3.2027 \(\int \frac {(a+b x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e (a+b x) (b d-a e) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {768, 640, 608, 31} \[ -\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e (a+b x) (b d-a e) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + e*x)^2/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (2*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (2*e*(b*d - a*e
)*(a + b*x)*Log[a + b*x])/(b^3*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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(2 e) \int \frac {d+e x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b}\\ &=-\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}{b^3}+\frac {(2 e (b d-a e)) \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b^2}\\ &=-\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}{b^3}+\frac {\left (2 e (b d-a e) \left (a b+b^2 x\right )\right ) \int \frac {1}{a b+b^2 x} \, dx}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^2}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}{b^3}+\frac {2 e (b d-a 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 = 0.03, size = 74, normalized size = 0.70 \[ \frac {-a^2 e^2+a b e (2 d+e x)-2 e (a+b x) (a e-b d) \log (a+b x)+b^2 \left (e^2 x^2-d^2\right )}{b^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 92, normalized size = 0.87 \[ \frac {b^{2} e^{2} x^{2} + a b e^{2} x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} + 2 \, {\left (a b d e - a^{2} e^{2} + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.36, size = 148, normalized size = 1.40 \[ -\frac {2 \, {\left (b d e - a e^{2}\right )} \log \left ({\left | -3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{2} a b - a^{3} b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{3} {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )} a^{2} {\left | b \right |} \right |}\right )}{3 \, b^{2} {\left | b \right |}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{2}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(b*d*e - a*e^2)*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^2*abs(b)) + sq
rt(b^2*x^2 + 2*a*b*x + a^2)*e^2/b^3

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maple [A]  time = 0.07, size = 116, normalized size = 1.09 \[ -\frac {\left (2 a b \,e^{2} x \ln \left (b x +a \right )-2 b^{2} d e x \ln \left (b x +a \right )-b^{2} e^{2} x^{2}+2 a^{2} e^{2} \ln \left (b x +a \right )-2 a b d e \ln \left (b x +a \right )-a b \,e^{2} x +a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.57, size = 272, normalized size = 2.57 \[ \frac {e^{2} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b} - \frac {3 \, a e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, a^{2} e^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} - \frac {6 \, a^{2} e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (2 \, b d e + a e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {b d^{2} + 2 \, a d e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {a d^{2}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, a^{3} e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, {\left (2 \, b d e + a e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, {\left (2 \, b d e + a e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (b d^{2} + 2 \, a d e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b) - 3*a*e^2*log(x + a/b)/b^3 + 2*a^2*e^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2
)*b^3) - 6*a^2*e^2*x/(b^4*(x + a/b)^2) + (2*b*d*e + a*e^2)*log(x + a/b)/b^3 - (b*d^2 + 2*a*d*e)/(sqrt(b^2*x^2
+ 2*a*b*x + a^2)*b^2) - 1/2*a*d^2/(b^3*(x + a/b)^2) - 11/2*a^3*e^2/(b^5*(x + a/b)^2) + 2*(2*b*d*e + a*e^2)*a*x
/(b^4*(x + a/b)^2) + 3/2*(2*b*d*e + a*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(b*d^2 + 2*a*d*e)*a/(b^4*(x + a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((a + b*x)*(d + e*x)^2)/(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 \[ \int \frac {\left (a + b x\right ) \left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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