Integrand size = 24, antiderivative size = 167 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 b^3 (b B-A c) \sqrt {b x+c x^2}}{c^5 \sqrt {x}}-\frac {2 b^2 (4 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{3 c^5 x^{3/2}}+\frac {6 b (2 b B-A c) \left (b x+c x^2\right )^{5/2}}{5 c^5 x^{5/2}}-\frac {2 (4 b B-A c) \left (b x+c x^2\right )^{7/2}}{7 c^5 x^{7/2}}+\frac {2 B \left (b x+c x^2\right )^{9/2}}{9 c^5 x^{9/2}} \] Output:
2*b^3*(-A*c+B*b)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)-2/3*b^2*(-3*A*c+4*B*b)*(c*x ^2+b*x)^(3/2)/c^5/x^(3/2)+6/5*b*(-A*c+2*B*b)*(c*x^2+b*x)^(5/2)/c^5/x^(5/2) -2/7*(-A*c+4*B*b)*(c*x^2+b*x)^(7/2)/c^5/x^(7/2)+2/9*B*(c*x^2+b*x)^(9/2)/c^ 5/x^(9/2)
Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {x (b+c x)} \left (128 b^4 B+24 b^2 c^2 x (3 A+2 B x)-16 b^3 c (9 A+4 B x)+5 c^4 x^3 (9 A+7 B x)-2 b c^3 x^2 (27 A+20 B x)\right )}{315 c^5 \sqrt {x}} \] Input:
Integrate[(x^(7/2)*(A + B*x))/Sqrt[b*x + c*x^2],x]
Output:
(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 = 0.49 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98, 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 |
\(\displaystyle \frac {2 B x^{7/2} \sqrt {b x+c x^2}}{9 c}-\frac {\left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right )}{5 c}\right )}{7 c}\right ) (8 b B-9 A c)}{9 c}\) |
Input:
Int[(x^(7/2)*(A + B*x))/Sqrt[b*x + c*x^2],x]
Output:
(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)
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 = 0.98 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.60
method | result | size |
default | \(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-35 B \,c^{4} x^{4}-45 A \,c^{4} x^{3}+40 B \,c^{3} x^{3} b +54 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 B \,b^{4}\right )}{315 \sqrt {x}\, c^{5}}\) | \(100\) |
risch | \(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-35 B \,c^{4} x^{4}-45 A \,c^{4} x^{3}+40 B \,c^{3} x^{3} b +54 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 B \,b^{4}\right )}{315 \sqrt {x \left (c x +b \right )}\, c^{5}}\) | \(105\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-35 B \,c^{4} x^{4}-45 A \,c^{4} x^{3}+40 B \,c^{3} x^{3} b +54 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 B \,b^{4}\right ) \sqrt {x}}{315 c^{5} \sqrt {c \,x^{2}+b x}}\) | \(107\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-35 B \,c^{4} x^{4}-45 A \,c^{4} x^{3}+40 B \,c^{3} x^{3} b +54 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}-72 A \,b^{2} c^{2} x +64 B \,b^{3} c x +144 A \,b^{3} c -128 B \,b^{4}\right ) \sqrt {x}}{315 c^{5} \sqrt {c \,x^{2}+b x}}\) | \(107\) |
Input:
int(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-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 = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.62 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (35 \, B c^{4} x^{4} + 128 \, B b^{4} - 144 \, A b^{3} c - 5 \, {\left (8 \, B b c^{3} - 9 \, A c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, c^{5} \sqrt {x}} \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")
Output:
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))
\[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{\frac {7}{2}} \left (A + B x\right )}{\sqrt {x \left (b + c x\right )}}\, dx \] Input:
integrate(x**(7/2)*(B*x+A)/(c*x**2+b*x)**(1/2),x)
Output:
Integral(x**(7/2)*(A + B*x)/sqrt(x*(b + c*x)), x)
Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.72 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )} A}{35 \, \sqrt {c x + b} c^{4}} + \frac {2 \, {\left (35 \, c^{5} x^{5} - 5 \, b c^{4} x^{4} + 8 \, b^{2} c^{3} x^{3} - 16 \, b^{3} c^{2} x^{2} + 64 \, b^{4} c x + 128 \, b^{5}\right )} B}{315 \, \sqrt {c x + b} c^{5}} \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")
Output:
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 = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.69 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \sqrt {c x + b}}{c^{5}} + \frac {2 \, {\left (35 \, {\left (c x + b\right )}^{\frac {9}{2}} B - 180 \, {\left (c x + b\right )}^{\frac {7}{2}} B b + 378 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} - 420 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} + 45 \, {\left (c x + b\right )}^{\frac {7}{2}} A c - 189 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c + 315 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c\right )}}{315 \, c^{5}} \] Input:
integrate(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="giac")
Output:
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
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{\sqrt {c\,x^2+b\,x}} \,d x \] Input:
int((x^(7/2)*(A + B*x))/(b*x + c*x^2)^(1/2),x)
Output:
int((x^(7/2)*(A + B*x))/(b*x + c*x^2)^(1/2), x)
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.54 \[ \int \frac {x^{7/2} (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {c x +b}\, \left (35 b \,c^{4} x^{4}+45 a \,c^{4} x^{3}-40 b^{2} c^{3} x^{3}-54 a b \,c^{3} x^{2}+48 b^{3} c^{2} x^{2}+72 a \,b^{2} c^{2} x -64 b^{4} c x -144 a \,b^{3} c +128 b^{5}\right )}{315 c^{5}} \] Input:
int(x^(7/2)*(B*x+A)/(c*x^2+b*x)^(1/2),x)
Output:
(2*sqrt(b + c*x)*( - 144*a*b**3*c + 72*a*b**2*c**2*x - 54*a*b*c**3*x**2 + 45*a*c**4*x**3 + 128*b**5 - 64*b**4*c*x + 48*b**3*c**2*x**2 - 40*b**2*c**3 *x**3 + 35*b*c**4*x**4))/(315*c**5)