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]

-2/5*(2*c*x+b)/b^2/(c*x^2+b*x)^(5/2)+32/15*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(3/2)-256/15*c^2*(2*c*x+b)/b^6/(c*x^2+b
*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 70, normalized size = 0.84 \[ -\frac {2 \left (3 b^5-10 b^4 c x+80 b^3 c^2 x^2+480 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )}{15 b^6 (x (b+c x))^{5/2}} \]

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))

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fricas [A]  time = 0.80, size = 105, normalized size = 1.27 \[ -\frac {2 \, {\left (256 \, c^{5} x^{5} + 640 \, b c^{4} x^{4} + 480 \, b^{2} c^{3} x^{3} + 80 \, b^{3} c^{2} x^{2} - 10 \, b^{4} c x + 3 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{6} c^{3} x^{6} + 3 \, b^{7} c^{2} x^{5} + 3 \, b^{8} c x^{4} + b^{9} x^{3}\right )}} \]

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)

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giac [A]  time = 0.49, size = 74, normalized size = 0.89 \[ -\frac {2 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, c^{5} x}{b^{6}} + \frac {5 \, c^{4}}{b^{5}}\right )} + \frac {15 \, c^{3}}{b^{4}}\right )} x + \frac {5 \, c^{2}}{b^{3}}\right )} x - \frac {5 \, c}{b^{2}}\right )} x + \frac {3}{b}\right )}}{15 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \]

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)

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maple [A]  time = 0.05, size = 75, normalized size = 0.90 \[ -\frac {2 \left (c x +b \right ) \left (256 c^{5} x^{5}+640 c^{4} x^{4} b +480 c^{3} x^{3} b^{2}+80 c^{2} x^{2} b^{3}-10 c x \,b^{4}+3 b^{5}\right ) x}{15 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(c*x+b)*x*(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)

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maxima [A]  time = 1.34, size = 111, normalized size = 1.34 \[ -\frac {4 \, c x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2}} + \frac {64 \, c^{2} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} - \frac {512 \, c^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{6}} - \frac {2}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b} + \frac {32 \, c}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} - \frac {256 \, c^{2}}{15 \, \sqrt {c x^{2} + b x} b^{5}} \]

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)

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mupad [B]  time = 0.27, size = 96, normalized size = 1.16 \[ -\frac {6\,b^5+256\,b\,c^2\,{\left (c\,x^2+b\,x\right )}^2+512\,c^3\,x\,{\left (c\,x^2+b\,x\right )}^2-32\,b^3\,c\,\left (c\,x^2+b\,x\right )+12\,b^4\,c\,x-64\,b^2\,c^2\,x\,\left (c\,x^2+b\,x\right )}{15\,b^6\,{\left (c\,x^2+b\,x\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(7/2),x)

[Out]

-(6*b^5 + 256*b*c^2*(b*x + c*x^2)^2 + 512*c^3*x*(b*x + c*x^2)^2 - 32*b^3*c*(b*x + c*x^2) + 12*b^4*c*x - 64*b^2
*c^2*x*(b*x + c*x^2))/(15*b^6*(b*x + c*x^2)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x + c x^{2}\right )^{\frac {7}{2}}}\, dx \]

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)

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