3.16 \(\int \frac{1}{(b x+c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{256 c^2 (b+2 c x)}{15 b^6 \sqrt{b x+c x^2}}+\frac{32 c (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}} \]

[Out]

(-2*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/2)) + (32*c*(b + 2*c*x))/(15*b^4*(b*x + c*x^2)^(3/2)) - (256*c^2*(b +
2*c*x))/(15*b^6*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0184755, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {614, 613} \[ -\frac{256 c^2 (b+2 c x)}{15 b^6 \sqrt{b x+c x^2}}+\frac{32 c (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-7/2),x]

[Out]

(-2*(b + 2*c*x))/(5*b^2*(b*x + c*x^2)^(5/2)) + (32*c*(b + 2*c*x))/(15*b^4*(b*x + c*x^2)^(3/2)) - (256*c^2*(b +
2*c*x))/(15*b^6*Sqrt[b*x + c*x^2])

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (b x+c x^2\right )^{7/2}} \, dx &=-\frac{2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}-\frac{(16 c) \int \frac{1}{\left (b x+c x^2\right )^{5/2}} \, dx}{5 b^2}\\ &=-\frac{2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}+\frac{32 c (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}+\frac{\left (128 c^2\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 b^4}\\ &=-\frac{2 (b+2 c x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}+\frac{32 c (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{256 c^2 (b+2 c x)}{15 b^6 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0248406, size = 70, normalized size = 0.84 \[ -\frac{2 \left (80 b^3 c^2 x^2+480 b^2 c^3 x^3-10 b^4 c x+3 b^5+640 b c^4 x^4+256 c^5 x^5\right )}{15 b^6 (x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-7/2),x]

[Out]

(-2*(3*b^5 - 10*b^4*c*x + 80*b^3*c^2*x^2 + 480*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x^5))/(15*b^6*(x*(b + c*x
))^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 75, normalized size = 0.9 \begin{align*} -{\frac{2\,x \left ( cx+b \right ) \left ( 256\,{c}^{5}{x}^{5}+640\,{c}^{4}{x}^{4}b+480\,{c}^{3}{x}^{3}{b}^{2}+80\,{c}^{2}{x}^{2}{b}^{3}-10\,cx{b}^{4}+3\,{b}^{5} \right ) }{15\,{b}^{6}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^(7/2),x)

[Out]

-2/15*x*(c*x+b)*(256*c^5*x^5+640*b*c^4*x^4+480*b^2*c^3*x^3+80*b^3*c^2*x^2-10*b^4*c*x+3*b^5)/b^6/(c*x^2+b*x)^(7
/2)

________________________________________________________________________________________

Maxima [A]  time = 1.30025, size = 150, normalized size = 1.81 \begin{align*} -\frac{4 \, c x}{5 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}} b^{2}} + \frac{64 \, c^{2} x}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{4}} - \frac{512 \, c^{3} x}{15 \, \sqrt{c x^{2} + b x} b^{6}} - \frac{2}{5 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}} b} + \frac{32 \, c}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{3}} - \frac{256 \, c^{2}}{15 \, \sqrt{c x^{2} + b x} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="maxima")

[Out]

-4/5*c*x/((c*x^2 + b*x)^(5/2)*b^2) + 64/15*c^2*x/((c*x^2 + b*x)^(3/2)*b^4) - 512/15*c^3*x/(sqrt(c*x^2 + b*x)*b
^6) - 2/5/((c*x^2 + b*x)^(5/2)*b) + 32/15*c/((c*x^2 + b*x)^(3/2)*b^3) - 256/15*c^2/(sqrt(c*x^2 + b*x)*b^5)

________________________________________________________________________________________

Fricas [A]  time = 2.22589, size = 223, normalized size = 2.69 \begin{align*} -\frac{2 \,{\left (256 \, c^{5} x^{5} + 640 \, b c^{4} x^{4} + 480 \, b^{2} c^{3} x^{3} + 80 \, b^{3} c^{2} x^{2} - 10 \, b^{4} c x + 3 \, b^{5}\right )} \sqrt{c x^{2} + b x}}{15 \,{\left (b^{6} c^{3} x^{6} + 3 \, b^{7} c^{2} x^{5} + 3 \, b^{8} c x^{4} + b^{9} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(256*c^5*x^5 + 640*b*c^4*x^4 + 480*b^2*c^3*x^3 + 80*b^3*c^2*x^2 - 10*b^4*c*x + 3*b^5)*sqrt(c*x^2 + b*x)/
(b^6*c^3*x^6 + 3*b^7*c^2*x^5 + 3*b^8*c*x^4 + b^9*x^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b x + c x^{2}\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**(7/2),x)

[Out]

Integral((b*x + c*x**2)**(-7/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.30193, size = 100, normalized size = 1.2 \begin{align*} -\frac{2 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (4 \, x{\left (\frac{2 \, c^{5} x}{b^{6}} + \frac{5 \, c^{4}}{b^{5}}\right )} + \frac{15 \, c^{3}}{b^{4}}\right )} x + \frac{5 \, c^{2}}{b^{3}}\right )} x - \frac{5 \, c}{b^{2}}\right )} x + \frac{3}{b}\right )}}{15 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(7/2),x, algorithm="giac")

[Out]

-2/15*(2*(8*(2*(4*x*(2*c^5*x/b^6 + 5*c^4/b^5) + 15*c^3/b^4)*x + 5*c^2/b^3)*x - 5*c/b^2)*x + 3/b)/(c*x^2 + b*x)
^(5/2)