3.16.50 \(\int \frac {(A+B x) (d+e x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=127 \[ -\frac {(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {-2 a B e+A b e+b B d}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.10, antiderivative size = 127, 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 {(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {-2 a B e+A b e+b B d}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e (a+b x) \log (a+b x)}{b^3 \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)^(3/2),x]
[Out]
-((b*B*d + A*b*e - 2*a*B*e)/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - ((A*b - a*B)*(b*d - a*e))/(2*b^3*(a + b*x)*
Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*e*(a + b*x)*Log[a + b*x])/(b^3*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 )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(A+B x) (d+e x)}{\left (a b+b^2 x\right )^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 d-a e)}{b^5 (a+b x)^3}+\frac {b B d+A b e-2 a B e}{b^5 (a+b x)^2}+\frac {B e}{b^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {b B d+A b e-2 a B e}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 86, normalized size = 0.68 \begin {gather*} \frac {B \left (3 a^2 e-a b d+4 a b e x-2 b^2 d x\right )-A b (a e+b d+2 b e x)+2 B e (a+b x)^2 \log (a+b x)}{2 b^3 (a+b x) \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)^(3/2),x]
[Out]
(-(A*b*(b*d + a*e + 2*b*e*x)) + B*(-(a*b*d) + 3*a^2*e - 2*b^2*d*x + 4*a*b*e*x) + 2*B*e*(a + b*x)^2*Log[a + b*x
])/(2*b^3*(a + b*x)*Sqrt[(a + b*x)^2])
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IntegrateAlgebraic [B] time = 2.14, size = 1723, normalized size = 13.57
result too large to display
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(a^2*A*b*d - a^3*B*d - a^3*A*e + a*A*b^2*d*x - 2*a^2*b*B*d*x - 2*a^2*A*b*e*x + A*b^3*d*x^2 - 2*a*b^2*B*d*x^2 -
2*a*A*b^2*e*x^2 + Sqrt[b^2]*(-(A*b*d*x) + 2*a*B*d*x + 2*a*A*e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b*x^2*(-2*a*
b^3 - 2*b^4*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + b*Sqrt[b^2]*x^2*(2*a^2*b^2 + 4*a*b^3*x + 2*b^4*x^2)) + ((-4*a*A
*b*d*x)/Sqrt[b^2] - (12*a^3*B*e*x)/(b*Sqrt[b^2]) - (24*a^2*B*e*x^2)/Sqrt[b^2] - (16*a*b*B*e*x^3)/Sqrt[b^2] + (
4*a*A*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (4*a^3*B*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (8*a^2*B*e*x*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/b^2 + (16*a*B*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (8*a^2*B*e*x^2*ArcTanh[(-(Sqrt
[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 16*a*B*e*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/a] + 8*b*B*e*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (8*a*B*e*x^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^2] - (8*b*B*e*x^3*Sq
rt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^2])/((-a - Sqr
t[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((4*a^4*B*e
)/(b^2)^(3/2) + (8*a^2*B*d*x)/Sqrt[b^2] + (8*a^2*A*e*x)/Sqrt[b^2] + (12*a^3*b*B*e*x)/(b^2)^(3/2) + (12*a*b^3*B
*d*x^2)/(b^2)^(3/2) + (12*a*A*b^3*e*x^2)/(b^2)^(3/2) + (12*a^2*B*e*x^2)/Sqrt[b^2] + (8*b^4*B*d*x^3)/(b^2)^(3/2
) + (8*A*b^4*e*x^3)/(b^2)^(3/2) - (4*a^2*B*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (4*a^2*A*e*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/b^2 - (4*a*B*d*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (4*a*A*e*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b
- (12*a^2*B*e*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - 8*B*d*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 8*A*e*x^2*Sqrt[
a^2 + 2*a*b*x + b^2*x^2] - (4*a^2*B*e*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (
8*a*b^3*B*e*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (4*b^4*B*e*x^4*Log[-a - S
qrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (4*a*B*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 4*B*e*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*
x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (4*a^2*B*e*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt
[b^2] - (8*a*b^3*B*e*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (4*b^4*B*e*x^4*Lo
g[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (4*a*B*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*L
og[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 4*B*e*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[
b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*
x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)
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fricas [A] time = 0.43, size = 110, normalized size = 0.87 \begin {gather*} -\frac {{\left (B a b + A b^{2}\right )} d - {\left (3 \, B a^{2} - A a b\right )} e + 2 \, {\left (B b^{2} d - {\left (2 \, B a b - A b^{2}\right )} e\right )} x - 2 \, {\left (B b^{2} e x^{2} + 2 \, B a b e x + B a^{2} e\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} 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)^(3/2),x, algorithm="fricas")
[Out]
-1/2*((B*a*b + A*b^2)*d - (3*B*a^2 - A*a*b)*e + 2*(B*b^2*d - (2*B*a*b - A*b^2)*e)*x - 2*(B*b^2*e*x^2 + 2*B*a*b
*e*x + B*a^2*e)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*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)^(3/2),x, algorithm="giac")
[Out]
sage0*x
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maple [A] time = 0.06, size = 109, normalized size = 0.86 \begin {gather*} -\frac {\left (-2 B \,b^{2} e \,x^{2} \ln \left (b x +a \right )-4 B a b e x \ln \left (b x +a \right )+2 A \,b^{2} e x -2 B \,a^{2} e \ln \left (b x +a \right )-4 B a b e x +2 B \,b^{2} d x +A a b e +A \,b^{2} d -3 B \,a^{2} e +B a b d \right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{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)^(3/2),x)
[Out]
-1/2*(-2*B*ln(b*x+a)*x^2*b^2*e-4*B*ln(b*x+a)*x*a*b*e+2*A*b^2*e*x-2*B*a^2*e*ln(b*x+a)-4*B*a*b*e*x+2*B*b^2*d*x+a
*A*e*b+A*b^2*d-3*B*e*a^2+a*B*d*b)*(b*x+a)/b^3/((b*x+a)^2)^(3/2)
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maxima [A] time = 0.57, size = 120, normalized size = 0.94 \begin {gather*} \frac {B e \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {B d + A e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, B a e x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A d}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, B a^{2} e}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (B d + A e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \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)^(3/2),x, algorithm="maxima")
[Out]
B*e*log(x + a/b)/b^3 - (B*d + A*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 2*B*a*e*x/(b^4*(x + a/b)^2) - 1/2*A*d
/(b^3*(x + a/b)^2) + 3/2*B*a^2*e/(b^5*(x + a/b)^2) + 1/2*(B*d + A*e)*a/(b^4*(x + a/b)^2)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\left (d+e\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \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)^(3/2),x)
[Out]
int(((A + B*x)*(d + e*x))/(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 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{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)**(3/2),x)
[Out]
Integral((A + B*x)*(d + e*x)/((a + b*x)**2)**(3/2), x)
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