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

Optimal. Leaf size=152 \[ \frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \]

[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

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Rubi [A]
time = 0.06, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 43, 52, 65, 214} \begin {gather*} -\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}}-\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 &=\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*}

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Mathematica [A]
time = 0.53, size = 149, normalized size = 0.98 \begin {gather*} -\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} \tan ^{-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 {gather*}

Antiderivative was successfully verified.

[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))

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Maple [A]
time = 1.46, size = 173, normalized size = 1.14

method	result	size
derivativedivides	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} a d \,e^{2} c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\)	\(173\)
default	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} a d \,e^{2} c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {15 \left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{c^{3} d^{3}}\right )\)	\(173\)
risch	\(\frac {2 e^{2} \textit {\_O1} \sqrt {e x +d}}{d^{3}}-\frac {e^{4} \left (\munderset {\textit {\_R} =\RootOf \left (c^{2} d^{3} \textit {\_Z}^{6}+\left (3 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \textit {\_Z}^{4}+\left (3 a^{2} d \,e^{4}-6 a c \,d^{3} e^{2}+3 c^{2} d^{5}\right ) \textit {\_Z}^{2}+a^{3} c^{2} e^{6} \textit {\_O1} -3 a^{2} d^{2} e^{4}+3 a \,d^{4} e^{2} c -d^{6} c^{2}\right )}{\sum }\frac {\left (3 \textit {\_R}^{4} c^{2} d^{2}+3 c d \left (e^{2} a -c \,d^{2}\right ) \textit {\_R}^{2}+a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (\sqrt {e x +d}-\textit {\_R} \right )}{c^{2} d^{2} \textit {\_R}^{5}+2 a d \,e^{2} c \,\textit {\_R}^{3}-2 c^{2} d^{3} \textit {\_R}^{3}+a^{2} e^{4} \textit {\_R} -2 a c \,d^{2} e^{2} \textit {\_R} +c^{2} d^{4} \textit {\_R}}\right ) a}{3 c^{4} d^{4}}+\frac {e^{2} \left (\munderset {\textit {\_R} =\RootOf \left (c^{2} d^{3} \textit {\_Z}^{6}+\left (3 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \textit {\_Z}^{4}+\left (3 a^{2} d \,e^{4}-6 a c \,d^{3} e^{2}+3 c^{2} d^{5}\right ) \textit {\_Z}^{2}+a^{3} c^{2} e^{6} \textit {\_O1} -3 a^{2} d^{2} e^{4}+3 a \,d^{4} e^{2} c -d^{6} c^{2}\right )}{\sum }\frac {\left (3 \textit {\_R}^{4} c^{2} d^{2}+3 c d \left (e^{2} a -c \,d^{2}\right ) \textit {\_R}^{2}+a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (\sqrt {e x +d}-\textit {\_R} \right )}{c^{2} d^{2} \textit {\_R}^{5}+2 a d \,e^{2} c \,\textit {\_R}^{3}-2 c^{2} d^{3} \textit {\_R}^{3}+a^{2} e^{4} \textit {\_R} -2 a c \,d^{2} e^{2} \textit {\_R} +c^{2} d^{4} \textit {\_R}}\right )}{3 c^{3} d^{2}}\)	\(521\)

Verification of antiderivative is not currently implemented for this CAS.

[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]

2*e^2*(1/c^3/d^3*(e*x+d)^(1/2)-1/c^3/d^3*(((-9/8*a*d*e^2*c+9/8*c^2*d^3)*(e*x+d)^(3/2)+(-7/8*a^2*e^4+7/4*a*c*d^
2*e^2-7/8*c^2*d^4)*(e*x+d)^(1/2))/(c*d*(e*x+d)+e^2*a-c*d^2)^2+15/8*(a*e^2-c*d^2)/((a*e^2-c*d^2)*c*d)^(1/2)*arc
tan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))))

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

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Fricas [A]
time = 2.91, size = 425, normalized size = 2.80 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + a^{2} e^{4}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d x e + 2 \, c d^{2} - 2 \, \sqrt {x e + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}} - a e^{2}}{c d x + a e}\right ) - 2 \, {\left (9 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - 25 \, a c d x e^{3} - 15 \, a^{2} e^{4} - {\left (8 \, c^{2} d^{2} x^{2} - 5 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}}{8 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} x e + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + a^{2} e^{4}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {x e + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) + {\left (9 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - 25 \, a c d x e^{3} - 15 \, a^{2} e^{4} - {\left (8 \, c^{2} d^{2} x^{2} - 5 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}}{4 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} x e + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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

[Out]

Timed out

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Giac [A]
time = 1.08, size = 200, normalized size = 1.32 \begin {gather*} \frac {15 \, {\left (c d^{2} e^{2} - a e^{4}\right )} \arctan \left (\frac {\sqrt {x e + 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 {2 \, \sqrt {x e + d} e^{2}}{c^{3} d^{3}} - \frac {9 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{2} d^{3} e^{2} - 7 \, \sqrt {x e + d} c^{2} d^{4} e^{2} - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} a c d e^{4} + 14 \, \sqrt {x e + d} a c d^{2} e^{4} - 7 \, \sqrt {x e + d} a^{2} e^{6}}{4 \, {\left ({\left (x e + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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Mupad [B]
time = 0.17, size = 240, normalized size = 1.58 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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