3.21.21 \(\int \frac {(d+e x)^{9/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [2021]

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

[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 = 0.05, 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 = {640, 43, 65, 214} \begin {gather*} -\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} \end {gather*}

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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) \text {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 = 0.39, size = 118, normalized size = 0.91 \begin {gather*} -\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}+\frac {3 e^2 \tan ^{-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 {gather*}

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])

________________________________________________________________________________________

Maple [A]
time = 0.73, size = 126, normalized size = 0.97

method	result	size
derivativedivides	\(2 e^{2} \left (\frac {-\frac {5 \left (e x +d \right )^{\frac {3}{2}}}{8 c d}-\frac {3 \left (e^{2} a -c \,d^{2}\right ) \sqrt {e x +d}}{8 c^{2} d^{2}}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )\)	\(126\)
default	\(2 e^{2} \left (\frac {-\frac {5 \left (e x +d \right )^{\frac {3}{2}}}{8 c d}-\frac {3 \left (e^{2} a -c \,d^{2}\right ) \sqrt {e x +d}}{8 c^{2} d^{2}}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^2*((-5/8*(e*x+d)^(3/2)/c/d-3/8*(a*e^2-c*d^2)/c^2/d^2*(e*x+d)^(1/2))/(c*d*(e*x+d)+e^2*a-c*d^2)^2+3/8/c^2/d^
2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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(c*d^2-%e^2*a>0)', see `assume?
` for more d

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (109) = 218\).
time = 3.57, size = 478, normalized size = 3.68 \begin {gather*} \left [\frac {3 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d x e + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {x e + d}}{c d x + a e}\right ) - 2 \, {\left (5 \, c^{3} d^{4} x e + 2 \, c^{3} d^{5} - 5 \, a c^{2} d^{2} x e^{3} + a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4}\right )} \sqrt {x e + d}}{8 \, {\left (c^{6} d^{7} x^{2} + 2 \, a c^{5} d^{6} x e - 2 \, a^{2} c^{4} d^{4} x e^{3} - a^{3} c^{3} d^{3} e^{4} - {\left (a c^{5} d^{5} x^{2} - a^{2} c^{4} d^{5}\right )} e^{2}\right )}}, \frac {3 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + 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 {x e + d}}{c d x e + c d^{2}}\right ) - {\left (5 \, c^{3} d^{4} x e + 2 \, c^{3} d^{5} - 5 \, a c^{2} d^{2} x e^{3} + a c^{2} d^{3} e^{2} - 3 \, a^{2} c d e^{4}\right )} \sqrt {x e + d}}{4 \, {\left (c^{6} d^{7} x^{2} + 2 \, a c^{5} d^{6} x e - 2 \, a^{2} c^{4} d^{4} x e^{3} - a^{3} c^{3} d^{3} e^{4} - {\left (a c^{5} d^{5} x^{2} - a^{2} c^{4} d^{5}\right )} e^{2}\right )}}\right ] \end {gather*}

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Giac [A]
time = 1.18, size = 131, normalized size = 1.01 \begin {gather*} \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right ) e^{2}}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{2} d^{2}} - \frac {5 \, {\left (x e + d\right )}^{\frac {3}{2}} c d e^{2} - 3 \, \sqrt {x e + d} c d^{2} e^{2} + 3 \, \sqrt {x e + d} a e^{4}}{4 \, {\left ({\left (x e + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{2} d^{2}} \end {gather*}

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]

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

________________________________________________________________________________________

Mupad [B]
time = 0.71, size = 171, normalized size = 1.32 \begin {gather*} \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} \end {gather*}

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)

________________________________________________________________________________________