3.2025 \(\int \frac{\sqrt{d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=208 \[ -\frac{35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}+\frac{35 c d e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{35 e^2}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}+\frac{7 e}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)^2} \]

[Out]

(35*e^2)/(12*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(3/2)) + (7*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(3/2)) + (35*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (35
*c^(3/2)*d^(3/2)*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(9/2))

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Rubi [A]  time = 0.137705, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \[ -\frac{35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}+\frac{35 c d e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{35 e^2}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}+\frac{7 e}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)^2} \]

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)^3,x]

[Out]

(35*e^2)/(12*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(3/2)) + (7*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(3/2)) + (35*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (35
*c^(3/2)*d^(3/2)*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(9/2))

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 )^3} \, dx &=\int \frac{1}{(a e+c d x)^3 (d+e x)^{5/2}} \, dx\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}-\frac{(7 e) \int \frac{1}{(a e+c d x)^2 (d+e x)^{5/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac{7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac{\left (35 e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac{7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac{\left (35 c d e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^3}\\ &=\frac{35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac{7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac{35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt{d+e x}}+\frac{\left (35 c^2 d^2 e^2\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^4}\\ &=\frac{35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac{7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac{35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt{d+e x}}+\frac{\left (35 c^2 d^2 e\right ) \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 )}{4 \left (c d^2-a e^2\right )^4}\\ &=\frac{35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac{7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac{35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt{d+e x}}-\frac{35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0164422, size = 61, normalized size = 0.29 \[ -\frac{2 e^2 \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{3 (d+e x)^{3/2} \left (a e^2-c d^2\right )^3} \]

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)^3,x]

[Out]

(-2*e^2*Hypergeometric2F1[-3/2, 3, -1/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(3*(-(c*d^2) + a*e^2)^3*(d +
e*x)^(3/2))

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Maple [A]  time = 0.206, size = 262, normalized size = 1.3 \begin{align*} -{\frac{2\,{e}^{2}}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+6\,{\frac{{e}^{2}cd}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}\sqrt{ex+d}}}+{\frac{11\,{c}^{3}{d}^{3}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{13\,{e}^{4}{c}^{2}{d}^{2}a}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{13\,{c}^{3}{d}^{4}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{35\,{e}^{2}{c}^{2}{d}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}}\arctan \left ({cd\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \right ){\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \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)^3,x)

[Out]

-2/3*e^2/(a*e^2-c*d^2)^3/(e*x+d)^(3/2)+6*e^2/(a*e^2-c*d^2)^4*c*d/(e*x+d)^(1/2)+11/4*e^2/(a*e^2-c*d^2)^4*c^3*d^
3/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)+13/4*e^4/(a*e^2-c*d^2)^4*c^2*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a-13/4*e^2/
(a*e^2-c*d^2)^4*c^3*d^4/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)+35/4*e^2/(a*e^2-c*d^2)^4*c^2*d^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))

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.15021, size = 2707, normalized size = 13.01 \begin{align*} \text{result too large to display} \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)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^3*d^5*e^2 + 4*a*c^2*d^3
*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*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*(105*c^3*d^3*e^3*x^3
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*
(3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a
^4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*
c^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7
- 3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e
^8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3
*e^9 + a^6*d*e^11)*x), -1/12*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^
3*d^5*e^2 + 4*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*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)) - (105*c^3*d^3*e^3*x^3
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*(
3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a^
4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*c
^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7 -
3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e^
8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3*
e^9 + a^6*d*e^11)*x)]

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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)**3,x)

[Out]

Timed out

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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)^3,x, algorithm="giac")

[Out]

Timed out