3.13.96 \(\int \frac {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=182 \[ \frac {b (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}+\frac {a+b x}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac {b^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 182, 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, 44} \begin {gather*} \frac {b (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}+\frac {a+b x}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac {b^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 {1}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b^2}{(b d-a e)^3 (a+b x)}-\frac {e}{b (b d-a e) (d+e x)^3}-\frac {e}{(b d-a e)^2 (d+e x)^2}-\frac {b e}{(b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {a+b x}{2 (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 97, normalized size = 0.53 \begin {gather*} \frac {(a+b x) \left (2 b^2 (d+e x)^2 \log (a+b x)+(b d-a e) (-a e+3 b d+2 b e x)-2 b^2 (d+e x)^2 \log (d+e x)\right )}{2 \sqrt {(a+b x)^2} (d+e x)^2 (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 5.79, size = 1022, normalized size = 5.62 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 b x a+b^2 x^2}}{a}\right ) b^2}{(b d-a e)^3}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b^2} e x}{2 b d-a e}-\frac {e \sqrt {a^2+2 b x a+b^2 x^2}}{2 b d-a e}\right ) b^2}{(b d-a e)^3}-\frac {\sqrt {b^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b}{2 (b d-a e)^3}-\frac {\sqrt {b^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b}{2 (b d-a e)^3}+\frac {\sqrt {b^2} \log \left (2 b d-a e+\sqrt {b^2} e x-e \sqrt {a^2+2 b x a+b^2 x^2}\right ) b}{2 (b d-a e)^3}+\frac {\sqrt {b^2} \log \left (2 b d-a e-\sqrt {b^2} e x+e \sqrt {a^2+2 b x a+b^2 x^2}\right ) b}{2 (b d-a e)^3}+\frac {-8 \sqrt {a^2+2 b x a+b^2 x^2} \left (-3 d^3 b^4-2 e^3 x^3 b^4+d e^2 x^2 b^4+4 d^2 e x b^4-3 a e^3 x^2 b^3+7 a d^2 e b^3-4 a d e^2 x b^3-5 a^2 d e^2 b^2+a^3 e^3 b\right ) b^2-8 \left (b^2\right )^{3/2} \left (-e^3 a^4+5 b d e^2 a^3-b e^3 x a^3+3 b^2 e^3 x^2 a^2-7 b^2 d^2 e a^2+9 b^2 d e^2 x a^2+3 b^3 d^3 a+5 b^3 e^3 x^3 a+3 b^3 d e^2 x^2 a-11 b^3 d^2 e x a+2 b^4 e^3 x^4-b^4 d e^2 x^3-4 b^4 d^2 e x^2+3 b^4 d^3 x\right )}{\sqrt {b^2} \sqrt {a^2+2 b x a+b^2 x^2} \left (-16 d^4 b^4-16 e^4 x^4 b^4+32 d^2 e^2 x^2 b^4-32 a e^4 x^3 b^3-32 a d e^3 x^2 b^3+32 a d^3 e b^3+32 a d^2 e^2 x b^3-16 a^2 d^2 e^2 b^2-16 a^2 e^4 x^2 b^2-32 a^2 d e^3 x b^2\right ) (b d-a e)^2+\left (16 e^4 x^5 b^6-32 d^2 e^2 x^3 b^6+16 d^4 x b^6+16 a d^4 b^5+48 a e^4 x^4 b^5+32 a d e^3 x^3 b^5-64 a d^2 e^2 x^2 b^5-32 a d^3 e x b^5+48 a^2 e^4 x^3 b^4+64 a^2 d e^3 x^2 b^4-32 a^2 d^3 e b^4-16 a^2 d^2 e^2 x b^4+16 a^3 d^2 e^2 b^3+16 a^3 e^4 x^2 b^3+32 a^3 d e^3 x b^3\right ) (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

(-8*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-3*b^4*d^3 + 7*a*b^3*d^2*e - 5*a^2*b^2*d*e^2 + a^3*b*e^3 + 4*b^4*d^2*e*
x - 4*a*b^3*d*e^2*x + b^4*d*e^2*x^2 - 3*a*b^3*e^3*x^2 - 2*b^4*e^3*x^3) - 8*(b^2)^(3/2)*(3*a*b^3*d^3 - 7*a^2*b^
2*d^2*e + 5*a^3*b*d*e^2 - a^4*e^3 + 3*b^4*d^3*x - 11*a*b^3*d^2*e*x + 9*a^2*b^2*d*e^2*x - a^3*b*e^3*x - 4*b^4*d
^2*e*x^2 + 3*a*b^3*d*e^2*x^2 + 3*a^2*b^2*e^3*x^2 - b^4*d*e^2*x^3 + 5*a*b^3*e^3*x^3 + 2*b^4*e^3*x^4))/(Sqrt[b^2
]*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-16*b^4*d^4 + 32*a*b^3*d^3*e - 16*a^2*b^2*d^2*e^2 + 32*a*b^3*d^
2*e^2*x - 32*a^2*b^2*d*e^3*x + 32*b^4*d^2*e^2*x^2 - 32*a*b^3*d*e^3*x^2 - 16*a^2*b^2*e^4*x^2 - 32*a*b^3*e^4*x^3
- 16*b^4*e^4*x^4) + (b*d - a*e)^2*(16*a*b^5*d^4 - 32*a^2*b^4*d^3*e + 16*a^3*b^3*d^2*e^2 + 16*b^6*d^4*x - 32*a
*b^5*d^3*e*x - 16*a^2*b^4*d^2*e^2*x + 32*a^3*b^3*d*e^3*x - 64*a*b^5*d^2*e^2*x^2 + 64*a^2*b^4*d*e^3*x^2 + 16*a^
3*b^3*e^4*x^2 - 32*b^6*d^2*e^2*x^3 + 32*a*b^5*d*e^3*x^3 + 48*a^2*b^4*e^4*x^3 + 48*a*b^5*e^4*x^4 + 16*b^6*e^4*x
^5)) - (b^2*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/(b*d - a*e)^3 + (b^2*ArcTanh[(Sqrt[b^2
]*e*x)/(2*b*d - a*e) - (e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(b*d - a*e)^3 - (b*Sqrt[b^2]*Log[-a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(b*d - a*e)^3) - (b*Sqrt[b^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(2*(b*d - a*e)^3) + (b*Sqrt[b^2]*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*
x + b^2*x^2]])/(2*(b*d - a*e)^3) + (b*Sqrt[b^2]*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]])/(2*(b*d - a*e)^3)

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fricas [A]  time = 0.40, size = 242, normalized size = 1.33 \begin {gather*} \frac {3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (b^{3} d^{5} - 3 \, a b^{2} d^{4} e + 3 \, a^{2} b d^{3} e^{2} - a^{3} d^{2} e^{3} + {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} x^{2} + 2 \, {\left (b^{3} d^{4} e - 3 \, a b^{2} d^{3} e^{2} + 3 \, a^{2} b d^{2} e^{3} - a^{3} d e^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 174, normalized size = 0.96 \begin {gather*} \frac {1}{2} \, {\left (\frac {2 \, b^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}} - \frac {2 \, b^{2} e \log \left ({\left | x e + d \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} + \frac {3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x}{{\left (b d - a e\right )}^{3} {\left (x e + d\right )}^{2}}\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 162, normalized size = 0.89 \begin {gather*} -\frac {\left (b x +a \right ) \left (2 b^{2} e^{2} x^{2} \ln \left (b x +a \right )-2 b^{2} e^{2} x^{2} \ln \left (e x +d \right )+4 b^{2} d e x \ln \left (b x +a \right )-4 b^{2} d e x \ln \left (e x +d \right )-2 a b \,e^{2} x +2 b^{2} d^{2} \ln \left (b x +a \right )-2 b^{2} d^{2} \ln \left (e x +d \right )+2 b^{2} d e x +a^{2} e^{2}-4 a b d e +3 b^{2} d^{2}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \left (e x +d \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(1/(((a + b*x)^2)^(1/2)*(d + e*x)^3), x)

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sympy [B]  time = 1.12, size = 381, normalized size = 2.09 \begin {gather*} \frac {b^{2} \log {\left (x + \frac {- \frac {a^{4} b^{2} e^{4}}{\left (a e - b d\right )^{3}} + \frac {4 a^{3} b^{3} d e^{3}}{\left (a e - b d\right )^{3}} - \frac {6 a^{2} b^{4} d^{2} e^{2}}{\left (a e - b d\right )^{3}} + \frac {4 a b^{5} d^{3} e}{\left (a e - b d\right )^{3}} + a b^{2} e - \frac {b^{6} d^{4}}{\left (a e - b d\right )^{3}} + b^{3} d}{2 b^{3} e} \right )}}{\left (a e - b d\right )^{3}} - \frac {b^{2} \log {\left (x + \frac {\frac {a^{4} b^{2} e^{4}}{\left (a e - b d\right )^{3}} - \frac {4 a^{3} b^{3} d e^{3}}{\left (a e - b d\right )^{3}} + \frac {6 a^{2} b^{4} d^{2} e^{2}}{\left (a e - b d\right )^{3}} - \frac {4 a b^{5} d^{3} e}{\left (a e - b d\right )^{3}} + a b^{2} e + \frac {b^{6} d^{4}}{\left (a e - b d\right )^{3}} + b^{3} d}{2 b^{3} e} \right )}}{\left (a e - b d\right )^{3}} + \frac {- a e + 3 b d + 2 b e x}{2 a^{2} d^{2} e^{2} - 4 a b d^{3} e + 2 b^{2} d^{4} + x^{2} \left (2 a^{2} e^{4} - 4 a b d e^{3} + 2 b^{2} d^{2} e^{2}\right ) + x \left (4 a^{2} d e^{3} - 8 a b d^{2} e^{2} + 4 b^{2} d^{3} e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

b**2*log(x + (-a**4*b**2*e**4/(a*e - b*d)**3 + 4*a**3*b**3*d*e**3/(a*e - b*d)**3 - 6*a**2*b**4*d**2*e**2/(a*e
- b*d)**3 + 4*a*b**5*d**3*e/(a*e - b*d)**3 + a*b**2*e - b**6*d**4/(a*e - b*d)**3 + b**3*d)/(2*b**3*e))/(a*e -
b*d)**3 - b**2*log(x + (a**4*b**2*e**4/(a*e - b*d)**3 - 4*a**3*b**3*d*e**3/(a*e - b*d)**3 + 6*a**2*b**4*d**2*e
**2/(a*e - b*d)**3 - 4*a*b**5*d**3*e/(a*e - b*d)**3 + a*b**2*e + b**6*d**4/(a*e - b*d)**3 + b**3*d)/(2*b**3*e)
)/(a*e - b*d)**3 + (-a*e + 3*b*d + 2*b*e*x)/(2*a**2*d**2*e**2 - 4*a*b*d**3*e + 2*b**2*d**4 + x**2*(2*a**2*e**4
- 4*a*b*d*e**3 + 2*b**2*d**2*e**2) + x*(4*a**2*d*e**3 - 8*a*b*d**2*e**2 + 4*b**2*d**3*e))

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