\(\int (a+b \sqrt {x})^4 x^m \, dx\) [2257]

Optimal result

Integrand size = 15, antiderivative size = 87 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {a^4 x^{1+m}}{1+m}+\frac {8 a^3 b x^{\frac {3}{2}+m}}{3+2 m}+\frac {6 a^2 b^2 x^{2+m}}{2+m}+\frac {8 a b^3 x^{\frac {5}{2}+m}}{5+2 m}+\frac {b^4 x^{3+m}}{3+m} \]

[Out]

a^4*x^(1+m)/(1+m)+8*a^3*b*x^(3/2+m)/(3+2*m)+6*a^2*b^2*x^(2+m)/(2+m)+8*a*b^3*x^(5/2+m)/(5+2*m)+b^4*x^(3+m)/(3+m
)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {276} \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {a^4 x^{m+1}}{m+1}+\frac {8 a^3 b x^{m+\frac {3}{2}}}{2 m+3}+\frac {6 a^2 b^2 x^{m+2}}{m+2}+\frac {8 a b^3 x^{m+\frac {5}{2}}}{2 m+5}+\frac {b^4 x^{m+3}}{m+3} \]

[In]

Int[(a + b*Sqrt[x])^4*x^m,x]

[Out]

(a^4*x^(1 + m))/(1 + m) + (8*a^3*b*x^(3/2 + m))/(3 + 2*m) + (6*a^2*b^2*x^(2 + m))/(2 + m) + (8*a*b^3*x^(5/2 +
m))/(5 + 2*m) + (b^4*x^(3 + m))/(3 + m)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^4 x^m+4 a^3 b x^{\frac {1}{2}+m}+6 a^2 b^2 x^{1+m}+4 a b^3 x^{\frac {3}{2}+m}+b^4 x^{2+m}\right ) \, dx \\ & = \frac {a^4 x^{1+m}}{1+m}+\frac {8 a^3 b x^{\frac {3}{2}+m}}{3+2 m}+\frac {6 a^2 b^2 x^{2+m}}{2+m}+\frac {8 a b^3 x^{\frac {5}{2}+m}}{5+2 m}+\frac {b^4 x^{3+m}}{3+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=x^{1+m} \left (\frac {a^4}{1+m}+\frac {8 a^3 b \sqrt {x}}{3+2 m}+\frac {6 a^2 b^2 x}{2+m}+\frac {8 a b^3 x^{3/2}}{5+2 m}+\frac {b^4 x^2}{3+m}\right ) \]

[In]

Integrate[(a + b*Sqrt[x])^4*x^m,x]

[Out]

x^(1 + m)*(a^4/(1 + m) + (8*a^3*b*Sqrt[x])/(3 + 2*m) + (6*a^2*b^2*x)/(2 + m) + (8*a*b^3*x^(3/2))/(5 + 2*m) + (
b^4*x^2)/(3 + m))

Maple [F]

\[\int x^{m} \left (a +b \sqrt {x}\right )^{4}d x\]

[In]

int(x^m*(a+b*x^(1/2))^4,x)

[Out]

int(x^m*(a+b*x^(1/2))^4,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (83) = 166\).

Time = 0.26 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.99 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {{\left ({\left (4 \, b^{4} m^{4} + 28 \, b^{4} m^{3} + 71 \, b^{4} m^{2} + 77 \, b^{4} m + 30 \, b^{4}\right )} x^{3} + 6 \, {\left (4 \, a^{2} b^{2} m^{4} + 32 \, a^{2} b^{2} m^{3} + 91 \, a^{2} b^{2} m^{2} + 108 \, a^{2} b^{2} m + 45 \, a^{2} b^{2}\right )} x^{2} + {\left (4 \, a^{4} m^{4} + 36 \, a^{4} m^{3} + 119 \, a^{4} m^{2} + 171 \, a^{4} m + 90 \, a^{4}\right )} x + 8 \, {\left ({\left (2 \, a b^{3} m^{4} + 15 \, a b^{3} m^{3} + 40 \, a b^{3} m^{2} + 45 \, a b^{3} m + 18 \, a b^{3}\right )} x^{2} + {\left (2 \, a^{3} b m^{4} + 17 \, a^{3} b m^{3} + 52 \, a^{3} b m^{2} + 67 \, a^{3} b m + 30 \, a^{3} b\right )} x\right )} \sqrt {x}\right )} x^{m}}{4 \, m^{5} + 40 \, m^{4} + 155 \, m^{3} + 290 \, m^{2} + 261 \, m + 90} \]

[In]

integrate(x^m*(a+b*x^(1/2))^4,x, algorithm="fricas")

[Out]

((4*b^4*m^4 + 28*b^4*m^3 + 71*b^4*m^2 + 77*b^4*m + 30*b^4)*x^3 + 6*(4*a^2*b^2*m^4 + 32*a^2*b^2*m^3 + 91*a^2*b^
2*m^2 + 108*a^2*b^2*m + 45*a^2*b^2)*x^2 + (4*a^4*m^4 + 36*a^4*m^3 + 119*a^4*m^2 + 171*a^4*m + 90*a^4)*x + 8*((
2*a*b^3*m^4 + 15*a*b^3*m^3 + 40*a*b^3*m^2 + 45*a*b^3*m + 18*a*b^3)*x^2 + (2*a^3*b*m^4 + 17*a^3*b*m^3 + 52*a^3*
b*m^2 + 67*a^3*b*m + 30*a^3*b)*x)*sqrt(x))*x^m/(4*m^5 + 40*m^4 + 155*m^3 + 290*m^2 + 261*m + 90)

Sympy [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.72 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=a^{4} \left (\begin {cases} \frac {x^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 8 a^{3} b \left (\begin {cases} \frac {x^{\frac {3}{2}} x^{m}}{2 m + 3} & \text {for}\: m \neq - \frac {3}{2} \\x^{\frac {3}{2}} x^{m} \log {\left (\sqrt {x} \right )} & \text {otherwise} \end {cases}\right ) + 6 a^{2} b^{2} \left (\begin {cases} \frac {x^{2} x^{m}}{m + 2} & \text {for}\: m \neq -2 \\x^{2} x^{m} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + 8 a b^{3} \left (\begin {cases} \frac {x^{\frac {5}{2}} x^{m}}{2 m + 5} & \text {for}\: m \neq - \frac {5}{2} \\x^{\frac {5}{2}} x^{m} \log {\left (\sqrt {x} \right )} & \text {otherwise} \end {cases}\right ) + b^{4} \left (\begin {cases} \frac {x^{3} x^{m}}{m + 3} & \text {for}\: m \neq -3 \\x^{3} x^{m} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(x**m*(a+b*x**(1/2))**4,x)

[Out]

a**4*Piecewise((x**(m + 1)/(m + 1), Ne(m, -1)), (log(x), True)) + 8*a**3*b*Piecewise((x**(3/2)*x**m/(2*m + 3),
Ne(m, -3/2)), (x**(3/2)*x**m*log(sqrt(x)), True)) + 6*a**2*b**2*Piecewise((x**2*x**m/(m + 2), Ne(m, -2)), (x*
*2*x**m*log(x), True)) + 8*a*b**3*Piecewise((x**(5/2)*x**m/(2*m + 5), Ne(m, -5/2)), (x**(5/2)*x**m*log(sqrt(x)
), True)) + b**4*Piecewise((x**3*x**m/(m + 3), Ne(m, -3)), (x**3*x**m*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^{4} x^{m + 3}}{m + 3} + \frac {8 \, a b^{3} x^{m + \frac {5}{2}}}{2 \, m + 5} + \frac {6 \, a^{2} b^{2} x^{m + 2}}{m + 2} + \frac {8 \, a^{3} b x^{m + \frac {3}{2}}}{2 \, m + 3} + \frac {a^{4} x^{m + 1}}{m + 1} \]

[In]

integrate(x^m*(a+b*x^(1/2))^4,x, algorithm="maxima")

[Out]

b^4*x^(m + 3)/(m + 3) + 8*a*b^3*x^(m + 5/2)/(2*m + 5) + 6*a^2*b^2*x^(m + 2)/(m + 2) + 8*a^3*b*x^(m + 3/2)/(2*m
+ 3) + a^4*x^(m + 1)/(m + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^{4} x^{3} \sqrt {x}^{2 \, m}}{m + 3} + \frac {8 \, a b^{3} x^{\frac {5}{2}} \sqrt {x}^{2 \, m}}{2 \, m + 5} + \frac {6 \, a^{2} b^{2} x^{2} \sqrt {x}^{2 \, m}}{m + 2} + \frac {8 \, a^{3} b x^{\frac {3}{2}} \sqrt {x}^{2 \, m}}{2 \, m + 3} + \frac {a^{4} x \sqrt {x}^{2 \, m}}{m + 1} \]

[In]

integrate(x^m*(a+b*x^(1/2))^4,x, algorithm="giac")

[Out]

b^4*x^3*sqrt(x)^(2*m)/(m + 3) + 8*a*b^3*x^(5/2)*sqrt(x)^(2*m)/(2*m + 5) + 6*a^2*b^2*x^2*sqrt(x)^(2*m)/(m + 2)
+ 8*a^3*b*x^(3/2)*sqrt(x)^(2*m)/(2*m + 3) + a^4*x*sqrt(x)^(2*m)/(m + 1)

Mupad [B] (verification not implemented)

Time = 6.05 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.36 \[ \int \left (a+b \sqrt {x}\right )^4 x^m \, dx=\frac {b^4\,x^m\,x^3\,\left (4\,m^4+28\,m^3+71\,m^2+77\,m+30\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {a^4\,x\,x^m\,\left (4\,m^4+36\,m^3+119\,m^2+171\,m+90\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {8\,a\,b^3\,x^m\,x^{5/2}\,\left (2\,m^4+15\,m^3+40\,m^2+45\,m+18\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {8\,a^3\,b\,x^m\,x^{3/2}\,\left (2\,m^4+17\,m^3+52\,m^2+67\,m+30\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90}+\frac {6\,a^2\,b^2\,x^m\,x^2\,\left (4\,m^4+32\,m^3+91\,m^2+108\,m+45\right )}{4\,m^5+40\,m^4+155\,m^3+290\,m^2+261\,m+90} \]

[In]

int(x^m*(a + b*x^(1/2))^4,x)

[Out]

(b^4*x^m*x^3*(77*m + 71*m^2 + 28*m^3 + 4*m^4 + 30))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (a^4*x
*x^m*(171*m + 119*m^2 + 36*m^3 + 4*m^4 + 90))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (8*a*b^3*x^m
*x^(5/2)*(45*m + 40*m^2 + 15*m^3 + 2*m^4 + 18))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (8*a^3*b*x
^m*x^(3/2)*(67*m + 52*m^2 + 17*m^3 + 2*m^4 + 30))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90) + (6*a^2*b
^2*x^m*x^2*(108*m + 91*m^2 + 32*m^3 + 4*m^4 + 45))/(261*m + 290*m^2 + 155*m^3 + 40*m^4 + 4*m^5 + 90)