3.1769 \(\int \frac{(A+B x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=186 \[ -\frac{(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^2 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.149695, antiderivative size = 186, normalized size of antiderivative = 1., 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} \[ -\frac{(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^2 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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]

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

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]

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]))))

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{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)^2}{\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{B e^2}{b^6}+\frac{(A b-a B) (b d-a e)^2}{b^6 (a+b x)^3}+\frac{(b d-a e) (b B d+2 A b e-3 a B e)}{b^6 (a+b x)^2}+\frac{e (2 b B d+A b e-3 a B e)}{b^6 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(b d-a e) (b B d+2 A b e-3 a B e)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^2 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (2 b B d+A b e-3 a B e) (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.104991, size = 151, normalized size = 0.81 \[ \frac{B \left (2 a^2 b e (3 d-2 e x)-5 a^3 e^2+a b^2 \left (-d^2+8 d e x+4 e^2 x^2\right )+2 b^3 x \left (e^2 x^2-d^2\right )\right )+2 e (a+b x)^2 \log (a+b x) (-3 a B e+A b e+2 b B d)-A b (b d-a e) (3 a e+b (d+4 e x))}{2 b^4 (a+b x) \sqrt{(a+b x)^2}} \]

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*b*(b*d - a*e)*(3*a*e + b*(d + 4*e*x))) + B*(-5*a^3*e^2 + 2*a^2*b*e*(3*d - 2*e*x) + 2*b^3*x*(-d^2 + e^2*x^
2) + a*b^2*(-d^2 + 8*d*e*x + 4*e^2*x^2)) + 2*e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a
+ b*x)*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.015, size = 303, normalized size = 1.6 \begin{align*}{\frac{ \left ( 2\,A\ln \left ( bx+a \right ){x}^{2}{b}^{3}{e}^{2}-6\,B\ln \left ( bx+a \right ){x}^{2}a{b}^{2}{e}^{2}+4\,B\ln \left ( bx+a \right ){x}^{2}{b}^{3}de+2\,B{x}^{3}{b}^{3}{e}^{2}+4\,A\ln \left ( bx+a \right ) xa{b}^{2}{e}^{2}-12\,B\ln \left ( bx+a \right ) x{a}^{2}b{e}^{2}+8\,B\ln \left ( bx+a \right ) xa{b}^{2}de+4\,B{x}^{2}a{b}^{2}{e}^{2}+2\,A\ln \left ( bx+a \right ){a}^{2}b{e}^{2}+4\,Axa{b}^{2}{e}^{2}-4\,Ax{b}^{3}de-6\,B\ln \left ( bx+a \right ){a}^{3}{e}^{2}+4\,B\ln \left ( bx+a \right ){a}^{2}bde-4\,Bx{a}^{2}b{e}^{2}+8\,Bxa{b}^{2}de-2\,Bx{b}^{3}{d}^{2}+3\,Ab{a}^{2}{e}^{2}-2\,Aa{b}^{2}de-A{b}^{3}{d}^{2}-5\,B{a}^{3}{e}^{2}+6\,B{a}^{2}bde-Ba{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) }{2\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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]

1/2*(2*A*ln(b*x+a)*x^2*b^3*e^2-6*B*ln(b*x+a)*x^2*a*b^2*e^2+4*B*ln(b*x+a)*x^2*b^3*d*e+2*B*x^3*b^3*e^2+4*A*ln(b*
x+a)*x*a*b^2*e^2-12*B*ln(b*x+a)*x*a^2*b*e^2+8*B*ln(b*x+a)*x*a*b^2*d*e+4*B*x^2*a*b^2*e^2+2*A*ln(b*x+a)*a^2*b*e^
2+4*A*x*a*b^2*e^2-4*A*x*b^3*d*e-6*B*ln(b*x+a)*a^3*e^2+4*B*ln(b*x+a)*a^2*b*d*e-4*B*x*a^2*b*e^2+8*B*x*a*b^2*d*e-
2*B*x*b^3*d^2+3*A*b*a^2*e^2-2*A*a*b^2*d*e-A*b^3*d^2-5*B*a^3*e^2+6*B*a^2*b*d*e-B*a*b^2*d^2)*(b*x+a)/b^4/((b*x+a
)^2)^(3/2)

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Maxima [B]  time = 0.977602, size = 446, normalized size = 2.4 \begin{align*} \frac{B e^{2} x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{3 \, B a e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} - \frac{9 \, B a^{3} b e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, B a^{2} e^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, B a^{2} e^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac{{\left (2 \, B d e + A e^{2}\right )} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{3 \,{\left (2 \, B d e + A e^{2}\right )} a^{2} b^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{A d^{2}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{B a^{3} e^{2}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \,{\left (2 \, B d e + A e^{2}\right )} a b x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{B d^{2} + 2 \, A d e}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{{\left (B d^{2} + 2 \, A d e\right )} a}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

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]

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

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Fricas [A]  time = 1.3562, size = 521, normalized size = 2.8 \begin{align*} \frac{2 \, B b^{3} e^{2} x^{3} + 4 \, B a b^{2} e^{2} x^{2} -{\left (B a b^{2} + A b^{3}\right )} d^{2} + 2 \,{\left (3 \, B a^{2} b - A a b^{2}\right )} d e -{\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \,{\left (B b^{3} d^{2} - 2 \,{\left (2 \, B a b^{2} - A b^{3}\right )} d e + 2 \,{\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \,{\left (2 \, B a^{2} b d e -{\left (3 \, B a^{3} - A a^{2} b\right )} e^{2} +{\left (2 \, B b^{3} d e -{\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \,{\left (2 \, B a b^{2} d e -{\left (3 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end{align*}

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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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]

sage0*x