3.2016 \(\int \frac {1}{\sqrt {d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=158 \[ \frac {5 c^{3/2} d^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}-\frac {5 c d e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^3}-\frac {1}{(d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {5 e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2} \]

[Out]

-5/3*e/(-a*e^2+c*d^2)^2/(e*x+d)^(3/2)-1/(-a*e^2+c*d^2)/(c*d*x+a*e)/(e*x+d)^(3/2)+5*c^(3/2)*d^(3/2)*e*arctanh(c
^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/(-a*e^2+c*d^2)^(7/2)-5*c*d*e/(-a*e^2+c*d^2)^3/(e*x+d)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \[ \frac {5 c^{3/2} d^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}-\frac {5 c d e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^3}-\frac {1}{(d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {5 e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2} \]

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)^2),x]

[Out]

(-5*e)/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*(d + e*x)^(3/2)) - (5*c*d*e)/(
(c*d^2 - a*e^2)^3*Sqrt[d + e*x]) + (5*c^(3/2)*d^(3/2)*e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a
*e^2]])/(c*d^2 - a*e^2)^(7/2)

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

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Mathematica [C]  time = 0.02, size = 59, normalized size = 0.37 \[ -\frac {2 e \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{3 (d+e x)^{3/2} \left (a e^2-c d^2\right )^2} \]

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)^2),x]

[Out]

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

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fricas [B]  time = 0.98, size = 881, normalized size = 5.58 \[ \left [-\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + a c d^{3} e^{2} + {\left (2 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x^{2} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} 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 (15 \, c^{2} d^{2} e^{2} x^{2} + 3 \, c^{2} d^{4} + 14 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + 10 \, {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (a c^{3} d^{8} e - 3 \, a^{2} c^{2} d^{6} e^{3} + 3 \, a^{3} c d^{4} e^{5} - a^{4} d^{2} e^{7} + {\left (c^{4} d^{7} e^{2} - 3 \, a c^{3} d^{5} e^{4} + 3 \, a^{2} c^{2} d^{3} e^{6} - a^{3} c d e^{8}\right )} x^{3} + {\left (2 \, c^{4} d^{8} e - 5 \, a c^{3} d^{6} e^{3} + 3 \, a^{2} c^{2} d^{4} e^{5} + a^{3} c d^{2} e^{7} - a^{4} e^{9}\right )} x^{2} + {\left (c^{4} d^{9} - a c^{3} d^{7} e^{2} - 3 \, a^{2} c^{2} d^{5} e^{4} + 5 \, a^{3} c d^{3} e^{6} - 2 \, a^{4} d e^{8}\right )} x\right )}}, \frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + a c d^{3} e^{2} + {\left (2 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} x^{2} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} 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 (15 \, c^{2} d^{2} e^{2} x^{2} + 3 \, c^{2} d^{4} + 14 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + 10 \, {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (a c^{3} d^{8} e - 3 \, a^{2} c^{2} d^{6} e^{3} + 3 \, a^{3} c d^{4} e^{5} - a^{4} d^{2} e^{7} + {\left (c^{4} d^{7} e^{2} - 3 \, a c^{3} d^{5} e^{4} + 3 \, a^{2} c^{2} d^{3} e^{6} - a^{3} c d e^{8}\right )} x^{3} + {\left (2 \, c^{4} d^{8} e - 5 \, a c^{3} d^{6} e^{3} + 3 \, a^{2} c^{2} d^{4} e^{5} + a^{3} c d^{2} e^{7} - a^{4} e^{9}\right )} x^{2} + {\left (c^{4} d^{9} - a c^{3} d^{7} e^{2} - 3 \, a^{2} c^{2} d^{5} e^{4} + 5 \, a^{3} c d^{3} e^{6} - 2 \, a^{4} d e^{8}\right )} x\right )}}\right ] \]

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)^2,x, algorithm="fricas")

[Out]

[-1/6*(15*(c^2*d^2*e^3*x^3 + a*c*d^3*e^2 + (2*c^2*d^3*e^2 + a*c*d*e^4)*x^2 + (c^2*d^4*e + 2*a*c*d^2*e^3)*x)*sq
rt(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*(15*c^2*d^2*e^2*x^2 + 3*c^2*d^4 + 14*a*c*d^2*e^2 - 2*a^2*e^4 + 10*(2*c^2*d^3*e + a*c*
d*e^3)*x)*sqrt(e*x + d))/(a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a
*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3
*c*d^2*e^7 - a^4*e^9)*x^2 + (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8)*x),
1/3*(15*(c^2*d^2*e^3*x^3 + a*c*d^3*e^2 + (2*c^2*d^3*e^2 + a*c*d*e^4)*x^2 + (c^2*d^4*e + 2*a*c*d^2*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)) - (
15*c^2*d^2*e^2*x^2 + 3*c^2*d^4 + 14*a*c*d^2*e^2 - 2*a^2*e^4 + 10*(2*c^2*d^3*e + a*c*d*e^3)*x)*sqrt(e*x + d))/(
a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a*c^3*d^5*e^4 + 3*a^2*c^2*d
^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3*c*d^2*e^7 - a^4*e^9)*x^2
+ (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8)*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 162, normalized size = 1.03 \[ \frac {5 c^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}+\frac {\sqrt {e x +d}\, c^{2} d^{2} e}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \left (c d e x +a \,e^{2}\right )}+\frac {4 c d e}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}-\frac {2 e}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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)^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 = 0.78, size = 200, normalized size = 1.27 \[ \frac {\frac {10\,c\,d\,e\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}-\frac {2\,e}{3\,\left (a\,e^2-c\,d^2\right )}+\frac {5\,c^2\,d^2\,e\,{\left (d+e\,x\right )}^2}{{\left (a\,e^2-c\,d^2\right )}^3}}{\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{3/2}+c\,d\,{\left (d+e\,x\right )}^{5/2}}+\frac {5\,c^{3/2}\,d^{3/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((10*c*d*e*(d + e*x))/(3*(a*e^2 - c*d^2)^2) - (2*e)/(3*(a*e^2 - c*d^2)) + (5*c^2*d^2*e*(d + e*x)^2)/(a*e^2 - c
*d^2)^3)/((a*e^2 - c*d^2)*(d + e*x)^(3/2) + c*d*(d + e*x)^(5/2)) + (5*c^(3/2)*d^(3/2)*e*atan((c^(1/2)*d^(1/2)*
(d + e*x)^(1/2)*(a^3*e^6 - c^3*d^6 + 3*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4))/(a*e^2 - c*d^2)^(7/2)))/(a*e^2 - c*d^
2)^(7/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \left (a e + c d x\right )^{2}}\, dx \]

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)**2,x)

[Out]

Integral(1/((d + e*x)**(5/2)*(a*e + c*d*x)**2), x)

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