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

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Rubi [A]  time = 0.0222112, antiderivative size = 48, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.0220034, 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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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sqrt{x} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99831, size = 61, normalized size = 1.27 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.43857, size = 120, normalized size = 2.5 \begin{align*} \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}} \end{align*}

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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Sympy [B]  time = 1.93248, size = 823, normalized size = 17.15 \begin{align*} - \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}}} \end{align*}

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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Giac [B]  time = 1.08253, size = 127, normalized size = 2.65 \begin{align*} \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}} \end{align*}

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)