\(\int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx\) [365]

Optimal result

Integrand size = 18, antiderivative size = 123 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=-\frac {3 (A b-5 a B) \sqrt {x}}{4 a b^3}+\frac {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}+\frac {(A b-5 a B) x^{3/2}}{4 a b^2 (a+b x)}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \]

[Out]

1/2*(A*b-B*a)*x^(5/2)/a/b/(b*x+a)^2+1/4*(A*b-5*B*a)*x^(3/2)/a/b^2/(b*x+a)+3/4*(A*b-5*B*a)*arctan(b^(1/2)*x^(1/
2)/a^(1/2))/b^(7/2)/a^(1/2)-3/4*(A*b-5*B*a)*x^(1/2)/a/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 43, 52, 65, 211} \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}}-\frac {3 \sqrt {x} (A b-5 a B)}{4 a b^3}+\frac {x^{3/2} (A b-5 a B)}{4 a b^2 (a+b x)}+\frac {x^{5/2} (A b-a B)}{2 a b (a+b x)^2} \]

[In]

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

[Out]

(-3*(A*b - 5*a*B)*Sqrt[x])/(4*a*b^3) + ((A*b - a*B)*x^(5/2))/(2*a*b*(a + b*x)^2) + ((A*b - 5*a*B)*x^(3/2))/(4*
a*b^2*(a + b*x)) + (3*(A*b - 5*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(7/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}-\frac {\left (\frac {A b}{2}-\frac {5 a B}{2}\right ) \int \frac {x^{3/2}}{(a+b x)^2} \, dx}{2 a b} \\ & = \frac {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}+\frac {(A b-5 a B) x^{3/2}}{4 a b^2 (a+b x)}-\frac {(3 (A b-5 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{8 a b^2} \\ & = -\frac {3 (A b-5 a B) \sqrt {x}}{4 a b^3}+\frac {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}+\frac {(A b-5 a B) x^{3/2}}{4 a b^2 (a+b x)}+\frac {(3 (A b-5 a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 b^3} \\ & = -\frac {3 (A b-5 a B) \sqrt {x}}{4 a b^3}+\frac {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}+\frac {(A b-5 a B) x^{3/2}}{4 a b^2 (a+b x)}+\frac {(3 (A b-5 a B)) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = -\frac {3 (A b-5 a B) \sqrt {x}}{4 a b^3}+\frac {(A b-a B) x^{5/2}}{2 a b (a+b x)^2}+\frac {(A b-5 a B) x^{3/2}}{4 a b^2 (a+b x)}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.74 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\frac {\sqrt {x} \left (15 a^2 B+b^2 x (-5 A+8 B x)+a (-3 A b+25 b B x)\right )}{4 b^3 (a+b x)^2}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2}} \]

[In]

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

[Out]

(Sqrt[x]*(15*a^2*B + b^2*x*(-5*A + 8*B*x) + a*(-3*A*b + 25*b*B*x)))/(4*b^3*(a + b*x)^2) + (3*(A*b - 5*a*B)*Arc
Tan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*Sqrt[a]*b^(7/2))

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{b^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (3 A b -7 B a \right ) \sqrt {x}}{8}\right )}{\left (b x +a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(83\)
default	\(\frac {2 B \sqrt {x}}{b^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (3 A b -7 B a \right ) \sqrt {x}}{8}\right )}{\left (b x +a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(83\)
risch	\(\frac {2 B \sqrt {x}}{b^{3}}+\frac {\frac {2 \left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (3 A b -7 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{b^{3}}\)	\(83\)

[In]

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

[Out]

2*B/b^3*x^(1/2)+2/b^3*(((-5/8*b^2*A+9/8*a*b*B)*x^(3/2)-1/8*a*(3*A*b-7*B*a)*x^(1/2))/(b*x+a)^2+3/8*(A*b-5*B*a)/
(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.59 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\left [\frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \]

[In]

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

[Out]

[1/8*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*
sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^2 - A*a*b^3)*x)*sq
rt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4), 1/4*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b
- A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^
2*b^2 - A*a*b^3)*x)*sqrt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1333 vs. \(2 (116) = 232\).

Time = 9.74 (sec) , antiderivative size = 1333, normalized size of antiderivative = 10.84 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**3, Eq
(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**3, Eq(a, 0)), (3*A*a**2*b*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqr
t(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 3*A*a**2*b*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4
*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 6*A*a*b**2*sqrt(x)*sqrt(-a/b)/(8*a**2*b**4*sq
rt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 6*A*a*b**2*x*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b
**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 6*A*a*b**2*x*log(sqrt(x) + sqrt(-a/b))/(8*
a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 10*A*b**3*x**(3/2)*sqrt(-a/b)/(8*a**
2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 3*A*b**3*x**2*log(sqrt(x) - sqrt(-a/b))
/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 3*A*b**3*x**2*log(sqrt(x) + sqrt
(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 15*B*a**3*log(sqrt(x) - s
qrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 15*B*a**3*log(sqrt(x)
+ sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 30*B*a**2*b*sqrt(x)
*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 30*B*a**2*b*x*log(sqr
t(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 30*B*a**2*b*x*
log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) + 50*B*a*
b**2*x**(3/2)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 15*B*a*b
**2*x**2*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b))
+ 15*B*a*b**2*x**2*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sq
rt(-a/b)) + 16*B*b**3*x**(5/2)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(
-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} + {\left (7 \, B a^{2} - 3 \, A a b\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {2 \, B \sqrt {x}}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} \]

[In]

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

[Out]

1/4*((9*B*a*b - 5*A*b^2)*x^(3/2) + (7*B*a^2 - 3*A*a*b)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) + 2*B*sqrt(x)/
b^3 - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\frac {2 \, B \sqrt {x}}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{\frac {3}{2}} - 5 \, A b^{2} x^{\frac {3}{2}} + 7 \, B a^{2} \sqrt {x} - 3 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \]

[In]

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

[Out]

2*B*sqrt(x)/b^3 - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/4*(9*B*a*b*x^(3/2) - 5*A*b
^2*x^(3/2) + 7*B*a^2*sqrt(x) - 3*A*a*b*sqrt(x))/((b*x + a)^2*b^3)

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.78 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^3} \, dx=\frac {\sqrt {x}\,\left (\frac {7\,B\,a^2}{4}-\frac {3\,A\,a\,b}{4}\right )-x^{3/2}\,\left (\frac {5\,A\,b^2}{4}-\frac {9\,B\,a\,b}{4}\right )}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,B\,\sqrt {x}}{b^3}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-5\,B\,a\right )}{4\,\sqrt {a}\,b^{7/2}} \]

[In]

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

[Out]

(x^(1/2)*((7*B*a^2)/4 - (3*A*a*b)/4) - x^(3/2)*((5*A*b^2)/4 - (9*B*a*b)/4))/(a^2*b^3 + b^5*x^2 + 2*a*b^4*x) +
(2*B*x^(1/2))/b^3 + (3*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b - 5*B*a))/(4*a^(1/2)*b^(7/2))