\(\int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx\) [445]

Optimal result

Integrand size = 18, antiderivative size = 118 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 a^3 (A b-a B)}{3 b^5 (a+b x)^{3/2}}-\frac {2 a^2 (3 A b-4 a B)}{b^5 \sqrt {a+b x}}-\frac {6 a (A b-2 a B) \sqrt {a+b x}}{b^5}+\frac {2 (A b-4 a B) (a+b x)^{3/2}}{3 b^5}+\frac {2 B (a+b x)^{5/2}}{5 b^5} \]

[Out]

2/3*a^3*(A*b-B*a)/b^5/(b*x+a)^(3/2)+2/3*(A*b-4*B*a)*(b*x+a)^(3/2)/b^5+2/5*B*(b*x+a)^(5/2)/b^5-2*a^2*(3*A*b-4*B
*a)/b^5/(b*x+a)^(1/2)-6*a*(A*b-2*B*a)*(b*x+a)^(1/2)/b^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 a^3 (A b-a B)}{3 b^5 (a+b x)^{3/2}}-\frac {2 a^2 (3 A b-4 a B)}{b^5 \sqrt {a+b x}}-\frac {6 a \sqrt {a+b x} (A b-2 a B)}{b^5}+\frac {2 (a+b x)^{3/2} (A b-4 a B)}{3 b^5}+\frac {2 B (a+b x)^{5/2}}{5 b^5} \]

[In]

Int[(x^3*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*a^3*(A*b - a*B))/(3*b^5*(a + b*x)^(3/2)) - (2*a^2*(3*A*b - 4*a*B))/(b^5*Sqrt[a + b*x]) - (6*a*(A*b - 2*a*B)
*Sqrt[a + b*x])/b^5 + (2*(A*b - 4*a*B)*(a + b*x)^(3/2))/(3*b^5) + (2*B*(a + b*x)^(5/2))/(5*b^5)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^3 (-A b+a B)}{b^4 (a+b x)^{5/2}}-\frac {a^2 (-3 A b+4 a B)}{b^4 (a+b x)^{3/2}}+\frac {3 a (-A b+2 a B)}{b^4 \sqrt {a+b x}}+\frac {(A b-4 a B) \sqrt {a+b x}}{b^4}+\frac {B (a+b x)^{3/2}}{b^4}\right ) \, dx \\ & = \frac {2 a^3 (A b-a B)}{3 b^5 (a+b x)^{3/2}}-\frac {2 a^2 (3 A b-4 a B)}{b^5 \sqrt {a+b x}}-\frac {6 a (A b-2 a B) \sqrt {a+b x}}{b^5}+\frac {2 (A b-4 a B) (a+b x)^{3/2}}{3 b^5}+\frac {2 B (a+b x)^{5/2}}{5 b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \left (128 a^4 B+24 a^2 b^2 x (-5 A+2 B x)+b^4 x^3 (5 A+3 B x)-2 a b^3 x^2 (15 A+4 B x)+a^3 (-80 A b+192 b B x)\right )}{15 b^5 (a+b x)^{3/2}} \]

[In]

Integrate[(x^3*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*(128*a^4*B + 24*a^2*b^2*x*(-5*A + 2*B*x) + b^4*x^3*(5*A + 3*B*x) - 2*a*b^3*x^2*(15*A + 4*B*x) + a^3*(-80*A*
b + 192*b*B*x)))/(15*b^5*(a + b*x)^(3/2))

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(-\frac {32 \left (-\frac {\left (\frac {3 B x}{5}+A \right ) x^{3} b^{4}}{16}+\frac {3 x^{2} \left (\frac {4 B x}{15}+A \right ) a \,b^{3}}{8}+\frac {3 x \left (-\frac {2 B x}{5}+A \right ) a^{2} b^{2}}{2}+a^{3} \left (-\frac {12 B x}{5}+A \right ) b -\frac {8 B \,a^{4}}{5}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{5}}\)	\(75\)
risch	\(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +14 B a b x +40 a b A -73 a^{2} B \right ) \sqrt {b x +a}}{15 b^{5}}-\frac {2 a^{2} \left (9 A \,b^{2} x -12 B a b x +8 a b A -11 a^{2} B \right )}{3 b^{5} \left (b x +a \right )^{\frac {3}{2}}}\)	\(88\)
gosper	\(-\frac {2 \left (-3 B \,x^{4} b^{4}-5 A \,x^{3} b^{4}+8 B \,x^{3} a \,b^{3}+30 A \,x^{2} a \,b^{3}-48 B \,x^{2} a^{2} b^{2}+120 A x \,a^{2} b^{2}-192 B x \,a^{3} b +80 A \,a^{3} b -128 B \,a^{4}\right )}{15 \left (b x +a \right )^{\frac {3}{2}} b^{5}}\)	\(95\)
trager	\(-\frac {2 \left (-3 B \,x^{4} b^{4}-5 A \,x^{3} b^{4}+8 B \,x^{3} a \,b^{3}+30 A \,x^{2} a \,b^{3}-48 B \,x^{2} a^{2} b^{2}+120 A x \,a^{2} b^{2}-192 B x \,a^{3} b +80 A \,a^{3} b -128 B \,a^{4}\right )}{15 \left (b x +a \right )^{\frac {3}{2}} b^{5}}\)	\(95\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {8 B a \left (b x +a \right )^{\frac {3}{2}}}{3}-6 A a b \sqrt {b x +a}+12 B \,a^{2} \sqrt {b x +a}-\frac {2 a^{2} \left (3 A b -4 B a \right )}{\sqrt {b x +a}}+\frac {2 a^{3} \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{5}}\)	\(105\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {8 B a \left (b x +a \right )^{\frac {3}{2}}}{3}-6 A a b \sqrt {b x +a}+12 B \,a^{2} \sqrt {b x +a}-\frac {2 a^{2} \left (3 A b -4 B a \right )}{\sqrt {b x +a}}+\frac {2 a^{3} \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{5}}\)	\(105\)

[In]

int(x^3*(B*x+A)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-32/3*(-1/16*(3/5*B*x+A)*x^3*b^4+3/8*x^2*(4/15*B*x+A)*a*b^3+3/2*x*(-2/5*B*x+A)*a^2*b^2+a^3*(-12/5*B*x+A)*b-8/5
*B*a^4)/(b*x+a)^(3/2)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b^{4} x^{4} + 128 \, B a^{4} - 80 \, A a^{3} b - {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 6 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 24 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]

[In]

integrate(x^3*(B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b^4*x^4 + 128*B*a^4 - 80*A*a^3*b - (8*B*a*b^3 - 5*A*b^4)*x^3 + 6*(8*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 24*
(8*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(b*x + a)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

Sympy [A] (verification not implemented)

Time = 1.49 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.13 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {5}{2}}}{5 b} - \frac {a^{3} \left (- A b + B a\right )}{3 b \left (a + b x\right )^{\frac {3}{2}}} + \frac {a^{2} \left (- 3 A b + 4 B a\right )}{b \sqrt {a + b x}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b - 4 B a\right )}{3 b} + \frac {\sqrt {a + b x} \left (- 3 A a b + 6 B a^{2}\right )}{b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{5}}{5}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(B*x+A)/(b*x+a)**(5/2),x)

[Out]

Piecewise((2*(B*(a + b*x)**(5/2)/(5*b) - a**3*(-A*b + B*a)/(3*b*(a + b*x)**(3/2)) + a**2*(-3*A*b + 4*B*a)/(b*s
qrt(a + b*x)) + (a + b*x)**(3/2)*(A*b - 4*B*a)/(3*b) + sqrt(a + b*x)*(-3*A*a*b + 6*B*a**2)/b)/b**4, Ne(b, 0)),
((A*x**4/4 + B*x**5/5)/a**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.90 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} B - 5 \, {\left (4 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 45 \, {\left (2 \, B a^{2} - A a b\right )} \sqrt {b x + a}}{b} - \frac {5 \, {\left (B a^{4} - A a^{3} b - 3 \, {\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b}\right )}}{15 \, b^{4}} \]

[In]

integrate(x^3*(B*x+A)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(b*x + a)^(5/2)*B - 5*(4*B*a - A*b)*(b*x + a)^(3/2) + 45*(2*B*a^2 - A*a*b)*sqrt(b*x + a))/b - 5*(B*a^
4 - A*a^3*b - 3*(4*B*a^3 - 3*A*a^2*b)*(b*x + a))/((b*x + a)^(3/2)*b))/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (b x + a\right )} B a^{3} - B a^{4} - 9 \, {\left (b x + a\right )} A a^{2} b + A a^{3} b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B b^{20} - 20 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b^{20} + 90 \, \sqrt {b x + a} B a^{2} b^{20} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{21} - 45 \, \sqrt {b x + a} A a b^{21}\right )}}{15 \, b^{25}} \]

[In]

integrate(x^3*(B*x+A)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*(12*(b*x + a)*B*a^3 - B*a^4 - 9*(b*x + a)*A*a^2*b + A*a^3*b)/((b*x + a)^(3/2)*b^5) + 2/15*(3*(b*x + a)^(5/
2)*B*b^20 - 20*(b*x + a)^(3/2)*B*a*b^20 + 90*sqrt(b*x + a)*B*a^2*b^20 + 5*(b*x + a)^(3/2)*A*b^21 - 45*sqrt(b*x
+ a)*A*a*b^21)/b^25

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {\left (12\,B\,a^2-6\,A\,a\,b\right )\,\sqrt {a+b\,x}}{b^5}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5\,b^5}+\frac {\left (8\,B\,a^3-6\,A\,a^2\,b\right )\,\left (a+b\,x\right )-\frac {2\,B\,a^4}{3}+\frac {2\,A\,a^3\,b}{3}}{b^5\,{\left (a+b\,x\right )}^{3/2}}+\frac {\left (2\,A\,b-8\,B\,a\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,b^5} \]

[In]

int((x^3*(A + B*x))/(a + b*x)^(5/2),x)

[Out]

((12*B*a^2 - 6*A*a*b)*(a + b*x)^(1/2))/b^5 + (2*B*(a + b*x)^(5/2))/(5*b^5) + ((8*B*a^3 - 6*A*a^2*b)*(a + b*x)
- (2*B*a^4)/3 + (2*A*a^3*b)/3)/(b^5*(a + b*x)^(3/2)) + ((2*A*b - 8*B*a)*(a + b*x)^(3/2))/(3*b^5)