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

Optimal. Leaf size=128 \[ -\frac {3 e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^2}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}+\frac {3 \sqrt {c} \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \begin {gather*} -\frac {3 e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^2}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}+\frac {3 \sqrt {c} \sqrt {d} e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.45 \begin {gather*} -\frac {2 e \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{\sqrt {d+e x} \left (a e^2-c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 152, normalized size = 1.19 \begin {gather*} \frac {3 \sqrt {c} \sqrt {d} e \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e^2-c d^2}}{c d^2-a e^2}\right )}{\left (a e^2-c d^2\right )^{5/2}}+\frac {2 a e^3-2 c d^2 e+3 c d e (d+e x)}{\sqrt {d+e x} \left (c d^2-a e^2\right )^2 \left (-a e^2+c d^2-c d (d+e x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*d^2*e + 2*a*e^3 + 3*c*d*e*(d + e*x))/((c*d^2 - a*e^2)^2*Sqrt[d + e*x]*(c*d^2 - a*e^2 - c*d*(d + e*x))) +
(3*Sqrt[c]*Sqrt[d]*e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(-(c*d^2
) + a*e^2)^(5/2)

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fricas [B]  time = 0.42, size = 502, normalized size = 3.92 \begin {gather*} \left [\frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x\right )}}, \frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) - {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2
+ 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(c*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) - 2*(3*c*d*e*x + c*d^2 + 2*a*e^2)
*sqrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 +
(c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6)*x), (3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt
(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) - (
3*c*d*e*x + c*d^2 + 2*a*e^2)*sqrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*
d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6)*x)]

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

[Out]

Timed out

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maple [A]  time = 0.06, size = 129, normalized size = 1.01 \begin {gather*} -\frac {3 c d e \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}-\frac {\sqrt {e x +d}\, c d e}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \left (c d e x +a \,e^{2}\right )}-\frac {2 e}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e*c*d/(a*e^2-c*d^2)^2*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)-3*e*c*d/(a*e^2-c*d^2)^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan(
(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)-2*e/(a*e^2-c*d^2)^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((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 positive or negative?

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mupad [B]  time = 0.75, size = 155, normalized size = 1.21 \begin {gather*} -\frac {\frac {2\,e}{a\,e^2-c\,d^2}+\frac {3\,c\,d\,e\,\left (d+e\,x\right )}{{\left (a\,e^2-c\,d^2\right )}^2}}{\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}+c\,d\,{\left (d+e\,x\right )}^{3/2}}-\frac {3\,\sqrt {c}\,\sqrt {d}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 42.33, size = 452, normalized size = 3.53 \begin {gather*} \frac {c d e \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} \log {\left (- a^{2} e^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + 2 a c d^{2} e^{2} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} - c^{2} d^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + \sqrt {d + e x} \right )}}{2 a e^{2} - 2 c d^{2}} - \frac {c d e \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} \log {\left (a^{2} e^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} - 2 a c d^{2} e^{2} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + c^{2} d^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + \sqrt {d + e x} \right )}}{2 a e^{2} - 2 c d^{2}} - \frac {2 c d e \sqrt {d + e x}}{2 a^{3} e^{6} - 4 a^{2} c d^{2} e^{4} + 2 a^{2} c d e^{5} x + 2 a c^{2} d^{4} e^{2} - 4 a c^{2} d^{3} e^{3} x + 2 c^{3} d^{5} e x} - \frac {2 e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2}}{c d} - d}} \right )}}{\left (a e^{2} - c d^{2}\right )^{2} \sqrt {\frac {a e^{2}}{c d} - d}} - \frac {2 e}{\sqrt {d + e x} \left (a e^{2} - c d^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*d*e*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3))*log(-a**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + 2*a*c*d**2*e**2
*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) - c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + sqrt(d + e*x))/(2*a*e**
2 - 2*c*d**2) - c*d*e*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3))*log(a**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) -
2*a*c*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + sqrt(d +
e*x))/(2*a*e**2 - 2*c*d**2) - 2*c*d*e*sqrt(d + e*x)/(2*a**3*e**6 - 4*a**2*c*d**2*e**4 + 2*a**2*c*d*e**5*x + 2
*a*c**2*d**4*e**2 - 4*a*c**2*d**3*e**3*x + 2*c**3*d**5*e*x) - 2*e*atan(sqrt(d + e*x)/sqrt(a*e**2/(c*d) - d))/(
(a*e**2 - c*d**2)**2*sqrt(a*e**2/(c*d) - d)) - 2*e/(sqrt(d + e*x)*(a*e**2 - c*d**2)**2)

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