\(\int \frac {(d+e x)^{11/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [2020]

Optimal result

Integrand size = 37, antiderivative size = 152 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \]

[Out]

-5/4*e*(e*x+d)^(3/2)/c^2/d^2/(c*d*x+a*e)-1/2*(e*x+d)^(5/2)/c/d/(c*d*x+a*e)^2-15/4*e^2*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)^(1/2)/c^(7/2)/d^(7/2)+15/4*e^2*(e*x+d)^(1/2)/c^3/d^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 43, 52, 65, 214} \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {15 e^2 \sqrt {c d^2-a e^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3} \]

[In]

Int[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(15*e^2*Sqrt[d + e*x])/(4*c^3*d^3) - (5*e*(d + e*x)^(3/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(5/2)/(2*c*d*
(a*e + c*d*x)^2) - (15*e^2*Sqrt[c*d^2 - a*e^2]*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(
4*c^(7/2)*d^(7/2))

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 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)^{5/2}}{(a e+c d x)^3} \, dx \\ & = -\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{(a e+c d x)^2} \, dx}{4 c d} \\ & = -\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e^2\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{8 c^2 d^2} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e^2 \left (c d^2-a e^2\right )\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c^3 d^3} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}+\frac {\left (15 e \left (c d^2-a e^2\right )\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 )}{4 c^3 d^3} \\ & = \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (-15 a^2 e^4+5 a c d e^2 (d-5 e x)+c^2 d^2 \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )}{4 c^3 d^3 (a e+c d x)^2}-\frac {15 e^2 \sqrt {-c d^2+a e^2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{7/2} d^{7/2}} \]

[In]

Integrate[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

-1/4*(Sqrt[d + e*x]*(-15*a^2*e^4 + 5*a*c*d*e^2*(d - 5*e*x) + c^2*d^2*(2*d^2 + 9*d*e*x - 8*e^2*x^2)))/(c^3*d^3*
(a*e + c*d*x)^2) - (15*e^2*Sqrt[-(c*d^2) + a*e^2]*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2
]])/(4*c^(7/2)*d^(7/2))

Maple [A] (verified)

Time = 11.84 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {15 \left (e^{2} \left (e^{2} a -c \,d^{2}\right ) \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )-\left (-\frac {2 d^{2} \left (-4 x^{2} e^{2}+\frac {9}{2} d e x +d^{2}\right ) c^{2}}{15}-\frac {a d \,e^{2} \left (-5 e x +d \right ) c}{3}+a^{2} e^{4}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \sqrt {e x +d}\right )}{4 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, c^{3} d^{3} \left (c d x +a e \right )^{2}}\)	\(164\)
derivativedivides	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} d \,e^{2} a c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\)	\(173\)
default	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} d \,e^{2} a c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\)	\(173\)
risch	\(\frac {2 e^{2} \textit {\_O1} \sqrt {e x +d}}{d^{3}}-\frac {\left (e^{2} a -c \,d^{2}\right ) e^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} d^{3} \textit {\_Z}^{6}+\left (3 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \textit {\_Z}^{4}+\left (3 a^{2} d \,e^{4}-6 a \,d^{3} e^{2} c +3 d^{5} c^{2}\right ) \textit {\_Z}^{2}+a^{3} c^{2} e^{6} \textit {\_O1} -3 a^{2} d^{2} e^{4}+3 a \,d^{4} e^{2} c -d^{6} c^{2}\right )}{\sum }\frac {\left (d^{2} \left (3 \textit {\_R}^{4}-3 \textit {\_R}^{2} d +d^{2}\right ) c^{2}-d \,e^{2} a c \left (-3 \textit {\_R}^{2}+2 d \right )+a^{2} e^{4}\right ) \ln \left (\sqrt {e x +d}-\textit {\_R} \right )}{\textit {\_R} \left (c d \,\textit {\_R}^{2}+e^{2} a -c \,d^{2}\right )^{2}}\right )}{3 c^{4} d^{4}}\)	\(234\)

[In]

int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

-15/4/((a*e^2-c*d^2)*c*d)^(1/2)*(e^2*(a*e^2-c*d^2)*(c*d*x+a*e)^2*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^
(1/2))-(-2/15*d^2*(-4*x^2*e^2+9/2*d*e*x+d^2)*c^2-1/3*a*d*e^2*(-5*e*x+d)*c+a^2*e^4)*((a*e^2-c*d^2)*c*d)^(1/2)*(
e*x+d)^(1/2))/c^3/d^3/(c*d*x+a*e)^2

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.89 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \]

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*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*(8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c
*d^2*e^2 + 15*a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^
3*d^3*e^2), -1/4*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*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)) - (8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c*d^2*e^2 + 15*
a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**(11/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)^(11/2)/(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 de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.34 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2 \, \sqrt {e x + d} e^{2}}{c^{3} d^{3}} + \frac {15 \, {\left (c d^{2} e^{2} - a e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{3} d^{3}} - \frac {9 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{3} e^{2} - 7 \, \sqrt {e x + d} c^{2} d^{4} e^{2} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} a c d e^{4} + 14 \, \sqrt {e x + d} a c d^{2} e^{4} - 7 \, \sqrt {e x + d} a^{2} e^{6}}{4 \, {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{3} d^{3}} \]

[In]

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

2*sqrt(e*x + d)*e^2/(c^3*d^3) + 15/4*(c*d^2*e^2 - a*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^3*d^3) - 1/4*(9*(e*x + d)^(3/2)*c^2*d^3*e^2 - 7*sqrt(e*x + d)*c^2*d^4*e^2 - 9*(e
*x + d)^(3/2)*a*c*d*e^4 + 14*sqrt(e*x + d)*a*c*d^2*e^4 - 7*sqrt(e*x + d)*a^2*e^6)/(((e*x + d)*c*d - c*d^2 + a*
e^2)^2*c^3*d^3)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2\,e^2\,\sqrt {d+e\,x}}{c^3\,d^3}-\frac {\left (\frac {9\,c^2\,d^3\,e^2}{4}-\frac {9\,a\,c\,d\,e^4}{4}\right )\,{\left (d+e\,x\right )}^{3/2}-\sqrt {d+e\,x}\,\left (\frac {7\,a^2\,e^6}{4}-\frac {7\,a\,c\,d^2\,e^4}{2}+\frac {7\,c^2\,d^4\,e^2}{4}\right )}{c^5\,d^7-\left (2\,c^5\,d^6-2\,a\,c^4\,d^4\,e^2\right )\,\left (d+e\,x\right )+c^5\,d^5\,{\left (d+e\,x\right )}^2-2\,a\,c^4\,d^5\,e^2+a^2\,c^3\,d^3\,e^4}-\frac {15\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,\sqrt {a\,e^2-c\,d^2}\,\sqrt {d+e\,x}}{a\,e^4-c\,d^2\,e^2}\right )\,\sqrt {a\,e^2-c\,d^2}}{4\,c^{7/2}\,d^{7/2}} \]

[In]

int((d + e*x)^(11/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

(2*e^2*(d + e*x)^(1/2))/(c^3*d^3) - (((9*c^2*d^3*e^2)/4 - (9*a*c*d*e^4)/4)*(d + e*x)^(3/2) - (d + e*x)^(1/2)*(
(7*a^2*e^6)/4 + (7*c^2*d^4*e^2)/4 - (7*a*c*d^2*e^4)/2))/(c^5*d^7 - (2*c^5*d^6 - 2*a*c^4*d^4*e^2)*(d + e*x) + c
^5*d^5*(d + e*x)^2 - 2*a*c^4*d^5*e^2 + a^2*c^3*d^3*e^4) - (15*e^2*atan((c^(1/2)*d^(1/2)*e^2*(a*e^2 - c*d^2)^(1
/2)*(d + e*x)^(1/2))/(a*e^4 - c*d^2*e^2))*(a*e^2 - c*d^2)^(1/2))/(4*c^(7/2)*d^(7/2))