Integrand size = 22, antiderivative size = 114 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=-\frac {2 (B d-A e) \left (c d^2+a e^2\right ) \sqrt {d+e x}}{e^4}+\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right ) (d+e x)^{3/2}}{3 e^4}-\frac {2 c (3 B d-A e) (d+e x)^{5/2}}{5 e^4}+\frac {2 B c (d+e x)^{7/2}}{7 e^4} \]
2/3*(-2*A*c*d*e+B*a*e^2+3*B*c*d^2)*(e*x+d)^(3/2)/e^4-2/5*c*(-A*e+3*B*d)*(e *x+d)^(5/2)/e^4+2/7*B*c*(e*x+d)^(7/2)/e^4-2*(-A*e+B*d)*(a*e^2+c*d^2)*(e*x+ d)^(1/2)/e^4
Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (105 a A e^3+35 a B e^2 (-2 d+e x)+7 A c e \left (8 d^2-4 d e x+3 e^2 x^2\right )-3 B c \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right )\right )}{105 e^4} \]
(2*Sqrt[d + e*x]*(105*a*A*e^3 + 35*a*B*e^2*(-2*d + e*x) + 7*A*c*e*(8*d^2 - 4*d*e*x + 3*e^2*x^2) - 3*B*c*(16*d^3 - 8*d^2*e*x + 6*d*e^2*x^2 - 5*e^3*x^ 3)))/(105*e^4)
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {652, 2009}
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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) (A+B x)}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {\sqrt {d+e x} \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^3}+\frac {\left (a e^2+c d^2\right ) (A e-B d)}{e^3 \sqrt {d+e x}}+\frac {c (d+e x)^{3/2} (A e-3 B d)}{e^3}+\frac {B c (d+e x)^{5/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (d+e x)^{3/2} \left (a B e^2-2 A c d e+3 B c d^2\right )}{3 e^4}-\frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right ) (B d-A e)}{e^4}-\frac {2 c (d+e x)^{5/2} (3 B d-A e)}{5 e^4}+\frac {2 B c (d+e x)^{7/2}}{7 e^4}\) |
(-2*(B*d - A*e)*(c*d^2 + a*e^2)*Sqrt[d + e*x])/e^4 + (2*(3*B*c*d^2 - 2*A*c *d*e + a*B*e^2)*(d + e*x)^(3/2))/(3*e^4) - (2*c*(3*B*d - A*e)*(d + e*x)^(5 /2))/(5*e^4) + (2*B*c*(d + e*x)^(7/2))/(7*e^4)
3.15.31.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {\left (\frac {5 B x}{7}+A \right ) x^{2} c}{5}+a \left (\frac {B x}{3}+A \right )\right ) e^{3}-\frac {4 \left (x \left (\frac {9 B x}{14}+A \right ) c +\frac {5 B a}{2}\right ) d \,e^{2}}{15}+\frac {8 c \left (\frac {3 B x}{7}+A \right ) d^{2} e}{15}-\frac {16 B c \,d^{3}}{35}\right )}{e^{4}}\) | \(79\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (15 B c \,x^{3} e^{3}+21 A c \,e^{3} x^{2}-18 B \,x^{2} c d \,e^{2}-28 A c d \,e^{2} x +35 B x a \,e^{3}+24 B c \,d^{2} e x +105 A a \,e^{3}+56 A c \,d^{2} e -70 B a d \,e^{2}-48 B c \,d^{3}\right )}{105 e^{4}}\) | \(101\) |
trager | \(\frac {2 \sqrt {e x +d}\, \left (15 B c \,x^{3} e^{3}+21 A c \,e^{3} x^{2}-18 B \,x^{2} c d \,e^{2}-28 A c d \,e^{2} x +35 B x a \,e^{3}+24 B c \,d^{2} e x +105 A a \,e^{3}+56 A c \,d^{2} e -70 B a d \,e^{2}-48 B c \,d^{3}\right )}{105 e^{4}}\) | \(101\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (15 B c \,x^{3} e^{3}+21 A c \,e^{3} x^{2}-18 B \,x^{2} c d \,e^{2}-28 A c d \,e^{2} x +35 B x a \,e^{3}+24 B c \,d^{2} e x +105 A a \,e^{3}+56 A c \,d^{2} e -70 B a d \,e^{2}-48 B c \,d^{3}\right )}{105 e^{4}}\) | \(101\) |
derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \sqrt {e x +d}}{e^{4}}\) | \(105\) |
default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \sqrt {e x +d}}{e^{4}}\) | \(105\) |
2*(e*x+d)^(1/2)*((1/5*(5/7*B*x+A)*x^2*c+a*(1/3*B*x+A))*e^3-4/15*(x*(9/14*B *x+A)*c+5/2*B*a)*d*e^2+8/15*c*(3/7*B*x+A)*d^2*e-16/35*B*c*d^3)/e^4
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, B c e^{3} x^{3} - 48 \, B c d^{3} + 56 \, A c d^{2} e - 70 \, B a d e^{2} + 105 \, A a e^{3} - 3 \, {\left (6 \, B c d e^{2} - 7 \, A c e^{3}\right )} x^{2} + {\left (24 \, B c d^{2} e - 28 \, A c d e^{2} + 35 \, B a e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{4}} \]
2/105*(15*B*c*e^3*x^3 - 48*B*c*d^3 + 56*A*c*d^2*e - 70*B*a*d*e^2 + 105*A*a *e^3 - 3*(6*B*c*d*e^2 - 7*A*c*e^3)*x^2 + (24*B*c*d^2*e - 28*A*c*d*e^2 + 35 *B*a*e^3)*x)*sqrt(e*x + d)/e^4
Time = 0.71 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A c e - 3 B c d\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (A a e^{3} + A c d^{2} e - B a d e^{2} - B c d^{3}\right )}{e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(B*c*(d + e*x)**(7/2)/(7*e**3) + (d + e*x)**(5/2)*(A*c*e - 3* B*c*d)/(5*e**3) + (d + e*x)**(3/2)*(-2*A*c*d*e + B*a*e**2 + 3*B*c*d**2)/(3 *e**3) + sqrt(d + e*x)*(A*a*e**3 + A*c*d**2*e - B*a*d*e**2 - B*c*d**3)/e** 3)/e, Ne(e, 0)), ((A*a*x + A*c*x**3/3 + B*a*x**2/2 + B*c*x**4/4)/sqrt(d), True))
Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} B c - 21 \, {\left (3 \, B c d - A c e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 105 \, {\left (B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}\right )} \sqrt {e x + d}\right )}}{105 \, e^{4}} \]
2/105*(15*(e*x + d)^(7/2)*B*c - 21*(3*B*c*d - A*c*e)*(e*x + d)^(5/2) + 35* (3*B*c*d^2 - 2*A*c*d*e + B*a*e^2)*(e*x + d)^(3/2) - 105*(B*c*d^3 - A*c*d^2 *e + B*a*d*e^2 - A*a*e^3)*sqrt(e*x + d))/e^4
Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} A a + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A c}{e^{2}} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B c}{e^{3}}\right )}}{105 \, e} \]
2/105*(105*sqrt(e*x + d)*A*a + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B* a/e + 7*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)* A*c/e^2 + 3*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2) *d^2 - 35*sqrt(e*x + d)*d^3)*B*c/e^3)/e
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{\sqrt {d+e x}} \, dx=\frac {{\left (d+e\,x\right )}^{3/2}\,\left (6\,B\,c\,d^2-4\,A\,c\,d\,e+2\,B\,a\,e^2\right )}{3\,e^4}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,c\,\left (A\,e-3\,B\,d\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,\left (c\,d^2+a\,e^2\right )\,\left (A\,e-B\,d\right )\,\sqrt {d+e\,x}}{e^4} \]