\(\int \frac {x^3}{\sqrt {a+b x}} \, dx\) [335]
Optimal result
Integrand size = 13, antiderivative size = 68 \[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=-\frac {2 a^3 \sqrt {a+b x}}{b^4}+\frac {2 a^2 (a+b x)^{3/2}}{b^4}-\frac {6 a (a+b x)^{5/2}}{5 b^4}+\frac {2 (a+b x)^{7/2}}{7 b^4}
\]
[Out]
2*a^2*(b*x+a)^(3/2)/b^4-6/5*a*(b*x+a)^(5/2)/b^4+2/7*(b*x+a)^(7/2)/b^4-2*a^3*(b*x+a)^(1/2)/b^4
Rubi [A] (verified)
Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00,
number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45}
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=-\frac {2 a^3 \sqrt {a+b x}}{b^4}+\frac {2 a^2 (a+b x)^{3/2}}{b^4}+\frac {2 (a+b x)^{7/2}}{7 b^4}-\frac {6 a (a+b x)^{5/2}}{5 b^4}
\]
[In]
Int[x^3/Sqrt[a + b*x],x]
[Out]
(-2*a^3*Sqrt[a + b*x])/b^4 + (2*a^2*(a + b*x)^(3/2))/b^4 - (6*a*(a + b*x)^(5/2))/(5*b^4) + (2*(a + b*x)^(7/2))
/(7*b^4)
Rule 45
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps \begin{align*}
\text {integral}& = \int \left (-\frac {a^3}{b^3 \sqrt {a+b x}}+\frac {3 a^2 \sqrt {a+b x}}{b^3}-\frac {3 a (a+b x)^{3/2}}{b^3}+\frac {(a+b x)^{5/2}}{b^3}\right ) \, dx \\ & = -\frac {2 a^3 \sqrt {a+b x}}{b^4}+\frac {2 a^2 (a+b x)^{3/2}}{b^4}-\frac {6 a (a+b x)^{5/2}}{5 b^4}+\frac {2 (a+b x)^{7/2}}{7 b^4} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \left (-16 a^3+8 a^2 b x-6 a b^2 x^2+5 b^3 x^3\right )}{35 b^4}
\]
[In]
Integrate[x^3/Sqrt[a + b*x],x]
[Out]
(2*Sqrt[a + b*x]*(-16*a^3 + 8*a^2*b*x - 6*a*b^2*x^2 + 5*b^3*x^3))/(35*b^4)
Maple [A] (verified)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63
| | |
method | result | size |
| | |
gosper |
\(-\frac {2 \sqrt {b x +a}\, \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right )}{35 b^{4}}\) |
\(43\) |
trager |
\(-\frac {2 \sqrt {b x +a}\, \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right )}{35 b^{4}}\) |
\(43\) |
risch |
\(-\frac {2 \sqrt {b x +a}\, \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right )}{35 b^{4}}\) |
\(43\) |
pseudoelliptic |
\(-\frac {2 \sqrt {b x +a}\, \left (-5 b^{3} x^{3}+6 a \,b^{2} x^{2}-8 a^{2} b x +16 a^{3}\right )}{35 b^{4}}\) |
\(43\) |
derivativedivides |
\(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {6 a \left (b x +a \right )^{\frac {5}{2}}}{5}+2 a^{2} \left (b x +a \right )^{\frac {3}{2}}-2 a^{3} \sqrt {b x +a}}{b^{4}}\) |
\(49\) |
default |
\(\frac {\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {6 a \left (b x +a \right )^{\frac {5}{2}}}{5}+2 a^{2} \left (b x +a \right )^{\frac {3}{2}}-2 a^{3} \sqrt {b x +a}}{b^{4}}\) |
\(49\) |
| | |
|
|
|
[In]
int(x^3/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2/35*(b*x+a)^(1/2)*(-5*b^3*x^3+6*a*b^2*x^2-8*a^2*b*x+16*a^3)/b^4
Fricas [A] (verification not implemented)
none
Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.62
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 8 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{4}}
\]
[In]
integrate(x^3/(b*x+a)^(1/2),x, algorithm="fricas")
[Out]
2/35*(5*b^3*x^3 - 6*a*b^2*x^2 + 8*a^2*b*x - 16*a^3)*sqrt(b*x + a)/b^4
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1640 vs. \(2 (65) = 130\).
Time = 2.00 (sec) , antiderivative size = 1640, normalized size of antiderivative = 24.12
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\text {Too large to display}
\]
[In]
integrate(x**3/(b*x+a)**(1/2),x)
[Out]
-32*a**(47/2)*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 +
525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 32*a**(47/2)/(35*a**20*b**4 + 210*a**19*b**
5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x
**6) - 176*a**(45/2)*b*x*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b
**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 192*a**(45/2)*b*x/(35*a**20*b**4
+ 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 +
35*a**14*b**10*x**6) - 396*a**(43/2)*b**2*x**2*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b
**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 480*a**(43
/2)*b**2*x**2/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x
**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) - 462*a**(41/2)*b**3*x**3*sqrt(1 + b*x/a)/(35*a**20*b**4 + 21
0*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a*
*14*b**10*x**6) + 640*a**(41/2)*b**3*x**3/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*
b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) - 280*a**(39/2)*b**4*x**4*sqrt(1
+ b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 +
210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 480*a**(39/2)*b**4*x**4/(35*a**20*b**4 + 210*a**19*b**5*x + 525*
a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) - 42*
a**(37/2)*b**5*x**5*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x
**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 192*a**(37/2)*b**5*x**5/(35*a**20*b**
4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 +
35*a**14*b**10*x**6) + 84*a**(35/2)*b**6*x**6*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b
**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 32*a**(35/
2)*b**6*x**6/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x*
*4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 94*a**(33/2)*b**7*x**7*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*
a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**1
4*b**10*x**6) + 48*a**(31/2)*b**8*x**8*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2
+ 700*a**17*b**7*x**3 + 525*a**16*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6) + 10*a**(29/2)*b**9*
x**9*sqrt(1 + b*x/a)/(35*a**20*b**4 + 210*a**19*b**5*x + 525*a**18*b**6*x**2 + 700*a**17*b**7*x**3 + 525*a**16
*b**8*x**4 + 210*a**15*b**9*x**5 + 35*a**14*b**10*x**6)
Maxima [A] (verification not implemented)
none
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}}}{7 \, b^{4}} - \frac {6 \, {\left (b x + a\right )}^{\frac {5}{2}} a}{5 \, b^{4}} + \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}}{b^{4}} - \frac {2 \, \sqrt {b x + a} a^{3}}{b^{4}}
\]
[In]
integrate(x^3/(b*x+a)^(1/2),x, algorithm="maxima")
[Out]
2/7*(b*x + a)^(7/2)/b^4 - 6/5*(b*x + a)^(5/2)*a/b^4 + 2*(b*x + a)^(3/2)*a^2/b^4 - 2*sqrt(b*x + a)*a^3/b^4
Giac [A] (verification not implemented)
none
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{35 \, b^{4}}
\]
[In]
integrate(x^3/(b*x+a)^(1/2),x, algorithm="giac")
[Out]
2/35*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)/b^4
Mupad [B] (verification not implemented)
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82
\[
\int \frac {x^3}{\sqrt {a+b x}} \, dx=\frac {2\,{\left (a+b\,x\right )}^{7/2}}{7\,b^4}-\frac {2\,a^3\,\sqrt {a+b\,x}}{b^4}+\frac {2\,a^2\,{\left (a+b\,x\right )}^{3/2}}{b^4}-\frac {6\,a\,{\left (a+b\,x\right )}^{5/2}}{5\,b^4}
\]
[In]
int(x^3/(a + b*x)^(1/2),x)
[Out]
(2*(a + b*x)^(7/2))/(7*b^4) - (2*a^3*(a + b*x)^(1/2))/b^4 + (2*a^2*(a + b*x)^(3/2))/b^4 - (6*a*(a + b*x)^(5/2)
)/(5*b^4)