\(\int (a+b x)^{p/2} \, dx\) [187]

Optimal result

Integrand size = 11, antiderivative size = 23 \[ \int (a+b x)^{p/2} \, dx=\frac {2 (a+b x)^{\frac {2+p}{2}}}{b (2+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {32} \[ \int (a+b x)^{p/2} \, dx=\frac {2 (a+b x)^{\frac {p+2}{2}}}{b (p+2)} \]

[In]

Int[(a + b*x)^(p/2),x]

[Out]

(2*(a + b*x)^((2 + p)/2))/(b*(2 + p))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (a+b x)^{\frac {2+p}{2}}}{b (2+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int (a+b x)^{p/2} \, dx=\frac {(a+b x)^{1+\frac {p}{2}}}{b \left (1+\frac {p}{2}\right )} \]

[In]

Integrate[(a + b*x)^(p/2),x]

[Out]

(a + b*x)^(1 + p/2)/(b*(1 + p/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (b x +a \right )^{\frac {p}{2}}}{b \left (2+p \right )}\)	\(25\)
risch	\(\frac {2 \left (b x +a \right ) \left (b x +a \right )^{\frac {p}{2}}}{b \left (2+p \right )}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int (a+b x)^{p/2} \, dx=\frac {2 \, {\left (b x + a\right )} \sqrt {b x + a}^{p}}{b p + 2 \, b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (a+b x)^{p/2} \, dx=\frac {\begin {cases} \frac {\left (a + b x\right )^{\frac {p}{2} + 1}}{\frac {p}{2} + 1} & \text {for}\: p \neq -2 \\\log {\left (a + b x \right )} & \text {otherwise} \end {cases}}{b} \]

[In]

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

[Out]

Piecewise(((a + b*x)**(p/2 + 1)/(p/2 + 1), Ne(p, -2)), (log(a + b*x), True))/b

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int (a+b x)^{p/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {1}{2} \, p + 1}}{b {\left (p + 2\right )}} \]

[In]

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

[Out]

2*(b*x + a)^(1/2*p + 1)/(b*(p + 2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int (a+b x)^{p/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {1}{2} \, p + 1}}{b {\left (p + 2\right )}} \]

[In]

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

[Out]

2*(b*x + a)^(1/2*p + 1)/(b*(p + 2))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int (a+b x)^{p/2} \, dx=\frac {2\,{\left (a+b\,x\right )}^{\frac {p}{2}+1}}{b\,\left (p+2\right )} \]

[In]

int((a + b*x)^(p/2),x)

[Out]

(2*(a + b*x)^(p/2 + 1))/(b*(p + 2))