3.1774 \(\int \frac {A+B x}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=332 \[ -\frac {b (A b-a B)}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b (2 a B e-3 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {e (a+b x) (a B e-3 A b e+2 b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^4}-\frac {e (a+b x) (B d-A e)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^3}-\frac {3 b e (a+b x) \log (a+b x) (a B e-2 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {3 b e (a+b x) \log (d+e x) (a B e-2 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 77} \[ -\frac {b (A b-a B)}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b (2 a B e-3 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {e (a+b x) (a B e-3 A b e+2 b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^4}-\frac {e (a+b x) (B d-A e)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^3}-\frac {3 b e (a+b x) \log (a+b x) (a B e-2 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {3 b e (a+b x) \log (d+e x) (a B e-2 A b e+b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

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

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Mathematica [A]  time = 0.27, size = 220, normalized size = 0.66 \[ \frac {(a+b x) \left (\frac {e (a+b x)^2 (b d-a e)^2 (A e-B d)}{(d+e x)^2}-2 b (a+b x) (b d-a e) (2 a B e-3 A b e+b B d)+\frac {2 e (a+b x)^2 (b d-a e) (-a B e+3 A b e-2 b B d)}{d+e x}-6 b e (a+b x)^2 \log (a+b x) (a B e-2 A b e+b B d)+6 b e (a+b x)^2 \log (d+e x) (a B e-2 A b e+b B d)-b (A b-a B) (b d-a e)^2\right )}{2 \left ((a+b x)^2\right )^{3/2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

((a + b*x)*(-(b*(A*b - a*B)*(b*d - a*e)^2) - 2*b*(b*d - a*e)*(b*B*d - 3*A*b*e + 2*a*B*e)*(a + b*x) + (e*(b*d -
a*e)^2*(-(B*d) + A*e)*(a + b*x)^2)/(d + e*x)^2 + (2*e*(b*d - a*e)*(-2*b*B*d + 3*A*b*e - a*B*e)*(a + b*x)^2)/(
d + e*x) - 6*b*e*(b*B*d - 2*A*b*e + a*B*e)*(a + b*x)^2*Log[a + b*x] + 6*b*e*(b*B*d - 2*A*b*e + a*B*e)*(a + b*x
)^2*Log[d + e*x]))/(2*(b*d - a*e)^5*((a + b*x)^2)^(3/2))

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fricas [B]  time = 0.75, size = 1215, normalized size = 3.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(9*B*a^3*b*d^2*e^2 + A*a^4*e^4 - (B*a*b^3 + A*b^4)*d^4 - (9*B*a^2*b^2 - 8*A*a*b^3)*d^3*e + (B*a^4 - 8*A*a^
3*b)*d*e^3 - 6*(B*b^4*d^2*e^2 - 2*A*b^4*d*e^3 - (B*a^2*b^2 - 2*A*a*b^3)*e^4)*x^3 - 9*(B*b^4*d^3*e - B*a^2*b^2*
d*e^3 + (B*a*b^3 - 2*A*b^4)*d^2*e^2 - (B*a^3*b - 2*A*a^2*b^2)*e^4)*x^2 - 2*(B*b^4*d^4 - 12*A*a*b^3*d^2*e^2 + (
7*B*a*b^3 - 2*A*b^4)*d^3*e - (7*B*a^3*b - 12*A*a^2*b^2)*d*e^3 - (B*a^4 - 2*A*a^3*b)*e^4)*x - 6*(B*a^2*b^2*d^3*
e + (B*a^3*b - 2*A*a^2*b^2)*d^2*e^2 + (B*b^4*d*e^3 + (B*a*b^3 - 2*A*b^4)*e^4)*x^4 + 2*(B*b^4*d^2*e^2 + 2*(B*a*
b^3 - A*b^4)*d*e^3 + (B*a^2*b^2 - 2*A*a*b^3)*e^4)*x^3 + (B*b^4*d^3*e + (5*B*a*b^3 - 2*A*b^4)*d^2*e^2 + (5*B*a^
2*b^2 - 8*A*a*b^3)*d*e^3 + (B*a^3*b - 2*A*a^2*b^2)*e^4)*x^2 + 2*(B*a*b^3*d^3*e + 2*(B*a^2*b^2 - A*a*b^3)*d^2*e
^2 + (B*a^3*b - 2*A*a^2*b^2)*d*e^3)*x)*log(b*x + a) + 6*(B*a^2*b^2*d^3*e + (B*a^3*b - 2*A*a^2*b^2)*d^2*e^2 + (
B*b^4*d*e^3 + (B*a*b^3 - 2*A*b^4)*e^4)*x^4 + 2*(B*b^4*d^2*e^2 + 2*(B*a*b^3 - A*b^4)*d*e^3 + (B*a^2*b^2 - 2*A*a
*b^3)*e^4)*x^3 + (B*b^4*d^3*e + (5*B*a*b^3 - 2*A*b^4)*d^2*e^2 + (5*B*a^2*b^2 - 8*A*a*b^3)*d*e^3 + (B*a^3*b - 2
*A*a^2*b^2)*e^4)*x^2 + 2*(B*a*b^3*d^3*e + 2*(B*a^2*b^2 - A*a*b^3)*d^2*e^2 + (B*a^3*b - 2*A*a^2*b^2)*d*e^3)*x)*
log(e*x + d))/(a^2*b^5*d^7 - 5*a^3*b^4*d^6*e + 10*a^4*b^3*d^5*e^2 - 10*a^5*b^2*d^4*e^3 + 5*a^6*b*d^3*e^4 - a^7
*d^2*e^5 + (b^7*d^5*e^2 - 5*a*b^6*d^4*e^3 + 10*a^2*b^5*d^3*e^4 - 10*a^3*b^4*d^2*e^5 + 5*a^4*b^3*d*e^6 - a^5*b^
2*e^7)*x^4 + 2*(b^7*d^6*e - 4*a*b^6*d^5*e^2 + 5*a^2*b^5*d^4*e^3 - 5*a^4*b^3*d^2*e^5 + 4*a^5*b^2*d*e^6 - a^6*b*
e^7)*x^3 + (b^7*d^7 - a*b^6*d^6*e - 9*a^2*b^5*d^5*e^2 + 25*a^3*b^4*d^4*e^3 - 25*a^4*b^3*d^3*e^4 + 9*a^5*b^2*d^
2*e^5 + a^6*b*d*e^6 - a^7*e^7)*x^2 + 2*(a*b^6*d^7 - 4*a^2*b^5*d^6*e + 5*a^3*b^4*d^5*e^2 - 5*a^5*b^2*d^3*e^4 +
4*a^6*b*d^2*e^5 - a^7*d*e^6)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.07, size = 1271, normalized size = 3.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/2*(-2*B*x*b^4*d^4+2*B*x*a^4*e^4-A*b^4*d^4+A*a^4*e^4-B*a*b^3*d^4+B*a^4*d*e^3+8*A*a*b^3*d^3*e+9*B*a^3*b*d^2*e
^2-9*B*a^2*b^2*d^3*e-8*A*a^3*b*d*e^3+12*B*ln(e*x+d)*x^3*a^2*b^2*e^4+12*B*ln(e*x+d)*x^3*b^4*d^2*e^2+12*A*ln(b*x
+a)*x^2*a^2*b^2*e^4+12*A*ln(b*x+a)*x^2*b^4*d^2*e^2-12*A*ln(e*x+d)*x^2*a^2*b^2*e^4-12*A*ln(e*x+d)*x^2*b^4*d^2*e
^2-6*B*ln(b*x+a)*x^2*a^3*b*e^4-6*B*ln(b*x+a)*x^2*b^4*d^3*e+6*B*ln(e*x+d)*x^2*a^3*b*e^4+6*B*ln(e*x+d)*x^2*b^4*d
^3*e-12*B*ln(b*x+a)*x^3*b^4*d^2*e^2+12*A*ln(b*x+a)*a^2*b^2*d^2*e^2-12*A*ln(e*x+d)*a^2*b^2*d^2*e^2-6*B*ln(b*x+a
)*a^3*b*d^2*e^2-6*B*ln(b*x+a)*a^2*b^2*d^3*e+6*B*ln(e*x+d)*a^3*b*d^2*e^2+6*B*ln(e*x+d)*a^2*b^2*d^3*e+9*B*x^2*a^
2*b^2*d*e^3-9*B*x^2*a*b^3*d^2*e^2-24*A*x*a^2*b^2*d*e^3+24*A*x*a*b^3*d^2*e^2+14*B*x*a^3*b*d*e^3-14*B*x*a*b^3*d^
3*e-6*B*ln(b*x+a)*x^4*a*b^3*e^4-6*B*ln(b*x+a)*x^4*b^4*d*e^3+6*B*ln(e*x+d)*x^4*a*b^3*e^4+6*B*ln(e*x+d)*x^4*b^4*
d*e^3+24*A*ln(b*x+a)*x^3*a*b^3*e^4+24*A*ln(b*x+a)*x^3*b^4*d*e^3-24*A*ln(e*x+d)*x^3*a*b^3*e^4-24*A*ln(e*x+d)*x^
3*b^4*d*e^3-12*B*ln(b*x+a)*x^3*a^2*b^2*e^4-12*A*x^3*a*b^3*e^4+12*A*x^3*b^4*d*e^3+6*B*x^3*a^2*b^2*e^4-6*B*x^3*b
^4*d^2*e^2-18*A*x^2*a^2*b^2*e^4+18*A*x^2*b^4*d^2*e^2+9*B*x^2*a^3*b*e^4-9*B*x^2*b^4*d^3*e+4*A*x*b^4*d^3*e-4*A*x
*a^3*b*e^4+12*A*ln(b*x+a)*x^4*b^4*e^4-12*A*ln(e*x+d)*x^4*b^4*e^4-12*B*ln(b*x+a)*x*a*b^3*d^3*e+12*B*ln(e*x+d)*x
*a^3*b*d*e^3+24*B*ln(e*x+d)*x*a^2*b^2*d^2*e^2+12*B*ln(e*x+d)*x*a*b^3*d^3*e+30*B*ln(e*x+d)*x^2*a^2*b^2*d*e^3+30
*B*ln(e*x+d)*x^2*a*b^3*d^2*e^2+24*A*ln(b*x+a)*x*a^2*b^2*d*e^3+24*A*ln(b*x+a)*x*a*b^3*d^2*e^2-24*A*ln(e*x+d)*x*
a^2*b^2*d*e^3-24*A*ln(e*x+d)*x*a*b^3*d^2*e^2-12*B*ln(b*x+a)*x*a^3*b*d*e^3-24*B*ln(b*x+a)*x^3*a*b^3*d*e^3+24*B*
ln(e*x+d)*x^3*a*b^3*d*e^3-24*B*ln(b*x+a)*x*a^2*b^2*d^2*e^2-48*A*ln(e*x+d)*x^2*a*b^3*d*e^3-30*B*ln(b*x+a)*x^2*a
^2*b^2*d*e^3-30*B*ln(b*x+a)*x^2*a*b^3*d^2*e^2+48*A*ln(b*x+a)*x^2*a*b^3*d*e^3)*(b*x+a)/(e*x+d)^2/(a*e-b*d)^5/((
b*x+a)^2)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{{\left (d+e\,x\right )}^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int((A + B*x)/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{\left (d + e x\right )^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((A + B*x)/((d + e*x)**3*((a + b*x)**2)**(3/2)), x)

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