Integrand size = 24, antiderivative size = 95 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 b (b B-A c) \left (b x+c x^2\right )^{3/2}}{3 c^3 x^{3/2}}-\frac {2 (2 b B-A c) \left (b x+c x^2\right )^{5/2}}{5 c^3 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{7 c^3 x^{7/2}} \] Output:
2/3*b*(-A*c+B*b)*(c*x^2+b*x)^(3/2)/c^3/x^(3/2)-2/5*(-A*c+2*B*b)*(c*x^2+b*x )^(5/2)/c^3/x^(5/2)+2/7*B*(c*x^2+b*x)^(7/2)/c^3/x^(7/2)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.59 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 (x (b+c x))^{3/2} \left (8 b^2 B+3 c^2 x (7 A+5 B x)-2 b c (7 A+6 B x)\right )}{105 c^3 x^{3/2}} \] Input:
Integrate[Sqrt[x]*(A + B*x)*Sqrt[b*x + c*x^2],x]
Output:
(2*(x*(b + c*x))^(3/2)*(8*b^2*B + 3*c^2*x*(7*A + 5*B*x) - 2*b*c*(7*A + 6*B *x)))/(105*c^3*x^(3/2))
Time = 0.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1221, 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.
\(\displaystyle \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {(4 b B-7 A c) \int \sqrt {x} \sqrt {c x^2+b x}dx}{7 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {(4 b B-7 A c) \left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {2 b \int \frac {\sqrt {c x^2+b x}}{\sqrt {x}}dx}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {\left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}\right ) (4 b B-7 A c)}{7 c}\) |
Input:
Int[Sqrt[x]*(A + B*x)*Sqrt[b*x + c*x^2],x]
Output:
(2*B*Sqrt[x]*(b*x + c*x^2)^(3/2))/(7*c) - ((4*b*B - 7*A*c)*((-4*b*(b*x + c *x^2)^(3/2))/(15*c^2*x^(3/2)) + (2*(b*x + c*x^2)^(3/2))/(5*c*Sqrt[x])))/(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 = 0.92 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.60
method | result | size |
default | \(-\frac {2 \left (c x +b \right ) \left (-15 B \,c^{2} x^{2}-21 A \,c^{2} x +12 B b c x +14 A b c -8 B \,b^{2}\right ) \sqrt {x \left (c x +b \right )}}{105 c^{3} \sqrt {x}}\) | \(57\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-15 B \,c^{2} x^{2}-21 A \,c^{2} x +12 B b c x +14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 c^{3} \sqrt {x}}\) | \(59\) |
orering | \(-\frac {2 \left (c x +b \right ) \left (-15 B \,c^{2} x^{2}-21 A \,c^{2} x +12 B b c x +14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 c^{3} \sqrt {x}}\) | \(59\) |
risch | \(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-15 B \,c^{3} x^{3}-21 A \,c^{3} x^{2}-3 B b \,c^{2} x^{2}-7 A b \,c^{2} x +4 B \,b^{2} c x +14 A \,b^{2} c -8 B \,b^{3}\right )}{105 \sqrt {x \left (c x +b \right )}\, c^{3}}\) | \(81\) |
Input:
int(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/105*(c*x+b)*(-15*B*c^2*x^2-21*A*c^2*x+12*B*b*c*x+14*A*b*c-8*B*b^2)*(x*( c*x+b))^(1/2)/c^3/x^(1/2)
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (15 \, B c^{3} x^{3} + 8 \, B b^{3} - 14 \, A b^{2} c + 3 \, {\left (B b c^{2} + 7 \, A c^{3}\right )} x^{2} - {\left (4 \, B b^{2} c - 7 \, A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, c^{3} \sqrt {x}} \] Input:
integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")
Output:
2/105*(15*B*c^3*x^3 + 8*B*b^3 - 14*A*b^2*c + 3*(B*b*c^2 + 7*A*c^3)*x^2 - ( 4*B*b^2*c - 7*A*b*c^2)*x)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))
\[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\int \sqrt {x} \sqrt {x \left (b + c x\right )} \left (A + B x\right )\, dx \] Input:
integrate(x**(1/2)*(B*x+A)*(c*x**2+b*x)**(1/2),x)
Output:
Integral(sqrt(x)*sqrt(x*(b + c*x))*(A + B*x), x)
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.79 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt {c x + b} A}{15 \, c^{2}} + \frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b} B}{105 \, c^{3}} \] Input:
integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")
Output:
2/15*(3*c^2*x^2 + b*c*x - 2*b^2)*sqrt(c*x + b)*A/c^2 + 2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x + b)*B/c^3
Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b\right )} A}{15 \, c^{2}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )} B}{105 \, c^{3}} \] Input:
integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")
Output:
2/15*(3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)*A/c^2 + 2/105*(15*(c*x + b) ^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)*B/c^3
Timed out. \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\int \sqrt {x}\,\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right ) \,d x \] Input:
int(x^(1/2)*(b*x + c*x^2)^(1/2)*(A + B*x),x)
Output:
int(x^(1/2)*(b*x + c*x^2)^(1/2)*(A + B*x), x)
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \sqrt {x} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \sqrt {c x +b}\, \left (15 b \,c^{3} x^{3}+21 a \,c^{3} x^{2}+3 b^{2} c^{2} x^{2}+7 a b \,c^{2} x -4 b^{3} c x -14 a \,b^{2} c +8 b^{4}\right )}{105 c^{3}} \] Input:
int(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x)
Output:
(2*sqrt(b + c*x)*( - 14*a*b**2*c + 7*a*b*c**2*x + 21*a*c**3*x**2 + 8*b**4 - 4*b**3*c*x + 3*b**2*c**2*x**2 + 15*b*c**3*x**3))/(105*c**3)