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*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)

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Rubi [A]  time = 0.0388309, antiderivative size = 100, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0136753, size = 51, normalized size = 0.51 $\frac{2 \sqrt{x (b+c x)} \left (6 b^2 c x-5 b^3-8 b c^2 x^2+16 c^3 x^3\right )}{35 b^4 x^4}$

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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Maple [A]  time = 0.046, size = 55, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -16\,{x}^{3}{c}^{3}+8\,b{x}^{2}{c}^{2}-6\,{b}^{2}xc+5\,{b}^{3} \right ) }{35\,{x}^{3}{b}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.88261, size = 109, normalized size = 1.09 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \sqrt{x \left (b + c x\right )}}\, dx \end{align*}

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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Giac [A]  time = 1.33767, size = 144, normalized size = 1.44 \begin{align*} \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}} \end{align*}

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