\(\int (a+b x)^{9/2} \, dx\) [316]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int (a+b x)^{9/2} \, dx=\frac {2 (a+b x)^{11/2}}{11 b} \]

[Out]

2/11*(b*x+a)^(11/2)/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} \[ \int (a+b x)^{9/2} \, dx=\frac {2 (a+b x)^{11/2}}{11 b} \]

[In]

Int[(a + b*x)^(9/2),x]

[Out]

(2*(a + b*x)^(11/2))/(11*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)^{11/2}}{11 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int (a+b x)^{9/2} \, dx=\frac {2 (a+b x)^{11/2}}{11 b} \]

[In]

Integrate[(a + b*x)^(9/2),x]

[Out]

(2*(a + b*x)^(11/2))/(11*b)

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {2 \left (b x +a \right )^{\frac {11}{2}}}{11 b}\)	\(13\)
derivativedivides	\(\frac {2 \left (b x +a \right )^{\frac {11}{2}}}{11 b}\)	\(13\)
default	\(\frac {2 \left (b x +a \right )^{\frac {11}{2}}}{11 b}\)	\(13\)
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{\frac {11}{2}}}{11 b}\)	\(13\)
trager	\(\frac {2 \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right ) \sqrt {b x +a}}{11 b}\)	\(62\)
risch	\(\frac {2 \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right ) \sqrt {b x +a}}{11 b}\)	\(62\)

[In]

int((b*x+a)^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/11*(b*x+a)^(11/2)/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (12) = 24\).

Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.81 \[ \int (a+b x)^{9/2} \, dx=\frac {2 \, {\left (b^{5} x^{5} + 5 \, a b^{4} x^{4} + 10 \, a^{2} b^{3} x^{3} + 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5}\right )} \sqrt {b x + a}}{11 \, b} \]

[In]

integrate((b*x+a)^(9/2),x, algorithm="fricas")

[Out]

2/11*(b^5*x^5 + 5*a*b^4*x^4 + 10*a^2*b^3*x^3 + 10*a^3*b^2*x^2 + 5*a^4*b*x + a^5)*sqrt(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{9/2} \, dx=\frac {2 \left (a + b x\right )^{\frac {11}{2}}}{11 b} \]

[In]

integrate((b*x+a)**(9/2),x)

[Out]

2*(a + b*x)**(11/2)/(11*b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{9/2} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {11}{2}}}{11 \, b} \]

[In]

integrate((b*x+a)^(9/2),x, algorithm="maxima")

[Out]

2/11*(b*x + a)^(11/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (12) = 24\).

Time = 0.30 (sec) , antiderivative size = 229, normalized size of antiderivative = 14.31 \[ \int (a+b x)^{9/2} \, dx=\frac {2 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} + 1155 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{4} + 462 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{3} + 198 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2} + 11 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a\right )}}{693 \, b} \]

[In]

integrate((b*x+a)^(9/2),x, algorithm="giac")

[Out]

2/693*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155
*(b*x + a)^(3/2)*a^4 + 1155*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^4 + 462*(3*(b*x + a)^(5/2) - 10*(b*x + a)^
(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^3 + 198*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 -
35*sqrt(b*x + a)*a^3)*a^2 + 11*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b
*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{9/2} \, dx=\frac {2\,{\left (a+b\,x\right )}^{11/2}}{11\,b} \]

[In]

int((a + b*x)^(9/2),x)

[Out]

(2*(a + b*x)^(11/2))/(11*b)