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

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 135, 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) (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}} \end {gather*}

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 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 {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 = 0.24, size = 119, normalized size = 0.88 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 2.18, size = 267, normalized size = 1.98 \begin {gather*} -\frac {2 \left (-96 a^3 c^2 e-48 a^2 b^2 c e+240 a^2 b c^2 d-240 a^2 b c^2 e x+480 a^2 c^3 d x+2 a b^4 e-40 a b^3 c d-120 a b^3 c e x+240 a b^2 c^2 d x-480 a b^2 c^2 e x^2+960 a b c^3 d x^2-320 a b c^3 e x^3+640 a c^4 d x^3+3 b^5 d+5 b^5 e x-10 b^4 c d x-40 b^4 c e x^2+80 b^3 c^2 d x^2-240 b^3 c^2 e x^3+480 b^2 c^3 d x^3-320 b^2 c^3 e x^4+640 b c^4 d x^4-128 b c^4 e x^5+256 c^5 d 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[(d + e*x)/(a + b*x + c*x^2)^(7/2),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 5.31, size = 550, normalized size = 4.07 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

giac [B]  time = 0.28, size = 448, normalized size = 3.32 \begin {gather*} -\frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, {\left (2 \, c^{5} d - b c^{4} e\right )} x}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}} + \frac {5 \, {\left (2 \, b c^{4} d - b^{2} c^{3} e\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 (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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\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} - 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((e*x+d)/(c*x^2+b*x+a)^(7/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.06, size = 288, normalized size = 2.13 \begin {gather*} -\frac {2 \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}+320 a b \,c^{3} e \,x^{3}-640 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 \right )}{15 \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((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)

________________________________________________________________________________________

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((e*x+d)/(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?

________________________________________________________________________________________

mupad [B]  time = 1.95, size = 395, normalized size = 2.93 \begin {gather*} \frac {x\,\left (\frac {4\,c^2\,d}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {2\,b\,c\,e}{5\,\left (4\,a\,c^2-b^2\,c\right )}\right )-\frac {4\,a\,c\,e}{5\,\left (4\,a\,c^2-b^2\,c\right )}+\frac {2\,b\,c\,d}{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 (20\,b\,e-32\,c\,d\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {8\,b\,c^2\,e}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}\right )+\frac {b\,c\,\left (20\,b\,e-32\,c\,d\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {16\,a\,c^2\,e}{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 {\frac {b\,c\,\left (256\,c^2\,d-128\,b\,c\,e\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\,c^2\,d-128\,b\,c\,e\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 {4\,e}{\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((d + e*x)/(a + b*x + c*x^2)^(7/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________