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

Optimal. Leaf size=135 \[ -\frac{128 c (b+2 c x) (2 c d-b e)}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}+\frac{16 (b+2 c x) (2 c d-b e)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 (-2 a e+x (2 c d-b e)+b d)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0383647, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {638, 614, 613} \[ -\frac{128 c (b+2 c x) (2 c d-b e)}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}+\frac{16 (b+2 c x) (2 c d-b e)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 (-2 a e+x (2 c d-b e)+b d)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.248946, size = 119, normalized size = 0.88 \[ \frac{2 \left (3 \left (b^2-4 a c\right )^2 (2 a e-b d+b e x-2 c d x)-8 \left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x)) (b e-2 c d)+64 c (b+2 c x) (a+x (b+c x))^2 (b e-2 c d)\right )}{15 \left (b^2-4 a c\right )^3 (a+x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(b^2 - 4*a*c)^2*(-(b*d) + 2*a*e - 2*c*d*x + b*e*x) - 8*(b^2 - 4*a*c)*(-2*c*d + b*e)*(b + 2*c*x)*(a + x*(
b + c*x)) + 64*c*(-2*c*d + b*e)*(b + 2*c*x)*(a + x*(b + c*x))^2))/(15*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^(5/2))

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Maple [B]  time = 0.007, size = 288, normalized size = 2.1 \begin{align*} -{\frac{256\,b{c}^{4}e{x}^{5}-512\,{c}^{5}d{x}^{5}+640\,{b}^{2}{c}^{3}e{x}^{4}-1280\,b{c}^{4}d{x}^{4}+640\,ab{c}^{3}e{x}^{3}-1280\,a{c}^{4}d{x}^{3}+480\,{b}^{3}{c}^{2}e{x}^{3}-960\,{b}^{2}{c}^{3}d{x}^{3}+960\,a{b}^{2}{c}^{2}e{x}^{2}-1920\,ab{c}^{3}d{x}^{2}+80\,{b}^{4}ce{x}^{2}-160\,{b}^{3}{c}^{2}d{x}^{2}+480\,{a}^{2}b{c}^{2}ex-960\,{a}^{2}{c}^{3}dx+240\,a{b}^{3}cex-480\,a{b}^{2}{c}^{2}dx-10\,{b}^{5}ex+20\,{b}^{4}cdx+192\,{a}^{3}{c}^{2}e+96\,{a}^{2}{b}^{2}ce-480\,{a}^{2}b{c}^{2}d-4\,a{b}^{4}e+80\,a{b}^{3}cd-6\,{b}^{5}d}{960\,{a}^{3}{c}^{3}-720\,{a}^{2}{b}^{2}{c}^{2}+180\,a{b}^{4}c-15\,{b}^{6}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)/(c*x^2+b*x+a)^(7/2),x)

[Out]

-2/15/(c*x^2+b*x+a)^(5/2)*(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+320*a*b*c^3*e*x^3-6
40*a*c^4*d*x^3+240*b^3*c^2*e*x^3-480*b^2*c^3*d*x^3+480*a*b^2*c^2*e*x^2-960*a*b*c^3*d*x^2+40*b^4*c*e*x^2-80*b^3
*c^2*d*x^2+240*a^2*b*c^2*e*x-480*a^2*c^3*d*x+120*a*b^3*c*e*x-240*a*b^2*c^2*d*x-5*b^5*e*x+10*b^4*c*d*x+96*a^3*c
^2*e+48*a^2*b^2*c*e-240*a^2*b*c^2*d-2*a*b^4*e+40*a*b^3*c*d-3*b^5*d)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 26.3252, size = 1170, normalized size = 8.67 \begin{align*} -\frac{2 \,{\left (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} + 80 \,{\left (2 \,{\left (3 \, b^{2} c^{3} + 4 \, a c^{4}\right )} d -{\left (3 \, b^{3} c^{2} + 4 \, a b c^{3}\right )} e\right )} x^{3} + 40 \,{\left (2 \,{\left (b^{3} c^{2} + 12 \, a b c^{3}\right )} d -{\left (b^{4} c + 12 \, a b^{2} c^{2}\right )} e\right )} x^{2} +{\left (3 \, b^{5} - 40 \, a b^{3} c + 240 \, a^{2} b c^{2}\right )} d + 2 \,{\left (a b^{4} - 24 \, a^{2} b^{2} c - 48 \, a^{3} c^{2}\right )} e - 5 \,{\left (2 \,{\left (b^{4} c - 24 \, a b^{2} c^{2} - 48 \, a^{2} c^{3}\right )} d -{\left (b^{5} - 24 \, a b^{3} c - 48 \, a^{2} b c^{2}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{15 \,{\left (a^{3} b^{6} - 12 \, a^{4} b^{4} c + 48 \, a^{5} b^{2} c^{2} - 64 \, a^{6} c^{3} +{\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{6} + 3 \,{\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{5} + 3 \,{\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} x^{4} +{\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} x^{3} + 3 \,{\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(128*(2*c^5*d - b*c^4*e)*x^5 + 320*(2*b*c^4*d - b^2*c^3*e)*x^4 + 80*(2*(3*b^2*c^3 + 4*a*c^4)*d - (3*b^3*
c^2 + 4*a*b*c^3)*e)*x^3 + 40*(2*(b^3*c^2 + 12*a*b*c^3)*d - (b^4*c + 12*a*b^2*c^2)*e)*x^2 + (3*b^5 - 40*a*b^3*c
+ 240*a^2*b*c^2)*d + 2*(a*b^4 - 24*a^2*b^2*c - 48*a^3*c^2)*e - 5*(2*(b^4*c - 24*a*b^2*c^2 - 48*a^2*c^3)*d - (
b^5 - 24*a*b^3*c - 48*a^2*b*c^2)*e)*x)*sqrt(c*x^2 + b*x + a)/(a^3*b^6 - 12*a^4*b^4*c + 48*a^5*b^2*c^2 - 64*a^6
*c^3 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^6 + 3*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4
- 64*a^3*b*c^5)*x^5 + 3*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*x^4 + (b^9 - 6*
a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*x^3 + 3*(a*b^8 - 11*a^2*b^6*c + 36*a^3*b^4*c^2 - 1
6*a^4*b^2*c^3 - 64*a^5*c^4)*x^2 + 3*(a^2*b^7 - 12*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.23579, size = 653, normalized size = 4.84 \begin{align*} -\frac{{\left (8 \,{\left (2 \,{\left (4 \,{\left (\frac{2 \,{\left (2 \, c^{5} d - b c^{4} e\right )} x}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}} + \frac{5 \,{\left (2 \, b c^{4} d - b^{2} c^{3} e\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{5 \,{\left (6 \, b^{2} c^{3} d + 8 \, a c^{4} d - 3 \, b^{3} c^{2} e - 4 \, a b c^{3} e\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{5 \,{\left (2 \, b^{3} c^{2} d + 24 \, a b c^{3} d - b^{4} c e - 12 \, a b^{2} c^{2} e\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x - \frac{5 \,{\left (2 \, b^{4} c d - 48 \, a b^{2} c^{2} d - 96 \, a^{2} c^{3} d - b^{5} e + 24 \, a b^{3} c e + 48 \, a^{2} b c^{2} e\right )}}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}\right )} x + \frac{3 \, b^{5} d - 40 \, a b^{3} c d + 240 \, a^{2} b c^{2} d + 2 \, a b^{4} e - 48 \, a^{2} b^{2} c e - 96 \, a^{3} c^{2} e}{b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}}}{15 \,{\left (c x^{2} + b x + a\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+a)^(7/2),x, algorithm="giac")

[Out]

-1/15*((8*(2*(4*(2*(2*c^5*d - b*c^4*e)*x/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6) + 5*(2*b*c^4*d
- b^2*c^3*e)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x + 5*(6*b^2*c^3*d + 8*a*c^4*d - 3*b^3*c
^2*e - 4*a*b*c^3*e)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x + 5*(2*b^3*c^2*d + 24*a*b*c^3*d
- b^4*c*e - 12*a*b^2*c^2*e)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))*x - 5*(2*b^4*c*d - 48*a*b^
2*c^2*d - 96*a^2*c^3*d - b^5*e + 24*a*b^3*c*e + 48*a^2*b*c^2*e)/(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*
a^3*c^6))*x + (3*b^5*d - 40*a*b^3*c*d + 240*a^2*b*c^2*d + 2*a*b^4*e - 48*a^2*b^2*c*e - 96*a^3*c^2*e)/(b^6*c^3
- 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6))/(c*x^2 + b*x + a)^(5/2)