\(\int (a+b x)^{5/2} \, dx\) [303]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int (a+b x)^{5/2} \, dx=\frac {2 (a+b x)^{7/2}}{7 b} \]

[Out]

2/7*(b*x+a)^(7/2)/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} \[ \int (a+b x)^{5/2} \, dx=\frac {2 (a+b x)^{7/2}}{7 b} \]

[In]

Int[(a + b*x)^(5/2),x]

[Out]

(2*(a + b*x)^(7/2))/(7*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)^{7/2}}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int (a+b x)^{5/2} \, dx=\frac {2 (a+b x)^{7/2}}{7 b} \]

[In]

Integrate[(a + b*x)^(5/2),x]

[Out]

(2*(a + b*x)^(7/2))/(7*b)

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7 b}\)	\(13\)
derivativedivides	\(\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7 b}\)	\(13\)
default	\(\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7 b}\)	\(13\)
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7 b}\)	\(13\)
trager	\(\frac {2 \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right ) \sqrt {b x +a}}{7 b}\)	\(40\)
risch	\(\frac {2 \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right ) \sqrt {b x +a}}{7 b}\)	\(40\)

[In]

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

[Out]

2/7*(b*x+a)^(7/2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (12) = 24\).

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.44 \[ \int (a+b x)^{5/2} \, dx=\frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b x + a}}{7 \, b} \]

[In]

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

[Out]

2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{5/2} \, dx=\frac {2 \left (a + b x\right )^{\frac {7}{2}}}{7 b} \]

[In]

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

[Out]

2*(a + b*x)**(7/2)/(7*b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{5/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}}}{7 \, b} \]

[In]

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

[Out]

2/7*(b*x + a)^(7/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (12) = 24\).

Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 5.94 \[ \int (a+b x)^{5/2} \, dx=\frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} + 35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} + 7 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a\right )}}{35 \, b} \]

[In]

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

[Out]

2/35*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 + 35*((b*x + a)^(3/2) - 3*sqrt(b*x + a
)*a)*a^2 + 7*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a)/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{5/2} \, dx=\frac {2\,{\left (a+b\,x\right )}^{7/2}}{7\,b} \]

[In]

int((a + b*x)^(5/2),x)

[Out]

(2*(a + b*x)^(7/2))/(7*b)