Integrand size = 30, antiderivative size = 83 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\frac {2 (3 f-2 g) (2-3 x)^{3/2}}{5 g (3 f+2 g) (f+g x)^{5/2}}-\frac {2 (9 f+14 g) (2-3 x)^{3/2}}{5 g (3 f+2 g)^2 (f+g x)^{3/2}} \] Output:
2/5*(3*f-2*g)*(2-3*x)^(3/2)/g/(3*f+2*g)/(g*x+f)^(5/2)-2/5*(9*f+14*g)*(2-3* x)^(3/2)/g/(3*f+2*g)^2/(g*x+f)^(3/2)
Time = 1.41 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=-\frac {2 (2-3 x)^2 \sqrt {2+3 x} (2 g (2+7 x)+f (14+9 x))}{5 (3 f+2 g)^2 (f+g x)^{5/2} \sqrt {4-9 x^2}} \] Input:
Integrate[(Sqrt[2 + 3*x]*Sqrt[4 - 9*x^2])/(f + g*x)^(7/2),x]
Output:
(-2*(2 - 3*x)^2*Sqrt[2 + 3*x]*(2*g*(2 + 7*x) + f*(14 + 9*x)))/(5*(3*f + 2* g)^2*(f + g*x)^(5/2)*Sqrt[4 - 9*x^2])
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {639, 87, 48}
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.
\(\displaystyle \int \frac {\sqrt {3 x+2} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 639 |
\(\displaystyle \int \frac {\sqrt {2-3 x} (3 x+2)}{(f+g x)^{7/2}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {3 (9 f+14 g) \int \frac {\sqrt {2-3 x}}{(f+g x)^{5/2}}dx}{5 g (3 f+2 g)}+\frac {2 (2-3 x)^{3/2} (3 f-2 g)}{5 g (3 f+2 g) (f+g x)^{5/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {2 (2-3 x)^{3/2} (3 f-2 g)}{5 g (3 f+2 g) (f+g x)^{5/2}}-\frac {2 (2-3 x)^{3/2} (9 f+14 g)}{5 g (3 f+2 g)^2 (f+g x)^{3/2}}\) |
Input:
Int[(Sqrt[2 + 3*x]*Sqrt[4 - 9*x^2])/(f + g*x)^(7/2),x]
Output:
(2*(3*f - 2*g)*(2 - 3*x)^(3/2))/(5*g*(3*f + 2*g)*(f + g*x)^(5/2)) - (2*(9* f + 14*g)*(2 - 3*x)^(3/2))/(5*g*(3*f + 2*g)^2*(f + g*x)^(3/2))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^ 2)^(p_.), x_Symbol] :> Int[(c + d*x)^(m + p)*(e + f*x)^n*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (I ntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[m]))
Time = 0.88 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.66
method | result | size |
default | \(\frac {2 \sqrt {-9 x^{2}+4}\, \left (-2+3 x \right ) \left (9 f x +14 g x +14 f +4 g \right )}{5 \sqrt {3 x +2}\, \left (g x +f \right )^{\frac {5}{2}} \left (3 f +2 g \right )^{2}}\) | \(55\) |
gosper | \(\frac {2 \left (-2+3 x \right ) \left (9 f x +14 g x +14 f +4 g \right ) \sqrt {-9 x^{2}+4}}{5 \left (g x +f \right )^{\frac {5}{2}} \left (9 f^{2}+12 f g +4 g^{2}\right ) \sqrt {3 x +2}}\) | \(63\) |
orering | \(\frac {2 \left (-2+3 x \right ) \left (9 f x +14 g x +14 f +4 g \right ) \sqrt {-9 x^{2}+4}}{5 \left (g x +f \right )^{\frac {5}{2}} \left (9 f^{2}+12 f g +4 g^{2}\right ) \sqrt {3 x +2}}\) | \(63\) |
Input:
int((3*x+2)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(7/2),x,method=_RETURNVERBOSE)
Output:
2/5/(3*x+2)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(5/2)*(-2+3*x)*(9*f*x+14*g*x+14 *f+4*g)/(3*f+2*g)^2
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (71) = 142\).
Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.39 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\frac {2 \, {\left (3 \, {\left (9 \, f + 14 \, g\right )} x^{2} + 8 \, {\left (3 \, f - 2 \, g\right )} x - 28 \, f - 8 \, g\right )} \sqrt {g x + f} \sqrt {-9 \, x^{2} + 4} \sqrt {3 \, x + 2}}{5 \, {\left (18 \, f^{5} + 24 \, f^{4} g + 8 \, f^{3} g^{2} + 3 \, {\left (9 \, f^{2} g^{3} + 12 \, f g^{4} + 4 \, g^{5}\right )} x^{4} + {\left (81 \, f^{3} g^{2} + 126 \, f^{2} g^{3} + 60 \, f g^{4} + 8 \, g^{5}\right )} x^{3} + 3 \, {\left (27 \, f^{4} g + 54 \, f^{3} g^{2} + 36 \, f^{2} g^{3} + 8 \, f g^{4}\right )} x^{2} + 3 \, {\left (9 \, f^{5} + 30 \, f^{4} g + 28 \, f^{3} g^{2} + 8 \, f^{2} g^{3}\right )} x\right )}} \] Input:
integrate((2+3*x)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(7/2),x, algorithm="frica s")
Output:
2/5*(3*(9*f + 14*g)*x^2 + 8*(3*f - 2*g)*x - 28*f - 8*g)*sqrt(g*x + f)*sqrt (-9*x^2 + 4)*sqrt(3*x + 2)/(18*f^5 + 24*f^4*g + 8*f^3*g^2 + 3*(9*f^2*g^3 + 12*f*g^4 + 4*g^5)*x^4 + (81*f^3*g^2 + 126*f^2*g^3 + 60*f*g^4 + 8*g^5)*x^3 + 3*(27*f^4*g + 54*f^3*g^2 + 36*f^2*g^3 + 8*f*g^4)*x^2 + 3*(9*f^5 + 30*f^ 4*g + 28*f^3*g^2 + 8*f^2*g^3)*x)
Timed out. \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((2+3*x)**(1/2)*(-9*x**2+4)**(1/2)/(g*x+f)**(7/2),x)
Output:
Timed out
\[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\int { \frac {\sqrt {-9 \, x^{2} + 4} \sqrt {3 \, x + 2}}{{\left (g x + f\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((2+3*x)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(7/2),x, algorithm="maxim a")
Output:
integrate(sqrt(-9*x^2 + 4)*sqrt(3*x + 2)/(g*x + f)^(7/2), x)
Time = 0.22 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\frac {6 \, {\left (\frac {{\left (9 \, \sqrt {3} f g^{2} + 14 \, \sqrt {3} g^{3}\right )} {\left (3 \, x - 2\right )}}{9 \, f^{2} g^{2} + 12 \, f g^{3} + 4 \, g^{4}} + \frac {20 \, {\left (3 \, \sqrt {3} f g^{2} + 2 \, \sqrt {3} g^{3}\right )}}{9 \, f^{2} g^{2} + 12 \, f g^{3} + 4 \, g^{4}}\right )} {\left (3 \, x - 2\right )} \sqrt {-3 \, x + 2}}{5 \, {\left (g {\left (3 \, x - 2\right )} + 3 \, f + 2 \, g\right )}^{\frac {5}{2}}} \] Input:
integrate((2+3*x)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(7/2),x, algorithm="giac" )
Output:
6/5*((9*sqrt(3)*f*g^2 + 14*sqrt(3)*g^3)*(3*x - 2)/(9*f^2*g^2 + 12*f*g^3 + 4*g^4) + 20*(3*sqrt(3)*f*g^2 + 2*sqrt(3)*g^3)/(9*f^2*g^2 + 12*f*g^3 + 4*g^ 4))*(3*x - 2)*sqrt(-3*x + 2)/(g*(3*x - 2) + 3*f + 2*g)^(5/2)
Time = 6.52 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.18 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\frac {\sqrt {f+g\,x}\,\left (\frac {x\,\sqrt {3\,x+2}\,\sqrt {4-9\,x^2}\,\left (48\,f-32\,g\right )}{15\,g^3\,{\left (3\,f+2\,g\right )}^2}-\frac {\sqrt {3\,x+2}\,\sqrt {4-9\,x^2}\,\left (56\,f+16\,g\right )}{15\,g^3\,{\left (3\,f+2\,g\right )}^2}+\frac {x^2\,\sqrt {3\,x+2}\,\sqrt {4-9\,x^2}\,\left (54\,f+84\,g\right )}{15\,g^3\,{\left (3\,f+2\,g\right )}^2}\right )}{x^4+\frac {2\,f^3}{3\,g^3}+\frac {x^3\,\left (9\,f+2\,g\right )}{3\,g}+\frac {f\,x^2\,\left (3\,f+2\,g\right )}{g^2}+\frac {f^2\,x\,\left (f+2\,g\right )}{g^3}} \] Input:
int(((3*x + 2)^(1/2)*(4 - 9*x^2)^(1/2))/(f + g*x)^(7/2),x)
Output:
((f + g*x)^(1/2)*((x*(3*x + 2)^(1/2)*(4 - 9*x^2)^(1/2)*(48*f - 32*g))/(15* g^3*(3*f + 2*g)^2) - ((3*x + 2)^(1/2)*(4 - 9*x^2)^(1/2)*(56*f + 16*g))/(15 *g^3*(3*f + 2*g)^2) + (x^2*(3*x + 2)^(1/2)*(4 - 9*x^2)^(1/2)*(54*f + 84*g) )/(15*g^3*(3*f + 2*g)^2)))/(x^4 + (2*f^3)/(3*g^3) + (x^3*(9*f + 2*g))/(3*g ) + (f*x^2*(3*f + 2*g))/g^2 + (f^2*x*(f + 2*g))/g^3)
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {2+3 x} \sqrt {4-9 x^2}}{(f+g x)^{7/2}} \, dx=\frac {2 \sqrt {-3 x +2}\, \left (27 f \,x^{2}+42 g \,x^{2}+24 f x -16 g x -28 f -8 g \right )}{5 \sqrt {g x +f}\, \left (9 f^{2} g^{2} x^{2}+12 f \,g^{3} x^{2}+4 g^{4} x^{2}+18 f^{3} g x +24 f^{2} g^{2} x +8 f \,g^{3} x +9 f^{4}+12 f^{3} g +4 f^{2} g^{2}\right )} \] Input:
int((2+3*x)^(1/2)*(-9*x^2+4)^(1/2)/(g*x+f)^(7/2),x)
Output:
(2*sqrt( - 3*x + 2)*(27*f*x**2 + 24*f*x - 28*f + 42*g*x**2 - 16*g*x - 8*g) )/(5*sqrt(f + g*x)*(9*f**4 + 18*f**3*g*x + 12*f**3*g + 9*f**2*g**2*x**2 + 24*f**2*g**2*x + 4*f**2*g**2 + 12*f*g**3*x**2 + 8*f*g**3*x + 4*g**4*x**2))