\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{9/2}} \, dx\) [2043]

Optimal result

Integrand size = 39, antiderivative size = 185 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {3 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}+\frac {3 c^2 d^2 \arctan \left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{4 e^{5/2} \sqrt {c d^2-a e^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {676, 674, 211} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {3 c^2 d^2 \arctan \left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{4 e^{5/2} \sqrt {c d^2-a e^2}}-\frac {3 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}} \]

[In]

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

[Out]

(-3*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e^2*(d + e*x)^(3/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*
d*e*x^2)^(3/2)/(2*e*(d + e*x)^(7/2)) + (3*c^2*d^2*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])
/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(4*e^(5/2)*Sqrt[c*d^2 - a*e^2])

Rule 211

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

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}+\frac {(3 c d) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{5/2}} \, dx}{4 e} \\ & = -\frac {3 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}+\frac {\left (3 c^2 d^2\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e^2} \\ & = -\frac {3 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}+\frac {\left (3 c^2 d^2\right ) \text {Subst}\left (\int \frac {1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{4 e} \\ & = -\frac {3 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}+\frac {3 c^2 d^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{4 e^{5/2} \sqrt {c d^2-a e^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {\sqrt {(a e+c d x) (d+e x)} \left (-\sqrt {e} \left (2 a e^2+c d (3 d+5 e x)\right )+\frac {3 c^2 d^2 (d+e x)^2 \arctan \left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c d^2-a e^2} \sqrt {a e+c d x}}\right )}{4 e^{5/2} (d+e x)^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {\sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c^{2} d^{2} e^{2} x^{2}+6 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c^{2} d^{3} e x +3 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c^{2} d^{4}+5 c d e x \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\, \sqrt {c d x +a e}+2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a \,e^{2}+3 \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, c \,d^{2}\right )}{4 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d x +a e}\, e^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\)	\(271\)

[In]

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

[Out]

-1/4*((c*d*x+a*e)*(e*x+d))^(1/2)*(3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^2*d^2*e^2*x^2+6*arc
tanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^2*d^3*e*x+3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^
(1/2))*c^2*d^4+5*c*d*e*x*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)+2*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)
*a*e^2+3*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c*d^2)/(e*x+d)^(5/2)/(c*d*x+a*e)^(1/2)/e^2/((a*e^2-c*d^2)*e
)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.51 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\left [-\frac {3 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {-c d^{2} e + a e^{3}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d^{2} e + a e^{3}} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} - 2 \, a^{2} e^{5} + 5 \, {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{8 \, {\left (c d^{5} e^{3} - a d^{3} e^{5} + {\left (c d^{2} e^{6} - a e^{8}\right )} x^{3} + 3 \, {\left (c d^{3} e^{5} - a d e^{7}\right )} x^{2} + 3 \, {\left (c d^{4} e^{4} - a d^{2} e^{6}\right )} x\right )}}, -\frac {3 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {c d^{2} e - a e^{3}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d^{2} e - a e^{3}} \sqrt {e x + d}}{c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x}\right ) + {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} - 2 \, a^{2} e^{5} + 5 \, {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{4 \, {\left (c d^{5} e^{3} - a d^{3} e^{5} + {\left (c d^{2} e^{6} - a e^{8}\right )} x^{3} + 3 \, {\left (c d^{3} e^{5} - a d e^{7}\right )} x^{2} + 3 \, {\left (c d^{4} e^{4} - a d^{2} e^{6}\right )} x\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {\frac {3 \, c^{3} d^{3} e {\left | e \right |} \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{\sqrt {c d^{2} e - a e^{3}}} - \frac {3 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{4} d^{5} e^{2} {\left | e \right |} - 3 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{3} d^{3} e^{4} {\left | e \right |} + 5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{3} d^{3} e {\left | e \right |}}{{\left (e x + d\right )}^{2} c^{2} d^{2} e^{2}}}{4 \, c d e^{4}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{{\left (d+e\,x\right )}^{9/2}} \,d x \]

[In]

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

[Out]

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