3.2015 $$\int \frac{\sqrt{d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx$$

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0640229, antiderivative size = 128, normalized size of antiderivative = 1., 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} $-\frac{3 e}{\sqrt{d+e x} \left (c d^2-a e^2\right )^2}-\frac{1}{\sqrt{d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}+\frac{3 \sqrt{c} \sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}$

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 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]

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{\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*}

Mathematica [C]  time = 0.0145002, size = 57, normalized size = 0.45 $-\frac{2 e \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{\sqrt{d+e x} \left (a e^2-c d^2\right )^2}$

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])

________________________________________________________________________________________

Maple [A]  time = 0.236, size = 129, normalized size = 1. \begin{align*} -2\,{\frac{e}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}\sqrt{ex+d}}}-{\frac{dec}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2} \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}-3\,{\frac{dec}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Fricas [B]  time = 2.06362, size = 1010, normalized size = 7.89 \begin{align*} \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{align*}

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)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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