Integrand size = 18, antiderivative size = 90 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 a (A b-a B) \sqrt {x}}{b^3}+\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{5/2}}{5 b}+\frac {2 a^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Output:
-2*a*(A*b-B*a)*x^(1/2)/b^3+2/3*(A*b-B*a)*x^(3/2)/b^2+2/5*B*x^(5/2)/b+2*a^( 3/2)*(A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(7/2)
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x} \left (15 a^2 B-5 a b (3 A+B x)+b^2 x (5 A+3 B x)\right )}{15 b^3}-\frac {2 a^{3/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \] Input:
Integrate[(x^(3/2)*(A + B*x))/(a + b*x),x]
Output:
(2*Sqrt[x]*(15*a^2*B - 5*a*b*(3*A + B*x) + b^2*x*(5*A + 3*B*x)))/(15*b^3) - (2*a^(3/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {90, 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.
\(\displaystyle \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(A b-a B) \int \frac {x^{3/2}}{a+b x}dx}{b}+\frac {2 B x^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
Input:
Int[(x^(3/2)*(A + B*x))/(a + b*x),x]
Output:
(2*B*x^(5/2))/(5*b) + ((A*b - 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))/b
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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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]
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +5 B a b x +15 a b A -15 a^{2} B \right ) \sqrt {x}}{15 b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(76\) |
derivativedivides | \(-\frac {2 \left (-\frac {b^{2} B \,x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {B a b \,x^{\frac {3}{2}}}{3}+a b A \sqrt {x}-a^{2} B \sqrt {x}\right )}{b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(82\) |
default | \(-\frac {2 \left (-\frac {b^{2} B \,x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {B a b \,x^{\frac {3}{2}}}{3}+a b A \sqrt {x}-a^{2} B \sqrt {x}\right )}{b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(82\) |
Input:
int(x^(3/2)*(B*x+A)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
-2/15*(-3*B*b^2*x^2-5*A*b^2*x+5*B*a*b*x+15*A*a*b-15*B*a^2)*x^(1/2)/b^3+2*a ^2*(A*b-B*a)/b^3/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.00 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\left [-\frac {15 \, {\left (B a^{2} - A a b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="fricas")
Output:
[-1/15*(15*(B*a^2 - A*a*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(3*B*b^2*x^2 + 15*B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x )*sqrt(x))/b^3, -2/15*(15*(B*a^2 - A*a*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt( a/b)/a) - (3*B*b^2*x^2 + 15*B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x)*sqrt(x ))/b^3]
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (87) = 174\).
Time = 1.55 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.89 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{a} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: a = 0 \\\frac {A a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {A a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 A a \sqrt {x}}{b^{2}} + \frac {2 A x^{\frac {3}{2}}}{3 b} - \frac {B a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {B a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {2 B a^{2} \sqrt {x}}{b^{3}} - \frac {2 B a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \] Input:
integrate(x**(3/2)*(B*x+A)/(b*x+a),x)
Output:
Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(b, 0)), (( 2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a, Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**( 5/2)/5)/b, Eq(a, 0)), (A*a**2*log(sqrt(x) - sqrt(-a/b))/(b**3*sqrt(-a/b)) - A*a**2*log(sqrt(x) + sqrt(-a/b))/(b**3*sqrt(-a/b)) - 2*A*a*sqrt(x)/b**2 + 2*A*x**(3/2)/(3*b) - B*a**3*log(sqrt(x) - sqrt(-a/b))/(b**4*sqrt(-a/b)) + B*a**3*log(sqrt(x) + sqrt(-a/b))/(b**4*sqrt(-a/b)) + 2*B*a**2*sqrt(x)/b* *3 - 2*B*a*x**(3/2)/(3*b**2) + 2*B*x**(5/2)/(5*b), True))
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (B a b - A b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (B a^{2} - A a b\right )} \sqrt {x}\right )}}{15 \, b^{3}} \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="maxima")
Output:
-2*(B*a^3 - A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3 *B*b^2*x^(5/2) - 5*(B*a*b - A*b^2)*x^(3/2) + 15*(B*a^2 - A*a*b)*sqrt(x))/b ^3
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{4} x^{\frac {5}{2}} - 5 \, B a b^{3} x^{\frac {3}{2}} + 5 \, A b^{4} x^{\frac {3}{2}} + 15 \, B a^{2} b^{2} \sqrt {x} - 15 \, A a b^{3} \sqrt {x}\right )}}{15 \, b^{5}} \] Input:
integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="giac")
Output:
-2*(B*a^3 - A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3 *B*b^4*x^(5/2) - 5*B*a*b^3*x^(3/2) + 5*A*b^4*x^(3/2) + 15*B*a^2*b^2*sqrt(x ) - 15*A*a*b^3*sqrt(x))/b^5
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{5/2}}{5\,b}-\frac {2\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^3-A\,a^2\,b}\right )\,\left (A\,b-B\,a\right )}{b^{7/2}}-\frac {a\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b} \] Input:
int((x^(3/2)*(A + B*x))/(a + b*x),x)
Output:
x^(3/2)*((2*A)/(3*b) - (2*B*a)/(3*b^2)) + (2*B*x^(5/2))/(5*b) - (2*a^(3/2) *atan((a^(3/2)*b^(1/2)*x^(1/2)*(A*b - B*a))/(B*a^3 - A*a^2*b))*(A*b - B*a) )/b^(7/2) - (a*x^(1/2)*((2*A)/b - (2*B*a)/b^2))/b
Time = 0.14 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.08 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x}\, x^{2}}{5} \] Input:
int(x^(3/2)*(B*x+A)/(b*x+a),x)
Output:
(2*sqrt(x)*x**2)/5