3.13.99 \(\int \frac {(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=161 \[ \frac {3 e^2 (a+b x) (b d-a e) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^3}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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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} \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 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 \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 )-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}} \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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IntegrateAlgebraic [B] time = 2.76, size = 2866, normalized size = 17.80 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(-(Sqrt[b^2]*d^2*(-(a*d) + b*d*x - 6*a*e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - d^2*(-(a^2*b*d) + 3*a^3*e + 6*a^2
*b*e*x - b^3*d*x^2 + 6*a*b^2*e*x^2))/(b*x^2*(-2*a*b^3 - 2*b^4*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + b*Sqrt[b^2]*x
^2*(2*a^2*b^2 + 4*a*b^3*x + 2*b^4*x^2)) + ((24*a^2*(b^2)^(3/2)*d^2*e*x)/b^4 + (16*a^4*Sqrt[b^2]*e^3*x)/b^4 + (
36*a*Sqrt[b^2]*d^2*e*x^2)/b + (32*a^3*Sqrt[b^2]*e^3*x^2)/b^3 + 24*Sqrt[b^2]*d^2*e*x^3 + (8*a^2*(b^2)^(3/2)*e^3
*x^3)/b^4 - (20*a*Sqrt[b^2]*e^3*x^4)/b - 8*Sqrt[b^2]*e^3*x^5 - (12*a^2*d^2*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^
2 - (4*a^4*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (12*a*d^2*e*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (12*a^3*e
^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - 24*d^2*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (20*a^2*e^3*x^2*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/b^2 + (12*a*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + 8*e^3*x^4*Sqrt[a^2 + 2*a*b*x +
b^2*x^2] - (24*a^3*e^3*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (48*a^2*e^3*x^3*
ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 24*a*e^3*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/a] + (24*a^2*Sqrt[b^2]*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x)
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^3 + (24*a*(b^2)^(3/2)*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(
Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4)/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(
a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((12*a^4*d*e^2)/(b^2)^(3/2) + (36*a^3*b*d*e^2*x)/(b^2)^(
3/2) + (36*a^2*d*e^2*x^2)/Sqrt[b^2] - (36*a^2*d*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (12*a^2*d*e^2*x^2*L
og[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (24*a*b^3*d*e^2*x^3*Log[-a - Sqrt[b^2]*x + S
qrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (12*b^4*d*e^2*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2]])/(b^2)^(3/2) + (12*a*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/b + 12*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^
2]] - (12*a^2*d*e^2*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (24*a*b^3*d*e^2*x^3*
Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (12*b^4*d*e^2*x^4*Log[a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (12*a*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 12*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^
2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2
*a*b*x + b^2*x^2])^2) + ((-4*a^5*e^3)/(b^3*Sqrt[b^2]) - (36*a^3*d*e^2*x)/(b*Sqrt[b^2]) - (16*a^4*e^3*x)/(b^2)^
(3/2) - (72*a^2*d*e^2*x^2)/Sqrt[b^2] - (16*a^3*e^3*x^2)/(b*Sqrt[b^2]) - (48*a*b*d*e^2*x^3)/Sqrt[b^2] + (12*a^3
*d*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (24*a^2*d*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (16*a^3*e^3*x
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (48*a*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (24*a^2*d*e^2*x^2*Arc
Tanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 48*a*d*e^2*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/a] + 24*b*d*e^2*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (24*
a*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^
2] - (24*b*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a]
)/Sqrt[b^2] + (12*a^3*e^3*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*e
^3*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (12*a*b*e^3*x^4*Log[-a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^2 - (12*a*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + (12*a^3*e^3*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*
Sqrt[b^2]) + (24*a^2*e^3*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (12*a*b*e^3*x^4
*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^2 - (12*a*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log
[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b)/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a
- Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)
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fricas [A] time = 0.39, size = 188, normalized size = 1.17 \begin {gather*} \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 {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*(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 \begin {gather*} \mathit {sage}_{0} x \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]
sage0*x
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maple [A] time = 0.06, size = 209, normalized size = 1.30 \begin {gather*} -\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}} \end {gather*}
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 \begin {gather*} \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}} \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]
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 \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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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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