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

Optimal. Leaf size=48 \[ \frac {4 b \left (a x+b x^2\right )^{7/2}}{63 a^2 x^7}-\frac {2 \left (a x+b x^2\right )^{7/2}}{9 a x^8} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {658, 650} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 650

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 658

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*} \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 {(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*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.75 \begin {gather*} \frac {2 (a+b x)^3 \sqrt {x (a+b x)} (2 b x-7 a)}{63 a^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^8,x]

[Out]

(2*(a + b*x)^3*Sqrt[x*(a + b*x)]*(-7*a + 2*b*x))/(63*a^2*x^5)

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IntegrateAlgebraic [A]  time = 0.32, size = 64, normalized size = 1.33 \begin {gather*} \frac {2 \sqrt {a x+b x^2} \left (-7 a^4-19 a^3 b x-15 a^2 b^2 x^2-a b^3 x^3+2 b^4 x^4\right )}{63 a^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*x + b*x^2)^(5/2)/x^8,x]

[Out]

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

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fricas [A]  time = 0.40, size = 60, normalized size = 1.25 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.21, size = 223, normalized size = 4.65 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.05, size = 33, normalized size = 0.69 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(b*x+a)*(-2*b*x+7*a)*(b*x^2+a*x)^(5/2)/a^2/x^7

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maxima [B]  time = 1.38, size = 134, normalized size = 2.79 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.01, size = 101, normalized size = 2.10 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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