\(\int (a+b x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [1997]

Optimal result

Integrand size = 26, antiderivative size = 27 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b} \]

[Out]

1/7*(b^2*x^2+2*a*b*x+a^2)^(7/2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {643} \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b} \]

[In]

Int[(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(a^2 + 2*a*b*x + b^2*x^2)^(7/2)/(7*b)

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {(a+b x)^6 \sqrt {(a+b x)^2}}{7 b} \]

[In]

Integrate[(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)^6*Sqrt[(a + b*x)^2])/(7*b)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{7}}{7 b}\)	\(19\)
default	\(\frac {\left (b x +a \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{7 b}\)	\(22\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b x +a \right )^{6}}{7 b}\)	\(22\)
gosper	\(\frac {x \left (b^{6} x^{6}+7 a \,b^{5} x^{5}+21 a^{2} b^{4} x^{4}+35 a^{3} b^{3} x^{3}+35 a^{4} b^{2} x^{2}+21 a^{5} b x +7 a^{6}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{7 \left (b x +a \right )^{5}}\)	\(82\)

[In]

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

[Out]

1/7*csgn(b*x+a)*(b*x+a)^7/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).

Time = 0.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.37 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{7} \, b^{6} x^{7} + a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{5} + 5 \, a^{3} b^{3} x^{4} + 5 \, a^{4} b^{2} x^{3} + 3 \, a^{5} b x^{2} + a^{6} x \]

[In]

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

[Out]

1/7*b^6*x^7 + a*b^5*x^6 + 3*a^2*b^4*x^5 + 5*a^3*b^3*x^4 + 5*a^4*b^2*x^3 + 3*a^5*b*x^2 + a^6*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (22) = 44\).

Time = 0.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 8.37 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\begin {cases} \frac {a^{6} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7 b} + \frac {6 a^{5} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} + \frac {15 a^{4} b x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} + \frac {20 a^{3} b^{2} x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} + \frac {15 a^{2} b^{3} x^{4} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} + \frac {6 a b^{4} x^{5} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} + \frac {b^{5} x^{6} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{7} & \text {for}\: b \neq 0 \\a x \left (a^{2}\right )^{\frac {5}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**6*sqrt(a**2 + 2*a*b*x + b**2*x**2)/(7*b) + 6*a**5*x*sqrt(a**2 + 2*a*b*x + b**2*x**2)/7 + 15*a**4
*b*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2)/7 + 20*a**3*b**2*x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2)/7 + 15*a**2*b*
*3*x**4*sqrt(a**2 + 2*a*b*x + b**2*x**2)/7 + 6*a*b**4*x**5*sqrt(a**2 + 2*a*b*x + b**2*x**2)/7 + b**5*x**6*sqrt
(a**2 + 2*a*b*x + b**2*x**2)/7, Ne(b, 0)), (a*x*(a**2)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{7 \, b} \]

[In]

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

[Out]

1/7*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.44 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{7} \, b^{6} x^{7} \mathrm {sgn}\left (b x + a\right ) + a b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{5} b x^{2} \mathrm {sgn}\left (b x + a\right ) + a^{6} x \mathrm {sgn}\left (b x + a\right ) + \frac {a^{7} \mathrm {sgn}\left (b x + a\right )}{7 \, b} \]

[In]

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

[Out]

1/7*b^6*x^7*sgn(b*x + a) + a*b^5*x^6*sgn(b*x + a) + 3*a^2*b^4*x^5*sgn(b*x + a) + 5*a^3*b^3*x^4*sgn(b*x + a) +
5*a^4*b^2*x^3*sgn(b*x + a) + 3*a^5*b*x^2*sgn(b*x + a) + a^6*x*sgn(b*x + a) + 1/7*a^7*sgn(b*x + a)/b

Mupad [B] (verification not implemented)

Time = 10.95 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left ({\left (a+b\,x\right )}^2\right )}^{7/2}}{7\,b} \]

[In]

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

[Out]

((a + b*x)^2)^(7/2)/(7*b)