 20 \, b c d\right )} x^{2} + {\left (6 \, a c d + 5 \, b d^{2}\right )} x\right )} \sqrt {\frac {c x + d}{x}}\right )}}{15 \, d x^{2}}\right ] \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^2,x, algorithm="fricas")
Output:
[1/15*(15*b*c^(3/2)*d*x^2*log(2*c*x + 2*sqrt(c)*x*sqrt((c*x + d)/x) + d) - 2*(3*a*d^2 + (3*a*c^2 + 20*b*c*d)*x^2 + (6*a*c*d + 5*b*d^2)*x)*sqrt((c*x + d)/x))/(d*x^2), -2/15*(15*b*sqrt(-c)*c*d*x^2*arctan(sqrt(-c)*x*sqrt((c*x + d)/x)/(c*x + d)) + (3*a*d^2 + (3*a*c^2 + 20*b*c*d)*x^2 + (6*a*c*d + 5*b *d^2)*x)*sqrt((c*x + d)/x))/(d*x^2)]
Time = 3.98 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=\begin {cases} - \frac {2 a \left (c + \frac {d}{x}\right )^{\frac {5}{2}}}{5 d} - \frac {2 b c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - 2 b c \sqrt {c + \frac {d}{x}} - \frac {2 b \left (c + \frac {d}{x}\right )^{\frac {3}{2}}}{3} & \text {for}\: d \neq 0 \\- \frac {a c^{\frac {3}{2}}}{x} - b c^{\frac {3}{2}} \log {\left (- \frac {a c^{\frac {3}{2}}}{x} \right )} & \text {otherwise} \end {cases} \] Input:
integrate((c+d/x)**(3/2)*(b*x+a)/x**2,x)
Output:
Piecewise((-2*a*(c + d/x)**(5/2)/(5*d) - 2*b*c**2*atan(sqrt(c + d/x)/sqrt( -c))/sqrt(-c) - 2*b*c*sqrt(c + d/x) - 2*b*(c + d/x)**(3/2)/3, Ne(d, 0)), ( -a*c**(3/2)/x - b*c**(3/2)*log(-a*c**(3/2)/x), True))
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=-\frac {2 \, a {\left (c + \frac {d}{x}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {1}{3} \, {\left (3 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x}} - \sqrt {c}}{\sqrt {c + \frac {d}{x}} + \sqrt {c}}\right ) + 2 \, {\left (c + \frac {d}{x}\right )}^{\frac {3}{2}} + 6 \, \sqrt {c + \frac {d}{x}} c\right )} b \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^2,x, algorithm="maxima")
Output:
-2/5*a*(c + d/x)^(5/2)/d - 1/3*(3*c^(3/2)*log((sqrt(c + d/x) - sqrt(c))/(s qrt(c + d/x) + sqrt(c))) + 2*(c + d/x)^(3/2) + 6*sqrt(c + d/x)*c)*b
Exception generated. \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c+d/x)^(3/2)*(b*x+a)/x^2,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 = 7.95 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=2\,b\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x}}}{\sqrt {c}}\right )-\frac {2\,b\,{\left (c+\frac {d}{x}\right )}^{3/2}}{3}-2\,b\,c\,\sqrt {c+\frac {d}{x}}-\frac {2\,a\,\sqrt {c+\frac {d}{x}}\,{\left (d+c\,x\right )}^2}{5\,d\,x^2} \] Input:
int(((c + d/x)^(3/2)*(a + b*x))/x^2,x)
Output:
2*b*c^(3/2)*atanh((c + d/x)^(1/2)/c^(1/2)) - (2*b*(c + d/x)^(3/2))/3 - 2*b *c*(c + d/x)^(1/2) - (2*a*(c + d/x)^(1/2)*(d + c*x)^2)/(5*d*x^2)
Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.74 \[ \int \frac {\left (c+\frac {d}{x}\right )^{3/2} (a+b x)}{x^2} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,c^{2} x^{2}}{5}-\frac {4 \sqrt {x}\, \sqrt {c x +d}\, a c d x}{5}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, a \,d^{2}}{5}-\frac {8 \sqrt {x}\, \sqrt {c x +d}\, b c d \,x^{2}}{3}-\frac {2 \sqrt {x}\, \sqrt {c x +d}\, b \,d^{2} x}{3}+2 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +d}+\sqrt {x}\, \sqrt {c}}{\sqrt {d}}\right ) b c d \,x^{3}-\frac {2 \sqrt {c}\, a \,c^{2} x^{3}}{5}+\frac {16 \sqrt {c}\, b c d \,x^{3}}{15}}{d \,x^{3}} \] Input:
int((c+d/x)^(3/2)*(b*x+a)/x^2,x)
Output:
(2*( - 3*sqrt(x)*sqrt(c*x + d)*a*c**2*x**2 - 6*sqrt(x)*sqrt(c*x + d)*a*c*d *x - 3*sqrt(x)*sqrt(c*x + d)*a*d**2 - 20*sqrt(x)*sqrt(c*x + d)*b*c*d*x**2 - 5*sqrt(x)*sqrt(c*x + d)*b*d**2*x + 15*sqrt(c)*log((sqrt(c*x + d) + sqrt( x)*sqrt(c))/sqrt(d))*b*c*d*x**3 - 3*sqrt(c)*a*c**2*x**3 + 8*sqrt(c)*b*c*d* x**3))/(15*d*x**3)