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

Optimal. Leaf size=110 \[ \frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2}}-\frac {e \sqrt {d+e x}}{4 b (a+b x) (b d-a e)}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 47, 51, 63, 208} \begin {gather*} \frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2}}-\frac {e \sqrt {d+e x}}{4 b (a+b x) (b d-a e)}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.47 \begin {gather*} \frac {2 e^2 (d+e x)^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (a e-b d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 125, normalized size = 1.14 \begin {gather*} -\frac {e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{3/2} (a e-b d)^{3/2}}-\frac {e^2 \sqrt {d+e x} (-a e+b (d+e x)+b d)}{4 b (b d-a e) (-a e-b (d+e x)+b d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(e^2*Sqrt[d + e*x]*(b*d - a*e + b*(d + e*x)))/(b*(b*d - a*e)*(b*d - a*e - b*(d + e*x))^2) - (e^2*ArcTan[(
Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(3/2)*(-(b*d) + a*e)^(3/2))

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fricas [B]  time = 0.44, size = 456, normalized size = 4.15 \begin {gather*} \left [-\frac {{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\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 (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} + {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} + {\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \, {\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}, -\frac {{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\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 (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} + {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} + {\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \, {\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*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*(2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d
))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5*d^2 - 2*a
^2*b^4*d*e + a^3*b^3*e^2)*x), -1/4*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^
2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sq
rt(e*x + d))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5
*d^2 - 2*a^2*b^4*d*e + a^3*b^3*e^2)*x)]

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giac [A]  time = 0.18, size = 132, normalized size = 1.20 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{2} d - a b e\right )} \sqrt {-b^{2} d + a b e}} - \frac {{\left (x e + d\right )}^{\frac {3}{2}} b e^{2} + \sqrt {x e + d} b d e^{2} - \sqrt {x e + d} a e^{3}}{4 \, {\left (b^{2} d - a b e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 111, normalized size = 1.01 \begin {gather*} \frac {e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} e^{2}}{4 \left (b e x +a e \right )^{2} \left (a e -b d \right )}-\frac {\sqrt {e x +d}\, e^{2}}{4 \left (b e x +a e \right )^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^2/(b*e*x+a*e)^2/(a*e-b*d)*(e*x+d)^(3/2)-1/4*e^2/(b*e*x+a*e)^2/b*(e*x+d)^(1/2)+1/4*e^2/(a*e-b*d)/b/((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((b*x+a)*(e*x+d)^(1/2)/(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 = 0.10, size = 135, normalized size = 1.23 \begin {gather*} \frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{4\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {e^2\,\sqrt {d+e\,x}}{4\,b}-\frac {e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,\left (a\,e-b\,d\right )}}{b^2\,{\left (d+e\,x\right )}^2-\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (d+e\,x\right )+a^2\,e^2+b^2\,d^2-2\,a\,b\,d\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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