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

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

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Rubi [A]  time = 0.04, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {638, 614, 613} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.20, size = 120, normalized size = 0.90 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 2.76, size = 267, normalized size = 2.01 \begin {gather*} -\frac {2 \left (-96 a^3 B c^2+240 a^2 A b c^2+480 a^2 A c^3 x-48 a^2 b^2 B c-240 a^2 b B c^2 x-40 a A b^3 c+240 a A b^2 c^2 x+960 a A b c^3 x^2+640 a A c^4 x^3+2 a b^4 B-120 a b^3 B c x-480 a b^2 B c^2 x^2-320 a b B c^3 x^3+3 A b^5-10 A b^4 c x+80 A b^3 c^2 x^2+480 A b^2 c^3 x^3+640 A b c^4 x^4+256 A c^5 x^5+5 b^5 B x-40 b^4 B c x^2-240 b^3 B c^2 x^3-320 b^2 B c^3 x^4-128 b B c^4 x^5\right )}{15 \left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.06, size = 543, normalized size = 4.08 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 0.28, size = 433, normalized size = 3.26 \begin {gather*} \frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} x}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}} + \frac {5 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )}}{15 \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \end {gather*}

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]

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

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maple [B]  time = 0.01, size = 288, normalized size = 2.17 \begin {gather*} \frac {\frac {512}{15} A \,c^{5} x^{5}-\frac {256}{15} B b \,c^{4} x^{5}+\frac {256}{3} A b \,c^{4} x^{4}-\frac {128}{3} B \,b^{2} c^{3} x^{4}+\frac {256}{3} A a \,c^{4} x^{3}+64 A \,b^{2} c^{3} x^{3}-\frac {128}{3} B a b \,c^{3} x^{3}-32 B \,b^{3} c^{2} x^{3}+128 A a b \,c^{3} x^{2}+\frac {32}{3} A \,b^{3} c^{2} x^{2}-64 B a \,b^{2} c^{2} x^{2}-\frac {16}{3} B \,b^{4} c \,x^{2}+64 A \,a^{2} c^{3} x +32 A a \,b^{2} c^{2} x -\frac {4}{3} A \,b^{4} c x -32 B \,a^{2} b \,c^{2} x -16 B a \,b^{3} c x +\frac {2}{3} B \,b^{5} x +32 A \,a^{2} b \,c^{2}-\frac {16}{3} A a \,b^{3} c +\frac {2}{5} A \,b^{5}-\frac {64}{5} B \,a^{3} c^{2}-\frac {32}{5} B \,a^{2} b^{2} c +\frac {4}{15} B a \,b^{4}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [B]  time = 3.12, size = 394, normalized size = 2.96 \begin {gather*} \frac {\frac {b\,c\,\left (256\,A\,c^2-128\,B\,b\,c\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}+\frac {2\,c^2\,x\,\left (256\,A\,c^2-128\,B\,b\,c\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {c\,x^2+b\,x+a}}+\frac {x\,\left (\frac {4\,A\,c^2}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {2\,B\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}\right )+\frac {2\,A\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {4\,B\,a\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}}+\frac {x\,\left (\frac {2\,c^2\,\left (32\,A\,c-20\,B\,b\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {8\,B\,b\,c^2}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}\right )+\frac {b\,c\,\left (32\,A\,c-20\,B\,b\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {16\,B\,a\,c^2}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}}-\frac {4\,B}{\left (60\,a\,c-15\,b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(a + b*x + c*x^2)^(7/2),x)

[Out]

((b*c*(256*A*c^2 - 128*B*b*c))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2) + (2*c^2*x*(256*A*c^2 - 128*B*b*c))/(15*
(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2))/(a + b*x + c*x^2)^(1/2) + (x*((4*A*c^2)/(5*(4*a*c^2 - b^2*c)) - (2*B*b*c)/
(5*(4*a*c^2 - b^2*c))) + (2*A*b*c)/(5*(4*a*c^2 - b^2*c)) - (4*B*a*c)/(5*(4*a*c^2 - b^2*c)))/(a + b*x + c*x^2)^
(5/2) + (x*((2*c^2*(32*A*c - 20*B*b))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) + (8*B*b*c^2)/(15*(4*a*c^2 - b^2*c)
*(4*a*c - b^2))) + (b*c*(32*A*c - 20*B*b))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) + (16*B*a*c^2)/(15*(4*a*c^2 -
b^2*c)*(4*a*c - b^2)))/(a + b*x + c*x^2)^(3/2) - (4*B)/((60*a*c - 15*b^2)*(a + b*x + c*x^2)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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