3.1.22 \(\int \frac {(b x+c x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=100 \[ \frac {32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac {4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8} \]

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Rubi [A]  time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac {4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^8,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(11*b*x^8) + (4*c*(b*x + c*x^2)^(5/2))/(33*b^2*x^7) - (16*c^2*(b*x + c*x^2)^(5/2))/(2
31*b^3*x^6) + (32*c^3*(b*x + c*x^2)^(5/2))/(1155*b^4*x^5)

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 (b x+c x^2\right )^{3/2}}{x^8} \, dx &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}-\frac {(6 c) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{11 b}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac {4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}+\frac {\left (8 c^2\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{33 b^2}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac {4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}-\frac {\left (16 c^3\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{231 b^3}\\ &=-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac {4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac {16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac {32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 0.51 \begin {gather*} \frac {2 (x (b+c x))^{5/2} \left (-105 b^3+70 b^2 c x-40 b c^2 x^2+16 c^3 x^3\right )}{1155 b^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^8,x]

[Out]

(2*(x*(b + c*x))^(5/2)*(-105*b^3 + 70*b^2*c*x - 40*b*c^2*x^2 + 16*c^3*x^3))/(1155*b^4*x^8)

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IntegrateAlgebraic [A]  time = 0.27, size = 75, normalized size = 0.75 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-105 b^5-140 b^4 c x-5 b^3 c^2 x^2+6 b^2 c^3 x^3-8 b c^4 x^4+16 c^5 x^5\right )}{1155 b^4 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x + c*x^2)^(3/2)/x^8,x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-105*b^5 - 140*b^4*c*x - 5*b^3*c^2*x^2 + 6*b^2*c^3*x^3 - 8*b*c^4*x^4 + 16*c^5*x^5))/(115
5*b^4*x^6)

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fricas [A]  time = 0.41, size = 71, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (16 \, c^{5} x^{5} - 8 \, b c^{4} x^{4} + 6 \, b^{2} c^{3} x^{3} - 5 \, b^{3} c^{2} x^{2} - 140 \, b^{4} c x - 105 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{1155 \, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="fricas")

[Out]

2/1155*(16*c^5*x^5 - 8*b*c^4*x^4 + 6*b^2*c^3*x^3 - 5*b^3*c^2*x^2 - 140*b^4*c*x - 105*b^5)*sqrt(c*x^2 + b*x)/(b
^4*x^6)

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giac [B]  time = 0.21, size = 223, normalized size = 2.23 \begin {gather*} \frac {2 \, {\left (2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} c^{\frac {7}{2}} + 10164 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} b c^{3} + 19635 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b^{2} c^{\frac {5}{2}} + 21285 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{3} c^{2} + 13860 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{4} c^{\frac {3}{2}} + 5390 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{5} c + 1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{6} \sqrt {c} + 105 \, b^{7}\right )}}{1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="giac")

[Out]

2/1155*(2310*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*c^(7/2) + 10164*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*b*c^3 + 19635
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^2*c^(5/2) + 21285*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^3*c^2 + 13860*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^3*b^4*c^(3/2) + 5390*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^5*c + 1155*(sqrt(c)*x - sq
rt(c*x^2 + b*x))*b^6*sqrt(c) + 105*b^7)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^11

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maple [A]  time = 0.05, size = 55, normalized size = 0.55 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-16 x^{3} c^{3}+40 b \,x^{2} c^{2}-70 b^{2} x c +105 b^{3}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{1155 b^{4} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^8,x)

[Out]

-2/1155*(c*x+b)*(-16*c^3*x^3+40*b*c^2*x^2-70*b^2*c*x+105*b^3)*(c*x^2+b*x)^(3/2)/x^7/b^4

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maxima [A]  time = 1.40, size = 139, normalized size = 1.39 \begin {gather*} \frac {32 \, \sqrt {c x^{2} + b x} c^{5}}{1155 \, b^{4} x} - \frac {16 \, \sqrt {c x^{2} + b x} c^{4}}{1155 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{2} + b x} c^{3}}{385 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{231 \, b x^{4}} + \frac {\sqrt {c x^{2} + b x} c}{132 \, x^{5}} + \frac {3 \, \sqrt {c x^{2} + b x} b}{44 \, x^{6}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{4 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="maxima")

[Out]

32/1155*sqrt(c*x^2 + b*x)*c^5/(b^4*x) - 16/1155*sqrt(c*x^2 + b*x)*c^4/(b^3*x^2) + 4/385*sqrt(c*x^2 + b*x)*c^3/
(b^2*x^3) - 2/231*sqrt(c*x^2 + b*x)*c^2/(b*x^4) + 1/132*sqrt(c*x^2 + b*x)*c/x^5 + 3/44*sqrt(c*x^2 + b*x)*b/x^6
- 1/4*(c*x^2 + b*x)^(3/2)/x^7

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mupad [B]  time = 1.01, size = 123, normalized size = 1.23 \begin {gather*} \frac {4\,c^3\,\sqrt {c\,x^2+b\,x}}{385\,b^2\,x^3}-\frac {8\,c\,\sqrt {c\,x^2+b\,x}}{33\,x^5}-\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}}{231\,b\,x^4}-\frac {2\,b\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {16\,c^4\,\sqrt {c\,x^2+b\,x}}{1155\,b^3\,x^2}+\frac {32\,c^5\,\sqrt {c\,x^2+b\,x}}{1155\,b^4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^8,x)

[Out]

(4*c^3*(b*x + c*x^2)^(1/2))/(385*b^2*x^3) - (8*c*(b*x + c*x^2)^(1/2))/(33*x^5) - (2*c^2*(b*x + c*x^2)^(1/2))/(
231*b*x^4) - (2*b*(b*x + c*x^2)^(1/2))/(11*x^6) - (16*c^4*(b*x + c*x^2)^(1/2))/(1155*b^3*x^2) + (32*c^5*(b*x +
c*x^2)^(1/2))/(1155*b^4*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**8,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**8, x)

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