Integrand size = 24, antiderivative size = 94 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 (b B-A c) x^{3/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 A \sqrt {x}}{b^2 \sqrt {b x+c x^2}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{5/2}} \] Output:
-2/3*(-A*c+B*b)*x^(3/2)/b/c/(c*x^2+b*x)^(3/2)+2*A*x^(1/2)/b^2/(c*x^2+b*x)^ (1/2)-2*A*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 x^{3/2} \left (\sqrt {b} \left (-b^2 B+4 A b c+3 A c^2 x\right )-3 A c (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{3 b^{5/2} c (x (b+c x))^{3/2}} \] Input:
Integrate[(x^(3/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(2*x^(3/2)*(Sqrt[b]*(-(b^2*B) + 4*A*b*c + 3*A*c^2*x) - 3*A*c*(b + c*x)^(3/ 2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(3*b^(5/2)*c*(x*(b + c*x))^(3/2))
Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1218, 1132, 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.
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1218 |
\(\displaystyle \frac {A \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{b}-\frac {2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1132 |
\(\displaystyle \frac {A \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{b}-\frac {2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {A \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{b}-\frac {2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {A \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{b}-\frac {2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
Input:
Int[(x^(3/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(-2*(b*B - A*c)*x^(3/2))/(3*b*c*(b*x + c*x^2)^(3/2)) + (A*((2*Sqrt[x])/(b* Sqrt[b*x + c*x^2]) - (2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3 /2)))/b
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[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ [0, m, 1] && 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*(c*d - b*e) + c*e*f)*(d + e*x)^m*(( a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Simp[e*((m*(g*(c* d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e))) I nt[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d , e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 1.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x \sqrt {c x +b}+3 A c \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \sqrt {c x +b}-3 A \sqrt {b}\, c^{2} x -4 A \,b^{\frac {3}{2}} c +B \,b^{\frac {5}{2}}\right )}{3 b^{\frac {5}{2}} \sqrt {x}\, \left (c x +b \right )^{2} c}\) | \(101\) |
Input:
int(x^(3/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(x*(c*x+b))^(1/2)/b^(5/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*c^2*x*( c*x+b)^(1/2)+3*A*c*arctanh((c*x+b)^(1/2)/b^(1/2))*b*(c*x+b)^(1/2)-3*A*b^(1 /2)*c^2*x-4*A*b^(3/2)*c+B*b^(5/2))/x^(1/2)/(c*x+b)^2/c
Time = 0.11 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.82 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (A c^{3} x^{3} + 2 \, A b c^{2} x^{2} + A b^{2} c x\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (3 \, A b c^{2} x - B b^{3} + 4 \, A b^{2} c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (b^{3} c^{3} x^{3} + 2 \, b^{4} c^{2} x^{2} + b^{5} c x\right )}}, \frac {2 \, {\left (3 \, {\left (A c^{3} x^{3} + 2 \, A b c^{2} x^{2} + A b^{2} c x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, A b c^{2} x - B b^{3} + 4 \, A b^{2} c\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{3 \, {\left (b^{3} c^{3} x^{3} + 2 \, b^{4} c^{2} x^{2} + b^{5} c x\right )}}\right ] \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
Output:
[1/3*(3*(A*c^3*x^3 + 2*A*b*c^2*x^2 + A*b^2*c*x)*sqrt(b)*log(-(c*x^2 + 2*b* x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(3*A*b*c^2*x - B*b^3 + 4 *A*b^2*c)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c^3*x^3 + 2*b^4*c^2*x^2 + b^5*c* x), 2/3*(3*(A*c^3*x^3 + 2*A*b*c^2*x^2 + A*b^2*c*x)*sqrt(-b)*arctan(sqrt(c* x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) + (3*A*b*c^2*x - B*b^3 + 4*A*b^2*c)*sqrt( c*x^2 + b*x)*sqrt(x))/(b^3*c^3*x^3 + 2*b^4*c^2*x^2 + b^5*c*x)]
\[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{\frac {3}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**(3/2)*(B*x+A)/(c*x**2+b*x)**(5/2),x)
Output:
Integral(x**(3/2)*(A + B*x)/(x*(b + c*x))**(5/2), x)
\[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {3}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
Output:
integrate((B*x + A)*x^(3/2)/(c*x^2 + b*x)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, A \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} - \frac {2 \, {\left (B b^{2} - 3 \, {\left (c x + b\right )} A c - A b c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c} \] Input:
integrate(x^(3/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
Output:
2*A*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) - 2/3*(B*b^2 - 3*(c*x + b)*A*c - A*b*c)/((c*x + b)^(3/2)*b^2*c)
Timed out. \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \] Input:
int((x^(3/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x)
Output:
int((x^(3/2)*(A + B*x))/(b*x + c*x^2)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.53 \[ \int \frac {x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a b c +3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{2} x -3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a b c -3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{2} x +8 a \,b^{2} c +6 a b \,c^{2} x -2 b^{4}}{3 \sqrt {c x +b}\, b^{3} c \left (c x +b \right )} \] Input:
int(x^(3/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x)
Output:
(3*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*b*c + 3*sqrt(b)*sq rt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*c**2*x - 3*sqrt(b)*sqrt(b + c*x )*log(sqrt(b + c*x) + sqrt(b))*a*b*c - 3*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*a*c**2*x + 8*a*b**2*c + 6*a*b*c**2*x - 2*b**4)/(3*sqrt(b + c*x)*b**3*c*(b + c*x))