3.138 \(\int \frac{d+e x}{(b x+c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=107 \[ -\frac{128 c (b+2 c x) (2 c d-b e)}{15 b^6 \sqrt{b x+c x^2}}+\frac{16 (b+2 c x) (2 c d-b e)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (x (2 c d-b e)+b d)}{5 b^2 \left (b x+c x^2\right )^{5/2}} \]

[Out]

(-2*(b*d + (2*c*d - b*e)*x))/(5*b^2*(b*x + c*x^2)^(5/2)) + (16*(2*c*d - b*e)*(b + 2*c*x))/(15*b^4*(b*x + c*x^2
)^(3/2)) - (128*c*(2*c*d - b*e)*(b + 2*c*x))/(15*b^6*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.0313922, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {638, 614, 613} \[ -\frac{128 c (b+2 c x) (2 c d-b e)}{15 b^6 \sqrt{b x+c x^2}}+\frac{16 (b+2 c x) (2 c d-b e)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{2 (x (2 c d-b e)+b d)}{5 b^2 \left (b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(b*x + c*x^2)^(7/2),x]

[Out]

(-2*(b*d + (2*c*d - b*e)*x))/(5*b^2*(b*x + c*x^2)^(5/2)) + (16*(2*c*d - b*e)*(b + 2*c*x))/(15*b^4*(b*x + c*x^2
)^(3/2)) - (128*c*(2*c*d - b*e)*(b + 2*c*x))/(15*b^6*Sqrt[b*x + c*x^2])

Rule 638

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

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{d+e x}{\left (b x+c x^2\right )^{7/2}} \, dx &=-\frac{2 (b d+(2 c d-b e) x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}-\frac{(8 (2 c d-b e)) \int \frac{1}{\left (b x+c x^2\right )^{5/2}} \, dx}{5 b^2}\\ &=-\frac{2 (b d+(2 c d-b e) x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}+\frac{16 (2 c d-b e) (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}+\frac{(64 c (2 c d-b e)) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{15 b^4}\\ &=-\frac{2 (b d+(2 c d-b e) x)}{5 b^2 \left (b x+c x^2\right )^{5/2}}+\frac{16 (2 c d-b e) (b+2 c x)}{15 b^4 \left (b x+c x^2\right )^{3/2}}-\frac{128 c (2 c d-b e) (b+2 c x)}{15 b^6 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0436897, size = 105, normalized size = 0.98 \[ -\frac{2 \left (80 b^3 c^2 x^2 (d-3 e x)+160 b^2 c^3 x^3 (3 d-2 e x)-10 b^4 c x (d+4 e x)+b^5 (3 d+5 e x)-128 b c^4 x^4 (e x-5 d)+256 c^5 d x^5\right )}{15 b^6 (x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(b*x + c*x^2)^(7/2),x]

[Out]

(-2*(256*c^5*d*x^5 + 80*b^3*c^2*x^2*(d - 3*e*x) + 160*b^2*c^3*x^3*(3*d - 2*e*x) - 128*b*c^4*x^4*(-5*d + e*x) -
10*b^4*c*x*(d + 4*e*x) + b^5*(3*d + 5*e*x)))/(15*b^6*(x*(b + c*x))^(5/2))

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Maple [A]  time = 0.004, size = 132, normalized size = 1.2 \begin{align*} -{\frac{2\,x \left ( cx+b \right ) \left ( -128\,b{c}^{4}e{x}^{5}+256\,{c}^{5}d{x}^{5}-320\,{b}^{2}{c}^{3}e{x}^{4}+640\,b{c}^{4}d{x}^{4}-240\,{b}^{3}{c}^{2}e{x}^{3}+480\,{b}^{2}{c}^{3}d{x}^{3}-40\,{b}^{4}ce{x}^{2}+80\,{b}^{3}{c}^{2}d{x}^{2}+5\,{b}^{5}ex-10\,{b}^{4}cdx+3\,d{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((e*x+d)/(c*x^2+b*x)^(7/2),x)

[Out]

-2/15*x*(c*x+b)*(-128*b*c^4*e*x^5+256*c^5*d*x^5-320*b^2*c^3*e*x^4+640*b*c^4*d*x^4-240*b^3*c^2*e*x^3+480*b^2*c^
3*d*x^3-40*b^4*c*e*x^2+80*b^3*c^2*d*x^2+5*b^5*e*x-10*b^4*c*d*x+3*b^5*d)/b^6/(c*x^2+b*x)^(7/2)

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Maxima [B]  time = 1.11062, size = 284, normalized size = 2.65 \begin{align*} -\frac{4 \, c d x}{5 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}} b^{2}} + \frac{64 \, c^{2} d x}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{4}} - \frac{512 \, c^{3} d x}{15 \, \sqrt{c x^{2} + b x} b^{6}} + \frac{2 \, e x}{5 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}} b} - \frac{32 \, c e x}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{3}} + \frac{256 \, c^{2} e x}{15 \, \sqrt{c x^{2} + b x} b^{5}} - \frac{2 \, d}{5 \,{\left (c x^{2} + b x\right )}^{\frac{5}{2}} b} + \frac{32 \, c d}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{3}} - \frac{256 \, c^{2} d}{15 \, \sqrt{c x^{2} + b x} b^{5}} - \frac{16 \, e}{15 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{2}} + \frac{128 \, c e}{15 \, \sqrt{c x^{2} + b x} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*c*d*x/((c*x^2 + b*x)^(5/2)*b^2) + 64/15*c^2*d*x/((c*x^2 + b*x)^(3/2)*b^4) - 512/15*c^3*d*x/(sqrt(c*x^2 +
b*x)*b^6) + 2/5*e*x/((c*x^2 + b*x)^(5/2)*b) - 32/15*c*e*x/((c*x^2 + b*x)^(3/2)*b^3) + 256/15*c^2*e*x/(sqrt(c*x
^2 + b*x)*b^5) - 2/5*d/((c*x^2 + b*x)^(5/2)*b) + 32/15*c*d/((c*x^2 + b*x)^(3/2)*b^3) - 256/15*c^2*d/(sqrt(c*x^
2 + b*x)*b^5) - 16/15*e/((c*x^2 + b*x)^(3/2)*b^2) + 128/15*c*e/(sqrt(c*x^2 + b*x)*b^4)

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Fricas [A]  time = 1.87097, size = 335, normalized size = 3.13 \begin{align*} -\frac{2 \,{\left (3 \, b^{5} d + 128 \,{\left (2 \, c^{5} d - b c^{4} e\right )} x^{5} + 320 \,{\left (2 \, b c^{4} d - b^{2} c^{3} e\right )} x^{4} + 240 \,{\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )} x^{3} + 40 \,{\left (2 \, b^{3} c^{2} d - b^{4} c e\right )} x^{2} - 5 \,{\left (2 \, b^{4} c d - b^{5} e\right )} x\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((e*x+d)/(c*x^2+b*x)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(3*b^5*d + 128*(2*c^5*d - b*c^4*e)*x^5 + 320*(2*b*c^4*d - b^2*c^3*e)*x^4 + 240*(2*b^2*c^3*d - b^3*c^2*e)
*x^3 + 40*(2*b^3*c^2*d - b^4*c*e)*x^2 - 5*(2*b^4*c*d - b^5*e)*x)*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x}{\left (x \left (b + c x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)/(x*(b + c*x))**(7/2), x)

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Giac [A]  time = 1.21587, size = 223, normalized size = 2.08 \begin{align*} -\frac{{\left (8 \,{\left (2 \,{\left (4 \, x{\left (\frac{2 \,{\left (2 \, c^{5} d - b c^{4} e\right )} x}{b^{6} c^{3}} + \frac{5 \,{\left (2 \, b c^{4} d - b^{2} c^{3} e\right )}}{b^{6} c^{3}}\right )} + \frac{15 \,{\left (2 \, b^{2} c^{3} d - b^{3} c^{2} e\right )}}{b^{6} c^{3}}\right )} x + \frac{5 \,{\left (2 \, b^{3} c^{2} d - b^{4} c e\right )}}{b^{6} c^{3}}\right )} x - \frac{5 \,{\left (2 \, b^{4} c d - b^{5} e\right )}}{b^{6} c^{3}}\right )} x + \frac{3 \, d}{b c^{3}}}{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((e*x+d)/(c*x^2+b*x)^(7/2),x, algorithm="giac")

[Out]

-1/15*((8*(2*(4*x*(2*(2*c^5*d - b*c^4*e)*x/(b^6*c^3) + 5*(2*b*c^4*d - b^2*c^3*e)/(b^6*c^3)) + 15*(2*b^2*c^3*d
- b^3*c^2*e)/(b^6*c^3))*x + 5*(2*b^3*c^2*d - b^4*c*e)/(b^6*c^3))*x - 5*(2*b^4*c*d - b^5*e)/(b^6*c^3))*x + 3*d/
(b*c^3))/(c*x^2 + b*x)^(5/2)