Integrand size = 24, antiderivative size = 141 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b B-A c) x^{7/2}}{b c \sqrt {b x+c x^2}}+\frac {16 b (6 b B-5 A c) \sqrt {b x+c x^2}}{15 c^4 \sqrt {x}}-\frac {8 (6 b B-5 A c) \sqrt {x} \sqrt {b x+c x^2}}{15 c^3}+\frac {2 (6 b B-5 A c) x^{3/2} \sqrt {b x+c x^2}}{5 b c^2} \]
-2*(-A*c+B*b)*x^(7/2)/b/c/(c*x^2+b*x)^(1/2)+2/5*(-5*A*c+6*B*b)*x^(3/2)*(c* x^2+b*x)^(1/2)/b/c^2+16/15*b*(-5*A*c+6*B*b)*(c*x^2+b*x)^(1/2)/c^4/x^(1/2)- 8/15*(-5*A*c+6*B*b)*x^(1/2)*(c*x^2+b*x)^(1/2)/c^3
Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.52 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (48 b^3 B-8 b^2 c (5 A-3 B x)+c^3 x^2 (5 A+3 B x)-2 b c^2 x (10 A+3 B x)\right )}{15 c^4 \sqrt {x (b+c x)}} \]
(2*Sqrt[x]*(48*b^3*B - 8*b^2*c*(5*A - 3*B*x) + c^3*x^2*(5*A + 3*B*x) - 2*b *c^2*x*(10*A + 3*B*x)))/(15*c^4*Sqrt[x*(b + c*x)])
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1218, 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.
\(\displaystyle \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1218 |
\(\displaystyle -\left (\frac {5 A}{b}-\frac {6 B}{c}\right ) \int \frac {x^{5/2}}{\sqrt {c x^2+b x}}dx-\frac {2 x^{7/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle -\left (\frac {5 A}{b}-\frac {6 B}{c}\right ) \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \int \frac {x^{3/2}}{\sqrt {c x^2+b x}}dx}{5 c}\right )-\frac {2 x^{7/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle -\left (\frac {5 A}{b}-\frac {6 B}{c}\right ) \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {c x^2+b x}}dx}{3 c}\right )}{5 c}\right )-\frac {2 x^{7/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle -\left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right )}{5 c}\right ) \left (\frac {5 A}{b}-\frac {6 B}{c}\right )-\frac {2 x^{7/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\) |
(-2*(b*B - A*c)*x^(7/2))/(b*c*Sqrt[b*x + c*x^2]) - ((5*A)/b - (6*B)/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))
3.3.34.3.1 Defintions of rubi rules used
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 = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.59
method | result | size |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-3 B \,c^{3} x^{3}-5 A \,c^{3} x^{2}+6 B b \,c^{2} x^{2}+20 A b \,c^{2} x -24 B \,b^{2} c x +40 A \,b^{2} c -48 B \,b^{3}\right ) x^{\frac {3}{2}}}{15 c^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(83\) |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-3 B \,c^{3} x^{3}-5 A \,c^{3} x^{2}+6 B b \,c^{2} x^{2}+20 A b \,c^{2} x -24 B \,b^{2} c x +40 A \,b^{2} c -48 B \,b^{3}\right )}{15 \sqrt {x}\, \left (c x +b \right ) c^{4}}\) | \(83\) |
risch | \(-\frac {2 \left (-3 B \,c^{2} x^{2}-5 A \,c^{2} x +9 B b c x +25 A b c -33 B \,b^{2}\right ) \left (c x +b \right ) \sqrt {x}}{15 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {2 b^{2} \left (A c -B b \right ) \sqrt {x}}{c^{4} \sqrt {x \left (c x +b \right )}}\) | \(86\) |
-2/15*(c*x+b)*(-3*B*c^3*x^3-5*A*c^3*x^2+6*B*b*c^2*x^2+20*A*b*c^2*x-24*B*b^ 2*c*x+40*A*b^2*c-48*B*b^3)*x^(3/2)/c^4/(c*x^2+b*x)^(3/2)
Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.65 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (3 \, B c^{3} x^{3} + 48 \, B b^{3} - 40 \, A b^{2} c - {\left (6 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{15 \, {\left (c^{5} x^{2} + b c^{4} x\right )}} \]
2/15*(3*B*c^3*x^3 + 48*B*b^3 - 40*A*b^2*c - (6*B*b*c^2 - 5*A*c^3)*x^2 + 4* (6*B*b^2*c - 5*A*b*c^2)*x)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^5*x^2 + b*c^4*x)
\[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{\frac {7}{2}} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {7}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]
2/15*((3*B*c^3*x^2 + B*b*c^2*x - 2*B*b^2*c)*x^3 - (4*B*b^3 + (4*B*b*c^2 - 5*A*c^3)*x^2 + (8*B*b^2*c - 5*A*b*c^2)*x)*x^2)*sqrt(c*x + b)/(c^5*x^3 + 2* b*c^4*x^2 + b^2*c^3*x) - integrate(-4/15*(2*B*b^4 + (7*B*b^2*c^2 - 5*A*b*c ^3)*x^2 + (9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x + b)*x^2/(c^6*x^5 + 3*b*c^ 5*x^4 + 3*b^2*c^4*x^3 + b^3*c^3*x^2), x)
Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.88 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (B b^{3} - A b^{2} c\right )}}{\sqrt {c x + b} c^{4}} - \frac {16 \, {\left (6 \, B b^{3} - 5 \, A b^{2} c\right )}}{15 \, \sqrt {b} c^{4}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} B c^{16} - 15 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{16} + 45 \, \sqrt {c x + b} B b^{2} c^{16} + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{17} - 30 \, \sqrt {c x + b} A b c^{17}\right )}}{15 \, c^{20}} \]
2*(B*b^3 - A*b^2*c)/(sqrt(c*x + b)*c^4) - 16/15*(6*B*b^3 - 5*A*b^2*c)/(sqr t(b)*c^4) + 2/15*(3*(c*x + b)^(5/2)*B*c^16 - 15*(c*x + b)^(3/2)*B*b*c^16 + 45*sqrt(c*x + b)*B*b^2*c^16 + 5*(c*x + b)^(3/2)*A*c^17 - 30*sqrt(c*x + b) *A*b*c^17)/c^20
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]