3.50 \(\int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=100 \[ \frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {2 \sqrt {b x+c x^2}}{7 b x^4} \]

[Out]

-2/7*(c*x^2+b*x)^(1/2)/b/x^4+12/35*c*(c*x^2+b*x)^(1/2)/b^2/x^3-16/35*c^2*(c*x^2+b*x)^(1/2)/b^3/x^2+32/35*c^3*(
c*x^2+b*x)^(1/2)/b^4/x

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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} \[ \frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {2 \sqrt {b x+c x^2}}{7 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[b*x + c*x^2])/(7*b*x^4) + (12*c*Sqrt[b*x + c*x^2])/(35*b^2*x^3) - (16*c^2*Sqrt[b*x + c*x^2])/(35*b^3*
x^2) + (32*c^3*Sqrt[b*x + c*x^2])/(35*b^4*x)

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 {1}{x^4 \sqrt {b x+c x^2}} \, dx &=-\frac {2 \sqrt {b x+c x^2}}{7 b x^4}-\frac {(6 c) \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx}{7 b}\\ &=-\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}+\frac {\left (24 c^2\right ) \int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx}{35 b^2}\\ &=-\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}-\frac {\left (16 c^3\right ) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{35 b^3}\\ &=-\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 51, normalized size = 0.51 \[ \frac {2 \sqrt {x (b+c x)} \left (-5 b^3+6 b^2 c x-8 b c^2 x^2+16 c^3 x^3\right )}{35 b^4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[b*x + c*x^2]),x]

[Out]

(2*Sqrt[x*(b + c*x)]*(-5*b^3 + 6*b^2*c*x - 8*b*c^2*x^2 + 16*c^3*x^3))/(35*b^4*x^4)

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fricas [A]  time = 0.67, size = 49, normalized size = 0.49 \[ \frac {2 \, {\left (16 \, c^{3} x^{3} - 8 \, b c^{2} x^{2} + 6 \, b^{2} c x - 5 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{35 \, b^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*c^3*x^3 - 8*b*c^2*x^2 + 6*b^2*c*x - 5*b^3)*sqrt(c*x^2 + b*x)/(b^4*x^4)

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giac [A]  time = 0.21, size = 107, normalized size = 1.07 \[ \frac {2 \, {\left (70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{\frac {3}{2}} + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + 5 \, b^{3}\right )}}{35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2/35*(70*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^(3/2) + 84*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c + 35*(sqrt(c)*x
- sqrt(c*x^2 + b*x))*b^2*sqrt(c) + 5*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^7

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maple [A]  time = 0.05, size = 55, normalized size = 0.55 \[ -\frac {2 \left (c x +b \right ) \left (-16 x^{3} c^{3}+8 b \,x^{2} c^{2}-6 b^{2} x c +5 b^{3}\right )}{35 \sqrt {c \,x^{2}+b x}\, b^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(c*x^2+b*x)^(1/2),x)

[Out]

-2/35*(c*x+b)*(-16*c^3*x^3+8*b*c^2*x^2-6*b^2*c*x+5*b^3)/x^3/b^4/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 1.37, size = 84, normalized size = 0.84 \[ \frac {32 \, \sqrt {c x^{2} + b x} c^{3}}{35 \, b^{4} x} - \frac {16 \, \sqrt {c x^{2} + b x} c^{2}}{35 \, b^{3} x^{2}} + \frac {12 \, \sqrt {c x^{2} + b x} c}{35 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x}}{7 \, b x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

32/35*sqrt(c*x^2 + b*x)*c^3/(b^4*x) - 16/35*sqrt(c*x^2 + b*x)*c^2/(b^3*x^2) + 12/35*sqrt(c*x^2 + b*x)*c/(b^2*x
^3) - 2/7*sqrt(c*x^2 + b*x)/(b*x^4)

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mupad [B]  time = 0.20, size = 84, normalized size = 0.84 \[ \frac {32\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x}-\frac {16\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b^3\,x^2}-\frac {2\,\sqrt {c\,x^2+b\,x}}{7\,b\,x^4}+\frac {12\,c\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(b*x + c*x^2)^(1/2)),x)

[Out]

(32*c^3*(b*x + c*x^2)^(1/2))/(35*b^4*x) - (16*c^2*(b*x + c*x^2)^(1/2))/(35*b^3*x^2) - (2*(b*x + c*x^2)^(1/2))/
(7*b*x^4) + (12*c*(b*x + c*x^2)^(1/2))/(35*b^2*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \sqrt {x \left (b + c x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(1/(x**4*sqrt(x*(b + c*x))), x)

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