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

Optimal. Leaf size=136 \[ \frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{3/2}}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (a+b x) (b d-a e)}-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 51, 63, 208} \begin {gather*} -\frac {e^2 \sqrt {d+e x}}{8 b^2 (a+b x) (b d-a e)}+\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{3/2}}-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*Sqrt[d + e*x])/(4*b^2*(a + b*x)^2) - (e^2*Sqrt[d + e*x])/(8*b^2*(b*d - a*e)*(a + b*x)) - (d + e*x)^(3/2)/(
3*b*(a + b*x)^3) + (e^3*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(5/2)*(b*d - a*e)^(3/2))

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 47

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

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 {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{3/2}}{(a+b x)^4} \, dx\\ &=-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}+\frac {e \int \frac {\sqrt {d+e x}}{(a+b x)^3} \, dx}{2 b}\\ &=-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}+\frac {e^2 \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{8 b^2}\\ &=-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (b d-a e) (a+b x)}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}-\frac {e^3 \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b^2 (b d-a e)}\\ &=-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (b d-a e) (a+b x)}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}-\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 b^2 (b d-a e)}\\ &=-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (b d-a e) (a+b x)}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}+\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.38 \begin {gather*} \frac {2 e^3 (d+e x)^{5/2} \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*(d + e*x)^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^4)

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IntegrateAlgebraic [A]  time = 0.68, size = 166, normalized size = 1.22 \begin {gather*} -\frac {e^3 \sqrt {d+e x} \left (3 a^2 e^2+8 a b e (d+e x)-6 a b d e+3 b^2 d^2-3 b^2 (d+e x)^2-8 b^2 d (d+e x)\right )}{24 b^2 (b d-a e) (-a e-b (d+e x)+b d)^3}-\frac {e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{5/2} (a e-b d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(e^3*Sqrt[d + e*x]*(3*b^2*d^2 - 6*a*b*d*e + 3*a^2*e^2 - 8*b^2*d*(d + e*x) + 8*a*b*e*(d + e*x) - 3*b^2*(d
+ e*x)^2))/(b^2*(b*d - a*e)*(b*d - a*e - b*(d + e*x))^3) - (e^3*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e
*x])/(b*d - a*e)])/(8*b^(5/2)*(-(b*d) + a*e)^(3/2))

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fricas [B]  time = 0.43, size = 666, normalized size = 4.90 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a
*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) + 2*(8*b^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b
*e^3 + 3*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(7*b^4*d^2*e - 11*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3
*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + (b^8*d^2 - 2*a*b^7*d*e + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*
d*e + a^3*b^5*e^2)*x^2 + 3*(a^2*b^6*d^2 - 2*a^3*b^5*d*e + a^4*b^4*e^2)*x), -1/24*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3
*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d))
+ (8*b^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b*e^3 + 3*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(7*b^4*d^2*e -
11*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + (b^8*d^2 - 2*a
*b^7*d*e + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*d*e + a^3*b^5*e^2)*x^2 + 3*(a^2*b^6*d^2 - 2*a^3*b^5*d*e
+ a^4*b^4*e^2)*x)]

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giac [A]  time = 0.19, size = 191, normalized size = 1.40 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{3} d - a b^{2} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} e^{3} + 8 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {x e + d} b^{2} d^{2} e^{3} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {x e + d} a b d e^{4} - 3 \, \sqrt {x e + d} a^{2} e^{5}}{24 \, {\left (b^{3} d - a b^{2} e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^3*d - a*b^2*e)*sqrt(-b^2*d + a*b*e)) - 1/24*(3*(x*e
+ d)^(5/2)*b^2*e^3 + 8*(x*e + d)^(3/2)*b^2*d*e^3 - 3*sqrt(x*e + d)*b^2*d^2*e^3 - 8*(x*e + d)^(3/2)*a*b*e^4 + 6
*sqrt(x*e + d)*a*b*d*e^4 - 3*sqrt(x*e + d)*a^2*e^5)/((b^3*d - a*b^2*e)*((x*e + d)*b - b*d + a*e)^3)

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maple [A]  time = 0.06, size = 163, normalized size = 1.20 \begin {gather*} -\frac {\sqrt {e x +d}\, a \,e^{4}}{8 \left (b e x +a e \right )^{3} b^{2}}+\frac {\sqrt {e x +d}\, d \,e^{3}}{8 \left (b e x +a e \right )^{3} b}+\frac {\left (e x +d \right )^{\frac {5}{2}} e^{3}}{8 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {3}{2}} e^{3}}{3 \left (b e x +a e \right )^{3} b}+\frac {e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(5/2)-1/3*e^3/(b*e*x+a*e)^3/b*(e*x+d)^(3/2)-1/8*e^4/(b*e*x+a*e)^3/b^2*
(e*x+d)^(1/2)*a+1/8*e^3/(b*e*x+a*e)^3/b*(e*x+d)^(1/2)*d+1/8*e^3/(a*e-b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+
d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)

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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((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^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 = 0.14, size = 209, normalized size = 1.54 \begin {gather*} \frac {e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{8\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,b}-\frac {e^3\,{\left (d+e\,x\right )}^{5/2}}{8\,\left (a\,e-b\,d\right )}+\frac {e^3\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{8\,b^2}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^3*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(8*b^(5/2)*(a*e - b*d)^(3/2)) - ((e^3*(d + e*x)^(3/2))
/(3*b) - (e^3*(d + e*x)^(5/2))/(8*(a*e - b*d)) + (e^3*(a*e - b*d)*(d + e*x)^(1/2))/(8*b^2))/((d + e*x)*(3*b^3*
d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e) + b^3*(d + e*x)^3 - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^2 + a^3*e^3 - b^3*d^3 + 3
*a*b^2*d^2*e - 3*a^2*b*d*e^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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