\(\int (a+b x)^{3/2} \, dx\) [295]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int (a+b x)^{3/2} \, dx=\frac {2 (a+b x)^{5/2}}{5 b} \]

[Out]

2/5*(b*x+a)^(5/2)/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {32} \[ \int (a+b x)^{3/2} \, dx=\frac {2 (a+b x)^{5/2}}{5 b} \]

[In]

Int[(a + b*x)^(3/2),x]

[Out]

(2*(a + b*x)^(5/2))/(5*b)

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)^{5/2}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int (a+b x)^{3/2} \, dx=\frac {2 (a+b x)^{5/2}}{5 b} \]

[In]

Integrate[(a + b*x)^(3/2),x]

[Out]

(2*(a + b*x)^(5/2))/(5*b)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\)	\(13\)
derivativedivides	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\)	\(13\)
default	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\)	\(13\)
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\)	\(13\)
trager	\(\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \sqrt {b x +a}}{5 b}\)	\(29\)
risch	\(\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \sqrt {b x +a}}{5 b}\)	\(29\)

[In]

int((b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*(b*x+a)^(5/2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (12) = 24\).

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75 \[ \int (a+b x)^{3/2} \, dx=\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{5 \, b} \]

[In]

integrate((b*x+a)^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{3/2} \, dx=\frac {2 \left (a + b x\right )^{\frac {5}{2}}}{5 b} \]

[In]

integrate((b*x+a)**(3/2),x)

[Out]

2*(a + b*x)**(5/2)/(5*b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{3/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}}}{5 \, b} \]

[In]

integrate((b*x+a)^(3/2),x, algorithm="maxima")

[Out]

2/5*(b*x + a)^(5/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (12) = 24\).

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.62 \[ \int (a+b x)^{3/2} \, dx=\frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 30 \, \sqrt {b x + a} a^{2} + 10 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a\right )}}{15 \, b} \]

[In]

integrate((b*x+a)^(3/2),x, algorithm="giac")

[Out]

2/15*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 30*sqrt(b*x + a)*a^2 + 10*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*
a)*a)/b

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{3/2} \, dx=\frac {2\,{\left (a+b\,x\right )}^{5/2}}{5\,b} \]

[In]

int((a + b*x)^(3/2),x)

[Out]

(2*(a + b*x)^(5/2))/(5*b)