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

Optimal. Leaf size=67 \[ \frac {e (a+b x) \log (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {d+e x}{b \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {768, 608, 31} \begin {gather*} \frac {e (a+b x) \log (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {d+e x}{b \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]

-((d + e*x)/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (e*(a + b*x)*Log[a + b*x])/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 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 {d+e x}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b}\\ &=-\frac {d+e x}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (e \left (a b+b^2 x\right )\right ) \int \frac {1}{a b+b^2 x} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {d+e x}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (a+b x) \log (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.54 \begin {gather*} \frac {e (a+b x) \log (a+b x)+a e-b d}{b^2 \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]

(-(b*d) + a*e + e*(a + b*x)*Log[a + b*x])/(b^2*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 0.76, size = 435, normalized size = 6.49 \begin {gather*} \frac {2 e x^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )+\frac {2 a e x \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )}{b}-\frac {2 e x \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )}{\sqrt {b^2}}-\frac {2 a d}{\sqrt {b^2}}}{\left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right ) \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}-\frac {b e \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 \left (b^2\right )^{3/2}}-\frac {b e \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 \left (b^2\right )^{3/2}}+\frac {\sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} (a e-b d)+a^2 b e-a b^2 e x+b^3 d x}{b^3 \sqrt {b^2} x (a+b x)-b^4 x \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

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*b*e + b^3*d*x - a*b^2*e*x + Sqrt[b^2]*(-(b*d) + a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b^3*Sqrt[b^2]*x*(a +
b*x) - b^4*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((-2*a*d)/Sqrt[b^2] + (2*a*e*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/a])/b + 2*e*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (2*e*x*
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])/((-a - S
qrt[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])) - (b*e*Log[-a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(b^2)^(3/2)) - (b*e*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/(2*(b^2)^(3/2))

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fricas [A]  time = 0.45, size = 39, normalized size = 0.58 \begin {gather*} -\frac {b d - a e - {\left (b e x + a e\right )} \log \left (b x + a\right )}{b^{3} x + a b^{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="fricas")

[Out]

-(b*d - a*e - (b*e*x + a*e)*log(b*x + a))/(b^3*x + a*b^2)

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giac [B]  time = 0.33, size = 114, normalized size = 1.70 \begin {gather*} -\frac {e \log \left ({\left | -3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{2} a b - a^{3} b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{3} {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )} a^{2} {\left | b \right |} \right |}\right )}{3 \, b {\left | b \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="giac")

[Out]

-1/3*e*log(abs(-3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))^2*a*b - a^3*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*
x + a^2))^3*abs(b) - 3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))*a^2*abs(b)))/(b*abs(b))

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maple [A]  time = 0.06, size = 48, normalized size = 0.72 \begin {gather*} \frac {\left (b e x \ln \left (b x +a \right )+a e \ln \left (b x +a \right )+a e -b d \right ) \left (b x +a \right )^{2}}{\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{2}} \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]

(ln(b*x+a)*x*b*e+ln(b*x+a)*a*e+a*e-b*d)*(b*x+a)^2/b^2/((b*x+a)^2)^(3/2)

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maxima [B]  time = 0.55, size = 117, normalized size = 1.75 \begin {gather*} \frac {e \log \left (x + \frac {a}{b}\right )}{b^{2}} - \frac {b d + a e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, a e x}{b^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {a d}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a^{2} e}{2 \, b^{4} {\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]

e*log(x + a/b)/b^2 - (b*d + a*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 2*a*e*x/(b^3*(x + a/b)^2) - 1/2*a*d/(b^
3*(x + a/b)^2) + 3/2*a^2*e/(b^4*(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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