\(\int \frac {(a x+b x^2)^{5/2}}{x^8} \, dx\) [34]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=-\frac {2 \left (a x+b x^2\right )^{7/2}}{9 a x^8}+\frac {4 b \left (a x+b x^2\right )^{7/2}}{63 a^2 x^7} \]

[Out]

-2/9*(b*x^2+a*x)^(7/2)/a/x^8+4/63*b*(b*x^2+a*x)^(7/2)/a^2/x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\frac {4 b \left (a x+b x^2\right )^{7/2}}{63 a^2 x^7}-\frac {2 \left (a x+b x^2\right )^{7/2}}{9 a x^8} \]

[In]

Int[(a*x + b*x^2)^(5/2)/x^8,x]

[Out]

(-2*(a*x + b*x^2)^(7/2))/(9*a*x^8) + (4*b*(a*x + b*x^2)^(7/2))/(63*a^2*x^7)

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a +
b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d -
b*e))), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a
*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a x+b x^2\right )^{7/2}}{9 a x^8}-\frac {(2 b) \int \frac {\left (a x+b x^2\right )^{5/2}}{x^7} \, dx}{9 a} \\ & = -\frac {2 \left (a x+b x^2\right )^{7/2}}{9 a x^8}+\frac {4 b \left (a x+b x^2\right )^{7/2}}{63 a^2 x^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=-\frac {2 (7 a-2 b x) (x (a+b x))^{7/2}}{63 a^2 x^8} \]

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^8,x]

[Out]

(-2*(7*a - 2*b*x)*(x*(a + b*x))^(7/2))/(63*a^2*x^8)

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {2 b x}{7}+a \right ) \left (b x +a \right )^{3} \sqrt {x \left (b x +a \right )}}{9 x^{5} a^{2}}\)	\(31\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-2 b x +7 a \right ) \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{63 a^{2} x^{7}}\)	\(33\)
default	\(-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{9 a \,x^{8}}+\frac {4 b \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{63 a^{2} x^{7}}\)	\(41\)
trager	\(-\frac {2 \left (-2 b^{4} x^{4}+a \,b^{3} x^{3}+15 a^{2} b^{2} x^{2}+19 a^{3} b x +7 a^{4}\right ) \sqrt {b \,x^{2}+a x}}{63 x^{5} a^{2}}\)	\(60\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-2 b^{4} x^{4}+a \,b^{3} x^{3}+15 a^{2} b^{2} x^{2}+19 a^{3} b x +7 a^{4}\right )}{63 x^{4} \sqrt {x \left (b x +a \right )}\, a^{2}}\)	\(63\)

[In]

int((b*x^2+a*x)^(5/2)/x^8,x,method=_RETURNVERBOSE)

[Out]

-2/9*(-2/7*b*x+a)*(b*x+a)^3*(x*(b*x+a))^(1/2)/x^5/a^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\frac {2 \, {\left (2 \, b^{4} x^{4} - a b^{3} x^{3} - 15 \, a^{2} b^{2} x^{2} - 19 \, a^{3} b x - 7 \, a^{4}\right )} \sqrt {b x^{2} + a x}}{63 \, a^{2} x^{5}} \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^8,x, algorithm="fricas")

[Out]

2/63*(2*b^4*x^4 - a*b^3*x^3 - 15*a^2*b^2*x^2 - 19*a^3*b*x - 7*a^4)*sqrt(b*x^2 + a*x)/(a^2*x^5)

Sympy [F]

\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{8}}\, dx \]

[In]

integrate((b*x**2+a*x)**(5/2)/x**8,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**8, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (40) = 80\).

Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.79 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\frac {4 \, \sqrt {b x^{2} + a x} b^{4}}{63 \, a^{2} x} - \frac {2 \, \sqrt {b x^{2} + a x} b^{3}}{63 \, a x^{2}} + \frac {\sqrt {b x^{2} + a x} b^{2}}{42 \, x^{3}} - \frac {5 \, \sqrt {b x^{2} + a x} a b}{252 \, x^{4}} - \frac {5 \, \sqrt {b x^{2} + a x} a^{2}}{36 \, x^{5}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{12 \, x^{6}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{2 \, x^{7}} \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^8,x, algorithm="maxima")

[Out]

4/63*sqrt(b*x^2 + a*x)*b^4/(a^2*x) - 2/63*sqrt(b*x^2 + a*x)*b^3/(a*x^2) + 1/42*sqrt(b*x^2 + a*x)*b^2/x^3 - 5/2
52*sqrt(b*x^2 + a*x)*a*b/x^4 - 5/36*sqrt(b*x^2 + a*x)*a^2/x^5 + 5/12*(b*x^2 + a*x)^(3/2)*a/x^6 - 1/2*(b*x^2 +
a*x)^(5/2)/x^7

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (40) = 80\).

Time = 0.29 (sec) , antiderivative size = 223, normalized size of antiderivative = 4.65 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\frac {2 \, {\left (63 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{7} b^{\frac {7}{2}} + 273 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{6} a b^{3} + 567 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5} a^{2} b^{\frac {5}{2}} + 693 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a^{3} b^{2} + 525 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a^{4} b^{\frac {3}{2}} + 243 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{5} b + 63 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{6} \sqrt {b} + 7 \, a^{7}\right )}}{63 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{9}} \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^8,x, algorithm="giac")

[Out]

2/63*(63*(sqrt(b)*x - sqrt(b*x^2 + a*x))^7*b^(7/2) + 273*(sqrt(b)*x - sqrt(b*x^2 + a*x))^6*a*b^3 + 567*(sqrt(b
)*x - sqrt(b*x^2 + a*x))^5*a^2*b^(5/2) + 693*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*a^3*b^2 + 525*(sqrt(b)*x - sqrt
(b*x^2 + a*x))^3*a^4*b^(3/2) + 243*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^5*b + 63*(sqrt(b)*x - sqrt(b*x^2 + a*x)
)*a^6*sqrt(b) + 7*a^7)/(sqrt(b)*x - sqrt(b*x^2 + a*x))^9

Mupad [B] (verification not implemented)

Time = 9.80 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.10 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^8} \, dx=\frac {4\,b^4\,\sqrt {b\,x^2+a\,x}}{63\,a^2\,x}-\frac {10\,b^2\,\sqrt {b\,x^2+a\,x}}{21\,x^3}-\frac {2\,b^3\,\sqrt {b\,x^2+a\,x}}{63\,a\,x^2}-\frac {2\,a^2\,\sqrt {b\,x^2+a\,x}}{9\,x^5}-\frac {38\,a\,b\,\sqrt {b\,x^2+a\,x}}{63\,x^4} \]

[In]

int((a*x + b*x^2)^(5/2)/x^8,x)

[Out]

(4*b^4*(a*x + b*x^2)^(1/2))/(63*a^2*x) - (10*b^2*(a*x + b*x^2)^(1/2))/(21*x^3) - (2*b^3*(a*x + b*x^2)^(1/2))/(
63*a*x^2) - (2*a^2*(a*x + b*x^2)^(1/2))/(9*x^5) - (38*a*b*(a*x + b*x^2)^(1/2))/(63*x^4)