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

Optimal. Leaf size=99 \[ -\frac {3 e}{\sqrt {d+e x} (b d-a e)^2}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}+\frac {3 \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \begin {gather*} -\frac {3 e}{\sqrt {d+e x} (b d-a e)^2}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}+\frac {3 \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*e)/((b*d - a*e)^2*Sqrt[d + e*x]) - 1/((b*d - a*e)*(a + b*x)*Sqrt[d + e*x]) + (3*Sqrt[b]*e*ArcTanh[(Sqrt[b]
*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(5/2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx\\ &=-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 e) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)}\\ &=-\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 b e) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 (b d-a e)^2}\\ &=-\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{(b d-a e)^2}\\ &=-\frac {3 e}{(b d-a e)^2 \sqrt {d+e x}}-\frac {1}{(b d-a e) (a+b x) \sqrt {d+e x}}+\frac {3 \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.48 \begin {gather*} -\frac {2 e \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {d+e x} (a e-b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*Hypergeometric2F1[-1/2, 2, 1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^2*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.31, size = 115, normalized size = 1.16 \begin {gather*} \frac {e (2 a e+3 b (d+e x)-2 b d)}{\sqrt {d+e x} (b d-a e)^2 (-a e-b (d+e x)+b d)}+\frac {3 \sqrt {b} e \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-2*b*d + 2*a*e + 3*b*(d + e*x)))/((b*d - a*e)^2*Sqrt[d + e*x]*(b*d - a*e - b*(d + e*x))) + (3*Sqrt[b]*e*Ar
cTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(-(b*d) + a*e)^(5/2)

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fricas [B]  time = 0.42, size = 423, normalized size = 4.27 \begin {gather*} \left [\frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*s
qrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d
^2*e + a^3*d*e^2 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^
3)*x), (3*(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(
-b/(b*d - a*e))/(b*e*x + b*d)) - (3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d^2*e + a^3*d*e^2
+ (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*x)]

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giac [A]  time = 0.18, size = 153, normalized size = 1.55 \begin {gather*} -\frac {3 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (x e + d\right )} b e - 2 \, b d e + 2 \, a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left ({\left (x e + d\right )}^{\frac {3}{2}} b - \sqrt {x e + d} b d + \sqrt {x e + d} a e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt(-b^2*d + a*b*e)) - (
3*(x*e + d)*b*e - 2*b*d*e + 2*a*e^2)/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*((x*e + d)^(3/2)*b - sqrt(x*e + d)*b*d +
sqrt(x*e + d)*a*e))

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maple [A]  time = 0.07, size = 101, normalized size = 1.02 \begin {gather*} -\frac {3 b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}-\frac {\sqrt {e x +d}\, b e}{\left (a e -b d \right )^{2} \left (b e x +a e \right )}-\frac {2 e}{\left (a e -b d \right )^{2} \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2)/(b^2*x^2+2*a*b*x+a^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 positive or negative?

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mupad [B]  time = 0.16, size = 123, normalized size = 1.24 \begin {gather*} -\frac {\frac {2\,e}{a\,e-b\,d}+\frac {3\,b\,e\,\left (d+e\,x\right )}{{\left (a\,e-b\,d\right )}^2}}{b\,{\left (d+e\,x\right )}^{3/2}+\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}-\frac {3\,\sqrt {b}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )}{{\left (a\,e-b\,d\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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