3.16.97 \(\int \frac {(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1597]

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

[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 = 0.13, 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 = {660, 45} \begin {gather*} \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}} \end {gather*}

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 45

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 660

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 = 0.04, size = 125, normalized size = 0.78 \begin {gather*} \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 )-b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )-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}} \end {gather*}

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 = 0.64, size = 209, normalized size = 1.30

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} x}{\left (b x +a \right ) b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 a^{2} e^{3}+6 a b d \,e^{2}-3 b^{2} d^{2} e \right ) x -\frac {5 e^{3} a^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}}{2 b}\right )}{\left (b x +a \right )^{3} b^{3}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (a e -b d \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\)	\(154\)
default	\(-\frac {\left (6 \ln \left (b x +a \right ) a \,b^{2} e^{3} x^{2}-6 \ln \left (b x +a \right ) b^{3} d \,e^{2} x^{2}-2 b^{3} e^{3} x^{3}+12 \ln \left (b x +a \right ) a^{2} b \,e^{3} x -12 \ln \left (b x +a \right ) a \,b^{2} d \,e^{2} x -4 a \,b^{2} e^{3} x^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}-6 \ln \left (b x +a \right ) a^{2} b d \,e^{2}+4 a^{2} b \,e^{3} x -12 a \,b^{2} d \,e^{2} x +6 b^{3} d^{2} e x +5 e^{3} a^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (b x +a \right )}{2 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(209\)

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,method=_RETURNVERBOSE)

[Out]

-1/2*(6*ln(b*x+a)*a*b^2*e^3*x^2-6*ln(b*x+a)*b^3*d*e^2*x^2-2*b^3*e^3*x^3+12*ln(b*x+a)*a^2*b*e^3*x-12*ln(b*x+a)*
a*b^2*d*e^2*x-4*a*b^2*e^3*x^2+6*ln(b*x+a)*a^3*e^3-6*ln(b*x+a)*a^2*b*d*e^2+4*a^2*b*e^3*x-12*a*b^2*d*e^2*x+6*b^3
*d^2*e*x+5*e^3*a^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 [A]
time = 0.28, size = 231, normalized size = 1.43 \begin {gather*} \frac {x^{2} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, d^{2} e}{\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 {2 \, a^{2} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, a d x e^{2}}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a d^{2} e}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, a^{2} x e^{3}}{b^{5} {\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}} \end {gather*}

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]

x^2*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*d^2*e/(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 + 2*a^2*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 1/2*d^3/(b^3*(x + a/b)^2)
+ 6*a*d*x*e^2/(b^4*(x + a/b)^2) + 3/2*a*d^2*e/(b^4*(x + a/b)^2) - 6*a^2*x*e^3/(b^5*(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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Fricas [A]
time = 3.01, size = 168, normalized size = 1.04 \begin {gather*} -\frac {b^{3} d^{3} - {\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3}\right )} e^{3} - 3 \, {\left (4 \, a b^{2} d x + 3 \, a^{2} b d\right )} e^{2} + 3 \, {\left (2 \, b^{3} d^{2} x + a b^{2} d^{2}\right )} e + 6 \, {\left ({\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} e^{3} - {\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} e^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 [A]
time = 1.29, size = 131, normalized size = 0.81 \begin {gather*} \frac {x e^{3}}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, {\left (b d e^{2} - a e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{3} d^{3} + 3 \, a b^{2} d^{2} e - 9 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 6 \, {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

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]

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

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

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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