Integrand size = 20, antiderivative size = 115 \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\frac {(b c-a d) \sqrt {c+d x}}{a b (a+b x)}-\frac {2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {\sqrt {b c-a d} (2 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}} \]
-2*c^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^2+(a*d+2*b*c)*arctanh(b^(1/2)* (d*x+c)^(1/2)/(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/a^2/b^(3/2)+(-a*d+b*c)*(d *x+c)^(1/2)/a/b/(b*x+a)
Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\frac {\frac {a (b c-a d) \sqrt {c+d x}}{b (a+b x)}+\frac {\sqrt {-b c+a d} (2 b c+a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}-2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \]
((a*(b*c - a*d)*Sqrt[c + d*x])/(b*(a + b*x)) + (Sqrt[-(b*c) + a*d]*(2*b*c + a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(3/2) - 2*c^( 3/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^2
Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 27, 174, 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.
| \(\displaystyle \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {\int \frac {2 b c^2+d (b c+a d) x}{2 x (a+b x) \sqrt {c+d x}}dx}{a b}+\frac {\sqrt {c+d x} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 b c^2+d (b c+a d) x}{x (a+b x) \sqrt {c+d x}}dx}{2 a b}+\frac {\sqrt {c+d x} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {\frac {2 b c^2 \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {(b c-a d) (a d+2 b c) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{2 a b}+\frac {\sqrt {c+d x} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {4 b c^2 \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {2 (b c-a d) (a d+2 b c) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{2 a b}+\frac {\sqrt {c+d x} (b c-a d)}{a b (a+b x)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 \sqrt {b c-a d} (a d+2 b c) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {4 b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}}{2 a b}+\frac {\sqrt {c+d x} (b c-a d)}{a b (a+b x)}\) |
((b*c - a*d)*Sqrt[c + d*x])/(a*b*(a + b*x)) + ((-4*b*c^(3/2)*ArcTanh[Sqrt[ c + d*x]/Sqrt[c]])/a + (2*Sqrt[b*c - a*d]*(2*b*c + a*d)*ArcTanh[(Sqrt[b]*S qrt[c + d*x])/Sqrt[b*c - a*d]])/(a*Sqrt[b]))/(2*a*b)
3.5.66.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
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]
Time = 0.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(2 d^{2} \left (-\frac {c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2} d^{2}}+\frac {\left (a d -b c \right ) \left (-\frac {a d \sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (a d +2 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )}{a^{2} d^{2}}\right )\) | \(124\) |
| default | \(2 d^{2} \left (-\frac {c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2} d^{2}}+\frac {\left (a d -b c \right ) \left (-\frac {a d \sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (a d +2 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )}{a^{2} d^{2}}\right )\) | \(124\) |
| pseudoelliptic | \(-\frac {-\left (a d +2 b c \right ) \left (a d -b c \right ) \left (b x +a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (2 b \,c^{\frac {3}{2}} \left (b x +a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+a \sqrt {d x +c}\, \left (a d -b c \right )\right ) \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, a^{2} b \left (b x +a \right )}\) | \(128\) |
2*d^2*(-1/a^2/d^2*c^(3/2)*arctanh((d*x+c)^(1/2)/c^(1/2))+(a*d-b*c)/a^2/d^2 *(-1/2*a*d/b*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+1/2*(a*d+2*b*c)/b/((a*d-b*c )*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
Time = 0.27 (sec) , antiderivative size = 624, normalized size of antiderivative = 5.43 \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\left [\frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (b^{2} c x + a b c\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (b^{2} c x + a b c\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{a^{2} b^{2} x + a^{3} b}, \frac {4 \, {\left (b^{2} c x + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{2 \, {\left (a^{2} b^{2} x + a^{3} b\right )}}, \frac {{\left (2 \, a b c + a^{2} d + {\left (2 \, b^{2} c + a b d\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + 2 \, {\left (b^{2} c x + a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (a b c - a^{2} d\right )} \sqrt {d x + c}}{a^{2} b^{2} x + a^{3} b}\right ] \]
[1/2*((2*a*b*c + a^2*d + (2*b^2*c + a*b*d)*x)*sqrt((b*c - a*d)/b)*log((b*d *x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*( b^2*c*x + a*b*c)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2* (a*b*c - a^2*d)*sqrt(d*x + c))/(a^2*b^2*x + a^3*b), ((2*a*b*c + a^2*d + (2 *b^2*c + a*b*d)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (b^2*c*x + a*b*c)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + (a*b*c - a^2*d)*sqrt(d*x + c))/(a^2*b^2*x + a^3*b ), 1/2*(4*(b^2*c*x + a*b*c)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (2 *a*b*c + a^2*d + (2*b^2*c + a*b*d)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b *c - a*d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(a*b*c - a^2*d)*sqrt(d*x + c))/(a^2*b^2*x + a^3*b), ((2*a*b*c + a^2*d + (2*b^2*c + a*b*d)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b )/(b*c - a*d)) + 2*(b^2*c*x + a*b*c)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c )/c) + (a*b*c - a^2*d)*sqrt(d*x + c))/(a^2*b^2*x + a^3*b)]
\[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}}}{x \left (a + b x\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=\frac {2 \, c^{2} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} - \frac {{\left (2 \, b^{2} c^{2} - a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b} + \frac {\sqrt {d x + c} b c d - \sqrt {d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b} \]
2*c^2*arctan(sqrt(d*x + c)/sqrt(-c))/(a^2*sqrt(-c)) - (2*b^2*c^2 - a*b*c*d - a^2*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a* b*d)*a^2*b) + (sqrt(d*x + c)*b*c*d - sqrt(d*x + c)*a*d^2)/(((d*x + c)*b - b*c + a*d)*a*b)
Time = 0.67 (sec) , antiderivative size = 482, normalized size of antiderivative = 4.19 \[ \int \frac {(c+d x)^{3/2}}{x (a+b x)^2} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {4\,d^6\,\sqrt {c^3}\,\sqrt {c+d\,x}}{4\,c^2\,d^6+\frac {8\,b\,c^3\,d^5}{a}-\frac {12\,b^2\,c^4\,d^4}{a^2}}+\frac {8\,c\,d^5\,\sqrt {c^3}\,\sqrt {c+d\,x}}{8\,c^3\,d^5+\frac {4\,a\,c^2\,d^6}{b}-\frac {12\,b\,c^4\,d^4}{a}}-\frac {12\,b\,c^2\,d^4\,\sqrt {c^3}\,\sqrt {c+d\,x}}{8\,a\,c^3\,d^5-12\,b\,c^4\,d^4+\frac {4\,a^2\,c^2\,d^6}{b}}\right )\,\sqrt {c^3}}{a^2}-\frac {\mathrm {atanh}\left (\frac {10\,c^2\,d^5\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{2\,a^2\,c\,d^7+2\,b^2\,c^3\,d^5-\frac {12\,b^3\,c^4\,d^4}{a}+8\,a\,b\,c^2\,d^6}+\frac {12\,c^3\,d^4\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{8\,a^2\,c^2\,d^6-12\,b^2\,c^4\,d^4+\frac {2\,a^3\,c\,d^7}{b}+2\,a\,b\,c^3\,d^5}+\frac {2\,c\,d^6\,\sqrt {b^4\,c-a\,b^3\,d}\,\sqrt {c+d\,x}}{8\,b^2\,c^2\,d^6+2\,a\,b\,c\,d^7+\frac {2\,b^3\,c^3\,d^5}{a}-\frac {12\,b^4\,c^4\,d^4}{a^2}}\right )\,\left (a\,d+2\,b\,c\right )\,\sqrt {-b^3\,\left (a\,d-b\,c\right )}}{a^2\,b^3}-\frac {d\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{a\,b\,\left (a\,d-b\,c+b\,\left (c+d\,x\right )\right )} \]
- (2*atanh((4*d^6*(c^3)^(1/2)*(c + d*x)^(1/2))/(4*c^2*d^6 + (8*b*c^3*d^5)/ a - (12*b^2*c^4*d^4)/a^2) + (8*c*d^5*(c^3)^(1/2)*(c + d*x)^(1/2))/(8*c^3*d ^5 + (4*a*c^2*d^6)/b - (12*b*c^4*d^4)/a) - (12*b*c^2*d^4*(c^3)^(1/2)*(c + d*x)^(1/2))/(8*a*c^3*d^5 - 12*b*c^4*d^4 + (4*a^2*c^2*d^6)/b))*(c^3)^(1/2)) /a^2 - (atanh((10*c^2*d^5*(b^4*c - a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(2*a^2* c*d^7 + 2*b^2*c^3*d^5 - (12*b^3*c^4*d^4)/a + 8*a*b*c^2*d^6) + (12*c^3*d^4* (b^4*c - a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(8*a^2*c^2*d^6 - 12*b^2*c^4*d^4 + (2*a^3*c*d^7)/b + 2*a*b*c^3*d^5) + (2*c*d^6*(b^4*c - a*b^3*d)^(1/2)*(c + d*x)^(1/2))/(8*b^2*c^2*d^6 + 2*a*b*c*d^7 + (2*b^3*c^3*d^5)/a - (12*b^4*c^4 *d^4)/a^2))*(a*d + 2*b*c)*(-b^3*(a*d - b*c))^(1/2))/(a^2*b^3) - (d*(a*d - b*c)*(c + d*x)^(1/2))/(a*b*(a*d - b*c + b*(c + d*x)))