Integrand size = 24, antiderivative size = 130 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=-\frac {2 b^2 (b B-A c) \sqrt {b x+c x^2}}{c^4 \sqrt {x}}+\frac {2 b (3 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{3 c^4 x^{3/2}}-\frac {2 (3 b B-A c) \left (b x+c x^2\right )^{5/2}}{5 c^4 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{7 c^4 x^{7/2}} \] Output:
-2*b^2*(-A*c+B*b)*(c*x^2+b*x)^(1/2)/c^4/x^(1/2)+2/3*b*(-2*A*c+3*B*b)*(c*x^ 2+b*x)^(3/2)/c^4/x^(3/2)-2/5*(-A*c+3*B*b)*(c*x^2+b*x)^(5/2)/c^4/x^(5/2)+2/ 7*B*(c*x^2+b*x)^(7/2)/c^4/x^(7/2)
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {x (b+c x)} \left (-48 b^3 B+8 b^2 c (7 A+3 B x)+3 c^3 x^2 (7 A+5 B x)-2 b c^2 x (14 A+9 B x)\right )}{105 c^4 \sqrt {x}} \] Input:
Integrate[(x^(5/2)*(A + B*x))/Sqrt[b*x + c*x^2],x]
Output:
(2*Sqrt[x*(b + c*x)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x) + 3*c^3*x^2*(7*A + 5*B*x) - 2*b*c^2*x*(14*A + 9*B*x)))/(105*c^4*Sqrt[x])
Time = 0.45 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1221, 1128, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {(6 b B-7 A c) \int \frac {x^{5/2}}{\sqrt {c x^2+b x}}dx}{7 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {(6 b B-7 A c) \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \int \frac {x^{3/2}}{\sqrt {c x^2+b x}}dx}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {(6 b B-7 A c) \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {c x^2+b x}}dx}{3 c}\right )}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {\left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right )}{5 c}\right ) (6 b B-7 A c)}{7 c}\) |
Input:
Int[(x^(5/2)*(A + B*x))/Sqrt[b*x + c*x^2],x]
Output:
(2*B*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c) - ((6*b*B - 7*A*c)*((2*x^(3/2)*Sqrt[ b*x + c*x^2])/(5*c) - (4*b*((-4*b*Sqrt[b*x + c*x^2])/(3*c^2*Sqrt[x]) + (2* Sqrt[x]*Sqrt[b*x + c*x^2])/(3*c)))/(5*c)))/(7*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*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x ] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
Time = 0.99 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-18 B b \,c^{2} x^{2}-28 A b \,c^{2} x +24 B \,b^{2} c x +56 A \,b^{2} c -48 B \,b^{3}\right )}{105 \sqrt {x}\, c^{4}}\) | \(76\) |
risch | \(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-18 B b \,c^{2} x^{2}-28 A b \,c^{2} x +24 B \,b^{2} c x +56 A \,b^{2} c -48 B \,b^{3}\right )}{105 \sqrt {x \left (c x +b \right )}\, c^{4}}\) | \(81\) |
gosper | \(\frac {2 \left (c x +b \right ) \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-18 B b \,c^{2} x^{2}-28 A b \,c^{2} x +24 B \,b^{2} c x +56 A \,b^{2} c -48 B \,b^{3}\right ) \sqrt {x}}{105 c^{4} \sqrt {c \,x^{2}+b x}}\) | \(83\) |
orering | \(\frac {2 \left (c x +b \right ) \left (15 B \,c^{3} x^{3}+21 A \,c^{3} x^{2}-18 B b \,c^{2} x^{2}-28 A b \,c^{2} x +24 B \,b^{2} c x +56 A \,b^{2} c -48 B \,b^{3}\right ) \sqrt {x}}{105 c^{4} \sqrt {c \,x^{2}+b x}}\) | \(83\) |
Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/105/x^(1/2)*(x*(c*x+b))^(1/2)*(15*B*c^3*x^3+21*A*c^3*x^2-18*B*b*c^2*x^2- 28*A*b*c^2*x+24*B*b^2*c*x+56*A*b^2*c-48*B*b^3)/c^4
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.61 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (15 \, B c^{3} x^{3} - 48 \, B b^{3} + 56 \, A b^{2} c - 3 \, {\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, c^{4} \sqrt {x}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")
Output:
2/105*(15*B*c^3*x^3 - 48*B*b^3 + 56*A*b^2*c - 3*(6*B*b*c^2 - 7*A*c^3)*x^2 + 4*(6*B*b^2*c - 7*A*b*c^2)*x)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))
\[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{\frac {5}{2}} \left (A + B x\right )}{\sqrt {x \left (b + c x\right )}}\, dx \] Input:
integrate(x**(5/2)*(B*x+A)/(c*x**2+b*x)**(1/2),x)
Output:
Integral(x**(5/2)*(A + B*x)/sqrt(x*(b + c*x)), x)
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (3 \, c^{3} x^{3} - b c^{2} x^{2} + 4 \, b^{2} c x + 8 \, b^{3}\right )} A}{15 \, \sqrt {c x + b} c^{3}} + \frac {2 \, {\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )} B}{35 \, \sqrt {c x + b} c^{4}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")
Output:
2/15*(3*c^3*x^3 - b*c^2*x^2 + 4*b^2*c*x + 8*b^3)*A/(sqrt(c*x + b)*c^3) + 2 /35*(5*c^4*x^4 - b*c^3*x^3 + 2*b^2*c^2*x^2 - 8*b^3*c*x - 16*b^4)*B/(sqrt(c *x + b)*c^4)
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=-\frac {2 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {c x + b}}{c^{4}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} B - 63 \, {\left (c x + b\right )}^{\frac {5}{2}} B b + 105 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} + 21 \, {\left (c x + b\right )}^{\frac {5}{2}} A c - 70 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c\right )}}{105 \, c^{4}} \] Input:
integrate(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="giac")
Output:
-2*(B*b^3 - A*b^2*c)*sqrt(c*x + b)/c^4 + 2/105*(15*(c*x + b)^(7/2)*B - 63* (c*x + b)^(5/2)*B*b + 105*(c*x + b)^(3/2)*B*b^2 + 21*(c*x + b)^(5/2)*A*c - 70*(c*x + b)^(3/2)*A*b*c)/c^4
Timed out. \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{5/2}\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \] Input:
int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(1/2),x)
Output:
int((x^(5/2)*(A + B*x))/(b*x + c*x^2)^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.52 \[ \int \frac {x^{5/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {c x +b}\, \left (15 b \,c^{3} x^{3}+21 a \,c^{3} x^{2}-18 b^{2} c^{2} x^{2}-28 a b \,c^{2} x +24 b^{3} c x +56 a \,b^{2} c -48 b^{4}\right )}{105 c^{4}} \] Input:
int(x^(5/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x)
Output:
(2*sqrt(b + c*x)*(56*a*b**2*c - 28*a*b*c**2*x + 21*a*c**3*x**2 - 48*b**4 + 24*b**3*c*x - 18*b**2*c**2*x**2 + 15*b*c**3*x**3))/(105*c**4)