3.2267 \(\int (a+b \sqrt {x})^p \, dx\)

Optimal. Leaf size=48 \[ \frac {2 \left (a+b \sqrt {x}\right )^{p+2}}{b^2 (p+2)}-\frac {2 a \left (a+b \sqrt {x}\right )^{p+1}}{b^2 (p+1)} \]

[Out]

-2*a*(a+b*x^(1/2))^(1+p)/b^2/(1+p)+2*(a+b*x^(1/2))^(2+p)/b^2/(2+p)

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {190, 43} \[ \frac {2 \left (a+b \sqrt {x}\right )^{p+2}}{b^2 (p+2)}-\frac {2 a \left (a+b \sqrt {x}\right )^{p+1}}{b^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^p,x]

[Out]

(-2*a*(a + b*Sqrt[x])^(1 + p))/(b^2*(1 + p)) + (2*(a + b*Sqrt[x])^(2 + p))/(b^2*(2 + p))

Rule 43

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 190

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*} \int \left (a+b \sqrt {x}\right )^p \, dx &=2 \operatorname {Subst}\left (\int x (a+b x)^p \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 a \left (a+b \sqrt {x}\right )^{1+p}}{b^2 (1+p)}+\frac {2 \left (a+b \sqrt {x}\right )^{2+p}}{b^2 (2+p)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.88 \[ \frac {2 \left (a+b \sqrt {x}\right )^{p+1} \left (b (p+1) \sqrt {x}-a\right )}{b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^p,x]

[Out]

(2*(a + b*Sqrt[x])^(1 + p)*(-a + b*(1 + p)*Sqrt[x]))/(b^2*(1 + p)*(2 + p))

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fricas [A]  time = 1.00, size = 56, normalized size = 1.17 \[ \frac {2 \, {\left (a b p \sqrt {x} - a^{2} + {\left (b^{2} p + b^{2}\right )} x\right )} {\left (b \sqrt {x} + a\right )}^{p}}{b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*b*p*sqrt(x) - a^2 + (b^2*p + b^2)*x)*(b*sqrt(x) + a)^p/(b^2*p^2 + 3*b^2*p + 2*b^2)

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giac [B]  time = 0.16, size = 94, normalized size = 1.96 \[ \frac {2 \, {\left ({\left (b \sqrt {x} + a\right )}^{2} {\left (b \sqrt {x} + a\right )}^{p} p - {\left (b \sqrt {x} + a\right )} {\left (b \sqrt {x} + a\right )}^{p} a p + {\left (b \sqrt {x} + a\right )}^{2} {\left (b \sqrt {x} + a\right )}^{p} - 2 \, {\left (b \sqrt {x} + a\right )} {\left (b \sqrt {x} + a\right )}^{p} a\right )}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b*sqrt(x) + a)^2*(b*sqrt(x) + a)^p*p - (b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a*p + (b*sqrt(x) + a)^2*(b*sqrt(x
) + a)^p - 2*(b*sqrt(x) + a)*(b*sqrt(x) + a)^p*a)/((p^2 + 3*p + 2)*b^2)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (b \sqrt {x}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.94, size = 45, normalized size = 0.94 \[ \frac {2 \, {\left (b^{2} {\left (p + 1\right )} x + a b p \sqrt {x} - a^{2}\right )} {\left (b \sqrt {x} + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*(p + 1)*x + a*b*p*sqrt(x) - a^2)*(b*sqrt(x) + a)^p/((p^2 + 3*p + 2)*b^2)

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mupad [B]  time = 1.18, size = 41, normalized size = 0.85 \[ \frac {x\,{\left (a+b\,\sqrt {x}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (2,-p;\ 3;\ -\frac {b\,\sqrt {x}}{a}\right )}{{\left (\frac {b\,\sqrt {x}}{a}+1\right )}^p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^p,x)

[Out]

(x*(a + b*x^(1/2))^p*hypergeom([2, -p], 3, -(b*x^(1/2))/a))/((b*x^(1/2))/a + 1)^p

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sympy [B]  time = 1.83, size = 823, normalized size = 17.15 \[ - \frac {2 a^{3} a^{p} x^{2} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a^{3} a^{p} x^{2}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a^{2} a^{p} b p x^{\frac {5}{2}} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} - \frac {2 a^{2} a^{p} b x^{\frac {5}{2}} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a^{2} a^{p} b x^{\frac {5}{2}}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {4 a a^{p} b^{2} p x^{3} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a a^{p} b^{2} x^{3} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a^{p} b^{3} p x^{\frac {7}{2}} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} + \frac {2 a^{p} b^{3} x^{\frac {7}{2}} \left (1 + \frac {b \sqrt {x}}{a}\right )^{p}}{a b^{2} p^{2} x^{2} + 3 a b^{2} p x^{2} + 2 a b^{2} x^{2} + b^{3} p^{2} x^{\frac {5}{2}} + 3 b^{3} p x^{\frac {5}{2}} + 2 b^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**3*a**p*x**2*(1 + b*sqrt(x)/a)**p/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**2*x**(5/2
) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 2*a**3*a**p*x**2/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**
2 + b**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 2*a**2*a**p*b*p*x**(5/2)*(1 + b*sqrt(x)/a)**p/
(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)
) - 2*a**2*a**p*b*x**(5/2)*(1 + b*sqrt(x)/a)**p/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**
2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 2*a**2*a**p*b*x**(5/2)/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2
+ 2*a*b**2*x**2 + b**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 4*a*a**p*b**2*p*x**3*(1 + b*sqr
t(x)/a)**p/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b*
*3*x**(5/2)) + 2*a*a**p*b**2*x**3*(1 + b*sqrt(x)/a)**p/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b
**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 2*a**p*b**3*p*x**(7/2)*(1 + b*sqrt(x)/a)**p/(a*b**2
*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**2*x**(5/2) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2)) + 2*a
**p*b**3*x**(7/2)*(1 + b*sqrt(x)/a)**p/(a*b**2*p**2*x**2 + 3*a*b**2*p*x**2 + 2*a*b**2*x**2 + b**3*p**2*x**(5/2
) + 3*b**3*p*x**(5/2) + 2*b**3*x**(5/2))

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