### 3.1597 $$\int \frac{(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=161 $\frac{3 e^2 (a+b x) (b d-a e) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^3 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

(-3*e*(b*d - a*e)^2)/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^3/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (e^3*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e^2*(b*d - a*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 = 0.0940398, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {646, 43} $\frac{3 e^2 (a+b x) (b d-a e) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^3 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{\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{(d+e x)^3}{\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{e^3}{b^6}+\frac{(b d-a e)^3}{b^6 (a+b x)^3}+\frac{3 e (b d-a e)^2}{b^6 (a+b x)^2}+\frac{3 e^2 (b d-a e)}{b^6 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 e (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^3}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^3 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e^2 (b d-a 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 = 0.056599, size = 125, normalized size = 0.78 $\frac{a^2 b e^2 (9 d-4 e x)-5 a^3 e^3+a b^2 e \left (-3 d^2+12 d e x+4 e^2 x^2\right )-6 e^2 (a+b x)^2 (a e-b d) \log (a+b x)+b^3 \left (-\left (6 d^2 e x+d^3-2 e^3 x^3\right )\right )}{2 b^4 (a+b x) \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-5*a^3*e^3 + a^2*b*e^2*(9*d - 4*e*x) + a*b^2*e*(-3*d^2 + 12*d*e*x + 4*e^2*x^2) - b^3*(d^3 + 6*d^2*e*x - 2*e^3
*x^3) - 6*e^2*(-(b*d) + a*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 [A]  time = 0.2, size = 209, normalized size = 1.3 \begin{align*} -{\frac{ \left ( 6\,\ln \left ( bx+a \right ){x}^{2}a{b}^{2}{e}^{3}-6\,\ln \left ( bx+a \right ){x}^{2}{b}^{3}d{e}^{2}-2\,{x}^{3}{b}^{3}{e}^{3}+12\,\ln \left ( bx+a \right ) x{a}^{2}b{e}^{3}-12\,\ln \left ( bx+a \right ) xa{b}^{2}d{e}^{2}-4\,{x}^{2}a{b}^{2}{e}^{3}+6\,\ln \left ( bx+a \right ){a}^{3}{e}^{3}-6\,\ln \left ( bx+a \right ){a}^{2}bd{e}^{2}+4\,x{a}^{2}b{e}^{3}-12\,xa{b}^{2}d{e}^{2}+6\,x{b}^{3}{d}^{2}e+5\,{a}^{3}{e}^{3}-9\,d{e}^{2}{a}^{2}b+3\,a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3} \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((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/2*(6*ln(b*x+a)*x^2*a*b^2*e^3-6*ln(b*x+a)*x^2*b^3*d*e^2-2*x^3*b^3*e^3+12*ln(b*x+a)*x*a^2*b*e^3-12*ln(b*x+a)*
x*a*b^2*d*e^2-4*x^2*a*b^2*e^3+6*ln(b*x+a)*a^3*e^3-6*ln(b*x+a)*a^2*b*d*e^2+4*x*a^2*b*e^3-12*x*a*b^2*d*e^2+6*x*b
^3*d^2*e+5*a^3*e^3-9*d*e^2*a^2*b+3*a*b^2*d^2*e+b^3*d^3)*(b*x+a)/b^4/((b*x+a)^2)^(3/2)

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Maxima [B]  time = 1.05174, size = 390, normalized size = 2.42 \begin{align*} \frac{e^{3} x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{3 \, d e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{3 \, a e^{3} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{9 \, a^{2} b^{2} d e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{9 \, a^{3} b e^{3}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{6 \, a b d e^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, a^{2} e^{3} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{3 \, d^{2} e}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{2 \, a^{2} e^{3}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac{d^{3}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{3 \, a d^{2} e}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{a^{3} e^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6106, size = 375, normalized size = 2.33 \begin{align*} \frac{2 \, b^{3} e^{3} x^{3} + 4 \, a b^{2} e^{3} x^{2} - b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - 2 \,{\left (3 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 2 \, a^{2} b e^{3}\right )} x + 6 \,{\left (a^{2} b d e^{2} - a^{3} e^{3} +{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (a b^{2} d e^{2} - a^{2} b e^{3}\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((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*b^3*e^3*x^3 + 4*a*b^2*e^3*x^2 - b^3*d^3 - 3*a*b^2*d^2*e + 9*a^2*b*d*e^2 - 5*a^3*e^3 - 2*(3*b^3*d^2*e -
6*a*b^2*d*e^2 + 2*a^2*b*e^3)*x + 6*(a^2*b*d*e^2 - a^3*e^3 + (b^3*d*e^2 - a*b^2*e^3)*x^2 + 2*(a*b^2*d*e^2 - a^2
*b*e^3)*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 (d + e x\right )^{3}}{\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((e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

[Out]

sage0*x