3.2021 \(\int \frac {(d+e x)^{9/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=130 \[ -\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{5/2} d^{5/2} \sqrt {c d^2-a e^2}}-\frac {3 e \sqrt {d+e x}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 130, 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, 47, 63, 208} \[ -\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{5/2} d^{5/2} \sqrt {c d^2-a e^2}}-\frac {3 e \sqrt {d+e x}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(-3*e*Sqrt[d + e*x])/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(3/2)/(2*c*d*(a*e + c*d*x)^2) - (3*e^2*ArcTanh[(Sqr
t[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(5/2)*d^(5/2)*Sqrt[c*d^2 - a*e^2])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac {(d+e x)^{3/2}}{(a e+c d x)^3} \, dx\\ &=-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2}+\frac {(3 e) \int \frac {\sqrt {d+e x}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac {3 e \sqrt {d+e x}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2}+\frac {\left (3 e^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c^2 d^2}\\ &=-\frac {3 e \sqrt {d+e x}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2}+\frac {(3 e) \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 c^2 d^2}\\ &=-\frac {3 e \sqrt {d+e x}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{3/2}}{2 c d (a e+c d x)^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{5/2} d^{5/2} \sqrt {c d^2-a e^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 118, normalized size = 0.91 \[ \frac {3 e^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {a e^2-c d^2}}\right )}{4 c^{5/2} d^{5/2} \sqrt {a e^2-c d^2}}-\frac {\sqrt {d+e x} \left (3 a e^2+c d (2 d+5 e x)\right )}{4 c^2 d^2 (a e+c d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

-1/4*(Sqrt[d + e*x]*(3*a*e^2 + c*d*(2*d + 5*e*x)))/(c^2*d^2*(a*e + c*d*x)^2) + (3*e^2*ArcTan[(Sqrt[c]*Sqrt[d]*
Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(4*c^(5/2)*d^(5/2)*Sqrt[-(c*d^2) + a*e^2])

________________________________________________________________________________________

fricas [B]  time = 0.98, size = 492, normalized size = 3.78 \[ \left [\frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{3} d^{5} + a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4} + 5 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} c^{4} d^{5} e^{2} - a^{3} c^{3} d^{3} e^{4} + {\left (c^{6} d^{7} - a c^{5} d^{5} e^{2}\right )} x^{2} + 2 \, {\left (a c^{5} d^{6} e - a^{2} c^{4} d^{4} e^{3}\right )} x\right )}}, \frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) - {\left (2 \, c^{3} d^{5} + a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4} + 5 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} c^{4} d^{5} e^{2} - a^{3} c^{3} d^{3} e^{4} + {\left (c^{6} d^{7} - a c^{5} d^{5} e^{2}\right )} x^{2} + 2 \, {\left (a c^{5} d^{6} e - a^{2} c^{4} d^{4} e^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*(3*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(c^2*d^3 - a*c*d*e^2)*log((c*d*e*x + 2*c*d^2 - a*e^2 -
2*sqrt(c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(c*d*x + a*e)) - 2*(2*c^3*d^5 + a*c^2*d^3*e^2 - 3*a^2*c*d*e^4 + 5*
(c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^4*d^5*e^2 - a^3*c^3*d^3*e^4 + (c^6*d^7 - a*c^5*d^5*e^2)*x
^2 + 2*(a*c^5*d^6*e - a^2*c^4*d^4*e^3)*x), 1/4*(3*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(-c^2*d^3 +
a*c*d*e^2)*arctan(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(e*x + d)/(c*d*e*x + c*d^2)) - (2*c^3*d^5 + a*c^2*d^3*e^2 - 3
*a^2*c*d*e^4 + 5*(c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^4*d^5*e^2 - a^3*c^3*d^3*e^4 + (c^6*d^7 -
a*c^5*d^5*e^2)*x^2 + 2*(a*c^5*d^6*e - a^2*c^4*d^4*e^3)*x)]

________________________________________________________________________________________

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((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.06, size = 149, normalized size = 1.15 \[ -\frac {3 \sqrt {e x +d}\, a \,e^{4}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{2} d^{2}}+\frac {3 \sqrt {e x +d}\, e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c}-\frac {5 \left (e x +d \right )^{\frac {3}{2}} e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c d}+\frac {3 e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{4 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(9/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^3,x)

[Out]

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

________________________________________________________________________________________

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((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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?

________________________________________________________________________________________

mupad [B]  time = 0.71, size = 171, normalized size = 1.32 \[ \frac {3\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{4\,c^{5/2}\,d^{5/2}\,\sqrt {a\,e^2-c\,d^2}}-\frac {\frac {5\,e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,c\,d}+\frac {3\,e^2\,\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}}{4\,c^2\,d^2}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(9/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

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

________________________________________________________________________________________

sympy [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((e*x+d)**(9/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________