Integrand size = 29, antiderivative size = 95 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {(3 A b-5 a B) \sqrt {x}}{b^3}+\frac {2 B x^{3/2}}{3 b^2}-\frac {(A b-a B) x^{3/2}}{b^2 (a+b x)}-\frac {\sqrt {a} (3 A b-5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Output:
(3*A*b-5*B*a)*x^(1/2)/b^3+2/3*B*x^(3/2)/b^2-(A*b-B*a)*x^(3/2)/b^2/(b*x+a)- a^(1/2)*(3*A*b-5*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(7/2)
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {x} \left (-15 a^2 B+a b (9 A-10 B x)+2 b^2 x (3 A+B x)\right )}{3 b^3 (a+b x)}+\frac {\sqrt {a} (-3 A b+5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Input:
Integrate[(x^(3/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
(Sqrt[x]*(-15*a^2*B + a*b*(9*A - 10*B*x) + 2*b^2*x*(3*A + B*x)))/(3*b^3*(a + b*x)) + (Sqrt[a]*(-3*A*b + 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^ (7/2)
Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1184, 27, 87, 60, 60, 73, 218}
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.
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {x^{3/2} (A+B x)}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {x^{5/2} (A b-a B)}{a b (a+b x)}-\frac {(3 A b-5 a B) \int \frac {x^{3/2}}{a+b x}dx}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2} (A b-a B)}{a b (a+b x)}-\frac {(3 A b-5 a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{5/2} (A b-a B)}{a b (a+b x)}-\frac {(3 A b-5 a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{5/2} (A b-a B)}{a b (a+b x)}-\frac {(3 A b-5 a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x^{5/2} (A b-a B)}{a b (a+b x)}-\frac {(3 A b-5 a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{2 a b}\) |
Input:
Int[(x^(3/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
((A*b - a*B)*x^(5/2))/(a*b*(a + b*x)) - ((3*A*b - 5*a*B)*((2*x^(3/2))/(3*b ) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3 /2)))/b))/(2*a*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {2 \left (B b x +3 A b -6 B a \right ) \sqrt {x}}{3 b^{3}}-\frac {a \left (\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (3 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b^{3}}\) | \(77\) |
derivativedivides | \(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-4 B a \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (3 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(82\) |
default | \(\frac {\frac {2 B b \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-4 B a \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (3 A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(82\) |
Input:
int(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
2/3*(B*b*x+3*A*b-6*B*a)*x^(1/2)/b^3-a/b^3*(2*(-1/2*A*b+1/2*B*a)*x^(1/2)/(b *x+a)+(3*A*b-5*B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.43 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {x}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \] Input:
integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
[-1/6*(3*(5*B*a^2 - 3*A*a*b + (5*B*a*b - 3*A*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(2*B*b^2*x^2 - 15*B*a^2 + 9*A* a*b - 2*(5*B*a*b - 3*A*b^2)*x)*sqrt(x))/(b^4*x + a*b^3), 1/3*(3*(5*B*a^2 - 3*A*a*b + (5*B*a*b - 3*A*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*B*b^2*x^2 - 15*B*a^2 + 9*A*a*b - 2*(5*B*a*b - 3*A*b^2)*x)*sqrt(x))/(b ^4*x + a*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 762 vs. \(2 (88) = 176\).
Time = 6.07 (sec) , antiderivative size = 762, normalized size of antiderivative = 8.02 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx =\text {Too large to display} \] Input:
integrate(x**(3/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2),x)
Output:
Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A *x**(5/2)/5 + 2*B*x**(7/2)/7)/a**2, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2 )/3)/b**2, Eq(a, 0)), (-9*A*a**2*b*log(sqrt(x) - sqrt(-a/b))/(6*a*b**4*sqr t(-a/b) + 6*b**5*x*sqrt(-a/b)) + 9*A*a**2*b*log(sqrt(x) + sqrt(-a/b))/(6*a *b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 18*A*a*b**2*sqrt(x)*sqrt(-a/b)/( 6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 9*A*a*b**2*x*log(sqrt(x) - sq rt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 9*A*a*b**2*x*log(s qrt(x) + sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 12*A*b* *3*x**(3/2)*sqrt(-a/b)/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 15*B* a**3*log(sqrt(x) - sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 15*B*a**3*log(sqrt(x) + sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqr t(-a/b)) - 30*B*a**2*b*sqrt(x)*sqrt(-a/b)/(6*a*b**4*sqrt(-a/b) + 6*b**5*x* sqrt(-a/b)) + 15*B*a**2*b*x*log(sqrt(x) - sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 15*B*a**2*b*x*log(sqrt(x) + sqrt(-a/b))/(6*a*b** 4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 20*B*a*b**2*x**(3/2)*sqrt(-a/b)/(6*a *b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 4*B*b**3*x**(5/2)*sqrt(-a/b)/(6* a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)), True))
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {{\left (B a^{2} - A a b\right )} \sqrt {x}}{b^{4} x + a b^{3}} + \frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (2 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{3}} \] Input:
integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-(B*a^2 - A*a*b)*sqrt(x)/(b^4*x + a*b^3) + (5*B*a^2 - 3*A*a*b)*arctan(b*sq rt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/3*(B*b*x^(3/2) - 3*(2*B*a - A*b)*sqrt (x))/b^3
Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {B a^{2} \sqrt {x} - A a b \sqrt {x}}{{\left (b x + a\right )} b^{3}} + \frac {2 \, {\left (B b^{4} x^{\frac {3}{2}} - 6 \, B a b^{3} \sqrt {x} + 3 \, A b^{4} \sqrt {x}\right )}}{3 \, b^{6}} \] Input:
integrate(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
(5*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - (B*a^2*s qrt(x) - A*a*b*sqrt(x))/((b*x + a)*b^3) + 2/3*(B*b^4*x^(3/2) - 6*B*a*b^3*s qrt(x) + 3*A*b^4*sqrt(x))/b^6
Time = 0.08 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b^2}-\frac {4\,B\,a}{b^3}\right )-\frac {\sqrt {x}\,\left (B\,a^2-A\,a\,b\right )}{x\,b^4+a\,b^3}+\frac {2\,B\,x^{3/2}}{3\,b^2}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (3\,A\,b-5\,B\,a\right )}{5\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-5\,B\,a\right )}{b^{7/2}} \] Input:
int((x^(3/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x),x)
Output:
x^(1/2)*((2*A)/b^2 - (4*B*a)/b^3) - (x^(1/2)*(B*a^2 - A*a*b))/(a*b^3 + b^4 *x) + (2*B*x^(3/2))/(3*b^2) + (a^(1/2)*atan((a^(1/2)*b^(1/2)*x^(1/2)*(3*A* b - 5*B*a))/(5*B*a^2 - 3*A*a*b))*(3*A*b - 5*B*a))/b^(7/2)
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.41 \[ \int \frac {x^{3/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a -2 \sqrt {x}\, a b +\frac {2 \sqrt {x}\, b^{2} x}{3}}{b^{3}} \] Input:
int(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x)
Output:
(2*(3*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a - 3*sqrt(x)*a* b + sqrt(x)*b**2*x))/(3*b**3)