Integrand size = 24, antiderivative size = 170 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {32 b^3 (8 b B-9 A c) \sqrt {b x+c x^2}}{315 c^5 \sqrt {x}}-\frac {16 b^2 (8 b B-9 A c) \sqrt {x} \sqrt {b x+c x^2}}{315 c^4}+\frac {4 b (8 b B-9 A c) x^{3/2} \sqrt {b x+c x^2}}{105 c^3}-\frac {2 (8 b B-9 A c) x^{5/2} \sqrt {b x+c x^2}}{63 c^2}+\frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c} \]
4/105*b*(-9*A*c+8*B*b)*x^(3/2)*(c*x^2+b*x)^(1/2)/c^3-2/63*(-9*A*c+8*B*b)*x ^(5/2)*(c*x^2+b*x)^(1/2)/c^2+2/9*B*x^(7/2)*(c*x^2+b*x)^(1/2)/c+32/315*b^3* (-9*A*c+8*B*b)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)-16/315*b^2*(-9*A*c+8*B*b)*x^( 1/2)*(c*x^2+b*x)^(1/2)/c^4
Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.55 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {x (b+c x)} \left (128 b^4 B+24 b^2 c^2 x (3 A+2 B x)-16 b^3 c (9 A+4 B x)+5 c^4 x^3 (9 A+7 B x)-2 b c^3 x^2 (27 A+20 B x)\right )}{315 c^5 \sqrt {x}} \]
(2*Sqrt[x*(b + c*x)]*(128*b^4*B + 24*b^2*c^2*x*(3*A + 2*B*x) - 16*b^3*c*(9 *A + 4*B*x) + 5*c^4*x^3*(9*A + 7*B*x) - 2*b*c^3*x^2*(27*A + 20*B*x)))/(315 *c^5*Sqrt[x])
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1221, 1128, 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.
\(\displaystyle \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {(8 b B-9 A c) \int \frac {x^{7/2}}{\sqrt {c x^2+b x}}dx}{9 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {(8 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \int \frac {x^{5/2}}{\sqrt {c x^2+b x}}dx}{7 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {(8 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \int \frac {x^{3/2}}{\sqrt {c x^2+b x}}dx}{5 c}\right )}{7 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {(8 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {c x^2+b x}}dx}{3 c}\right )}{5 c}\right )}{7 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {\left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right )}{5 c}\right )}{7 c}\right ) (8 b B-9 A c)}{9 c}\) |
(2*B*x^(7/2)*Sqrt[b*x + c*x^2])/(9*c) - ((8*b*B - 9*A*c)*((2*x^(5/2)*Sqrt[ b*x + c*x^2])/(7*c) - (6*b*((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)))/(9*c)
3.3.25.3.1 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*(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 = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.59
method | result | size |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-35 B \,x^{4} c^{4}-45 A \,c^{4} x^{3}+40 B b \,c^{3} x^{3}+54 A b \,c^{3} x^{2}-48 B \,b^{2} c^{2} x^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 b^{4} B \right )}{315 \sqrt {x}\, c^{5}}\) | \(100\) |
risch | \(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-35 B \,x^{4} c^{4}-45 A \,c^{4} x^{3}+40 B b \,c^{3} x^{3}+54 A b \,c^{3} x^{2}-48 B \,b^{2} c^{2} x^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 b^{4} B \right )}{315 \sqrt {x \left (c x +b \right )}\, c^{5}}\) | \(105\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-35 B \,x^{4} c^{4}-45 A \,c^{4} x^{3}+40 B b \,c^{3} x^{3}+54 A b \,c^{3} x^{2}-48 B \,b^{2} c^{2} x^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 b^{4} B \right ) \sqrt {x}}{315 c^{5} \sqrt {c \,x^{2}+b x}}\) | \(107\) |
-2/315/x^(1/2)*(x*(c*x+b))^(1/2)*(-35*B*c^4*x^4-45*A*c^4*x^3+40*B*b*c^3*x^ 3+54*A*b*c^3*x^2-48*B*b^2*c^2*x^2-72*A*b^2*c^2*x+64*B*b^3*c*x+144*A*b^3*c- 128*B*b^4)/c^5
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.61 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (35 \, B c^{4} x^{4} + 128 \, B b^{4} - 144 \, A b^{3} c - 5 \, {\left (8 \, B b c^{3} - 9 \, A c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{5} \sqrt {x}} \]
2/315*(35*B*c^4*x^4 + 128*B*b^4 - 144*A*b^3*c - 5*(8*B*b*c^3 - 9*A*c^4)*x^ 3 + 6*(8*B*b^2*c^2 - 9*A*b*c^3)*x^2 - 8*(8*B*b^3*c - 9*A*b^2*c^2)*x)*sqrt( c*x^2 + b*x)/(c^5*sqrt(x))
\[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{\frac {7}{2}} \left (A + B x\right )}{\sqrt {x \left (b + c x\right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )} A}{35 \, \sqrt {c x + b} c^{4}} + \frac {2 \, {\left (35 \, c^{5} x^{5} - 5 \, b c^{4} x^{4} + 8 \, b^{2} c^{3} x^{3} - 16 \, b^{3} c^{2} x^{2} + 64 \, b^{4} c x + 128 \, b^{5}\right )} B}{315 \, \sqrt {c x + b} c^{5}} \]
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)*A/(sqrt( c*x + b)*c^4) + 2/315*(35*c^5*x^5 - 5*b*c^4*x^4 + 8*b^2*c^3*x^3 - 16*b^3*c ^2*x^2 + 64*b^4*c*x + 128*b^5)*B/(sqrt(c*x + b)*c^5)
Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \sqrt {c x + b}}{c^{5}} + \frac {2 \, {\left (35 \, {\left (c x + b\right )}^{\frac {9}{2}} B - 180 \, {\left (c x + b\right )}^{\frac {7}{2}} B b + 378 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} - 420 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} + 45 \, {\left (c x + b\right )}^{\frac {7}{2}} A c - 189 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c + 315 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c\right )}}{315 \, c^{5}} - \frac {32 \, {\left (8 \, B b^{\frac {9}{2}} - 9 \, A b^{\frac {7}{2}} c\right )}}{315 \, c^{5}} \]
2*(B*b^4 - A*b^3*c)*sqrt(c*x + b)/c^5 + 2/315*(35*(c*x + b)^(9/2)*B - 180* (c*x + b)^(7/2)*B*b + 378*(c*x + b)^(5/2)*B*b^2 - 420*(c*x + b)^(3/2)*B*b^ 3 + 45*(c*x + b)^(7/2)*A*c - 189*(c*x + b)^(5/2)*A*b*c + 315*(c*x + b)^(3/ 2)*A*b^2*c)/c^5 - 32/315*(8*B*b^(9/2) - 9*A*b^(7/2)*c)/c^5
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \]