Integrand size = 24, antiderivative size = 91 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 b (b B-A c) x^{3/2}}{3 c^3 \left (b x+c x^2\right )^{3/2}}+\frac {2 (2 b B-A c) \sqrt {x}}{c^3 \sqrt {b x+c x^2}}+\frac {2 B \sqrt {b x+c x^2}}{c^3 \sqrt {x}} \] Output:
-2/3*b*(-A*c+B*b)*x^(3/2)/c^3/(c*x^2+b*x)^(3/2)+2*(-A*c+2*B*b)*x^(1/2)/c^3 /(c*x^2+b*x)^(1/2)+2*B*(c*x^2+b*x)^(1/2)/c^3/x^(1/2)
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 x^{3/2} \left (8 b^2 B-2 b c (A-6 B x)+3 c^2 x (-A+B x)\right )}{3 c^3 (x (b+c x))^{3/2}} \] Input:
Integrate[(x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(2*x^(3/2)*(8*b^2*B - 2*b*c*(A - 6*B*x) + 3*c^2*x*(-A + B*x)))/(3*c^3*(x*( b + c*x))^(3/2))
Time = 0.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14, 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.
\(\displaystyle \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1218 |
\(\displaystyle \frac {(4 b B-A c) \int \frac {x^{5/2}}{\left (c x^2+b x\right )^{3/2}}dx}{3 b c}-\frac {2 x^{7/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {(4 b B-A c) \left (\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}-\frac {2 b \int \frac {x^{3/2}}{\left (c x^2+b x\right )^{3/2}}dx}{c}\right )}{3 b c}-\frac {2 x^{7/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {\left (\frac {4 b \sqrt {x}}{c^2 \sqrt {b x+c x^2}}+\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}\right ) (4 b B-A c)}{3 b c}-\frac {2 x^{7/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\) |
Input:
Int[(x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
Output:
(-2*(b*B - A*c)*x^(7/2))/(3*b*c*(b*x + c*x^2)^(3/2)) + ((4*b*B - A*c)*((4* b*Sqrt[x])/(c^2*Sqrt[b*x + c*x^2]) + (2*x^(3/2))/(c*Sqrt[b*x + c*x^2])))/( 3*b*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 = 1.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-3 B \,c^{2} x^{2}+3 A \,c^{2} x -12 B b c x +2 A b c -8 B \,b^{2}\right ) x^{\frac {5}{2}}}{3 c^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\) | \(59\) |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-3 B \,c^{2} x^{2}+3 A \,c^{2} x -12 B b c x +2 A b c -8 B \,b^{2}\right )}{3 \sqrt {x}\, \left (c x +b \right )^{2} c^{3}}\) | \(59\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-3 B \,c^{2} x^{2}+3 A \,c^{2} x -12 B b c x +2 A b c -8 B \,b^{2}\right ) x^{\frac {5}{2}}}{3 c^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\) | \(59\) |
risch | \(\frac {2 B \left (c x +b \right ) \sqrt {x}}{c^{3} \sqrt {x \left (c x +b \right )}}-\frac {2 \left (3 A \,c^{2} x -6 B b c x +2 A b c -5 B \,b^{2}\right ) \sqrt {x}}{3 c^{3} \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}\) | \(74\) |
Input:
int(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(c*x+b)*(-3*B*c^2*x^2+3*A*c^2*x-12*B*b*c*x+2*A*b*c-8*B*b^2)*x^(5/2)/c ^3/(c*x^2+b*x)^(5/2)
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, B c^{2} x^{2} + 8 \, B b^{2} - 2 \, A b c + 3 \, {\left (4 \, B b c - A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (c^{5} x^{3} + 2 \, b c^{4} x^{2} + b^{2} c^{3} x\right )}} \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
Output:
2/3*(3*B*c^2*x^2 + 8*B*b^2 - 2*A*b*c + 3*(4*B*b*c - A*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^5*x^3 + 2*b*c^4*x^2 + b^2*c^3*x)
\[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{\frac {7}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**(7/2)*(B*x+A)/(c*x**2+b*x)**(5/2),x)
Output:
Integral(x**(7/2)*(A + B*x)/(x*(b + c*x))**(5/2), x)
\[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {7}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
Output:
integrate((B*x + A)*x^(7/2)/(c*x^2 + b*x)^(5/2), x)
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, \sqrt {c x + b} B}{c^{3}} + \frac {2 \, {\left (6 \, {\left (c x + b\right )} B b - B b^{2} - 3 \, {\left (c x + b\right )} A c + A b c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{3}} \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
Output:
2*sqrt(c*x + b)*B/c^3 + 2/3*(6*(c*x + b)*B*b - B*b^2 - 3*(c*x + b)*A*c + A *b*c)/((c*x + b)^(3/2)*c^3)
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \] Input:
int((x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x)
Output:
int((x^(7/2)*(A + B*x))/(b*x + c*x^2)^(5/2), x)
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.59 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 b \,c^{2} x^{2}-2 a \,c^{2} x +8 b^{2} c x -\frac {4}{3} a b c +\frac {16}{3} b^{3}}{\sqrt {c x +b}\, c^{3} \left (c x +b \right )} \] Input:
int(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x)
Output:
(2*( - 2*a*b*c - 3*a*c**2*x + 8*b**3 + 12*b**2*c*x + 3*b*c**2*x**2))/(3*sq rt(b + c*x)*c**3*(b + c*x))