\(\int \frac {x^{3/2} (A+B x)}{a+b x} \, dx\) [346]

Optimal result

Integrand size = 18, antiderivative size = 90 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 a (A b-a B) \sqrt {x}}{b^3}+\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{5/2}}{5 b}+\frac {2 a^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

[Out]

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)-2*a*(A*b
-B*a)*x^(1/2)/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 211} \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\frac {2 a^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}-\frac {2 a \sqrt {x} (A b-a B)}{b^3}+\frac {2 x^{3/2} (A b-a B)}{3 b^2}+\frac {2 B x^{5/2}}{5 b} \]

[In]

Int[(x^(3/2)*(A + B*x))/(a + b*x),x]

[Out]

(-2*a*(A*b - a*B)*Sqrt[x])/b^3 + (2*(A*b - a*B)*x^(3/2))/(3*b^2) + (2*B*x^(5/2))/(5*b) + (2*a^(3/2)*(A*b - a*B
)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B x^{5/2}}{5 b}+\frac {\left (2 \left (\frac {5 A b}{2}-\frac {5 a B}{2}\right )\right ) \int \frac {x^{3/2}}{a+b x} \, dx}{5 b} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{5/2}}{5 b}-\frac {(a (A b-a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{b^2} \\ & = -\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 {\left (a^2 (A b-a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{b^3} \\ & = -\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 {\left (2 a^2 (A b-a B)\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{b^3} \\ & = -\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) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x} \left (15 a^2 B-5 a b (3 A+B x)+b^2 x (5 A+3 B x)\right )}{15 b^3}-\frac {2 a^{3/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

[In]

Integrate[(x^(3/2)*(A + B*x))/(a + b*x),x]

[Out]

(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)

Maple [A] (verified)

Time = 1.29 (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	\(-\frac {2 \left (-\frac {b^{2} B \,x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {B a b \,x^{\frac {3}{2}}}{3}+a b A \sqrt {x}-a^{2} B \sqrt {x}\right )}{b^{3}}+\frac {2 a^{2} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\)	\(82\)

[In]

int(x^(3/2)*(B*x+A)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-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))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.00 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\left [-\frac {15 \, {\left (B a^{2} - A a b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \]

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="fricas")

[Out]

[-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)*sqr
t(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]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (87) = 174\).

Time = 1.47 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.89 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{a} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: a = 0 \\\frac {A a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {A a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 A a \sqrt {x}}{b^{2}} + \frac {2 A x^{\frac {3}{2}}}{3 b} - \frac {B a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {B a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {2 B a^{2} \sqrt {x}}{b^{3}} - \frac {2 B a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(3/2)*(B*x+A)/(b*x+a),x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (B a b - A b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (B a^{2} - A a b\right )} \sqrt {x}\right )}}{15 \, b^{3}} \]

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="maxima")

[Out]

-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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=-\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{4} x^{\frac {5}{2}} - 5 \, B a b^{3} x^{\frac {3}{2}} + 5 \, A b^{4} x^{\frac {3}{2}} + 15 \, B a^{2} b^{2} \sqrt {x} - 15 \, A a b^{3} \sqrt {x}\right )}}{15 \, b^{5}} \]

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a),x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {x^{3/2} (A+B x)}{a+b x} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{5/2}}{5\,b}-\frac {2\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^3-A\,a^2\,b}\right )\,\left (A\,b-B\,a\right )}{b^{7/2}}-\frac {a\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b} \]

[In]

int((x^(3/2)*(A + B*x))/(a + b*x),x)

[Out]

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