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

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Rubi [A]  time = 0.09, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.06, size = 125, normalized size = 0.78 \[ \frac {-5 a^3 e^3+a^2 b e^2 (9 d-4 e x)+a b^2 e \left (-3 d^2+12 d e x+4 e^2 x^2\right )-6 e^2 (a+b x)^2 (a e-b d) \log (a+b x)-\left (b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \]

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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fricas [A]  time = 0.96, size = 188, normalized size = 1.17 \[ \frac {2 \, b^{3} e^{3} x^{3} + 4 \, a b^{2} e^{3} x^{2} - b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} - 2 \, {\left (3 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 2 \, a^{2} b e^{3}\right )} x + 6 \, {\left (a^{2} b d e^{2} - a^{3} e^{3} + {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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

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maple [A]  time = 0.06, size = 209, normalized size = 1.30 \[ -\frac {\left (6 a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-6 b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )-2 b^{3} e^{3} x^{3}+12 a^{2} b \,e^{3} x \ln \left (b x +a \right )-12 a \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-4 a \,b^{2} e^{3} x^{2}+6 a^{3} e^{3} \ln \left (b x +a \right )-6 a^{2} b d \,e^{2} \ln \left (b x +a \right )+4 a^{2} b \,e^{3} x -12 a \,b^{2} d \,e^{2} x +6 b^{3} d^{2} e x +5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \]

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*b^3*d*e^2*x^2*ln(b*x+a)-2*b^3*e^3*x^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*a*b^2*e^3*x^2+6*a^3*e^3*ln(b*x+a)-6*a^2*b*d*e^2*ln(b*x+a)+4*a^2*b*e^3*x-12*a*b^2*d*e^2*x+6*b^3
*d^2*e*x+5*a^3*e^3-9*a^2*b*d*e^2+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 = 0.98, size = 237, normalized size = 1.47 \[ \frac {e^{3} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {3 \, d e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {3 \, a e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {3 \, d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, a^{2} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {6 \, a d e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, a^{2} e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a d^{2} e}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, a^{2} d e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, a^{3} e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \]

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^3 - 3*a*e^3*log(x + a/b)/b^4 - 3*d^2*e/(s
qrt(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) + 6*a*d*e^2*x/(b^4*(x + a/b)
^2) - 6*a^2*e^3*x/(b^5*(x + a/b)^2) - 1/2*d^3/(b^3*(x + a/b)^2) + 3/2*a*d^2*e/(b^4*(x + a/b)^2) + 9/2*a^2*d*e^
2/(b^5*(x + a/b)^2) - 11/2*a^3*e^3/(b^6*(x + a/b)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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