\(\int (a+b \sqrt {x})^5 \, dx\) [2145]

Optimal result

Integrand size = 11, antiderivative size = 38 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=-\frac {a \left (a+b \sqrt {x}\right )^6}{3 b^2}+\frac {2 \left (a+b \sqrt {x}\right )^7}{7 b^2} \]

[Out]

-1/3*a*(a+b*x^(1/2))^6/b^2+2/7*(a+b*x^(1/2))^7/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {196, 45} \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=\frac {2 \left (a+b \sqrt {x}\right )^7}{7 b^2}-\frac {a \left (a+b \sqrt {x}\right )^6}{3 b^2} \]

[In]

Int[(a + b*Sqrt[x])^5,x]

[Out]

-1/3*(a*(a + b*Sqrt[x])^6)/b^2 + (2*(a + b*Sqrt[x])^7)/(7*b^2)

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])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x (a+b x)^5 \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (-\frac {a (a+b x)^5}{b}+\frac {(a+b x)^6}{b}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {a \left (a+b \sqrt {x}\right )^6}{3 b^2}+\frac {2 \left (a+b \sqrt {x}\right )^7}{7 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=\frac {1}{21} \left (21 a^5 x+70 a^4 b x^{3/2}+105 a^3 b^2 x^2+84 a^2 b^3 x^{5/2}+35 a b^4 x^3+6 b^5 x^{7/2}\right ) \]

[In]

Integrate[(a + b*Sqrt[x])^5,x]

[Out]

(21*a^5*x + 70*a^4*b*x^(3/2) + 105*a^3*b^2*x^2 + 84*a^2*b^3*x^(5/2) + 35*a*b^4*x^3 + 6*b^5*x^(7/2))/21

Maple [A] (verified)

Time = 3.51 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(\frac {2 b^{5} x^{\frac {7}{2}}}{7}+\frac {5 a \,b^{4} x^{3}}{3}+4 a^{2} b^{3} x^{\frac {5}{2}}+5 a^{3} b^{2} x^{2}+\frac {10 a^{4} b \,x^{\frac {3}{2}}}{3}+a^{5} x\)	\(55\)
default	\(\frac {2 b^{5} x^{\frac {7}{2}}}{7}+\frac {5 a \,b^{4} x^{3}}{3}+4 a^{2} b^{3} x^{\frac {5}{2}}+5 a^{3} b^{2} x^{2}+\frac {10 a^{4} b \,x^{\frac {3}{2}}}{3}+a^{5} x\)	\(55\)
trager	\(\frac {a \left (5 b^{4} x^{2}+15 a^{2} b^{2} x +5 b^{4} x +3 a^{4}+15 a^{2} b^{2}+5 b^{4}\right ) \left (-1+x \right )}{3}+\frac {2 b \,x^{\frac {3}{2}} \left (3 b^{4} x^{2}+42 a^{2} b^{2} x +35 a^{4}\right )}{21}\)	\(79\)

[In]

int((a+b*x^(1/2))^5,x,method=_RETURNVERBOSE)

[Out]

2/7*b^5*x^(7/2)+5/3*a*b^4*x^3+4*a^2*b^3*x^(5/2)+5*a^3*b^2*x^2+10/3*a^4*b*x^(3/2)+a^5*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=\frac {5}{3} \, a b^{4} x^{3} + 5 \, a^{3} b^{2} x^{2} + a^{5} x + \frac {2}{21} \, {\left (3 \, b^{5} x^{3} + 42 \, a^{2} b^{3} x^{2} + 35 \, a^{4} b x\right )} \sqrt {x} \]

[In]

integrate((a+b*x^(1/2))^5,x, algorithm="fricas")

[Out]

5/3*a*b^4*x^3 + 5*a^3*b^2*x^2 + a^5*x + 2/21*(3*b^5*x^3 + 42*a^2*b^3*x^2 + 35*a^4*b*x)*sqrt(x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (32) = 64\).

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.74 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=a^{5} x + \frac {10 a^{4} b x^{\frac {3}{2}}}{3} + 5 a^{3} b^{2} x^{2} + 4 a^{2} b^{3} x^{\frac {5}{2}} + \frac {5 a b^{4} x^{3}}{3} + \frac {2 b^{5} x^{\frac {7}{2}}}{7} \]

[In]

integrate((a+b*x**(1/2))**5,x)

[Out]

a**5*x + 10*a**4*b*x**(3/2)/3 + 5*a**3*b**2*x**2 + 4*a**2*b**3*x**(5/2) + 5*a*b**4*x**3/3 + 2*b**5*x**(7/2)/7

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=\frac {2}{7} \, b^{5} x^{\frac {7}{2}} + \frac {5}{3} \, a b^{4} x^{3} + 4 \, a^{2} b^{3} x^{\frac {5}{2}} + 5 \, a^{3} b^{2} x^{2} + \frac {10}{3} \, a^{4} b x^{\frac {3}{2}} + a^{5} x \]

[In]

integrate((a+b*x^(1/2))^5,x, algorithm="maxima")

[Out]

2/7*b^5*x^(7/2) + 5/3*a*b^4*x^3 + 4*a^2*b^3*x^(5/2) + 5*a^3*b^2*x^2 + 10/3*a^4*b*x^(3/2) + a^5*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=\frac {2}{7} \, b^{5} x^{\frac {7}{2}} + \frac {5}{3} \, a b^{4} x^{3} + 4 \, a^{2} b^{3} x^{\frac {5}{2}} + 5 \, a^{3} b^{2} x^{2} + \frac {10}{3} \, a^{4} b x^{\frac {3}{2}} + a^{5} x \]

[In]

integrate((a+b*x^(1/2))^5,x, algorithm="giac")

[Out]

2/7*b^5*x^(7/2) + 5/3*a*b^4*x^3 + 4*a^2*b^3*x^(5/2) + 5*a^3*b^2*x^2 + 10/3*a^4*b*x^(3/2) + a^5*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \left (a+b \sqrt {x}\right )^5 \, dx=a^5\,x+\frac {2\,b^5\,x^{7/2}}{7}+\frac {5\,a\,b^4\,x^3}{3}+\frac {10\,a^4\,b\,x^{3/2}}{3}+5\,a^3\,b^2\,x^2+4\,a^2\,b^3\,x^{5/2} \]

[In]

int((a + b*x^(1/2))^5,x)

[Out]

a^5*x + (2*b^5*x^(7/2))/7 + (5*a*b^4*x^3)/3 + (10*a^4*b*x^(3/2))/3 + 5*a^3*b^2*x^2 + 4*a^2*b^3*x^(5/2)