3.16.59 \(\int \frac {(A+B x) (d+e x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac {-2 a B e+A b e+b B d}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B e}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 135, 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 = {770, 77} \begin {gather*} -\frac {-2 a B e+A b e+b B d}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B e}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 75, normalized size = 0.56 \begin {gather*} \frac {-B \left (a^2 e+a b (d+4 e x)+2 b^2 x (2 d+3 e x)\right )-A b (a e+3 b d+4 b e x)}{12 b^3 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-(A*b*(3*b*d + a*e + 4*b*e*x)) - B*(a^2*e + 2*b^2*x*(2*d + 3*e*x) + a*b*(d + 4*e*x)))/(12*b^3*(a + b*x)^3*Sqr
t[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 1.59, size = 497, normalized size = 3.68 \begin {gather*} \frac {-2 \left (-3 a^6 b B e+3 a^5 A b^2 e+3 a^5 b^2 B d-3 a^4 A b^3 d-a^2 b^5 B e x^4-a A b^6 e x^4-a b^6 B d x^4-4 a b^6 B e x^5-3 A b^7 d x^4-4 A b^7 e x^5-4 b^7 B d x^5-6 b^7 B e x^6\right )-2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (-3 a^5 B e+3 a^4 A b e+3 a^4 b B d+3 a^4 b B e x-3 a^3 A b^2 d-3 a^3 A b^2 e x-3 a^3 b^2 B d x-3 a^3 b^2 B e x^2+3 a^2 A b^3 d x+3 a^2 A b^3 e x^2+3 a^2 b^3 B d x^2+3 a^2 b^3 B e x^3-3 a A b^4 d x^2-3 a A b^4 e x^3-3 a b^4 B d x^3-2 a b^4 B e x^4+3 A b^5 d x^3+4 A b^5 e x^4+4 b^5 B d x^4+6 b^5 B e x^5\right )}{3 b^3 x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^5-24 a^2 b^6 x-24 a b^7 x^2-8 b^8 x^3\right )+3 b^3 \sqrt {b^2} x^4 \left (8 a^4 b^4+32 a^3 b^5 x+48 a^2 b^6 x^2+32 a b^7 x^3+8 b^8 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-3*a^3*A*b^2*d + 3*a^4*b*B*d + 3*a^4*A*b*e - 3*a^5*B*e + 3*a^2*A*
b^3*d*x - 3*a^3*b^2*B*d*x - 3*a^3*A*b^2*e*x + 3*a^4*b*B*e*x - 3*a*A*b^4*d*x^2 + 3*a^2*b^3*B*d*x^2 + 3*a^2*A*b^
3*e*x^2 - 3*a^3*b^2*B*e*x^2 + 3*A*b^5*d*x^3 - 3*a*b^4*B*d*x^3 - 3*a*A*b^4*e*x^3 + 3*a^2*b^3*B*e*x^3 + 4*b^5*B*
d*x^4 + 4*A*b^5*e*x^4 - 2*a*b^4*B*e*x^4 + 6*b^5*B*e*x^5) - 2*(-3*a^4*A*b^3*d + 3*a^5*b^2*B*d + 3*a^5*A*b^2*e -
3*a^6*b*B*e - 3*A*b^7*d*x^4 - a*b^6*B*d*x^4 - a*A*b^6*e*x^4 - a^2*b^5*B*e*x^4 - 4*b^7*B*d*x^5 - 4*A*b^7*e*x^5
- 4*a*b^6*B*e*x^5 - 6*b^7*B*e*x^6))/(3*b^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^5 - 24*a^2*b^6*x - 24*
a*b^7*x^2 - 8*b^8*x^3) + 3*b^3*Sqrt[b^2]*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*b^7*x^3 + 8*b^8
*x^4))

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fricas [A]  time = 0.44, size = 106, normalized size = 0.79 \begin {gather*} -\frac {6 \, B b^{2} e x^{2} + {\left (B a b + 3 \, A b^{2}\right )} d + {\left (B a^{2} + A a b\right )} e + 4 \, {\left (B b^{2} d + {\left (B a b + A b^{2}\right )} e\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(6*B*b^2*e*x^2 + (B*a*b + 3*A*b^2)*d + (B*a^2 + A*a*b)*e + 4*(B*b^2*d + (B*a*b + A*b^2)*e)*x)/(b^7*x^4 +
4*a*b^6*x^3 + 6*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 77, normalized size = 0.57 \begin {gather*} -\frac {\left (b x +a \right ) \left (6 B \,b^{2} e \,x^{2}+4 A \,b^{2} e x +4 B a b e x +4 B \,b^{2} d x +A a b e +3 A \,b^{2} d +B \,a^{2} e +B a b d \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(b*x+a)/b^3*(6*B*b^2*e*x^2+4*A*b^2*e*x+4*B*a*b*e*x+4*B*b^2*d*x+A*a*b*e+3*A*b^2*d+B*a^2*e+B*a*b*d)/((b*x+
a)^2)^(5/2)

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maxima [A]  time = 0.68, size = 121, normalized size = 0.90 \begin {gather*} -\frac {B d + A e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {B e}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a e}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {A d}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {B a^{2} e}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {{\left (B d + A e\right )} a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(B*d + A*e)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 1/2*B*e/(b^5*(x + a/b)^2) + 2/3*B*a*e/(b^6*(x + a/b)^
3) - 1/4*A*d/(b^5*(x + a/b)^4) - 1/4*B*a^2*e/(b^7*(x + a/b)^4) + 1/4*(B*d + A*e)*a/(b^6*(x + a/b)^4)

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mupad [B]  time = 2.24, size = 87, normalized size = 0.64 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (3\,A\,b^2\,d+B\,a^2\,e+4\,A\,b^2\,e\,x+4\,B\,b^2\,d\,x+6\,B\,b^2\,e\,x^2+A\,a\,b\,e+B\,a\,b\,d+4\,B\,a\,b\,e\,x\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(3*A*b^2*d + B*a^2*e + 4*A*b^2*e*x + 4*B*b^2*d*x + 6*B*b^2*e*x^2 + A*a*b*e +
B*a*b*d + 4*B*a*b*e*x))/(12*b^3*(a + b*x)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((A + B*x)*(d + e*x)/((a + b*x)**2)**(5/2), x)

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