### 3.2026 $$\int \frac{1}{\sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=244 $\frac{63 c^2 d^2 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^5}-\frac{63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}+\frac{21 c d e^2}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4}+\frac{9 e}{4 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{63 e^2}{20 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3}$

[Out]

(63*e^2)/(20*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(5/2)) + (9*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(5/2)) + (21*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*(d + e*x)^(3/2)) + (
63*c^2*d^2*e^2)/(4*(c*d^2 - a*e^2)^5*Sqrt[d + e*x]) - (63*c^(5/2)*d^(5/2)*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)^(11/2))

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Rubi [A]  time = 0.222973, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {626, 51, 63, 208} $\frac{63 c^2 d^2 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^5}-\frac{63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}+\frac{21 c d e^2}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4}+\frac{9 e}{4 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{63 e^2}{20 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(63*e^2)/(20*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(5/2)) + (9*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(5/2)) + (21*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*(d + e*x)^(3/2)) + (
63*c^2*d^2*e^2)/(4*(c*d^2 - a*e^2)^5*Sqrt[d + e*x]) - (63*c^(5/2)*d^(5/2)*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)^(11/2))

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{1}{\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)^{7/2}} \, dx\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}-\frac{(9 e) \int \frac{1}{(a e+c d x)^2 (d+e x)^{7/2}} \, 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)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{\left (63 e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{\left (63 c d 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 )^3}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{\left (63 c^2 d^2 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 )^4}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}+\frac{\left (63 c^3 d^3 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 )^5}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}+\frac{\left (63 c^3 d^3 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 )^5}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}-\frac{63 c^{5/2} d^{5/2} 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 )^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0207404, size = 61, normalized size = 0.25 $-\frac{2 e^2 \, _2F_1\left (-\frac{5}{2},3;-\frac{3}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{5 (d+e x)^{5/2} \left (a e^2-c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

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Maple [A]  time = 0.234, size = 294, normalized size = 1.2 \begin{align*} -{\frac{2\,{e}^{2}}{5\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-12\,{\frac{{e}^{2}{c}^{2}{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{5}\sqrt{ex+d}}}+2\,{\frac{{e}^{2}cd}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( ex+d \right ) ^{3/2}}}-{\frac{15\,{c}^{4}{d}^{4}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{17\,{e}^{4}{c}^{3}{d}^{3}a}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{17\,{c}^{4}{d}^{5}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{63\,{e}^{2}{c}^{3}{d}^{3}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5}}\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(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

-2/5*e^2/(a*e^2-c*d^2)^3/(e*x+d)^(5/2)-12*e^2/(a*e^2-c*d^2)^5*c^2*d^2/(e*x+d)^(1/2)+2*e^2/(a*e^2-c*d^2)^4*c*d/
(e*x+d)^(3/2)-15/4*e^2/(a*e^2-c*d^2)^5*c^4*d^4/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)-17/4*e^4/(a*e^2-c*d^2)^5*c^3*d^
3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+17/4*e^2/(a*e^2-c*d^2)^5*c^4*d^5/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)-63/4*e^2/
(a*e^2-c*d^2)^5*c^3*d^3/((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(1/(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 = 2.38589, size = 4073, normalized size = 16.69 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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]

[1/40*(315*(c^4*d^4*e^5*x^5 + a^2*c^2*d^5*e^4 + (3*c^4*d^5*e^4 + 2*a*c^3*d^3*e^6)*x^4 + (3*c^4*d^6*e^3 + 6*a*c
^3*d^4*e^5 + a^2*c^2*d^2*e^7)*x^3 + (c^4*d^7*e^2 + 6*a*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6)*x^2 + (2*a*c^3*d^6*e^3
+ 3*a^2*c^2*d^4*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*(315*c^4*d^4*e^4*x^4 - 10*c^4*d^8 + 85*a*c^3*d^6*e^2 + 288
*a^2*c^2*d^4*e^4 - 56*a^3*c*d^2*e^6 + 8*a^4*e^8 + 105*(7*c^4*d^5*e^3 + 5*a*c^3*d^3*e^5)*x^3 + 21*(23*c^4*d^6*e
^2 + 59*a*c^3*d^4*e^4 + 8*a^2*c^2*d^2*e^6)*x^2 + 3*(15*c^4*d^7*e + 277*a*c^3*d^5*e^3 + 136*a^2*c^2*d^3*e^5 - 8
*a^3*c*d*e^7)*x)*sqrt(e*x + d))/(a^2*c^5*d^13*e^2 - 5*a^3*c^4*d^11*e^4 + 10*a^4*c^3*d^9*e^6 - 10*a^5*c^2*d^7*e
^8 + 5*a^6*c*d^5*e^10 - a^7*d^3*e^12 + (c^7*d^12*e^3 - 5*a*c^6*d^10*e^5 + 10*a^2*c^5*d^8*e^7 - 10*a^3*c^4*d^6*
e^9 + 5*a^4*c^3*d^4*e^11 - a^5*c^2*d^2*e^13)*x^5 + (3*c^7*d^13*e^2 - 13*a*c^6*d^11*e^4 + 20*a^2*c^5*d^9*e^6 -
10*a^3*c^4*d^7*e^8 - 5*a^4*c^3*d^5*e^10 + 7*a^5*c^2*d^3*e^12 - 2*a^6*c*d*e^14)*x^4 + (3*c^7*d^14*e - 9*a*c^6*d
^12*e^3 + a^2*c^5*d^10*e^5 + 25*a^3*c^4*d^8*e^7 - 35*a^4*c^3*d^6*e^9 + 17*a^5*c^2*d^4*e^11 - a^6*c*d^2*e^13 -
a^7*e^15)*x^3 + (c^7*d^15 + a*c^6*d^13*e^2 - 17*a^2*c^5*d^11*e^4 + 35*a^3*c^4*d^9*e^6 - 25*a^4*c^3*d^7*e^8 - a
^5*c^2*d^5*e^10 + 9*a^6*c*d^3*e^12 - 3*a^7*d*e^14)*x^2 + (2*a*c^6*d^14*e - 7*a^2*c^5*d^12*e^3 + 5*a^3*c^4*d^10
*e^5 + 10*a^4*c^3*d^8*e^7 - 20*a^5*c^2*d^6*e^9 + 13*a^6*c*d^4*e^11 - 3*a^7*d^2*e^13)*x), -1/20*(315*(c^4*d^4*e
^5*x^5 + a^2*c^2*d^5*e^4 + (3*c^4*d^5*e^4 + 2*a*c^3*d^3*e^6)*x^4 + (3*c^4*d^6*e^3 + 6*a*c^3*d^4*e^5 + a^2*c^2*
d^2*e^7)*x^3 + (c^4*d^7*e^2 + 6*a*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6)*x^2 + (2*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*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)) - (315*c^4*d^4*e^4*x^4 - 10*c^4*d^8 + 85*a*c^3*d^6*e^2 + 288*a^2*c^2*d^4*e^4 - 56*a^3*c*d^2*e^6 + 8*a^4*e
^8 + 105*(7*c^4*d^5*e^3 + 5*a*c^3*d^3*e^5)*x^3 + 21*(23*c^4*d^6*e^2 + 59*a*c^3*d^4*e^4 + 8*a^2*c^2*d^2*e^6)*x^
2 + 3*(15*c^4*d^7*e + 277*a*c^3*d^5*e^3 + 136*a^2*c^2*d^3*e^5 - 8*a^3*c*d*e^7)*x)*sqrt(e*x + d))/(a^2*c^5*d^13
*e^2 - 5*a^3*c^4*d^11*e^4 + 10*a^4*c^3*d^9*e^6 - 10*a^5*c^2*d^7*e^8 + 5*a^6*c*d^5*e^10 - a^7*d^3*e^12 + (c^7*d
^12*e^3 - 5*a*c^6*d^10*e^5 + 10*a^2*c^5*d^8*e^7 - 10*a^3*c^4*d^6*e^9 + 5*a^4*c^3*d^4*e^11 - a^5*c^2*d^2*e^13)*
x^5 + (3*c^7*d^13*e^2 - 13*a*c^6*d^11*e^4 + 20*a^2*c^5*d^9*e^6 - 10*a^3*c^4*d^7*e^8 - 5*a^4*c^3*d^5*e^10 + 7*a
^5*c^2*d^3*e^12 - 2*a^6*c*d*e^14)*x^4 + (3*c^7*d^14*e - 9*a*c^6*d^12*e^3 + a^2*c^5*d^10*e^5 + 25*a^3*c^4*d^8*e
^7 - 35*a^4*c^3*d^6*e^9 + 17*a^5*c^2*d^4*e^11 - a^6*c*d^2*e^13 - a^7*e^15)*x^3 + (c^7*d^15 + a*c^6*d^13*e^2 -
17*a^2*c^5*d^11*e^4 + 35*a^3*c^4*d^9*e^6 - 25*a^4*c^3*d^7*e^8 - a^5*c^2*d^5*e^10 + 9*a^6*c*d^3*e^12 - 3*a^7*d*
e^14)*x^2 + (2*a*c^6*d^14*e - 7*a^2*c^5*d^12*e^3 + 5*a^3*c^4*d^10*e^5 + 10*a^4*c^3*d^8*e^7 - 20*a^5*c^2*d^6*e^
9 + 13*a^6*c*d^4*e^11 - 3*a^7*d^2*e^13)*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(1/(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(1/(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