Integrand size = 24, antiderivative size = 91 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 b (b B-A c) \sqrt {x}}{c^3 \sqrt {b x+c x^2}}-\frac {2 (2 b B-A c) \sqrt {b x+c x^2}}{c^3 \sqrt {x}}+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c^3 x^{3/2}} \] Output:
-2*b*(-A*c+B*b)*x^(1/2)/c^3/(c*x^2+b*x)^(1/2)-2*(-A*c+2*B*b)*(c*x^2+b*x)^( 1/2)/c^3/x^(1/2)+2/3*B*(c*x^2+b*x)^(3/2)/c^3/x^(3/2)
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.59 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (-8 b^2 B+c^2 x (3 A+B x)+b (6 A c-4 B c x)\right )}{3 c^3 \sqrt {x (b+c x)}} \] Input:
Integrate[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]
Output:
(2*Sqrt[x]*(-8*b^2*B + c^2*x*(3*A + B*x) + b*(6*A*c - 4*B*c*x)))/(3*c^3*Sq rt[x*(b + c*x)])
Time = 0.41 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1218, 1128, 1122}
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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1218 |
\(\displaystyle -\left (\frac {3 A}{b}-\frac {4 B}{c}\right ) \int \frac {x^{3/2}}{\sqrt {c x^2+b x}}dx-\frac {2 x^{5/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle -\left (\frac {3 A}{b}-\frac {4 B}{c}\right ) \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {c x^2+b x}}dx}{3 c}\right )-\frac {2 x^{5/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle -\left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right ) \left (\frac {3 A}{b}-\frac {4 B}{c}\right )-\frac {2 x^{5/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
Input:
Int[(x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x]
Output:
(-2*(b*B - A*c)*x^(5/2))/(b*c*Sqrt[b*x + c*x^2]) - ((3*A)/b - (4*B)/c)*((- 4*b*Sqrt[b*x + c*x^2])/(3*c^2*Sqrt[x]) + (2*Sqrt[x]*Sqrt[b*x + c*x^2])/(3* c))
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[m + p], 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 = 0.98 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(\frac {2 \left (c x +b \right ) \left (B \,c^{2} x^{2}+3 A \,c^{2} x -4 B b c x +6 A b c -8 B \,b^{2}\right ) x^{\frac {3}{2}}}{3 c^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(58\) |
default | \(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (B \,c^{2} x^{2}+3 A \,c^{2} x -4 B b c x +6 A b c -8 B \,b^{2}\right )}{3 \sqrt {x}\, \left (c x +b \right ) c^{3}}\) | \(58\) |
orering | \(\frac {2 \left (c x +b \right ) \left (B \,c^{2} x^{2}+3 A \,c^{2} x -4 B b c x +6 A b c -8 B \,b^{2}\right ) x^{\frac {3}{2}}}{3 c^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(58\) |
risch | \(\frac {2 \left (B c x +3 A c -5 B b \right ) \left (c x +b \right ) \sqrt {x}}{3 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {2 b \left (A c -B b \right ) \sqrt {x}}{c^{3} \sqrt {x \left (c x +b \right )}}\) | \(63\) |
Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3*(c*x+b)*(B*c^2*x^2+3*A*c^2*x-4*B*b*c*x+6*A*b*c-8*B*b^2)*x^(3/2)/c^3/(c *x^2+b*x)^(3/2)
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (B c^{2} x^{2} - 8 \, B b^{2} + 6 \, A b c - {\left (4 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (c^{4} x^{2} + b c^{3} x\right )}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
Output:
2/3*(B*c^2*x^2 - 8*B*b^2 + 6*A*b*c - (4*B*b*c - 3*A*c^2)*x)*sqrt(c*x^2 + b *x)*sqrt(x)/(c^4*x^2 + b*c^3*x)
\[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{\frac {5}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x)**(3/2),x)
Output:
Integral(x**(5/2)*(A + B*x)/(x*(b + c*x))**(3/2), x)
\[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {5}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
Output:
2/3*(B*c*x + B*b)*sqrt(c*x + b)*x^2/(c^3*x^2 + 2*b*c^2*x + b^2*c) + integr ate(1/3*(3*A*b*c*x^2 - (4*B*b^2 + (4*B*b*c - 3*A*c^2)*x)*x^2)*sqrt(c*x + b )/(c^4*x^4 + 3*b*c^3*x^3 + 3*b^2*c^2*x^2 + b^3*c*x), x)
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (B b^{2} - A b c\right )}}{\sqrt {c x + b} c^{3}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} B c^{6} - 6 \, \sqrt {c x + b} B b c^{6} + 3 \, \sqrt {c x + b} A c^{7}\right )}}{3 \, c^{9}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")
Output:
-2*(B*b^2 - A*b*c)/(sqrt(c*x + b)*c^3) + 2/3*((c*x + b)^(3/2)*B*c^6 - 6*sq rt(c*x + b)*B*b*c^6 + 3*sqrt(c*x + b)*A*c^7)/c^9
Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{5/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \] Input:
int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2),x)
Output:
int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.51 \[ \int \frac {x^{5/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {2}{3} b \,c^{2} x^{2}+2 a \,c^{2} x -\frac {8}{3} b^{2} c x +4 a b c -\frac {16}{3} b^{3}}{\sqrt {c x +b}\, c^{3}} \] Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(3/2),x)
Output:
(2*(6*a*b*c + 3*a*c**2*x - 8*b**3 - 4*b**2*c*x + b*c**2*x**2))/(3*sqrt(b + c*x)*c**3)