3.2493 \(\int \frac{A+B x}{(a+b x+c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac{128 c (b+2 c x) (b B-2 A c)}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}-\frac{16 (b+2 c x) (b B-2 A c)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 (-2 a B-x (b B-2 A c)+A b)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x))/(5*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5/2)) - (16*(b*B - 2*A*c)*(b + 2*c*x)
)/(15*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) + (128*c*(b*B - 2*A*c)*(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 = 0.0373278, antiderivative size = 133, 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} \[ \frac{128 c (b+2 c x) (b B-2 A c)}{15 \left (b^2-4 a c\right )^3 \sqrt{a+b x+c x^2}}-\frac{16 (b+2 c x) (b B-2 A c)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{2 (-2 a B-x (b B-2 A c)+A b)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x))/(5*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5/2)) - (16*(b*B - 2*A*c)*(b + 2*c*x)
)/(15*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) + (128*c*(b*B - 2*A*c)*(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{A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx &=-\frac{2 (A b-2 a B-(b B-2 A c) x)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}+\frac{(8 (b B-2 A c)) \int \frac{1}{\left (a+b x+c x^2\right )^{5/2}} \, dx}{5 \left (b^2-4 a c\right )}\\ &=-\frac{2 (A b-2 a B-(b B-2 A c) x)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac{16 (b B-2 A c) (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 (b B-2 A c)) \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 (A b-2 a B-(b B-2 A c) x)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac{16 (b B-2 A c) (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 (b B-2 A c) (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 = 0.211504, size = 120, normalized size = 0.9 \[ \frac{2 \left (3 \left (b^2-4 a c\right )^2 (B (2 a+b x)-A (b+2 c x))-8 \left (b^2-4 a c\right ) (b+2 c x) (a+x (b+c x)) (b B-2 A c)+64 c (b+2 c x) (a+x (b+c x))^2 (b B-2 A c)\right )}{15 \left (b^2-4 a c\right )^3 (a+x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-8*(b^2 - 4*a*c)*(b*B - 2*A*c)*(b + 2*c*x)*(a + x*(b + c*x)) + 64*c*(b*B - 2*A*c)*(b + 2*c*x)*(a + x*(b +
c*x))^2 + 3*(b^2 - 4*a*c)^2*(B*(2*a + b*x) - A*(b + 2*c*x))))/(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 = 2.2 \begin{align*}{\frac{512\,A{c}^{5}{x}^{5}-256\,Bb{c}^{4}{x}^{5}+1280\,Ab{c}^{4}{x}^{4}-640\,B{b}^{2}{c}^{3}{x}^{4}+1280\,Aa{c}^{4}{x}^{3}+960\,A{b}^{2}{c}^{3}{x}^{3}-640\,Bab{c}^{3}{x}^{3}-480\,B{b}^{3}{c}^{2}{x}^{3}+1920\,Aab{c}^{3}{x}^{2}+160\,A{b}^{3}{c}^{2}{x}^{2}-960\,Ba{b}^{2}{c}^{2}{x}^{2}-80\,B{b}^{4}c{x}^{2}+960\,A{a}^{2}{c}^{3}x+480\,Aa{b}^{2}{c}^{2}x-20\,A{b}^{4}cx-480\,B{a}^{2}b{c}^{2}x-240\,Ba{b}^{3}cx+10\,B{b}^{5}x+480\,A{a}^{2}b{c}^{2}-80\,Aa{b}^{3}c+6\,A{b}^{5}-192\,B{a}^{3}{c}^{2}-96\,B{a}^{2}{b}^{2}c+4\,Ba{b}^{4}}{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((B*x+A)/(c*x^2+b*x+a)^(7/2),x)

[Out]

2/15/(c*x^2+b*x+a)^(5/2)*(256*A*c^5*x^5-128*B*b*c^4*x^5+640*A*b*c^4*x^4-320*B*b^2*c^3*x^4+640*A*a*c^4*x^3+480*
A*b^2*c^3*x^3-320*B*a*b*c^3*x^3-240*B*b^3*c^2*x^3+960*A*a*b*c^3*x^2+80*A*b^3*c^2*x^2-480*B*a*b^2*c^2*x^2-40*B*
b^4*c*x^2+480*A*a^2*c^3*x+240*A*a*b^2*c^2*x-10*A*b^4*c*x-240*B*a^2*b*c^2*x-120*B*a*b^3*c*x+5*B*b^5*x+240*A*a^2
*b*c^2-40*A*a*b^3*c+3*A*b^5-96*B*a^3*c^2-48*B*a^2*b^2*c+2*B*a*b^4)/(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((B*x+A)/(c*x^2+b*x+a)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 47.8584, size = 1165, normalized size = 8.76 \begin{align*} -\frac{2 \,{\left (2 \, B a b^{4} + 3 \, A b^{5} - 128 \,{\left (B b c^{4} - 2 \, A c^{5}\right )} x^{5} - 320 \,{\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} x^{4} - 80 \,{\left (3 \, B b^{3} c^{2} - 8 \, A a c^{4} + 2 \,{\left (2 \, B a b - 3 \, A b^{2}\right )} c^{3}\right )} x^{3} - 48 \,{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} c^{2} - 40 \,{\left (B b^{4} c - 24 \, A a b c^{3} + 2 \,{\left (6 \, B a b^{2} - A b^{3}\right )} c^{2}\right )} x^{2} - 8 \,{\left (6 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} c + 5 \,{\left (B b^{5} + 96 \, A a^{2} c^{3} - 48 \,{\left (B a^{2} b - A a b^{2}\right )} c^{2} - 2 \,{\left (12 \, B a b^{3} + A b^{4}\right )} c\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((B*x+A)/(c*x^2+b*x+a)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(2*B*a*b^4 + 3*A*b^5 - 128*(B*b*c^4 - 2*A*c^5)*x^5 - 320*(B*b^2*c^3 - 2*A*b*c^4)*x^4 - 80*(3*B*b^3*c^2 -
8*A*a*c^4 + 2*(2*B*a*b - 3*A*b^2)*c^3)*x^3 - 48*(2*B*a^3 - 5*A*a^2*b)*c^2 - 40*(B*b^4*c - 24*A*a*b*c^3 + 2*(6
*B*a*b^2 - A*b^3)*c^2)*x^2 - 8*(6*B*a^2*b^2 + 5*A*a*b^3)*c + 5*(B*b^5 + 96*A*a^2*c^3 - 48*(B*a^2*b - A*a*b^2)*
c^2 - 2*(12*B*a*b^3 + A*b^4)*c)*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 - 6
4*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 - 16*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((B*x+A)/(c*x**2+b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.21119, size = 633, normalized size = 4.76 \begin{align*} \frac{{\left (8 \,{\left (2 \,{\left (4 \,{\left (\frac{2 \,{\left (B b c^{4} - 2 \, A c^{5}\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 (B b^{2} c^{3} - 2 \, A b c^{4}\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 (3 \, B b^{3} c^{2} + 4 \, B a b c^{3} - 6 \, A b^{2} c^{3} - 8 \, A a c^{4}\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 (B b^{4} c + 12 \, B a b^{2} c^{2} - 2 \, A b^{3} c^{2} - 24 \, A a b c^{3}\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 (B b^{5} - 24 \, B a b^{3} c - 2 \, A b^{4} c - 48 \, B a^{2} b c^{2} + 48 \, A a b^{2} c^{2} + 96 \, A a^{2} c^{3}\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{2 \, B a b^{4} + 3 \, A b^{5} - 48 \, B a^{2} b^{2} c - 40 \, A a b^{3} c - 96 \, B a^{3} c^{2} + 240 \, A a^{2} b c^{2}}{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((B*x+A)/(c*x^2+b*x+a)^(7/2),x, algorithm="giac")

[Out]

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