3.1760 \(\int \frac{(A+B x) (d+e x)^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=191 \[ \frac{(a+b x) (d+e x)^2 (A b-a B)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e x (a+b x) (A b-a B) (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-a B) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B (a+b x) (d+e x)^3}{3 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

((A*b - a*B)*e*(b*d - a*e)*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*(a + b*x)*(d + e*x)
^2)/(2*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*(a + b*x)*(d + e*x)^3)/(3*b*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
((A*b - a*B)*(b*d - a*e)^2*(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 = 0.107902, antiderivative size = 191, 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} \[ \frac{(a+b x) (d+e x)^2 (A b-a B)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e x (a+b x) (A b-a B) (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-a B) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B (a+b x) (d+e x)^3}{3 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*e*(b*d - a*e)*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*(a + b*x)*(d + e*x)
^2)/(2*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*(a + b*x)*(d + e*x)^3)/(3*b*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
((A*b - a*B)*(b*d - a*e)^2*(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}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(A+B x) (d+e x)^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (\frac{(A b-a B) e (b d-a e)}{b^4}+\frac{(A b-a B) (b d-a e)^2}{b^4 (a+b x)}+\frac{(A b-a B) e (d+e x)}{b^3}+\frac{B (d+e x)^2}{b^2}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(A b-a B) e (b d-a e) x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) (a+b x) (d+e x)^2}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B (a+b x) (d+e x)^3}{3 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) (b d-a e)^2 (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 = 0.0941073, size = 118, normalized size = 0.62 \[ \frac{(a+b x) \left (b x \left (6 a^2 B e^2-3 a b e (2 A e+4 B d+B e x)+b^2 \left (3 A e (4 d+e x)+2 B \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+6 (A b-a B) (b d-a e)^2 \log (a+b x)\right )}{6 b^4 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*x*(6*a^2*B*e^2 - 3*a*b*e*(4*B*d + 2*A*e + B*e*x) + b^2*(3*A*e*(4*d + e*x) + 2*B*(3*d^2 + 3*d*e*x
+ e^2*x^2))) + 6*(A*b - a*B)*(b*d - a*e)^2*Log[a + b*x]))/(6*b^4*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.011, size = 212, normalized size = 1.1 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 2\,B{x}^{3}{b}^{3}{e}^{2}+3\,A{x}^{2}{b}^{3}{e}^{2}-3\,B{x}^{2}a{b}^{2}{e}^{2}+6\,B{x}^{2}{b}^{3}de+6\,A\ln \left ( bx+a \right ){a}^{2}b{e}^{2}-12\,A\ln \left ( bx+a \right ) a{b}^{2}de+6\,A\ln \left ( bx+a \right ){b}^{3}{d}^{2}-6\,Axa{b}^{2}{e}^{2}+12\,Ax{b}^{3}de-6\,B\ln \left ( bx+a \right ){a}^{3}{e}^{2}+12\,B\ln \left ( bx+a \right ){a}^{2}bde-6\,B\ln \left ( bx+a \right ) a{b}^{2}{d}^{2}+6\,Bx{a}^{2}b{e}^{2}-12\,Bxa{b}^{2}de+6\,Bx{b}^{3}{d}^{2} \right ) }{6\,{b}^{4}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^2/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [B]  time = 0.974402, size = 400, normalized size = 2.09 \begin{align*} -\frac{5 \, B a^{3} b e^{2} \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{5 \, B a^{2} e^{2} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{5 \, B a e^{2} x^{2}}{6 \, \sqrt{b^{2}} b} + A \sqrt{\frac{1}{b^{2}}} d^{2} \log \left (x + \frac{a}{b}\right ) + \frac{2 \, B a^{3} \sqrt{\frac{1}{b^{2}}} e^{2} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B e^{2} x^{2}}{3 \, b^{2}} + \frac{{\left (2 \, B d e + A e^{2}\right )} a^{2} b^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{{\left (2 \, B d e + A e^{2}\right )} a b x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{{\left (2 \, B d e + A e^{2}\right )} x^{2}}{2 \, \sqrt{b^{2}}} - \frac{{\left (B d^{2} + 2 \, A d e\right )} a \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{b} - \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} e^{2}}{3 \, b^{4}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}{\left (B d^{2} + 2 \, A d e\right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.24659, size = 319, normalized size = 1.67 \begin{align*} \frac{2 \, B b^{3} e^{2} x^{3} + 3 \,{\left (2 \, B b^{3} d e -{\left (B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 6 \,{\left (B b^{3} d^{2} - 2 \,{\left (B a b^{2} - A b^{3}\right )} d e +{\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x - 6 \,{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \,{\left (B a^{2} b - A a b^{2}\right )} d e +{\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.675601, size = 117, normalized size = 0.61 \begin{align*} \frac{B e^{2} x^{3}}{3 b} - \frac{x^{2} \left (- A b e^{2} + B a e^{2} - 2 B b d e\right )}{2 b^{2}} + \frac{x \left (- A a b e^{2} + 2 A b^{2} d e + B a^{2} e^{2} - 2 B a b d e + B b^{2} d^{2}\right )}{b^{3}} - \frac{\left (- A b + B a\right ) \left (a e - b d\right )^{2} \log{\left (a + b x \right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**2/((b*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.13253, size = 343, normalized size = 1.8 \begin{align*} \frac{2 \, B b^{2} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, B b^{2} d x^{2} e \mathrm{sgn}\left (b x + a\right ) + 6 \, B b^{2} d^{2} x \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, A b^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 12 \, B a b d x e \mathrm{sgn}\left (b x + a\right ) + 12 \, A b^{2} d x e \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) - 6 \, A a b x e^{2} \mathrm{sgn}\left (b x + a\right )}{6 \, b^{3}} - \frac{{\left (B a b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm{sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm{sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*(2*B*b^2*x^3*e^2*sgn(b*x + a) + 6*B*b^2*d*x^2*e*sgn(b*x + a) + 6*B*b^2*d^2*x*sgn(b*x + a) - 3*B*a*b*x^2*e^
2*sgn(b*x + a) + 3*A*b^2*x^2*e^2*sgn(b*x + a) - 12*B*a*b*d*x*e*sgn(b*x + a) + 12*A*b^2*d*x*e*sgn(b*x + a) + 6*
B*a^2*x*e^2*sgn(b*x + a) - 6*A*a*b*x*e^2*sgn(b*x + a))/b^3 - (B*a*b^2*d^2*sgn(b*x + a) - A*b^3*d^2*sgn(b*x + a
) - 2*B*a^2*b*d*e*sgn(b*x + a) + 2*A*a*b^2*d*e*sgn(b*x + a) + B*a^3*e^2*sgn(b*x + a) - A*a^2*b*e^2*sgn(b*x + a
))*log(abs(b*x + a))/b^4