3.16.71 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=124 \[ \frac {2 b (-2 a B e-A b e+3 b B d)}{e^4 \sqrt {d+e x}}-\frac {2 (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac {2 b^2 B \sqrt {d+e x}}{e^4} \]

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Rubi [A]  time = 0.06, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 77} \begin {gather*} \frac {2 b (-2 a B e-A b e+3 b B d)}{e^4 \sqrt {d+e x}}-\frac {2 (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac {2 b^2 B \sqrt {d+e x}}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)^2*(B*d - A*e))/(5*e^4*(d + e*x)^(5/2)) - (2*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e))/(3*e^4*(d
+ e*x)^(3/2)) + (2*b*(3*b*B*d - A*b*e - 2*a*B*e))/(e^4*Sqrt[d + e*x]) + (2*b^2*B*Sqrt[d + e*x])/e^4

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx &=\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{7/2}} \, dx\\ &=\int \left (\frac {(-b d+a e)^2 (-B d+A e)}{e^3 (d+e x)^{7/2}}+\frac {(-b d+a e) (-3 b B d+2 A b e+a B e)}{e^3 (d+e x)^{5/2}}+\frac {b (-3 b B d+A b e+2 a B e)}{e^3 (d+e x)^{3/2}}+\frac {b^2 B}{e^3 \sqrt {d+e x}}\right ) \, dx\\ &=\frac {2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}-\frac {2 (b d-a e) (3 b B d-2 A b e-a B e)}{3 e^4 (d+e x)^{3/2}}+\frac {2 b (3 b B d-A b e-2 a B e)}{e^4 \sqrt {d+e x}}+\frac {2 b^2 B \sqrt {d+e x}}{e^4}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 107, normalized size = 0.86 \begin {gather*} \frac {2 \left (15 b (d+e x)^2 (-2 a B e-A b e+3 b B d)-5 (d+e x) (b d-a e) (-a B e-2 A b e+3 b B d)+3 (b d-a e)^2 (B d-A e)+15 b^2 B (d+e x)^3\right )}{15 e^4 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(b*d - a*e)^2*(B*d - A*e) - 5*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e)*(d + e*x) + 15*b*(3*b*B*d - A*b*e
- 2*a*B*e)*(d + e*x)^2 + 15*b^2*B*(d + e*x)^3))/(15*e^4*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 0.12, size = 193, normalized size = 1.56 \begin {gather*} \frac {2 \left (-3 a^2 A e^3-5 a^2 B e^2 (d+e x)+3 a^2 B d e^2-10 a A b e^2 (d+e x)+6 a A b d e^2-6 a b B d^2 e+20 a b B d e (d+e x)-30 a b B e (d+e x)^2-3 A b^2 d^2 e+10 A b^2 d e (d+e x)-15 A b^2 e (d+e x)^2+3 b^2 B d^3-15 b^2 B d^2 (d+e x)+45 b^2 B d (d+e x)^2+15 b^2 B (d+e x)^3\right )}{15 e^4 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*b^2*B*d^3 - 3*A*b^2*d^2*e - 6*a*b*B*d^2*e + 6*a*A*b*d*e^2 + 3*a^2*B*d*e^2 - 3*a^2*A*e^3 - 15*b^2*B*d^2*(
d + e*x) + 10*A*b^2*d*e*(d + e*x) + 20*a*b*B*d*e*(d + e*x) - 10*a*A*b*e^2*(d + e*x) - 5*a^2*B*e^2*(d + e*x) +
45*b^2*B*d*(d + e*x)^2 - 15*A*b^2*e*(d + e*x)^2 - 30*a*b*B*e*(d + e*x)^2 + 15*b^2*B*(d + e*x)^3))/(15*e^4*(d +
e*x)^(5/2))

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fricas [A]  time = 0.40, size = 188, normalized size = 1.52 \begin {gather*} \frac {2 \, {\left (15 \, B b^{2} e^{3} x^{3} + 48 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 8 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (6 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (24 \, B b^{2} d^{2} e - 4 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} - {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*B*b^2*e^3*x^3 + 48*B*b^2*d^3 - 3*A*a^2*e^3 - 8*(2*B*a*b + A*b^2)*d^2*e - 2*(B*a^2 + 2*A*a*b)*d*e^2 +
15*(6*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 5*(24*B*b^2*d^2*e - 4*(2*B*a*b + A*b^2)*d*e^2 - (B*a^2 + 2*A*
a*b)*e^3)*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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giac [A]  time = 0.21, size = 202, normalized size = 1.63 \begin {gather*} 2 \, \sqrt {x e + d} B b^{2} e^{\left (-4\right )} + \frac {2 \, {\left (45 \, {\left (x e + d\right )}^{2} B b^{2} d - 15 \, {\left (x e + d\right )} B b^{2} d^{2} + 3 \, B b^{2} d^{3} - 30 \, {\left (x e + d\right )}^{2} B a b e - 15 \, {\left (x e + d\right )}^{2} A b^{2} e + 20 \, {\left (x e + d\right )} B a b d e + 10 \, {\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (x e + d\right )} B a^{2} e^{2} - 10 \, {\left (x e + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )} e^{\left (-4\right )}}{15 \, {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*B*b^2*e^(-4) + 2/15*(45*(x*e + d)^2*B*b^2*d - 15*(x*e + d)*B*b^2*d^2 + 3*B*b^2*d^3 - 30*(x*e +
d)^2*B*a*b*e - 15*(x*e + d)^2*A*b^2*e + 20*(x*e + d)*B*a*b*d*e + 10*(x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A
*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 - 10*(x*e + d)*A*a*b*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A*a^2*e^3)*e^(
-4)/(x*e + d)^(5/2)

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maple [A]  time = 0.06, size = 169, normalized size = 1.36 \begin {gather*} -\frac {2 \left (-15 B \,b^{2} x^{3} e^{3}+15 A \,b^{2} e^{3} x^{2}+30 B a b \,e^{3} x^{2}-90 B \,b^{2} d \,e^{2} x^{2}+10 A a b \,e^{3} x +20 A \,b^{2} d \,e^{2} x +5 B \,a^{2} e^{3} x +40 B a b d \,e^{2} x -120 B \,b^{2} d^{2} e x +3 A \,a^{2} e^{3}+4 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e -48 B \,b^{2} d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-15*B*b^2*e^3*x^3+15*A*b^2*e^3*x^2+30*B*a*b*e^3*x^2-90*B*b^2*d*e^2*x^2+10*A*a*b*e^3*x+20*A*b^2*d*e^2*x+
5*B*a^2*e^3*x+40*B*a*b*d*e^2*x-120*B*b^2*d^2*e*x+3*A*a^2*e^3+4*A*a*b*d*e^2+8*A*b^2*d^2*e+2*B*a^2*d*e^2+16*B*a*
b*d^2*e-48*B*b^2*d^3)/(e*x+d)^(5/2)/e^4

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maxima [A]  time = 0.64, size = 164, normalized size = 1.32 \begin {gather*} \frac {2 \, {\left (\frac {15 \, \sqrt {e x + d} B b^{2}}{e^{3}} + \frac {3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{2} - 5 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {5}{2}} e^{3}}\right )}}{15 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(e*x + d)*B*b^2/e^3 + (3*B*b^2*d^3 - 3*A*a^2*e^3 - 3*(2*B*a*b + A*b^2)*d^2*e + 3*(B*a^2 + 2*A*a*b
)*d*e^2 + 15*(3*B*b^2*d - (2*B*a*b + A*b^2)*e)*(e*x + d)^2 - 5*(3*B*b^2*d^2 - 2*(2*B*a*b + A*b^2)*d*e + (B*a^2
+ 2*A*a*b)*e^2)*(e*x + d))/((e*x + d)^(5/2)*e^3))/e

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mupad [B]  time = 0.10, size = 168, normalized size = 1.35 \begin {gather*} -\frac {2\,\left (2\,B\,a^2\,d\,e^2+5\,B\,a^2\,e^3\,x+3\,A\,a^2\,e^3+16\,B\,a\,b\,d^2\,e+40\,B\,a\,b\,d\,e^2\,x+4\,A\,a\,b\,d\,e^2+30\,B\,a\,b\,e^3\,x^2+10\,A\,a\,b\,e^3\,x-48\,B\,b^2\,d^3-120\,B\,b^2\,d^2\,e\,x+8\,A\,b^2\,d^2\,e-90\,B\,b^2\,d\,e^2\,x^2+20\,A\,b^2\,d\,e^2\,x-15\,B\,b^2\,e^3\,x^3+15\,A\,b^2\,e^3\,x^2\right )}{15\,e^4\,{\left (d+e\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(3*A*a^2*e^3 - 48*B*b^2*d^3 + 8*A*b^2*d^2*e + 2*B*a^2*d*e^2 + 5*B*a^2*e^3*x + 15*A*b^2*e^3*x^2 - 15*B*b^2*
e^3*x^3 + 30*B*a*b*e^3*x^2 + 20*A*b^2*d*e^2*x - 120*B*b^2*d^2*e*x - 90*B*b^2*d*e^2*x^2 + 4*A*a*b*d*e^2 + 16*B*
a*b*d^2*e + 10*A*a*b*e^3*x + 40*B*a*b*d*e^2*x))/(15*e^4*(d + e*x)^(5/2))

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sympy [A]  time = 3.64, size = 1015, normalized size = 8.19 \begin {gather*} \begin {cases} - \frac {6 A a^{2} e^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {8 A a b d e^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {20 A a b e^{3} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {16 A b^{2} d^{2} e}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {40 A b^{2} d e^{2} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {30 A b^{2} e^{3} x^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {4 B a^{2} d e^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {10 B a^{2} e^{3} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {32 B a b d^{2} e}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {80 B a b d e^{2} x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} - \frac {60 B a b e^{3} x^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {96 B b^{2} d^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {240 B b^{2} d^{2} e x}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {180 B b^{2} d e^{2} x^{2}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} + \frac {30 B b^{2} e^{3} x^{3}}{15 d^{2} e^{4} \sqrt {d + e x} + 30 d e^{5} x \sqrt {d + e x} + 15 e^{6} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {A a^{2} x + A a b x^{2} + \frac {A b^{2} x^{3}}{3} + \frac {B a^{2} x^{2}}{2} + \frac {2 B a b x^{3}}{3} + \frac {B b^{2} x^{4}}{4}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-6*A*a**2*e**3/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)
) - 8*A*a*b*d*e**2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 20*
A*a*b*e**3*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 16*A*b**2
*d**2*e/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 40*A*b**2*d*e*
*2*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 30*A*b**2*e**3*x*
*2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 4*B*a**2*d*e**2/(15
*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 10*B*a**2*e**3*x/(15*d**2
*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 32*B*a*b*d**2*e/(15*d**2*e**4*
sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 80*B*a*b*d*e**2*x/(15*d**2*e**4*sqrt
(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 60*B*a*b*e**3*x**2/(15*d**2*e**4*sqrt(d
+ e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 96*B*b**2*d**3/(15*d**2*e**4*sqrt(d + e*x)
+ 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 240*B*b**2*d**2*e*x/(15*d**2*e**4*sqrt(d + e*x) +
30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 180*B*b**2*d*e**2*x**2/(15*d**2*e**4*sqrt(d + e*x) +
30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 30*B*b**2*e**3*x**3/(15*d**2*e**4*sqrt(d + e*x) + 3
0*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)), Ne(e, 0)), ((A*a**2*x + A*a*b*x**2 + A*b**2*x**3/3 + B
*a**2*x**2/2 + 2*B*a*b*x**3/3 + B*b**2*x**4/4)/d**(7/2), True))

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