\(\int \frac {(d+e x)^{5/2}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [2002]

Optimal result

Integrand size = 37, antiderivative size = 114 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^2 d^2}+\frac {2 (d+e x)^{3/2}}{3 c d}-\frac {2 \left (c d^2-a e^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 52, 65, 214} \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=-\frac {2 \left (c d^2-a e^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}+\frac {2 \sqrt {d+e x} \left (c d^2-a e^2\right )}{c^2 d^2}+\frac {2 (d+e x)^{3/2}}{3 c d} \]

[In]

Int[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2),x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \sqrt {d+e x} \left (-3 a e^2+c d (4 d+e x)\right )}{3 c^2 d^2}+\frac {2 \left (-c d^2+a e^2\right )^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{5/2} d^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\left (e^{2} a -c \,d^{2}\right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )+\sqrt {e x +d}\, \left (-\frac {4 \left (\frac {e x}{4}+d \right ) d c}{3}+e^{2} a \right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, c^{2} d^{2}}\)	\(112\)
risch	\(-\frac {2 \left (-x c d e +3 e^{2} a -4 c \,d^{2}\right ) \sqrt {e x +d}}{3 c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(114\)
derivativedivides	\(-\frac {2 \left (-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+a \,e^{2} \sqrt {e x +d}-c \,d^{2} \sqrt {e x +d}\right )}{c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(125\)
default	\(-\frac {2 \left (-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+a \,e^{2} \sqrt {e x +d}-c \,d^{2} \sqrt {e x +d}\right )}{c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(125\)

[In]

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.23 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\left [\frac {3 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt {e x + d}}{3 \, c^{2} d^{2}}, -\frac {2 \, {\left (3 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt {e x + d}\right )}}{3 \, c^{2} d^{2}}\right ] \]

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 5.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e \left (d + e x\right )^{\frac {3}{2}}}{3 c d} + \frac {\sqrt {d + e x} \left (- a e^{3} + c d^{2} e\right )}{c^{2} d^{2}} + \frac {e \left (a e^{2} - c d^{2}\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{c^{3} d^{3} \sqrt {\frac {a e^{2} - c d^{2}}{c d}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} \log {\left (x \right )}}{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(e*(d + e*x)**(3/2)/(3*c*d) + sqrt(d + e*x)*(-a*e**3 + c*d**2*e)/(c**2*d**2) + e*(a*e**2 - c*d**2
)**2*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d)))/(c**3*d**3*sqrt((a*e**2 - c*d**2)/(c*d))))/e, Ne(e, 0))
, (sqrt(d)*log(x)/c, True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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 de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c^{2} d^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{2} + 3 \, \sqrt {e x + d} c^{2} d^{3} - 3 \, \sqrt {e x + d} a c d e^{2}\right )}}{3 \, c^{3} d^{3}} \]

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}}{3\,c\,d}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{c^{5/2}\,d^{5/2}}-\frac {2\,\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}}{c^2\,d^2} \]

[In]

int((d + e*x)^(5/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2),x)

[Out]

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