3.20.65 \(\int \frac {(d+e x)^5}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [1965]

Optimal. Leaf size=231 \[ -\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {5 e^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{7/2} d^{7/2}} \]

[Out]

-2/3*(e*x+d)^4/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+5/2*e^(3/2)*(-a*e^2+c*d^2)*arctanh(1/2*(2*c*d*e*x+a
*e^2+c*d^2)/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/c^(7/2)/d^(7/2)-10/3*e*(e*x+d)^2/
c^2/d^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5*e^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3

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Rubi [A]
time = 0.10, antiderivative size = 231, 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 = {682, 654, 635, 212} \begin {gather*} \frac {5 e^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{7/2} d^{7/2}}+\frac {5 e^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^3 d^3}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^4}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^4)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (10*e*(d + e*x)^2)/(3*c^2*d^2*Sqrt[a*
d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*e^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c^3*d^3) + (5*e^(
3/2)*(c*d^2 - a*e^2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^
2)*x + c*d*e*x^2])])/(2*c^(7/2)*d^(7/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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

Rule 654

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

Rule 682

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {(5 e) \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 c d}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (5 e^2\right ) \int \frac {d+e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c^2 d^2}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {\left (5 e^2 \left (c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 c^3 d^3}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {\left (5 e^2 \left (c d^2-a e^2\right )\right ) \text {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{c^3 d^3}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {5 e^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{7/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 183, normalized size = 0.79 \begin {gather*} \frac {((a e+c d x) (d+e x))^{5/2} \left (\frac {c d \left (15 a^2 e^4-10 a c d e^2 (d-2 e x)+c^2 d^2 \left (-2 d^2-14 d e x+3 e^2 x^2\right )\right )}{(a e+c d x)^{3/2} (d+e x)^2}-\frac {15 \sqrt {\frac {c d}{e}} \left (c d^2 e^2-a e^4\right ) \log \left (\sqrt {a e+c d x}-\sqrt {\frac {c d}{e}} \sqrt {d+e x}\right )}{(d+e x)^{5/2}}\right )}{3 c^4 d^4 (a e+c d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*((c*d*(15*a^2*e^4 - 10*a*c*d*e^2*(d - 2*e*x) + c^2*d^2*(-2*d^2 - 14*d*e*x + 3
*e^2*x^2)))/((a*e + c*d*x)^(3/2)*(d + e*x)^2) - (15*Sqrt[(c*d)/e]*(c*d^2*e^2 - a*e^4)*Log[Sqrt[a*e + c*d*x] -
Sqrt[(c*d)/e]*Sqrt[d + e*x]])/(d + e*x)^(5/2)))/(3*c^4*d^4*(a*e + c*d*x)^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(4446\) vs. \(2(203)=406\).
time = 0.71, size = 4447, normalized size = 19.25

method	result	size
default	\(\text {Expression too large to display}\)	\(4447\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^5*(x^4/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-5/2*(a*e^2+c*d^2)/c/d/e*(-1/3*x^3/c/d/e/(a*d*e+(a*e^2+c
*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(a*e^
2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*
e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+
c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^
2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))+1/2*a/c*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+
c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^
2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))+2*a/c*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1
/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d
*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*
e*x^2)^(1/2))))+1/c/d/e*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/c/d/e/(a
*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)
^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*
e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)))-4*a/c*(-x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(
a*e^2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(
a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*
e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+
a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))+1/2*a/c*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*
e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+
a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))+2*a/c*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/
2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x
+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+
c*d*e*x^2)^(1/2)))))+5*d*e^4*(-1/3*x^3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(
-x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(a*e^2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*
x+c*d*e*x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^
2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+
16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))
+1/2*a/c*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+
16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)))
+2*a/c*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2
)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d
^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))))+1/c/d/e*(-x/c/d/e/(a*d*e+(a*e^2+c*
d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2
)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/c/d
/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)))+10*
d^2*e^3*(-x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(a*e^2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^
2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a
*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2)))+1/2*a/c*(2/3*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2)))+2*a/c*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a*
e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(
a*e^2+c*d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))))+10*d^3*e^2*(-1/2*x/c/d/e/
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^
(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(2*c*d*e*x+a...

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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)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 5.74, size = 626, normalized size = 2.71 \begin {gather*} \left [\frac {15 \, {\left (c^{3} d^{4} x^{2} e + 2 \, a c^{2} d^{3} x e^{2} - 2 \, a^{2} c d x e^{4} - a^{3} e^{5} - {\left (a c^{2} d^{2} x^{2} - a^{2} c d^{2}\right )} e^{3}\right )} \sqrt {\frac {1}{c d}} e^{\frac {1}{2}} \log \left (8 \, c^{2} d^{3} x e + c^{2} d^{4} + 8 \, a c d x e^{3} + a^{2} e^{4} + 4 \, {\left (2 \, c^{2} d^{2} x e + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {\frac {1}{c d}} e^{\frac {1}{2}} + 2 \, {\left (4 \, c^{2} d^{2} x^{2} + 3 \, a c d^{2}\right )} e^{2}\right ) - 4 \, {\left (14 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - 20 \, a c d x e^{3} - 15 \, a^{2} e^{4} - {\left (3 \, c^{2} d^{2} x^{2} - 10 \, a c d^{2}\right )} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{12 \, {\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^{3} d^{4} x^{2} e + 2 \, a c^{2} d^{3} x e^{2} - 2 \, a^{2} c d x e^{4} - a^{3} e^{5} - {\left (a c^{2} d^{2} x^{2} - a^{2} c d^{2}\right )} e^{3}\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d^{2} x e + a x e^{3} + {\left (c d x^{2} + a d\right )} e^{2}\right )}}\right ) + 2 \, {\left (14 \, c^{2} d^{3} x e + 2 \, c^{2} d^{4} - 20 \, a c d x e^{3} - 15 \, a^{2} e^{4} - {\left (3 \, c^{2} d^{2} x^{2} - 10 \, a c d^{2}\right )} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{6 \, {\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)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(c^3*d^4*x^2*e + 2*a*c^2*d^3*x*e^2 - 2*a^2*c*d*x*e^4 - a^3*e^5 - (a*c^2*d^2*x^2 - a^2*c*d^2)*e^3)*sq
rt(1/(c*d))*e^(1/2)*log(8*c^2*d^3*x*e + c^2*d^4 + 8*a*c*d*x*e^3 + a^2*e^4 + 4*(2*c^2*d^2*x*e + c^2*d^3 + a*c*d
*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(1/(c*d))*e^(1/2) + 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)*e^2) -
4*(14*c^2*d^3*x*e + 2*c^2*d^4 - 20*a*c*d*x*e^3 - 15*a^2*e^4 - (3*c^2*d^2*x^2 - 10*a*c*d^2)*e^2)*sqrt(c*d^2*x
+ a*x*e^2 + (c*d*x^2 + a*d)*e))/(c^5*d^5*x^2 + 2*a*c^4*d^4*x*e + a^2*c^3*d^3*e^2), -1/6*(15*(c^3*d^4*x^2*e + 2
*a*c^2*d^3*x*e^2 - 2*a^2*c*d*x*e^4 - a^3*e^5 - (a*c^2*d^2*x^2 - a^2*c*d^2)*e^3)*sqrt(-e/(c*d))*arctan(1/2*sqrt
(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(2*c*d*x*e + c*d^2 + a*e^2)*sqrt(-e/(c*d))/(c*d^2*x*e + a*x*e^3 + (c*d
*x^2 + a*d)*e^2)) + 2*(14*c^2*d^3*x*e + 2*c^2*d^4 - 20*a*c*d*x*e^3 - 15*a^2*e^4 - (3*c^2*d^2*x^2 - 10*a*c*d^2)
*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))/(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{5}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (206) = 412\).
time = 1.74, size = 621, normalized size = 2.69 \begin {gather*} \frac {\sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} e^{2}}{c^{3} d^{3}} - \frac {5 \, {\left (\sqrt {c d} c d^{2} e^{\frac {5}{2}} - \sqrt {c d} a e^{\frac {9}{2}}\right )} e^{\left (-1\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{2 \, c^{4} d^{4}} + \frac {2 \, {\left (c^{4} d^{8} + 3 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} c^{3} d^{6} e^{\frac {1}{2}} - a c^{3} d^{6} e^{2} + 9 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )}^{2} c^{3} d^{5} e + 9 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} a c^{2} d^{4} e^{\frac {5}{2}} + 6 \, a^{2} c^{2} d^{4} e^{4} - 18 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )}^{2} a c^{2} d^{3} e^{3} - 27 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} a^{2} c d^{2} e^{\frac {9}{2}} - 13 \, a^{3} c d^{2} e^{6} + 9 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )}^{2} a^{2} c d e^{5} + 15 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} a^{3} e^{\frac {13}{2}} + 7 \, a^{4} e^{8}\right )}}{3 \, {\left ({\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} c d + \sqrt {c d} a e^{\frac {3}{2}}\right )}^{3} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*e^2/(c^3*d^3) - 5/2*(sqrt(c*d)*c*d^2*e^(5/2) - sqrt(c*d)*a*e^(9/2)
)*e^(-1)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*
e))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^4*d^4) + 2/3*(c^4*d^8 + 3*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x
+ a*x*e^2 + a*d*e))*sqrt(c*d)*c^3*d^6*e^(1/2) - a*c^3*d^6*e^2 + 9*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d
^2*x + a*x*e^2 + a*d*e))^2*c^3*d^5*e + 9*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*s
qrt(c*d)*a*c^2*d^4*e^(5/2) + 6*a^2*c^2*d^4*e^4 - 18*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2
+ a*d*e))^2*a*c^2*d^3*e^3 - 27*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*sqrt(c*d)*a
^2*c*d^2*e^(9/2) - 13*a^3*c*d^2*e^6 + 9*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))^2*
a^2*c*d*e^5 + 15*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*sqrt(c*d)*a^3*e^(13/2) +
7*a^4*e^8)/(((sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*c*d + sqrt(c*d)*a*e^(3/2))^3*
c^2*d^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^5}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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