3.1937 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=261 \[ -\frac {5 \left (c d^2-a e^2\right )^4 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{3/2} d^{3/2} e^{7/2}}+\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e}+\frac {5 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c d e^3} \]

[Out]

5/24*(a-c*d^2/e^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/4*(c*d*x+a*e)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
3/2)/e-5/128*(-a*e^2+c*d^2)^4*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^(3/2)/d^(3/2)/e^(7/2)+5/64*(-a*e^2+c*d^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)/c/d/e^3

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Rubi [A]  time = 0.22, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {654, 670, 640, 612, 621, 206} \[ -\frac {5 \left (c d^2-a e^2\right )^4 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{3/2} d^{3/2} e^{7/2}}+\frac {5 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c d e^3}+\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(c*d^2 - a*e^2)^2*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(64*c*d*e^3) + (
5*(a - (c*d^2)/e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/24 + ((a*e + c*d*x)*(a*d*e + (c*d^2 + a*e^2
)*x + c*d*e*x^2)^(3/2))/(4*e) - (5*(c*d^2 - a*e^2)^4*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sq
rt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(128*c^(3/2)*d^(3/2)*e^(7/2))

Rule 206

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

Rule 612

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

Rule 621

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 640

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 654

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(a + b*x + c*x^2)^(m +
p)/(a/d + (c*x)/e)^m, 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]
&&  !IntegerQ[p] && IntegerQ[m] && RationalQ[p] && (LtQ[0, -m, p] || LtQ[p, -m, 0]) && NeQ[m, 2] && NeQ[m, -1
]

Rule 670

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*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\int (a e+c d x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx\\ &=\frac {(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e}+\frac {\left (5 \left (2 a c d e^2-c d \left (c d^2+a e^2\right )\right )\right ) \int (a e+c d x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{8 c d e}\\ &=\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e}+\frac {\left (5 \left (c d^2-a e^2\right )^2\right ) \int \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{16 e^2}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c d e^3}+\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e}-\frac {\left (5 \left (c d^2-a e^2\right )^4\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{128 c d e^3}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c d e^3}+\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e}-\frac {\left (5 \left (c d^2-a e^2\right )^4\right ) \operatorname {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 )}{64 c d e^3}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 c d e^3}+\frac {5}{24} \left (a-\frac {c d^2}{e^2}\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}+\frac {(a e+c d x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e}-\frac {5 \left (c d^2-a e^2\right )^4 \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 )}{128 c^{3/2} d^{3/2} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 316, normalized size = 1.21 \[ \frac {\sqrt {c} \sqrt {d} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d} (d+e x) \left (15 a^4 e^7+a^3 c d e^5 (73 d+133 e x)+a^2 c^2 d^2 e^3 \left (-55 d^2+109 d e x+254 e^2 x^2\right )+a c^3 d^3 e \left (15 d^3-65 d^2 e x+44 d e^2 x^2+184 e^3 x^3\right )+c^4 d^4 x \left (15 d^3-10 d^2 e x+8 d e^2 x^2+48 e^3 x^3\right )\right )-15 \left (c d^2-a e^2\right )^{9/2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )\right )}{192 e^{7/2} (c d)^{5/2} \sqrt {(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[d]*(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(15*a^4*e^7 + a^3*c*d*e^5*(73*d + 133*e*x) + a^2
*c^2*d^2*e^3*(-55*d^2 + 109*d*e*x + 254*e^2*x^2) + c^4*d^4*x*(15*d^3 - 10*d^2*e*x + 8*d*e^2*x^2 + 48*e^3*x^3)
+ a*c^3*d^3*e*(15*d^3 - 65*d^2*e*x + 44*d*e^2*x^2 + 184*e^3*x^3)) - 15*(c*d^2 - a*e^2)^(9/2)*Sqrt[a*e + c*d*x]
*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d
^2 - a*e^2])]))/(192*(c*d)^(5/2)*e^(7/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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fricas [A]  time = 1.07, size = 672, normalized size = 2.57 \[ \left [\frac {15 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \, {\left (48 \, c^{4} d^{4} e^{4} x^{3} + 15 \, c^{4} d^{7} e - 55 \, a c^{3} d^{5} e^{3} + 73 \, a^{2} c^{2} d^{3} e^{5} + 15 \, a^{3} c d e^{7} + 8 \, {\left (c^{4} d^{5} e^{3} + 17 \, a c^{3} d^{3} e^{5}\right )} x^{2} - 2 \, {\left (5 \, c^{4} d^{6} e^{2} - 18 \, a c^{3} d^{4} e^{4} - 59 \, a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{768 \, c^{2} d^{2} e^{4}}, \frac {15 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) + 2 \, {\left (48 \, c^{4} d^{4} e^{4} x^{3} + 15 \, c^{4} d^{7} e - 55 \, a c^{3} d^{5} e^{3} + 73 \, a^{2} c^{2} d^{3} e^{5} + 15 \, a^{3} c d e^{7} + 8 \, {\left (c^{4} d^{5} e^{3} + 17 \, a c^{3} d^{3} e^{5}\right )} x^{2} - 2 \, {\left (5 \, c^{4} d^{6} e^{2} - 18 \, a c^{3} d^{4} e^{4} - 59 \, a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{384 \, c^{2} d^{2} e^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(15*(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*sqrt(c*d*e)*log(8*c^2*d
^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*
d^2 + a*e^2)*sqrt(c*d*e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(48*c^4*d^4*e^4*x^3 + 15*c^4*d^7*e - 55*a*c^3*d^5*
e^3 + 73*a^2*c^2*d^3*e^5 + 15*a^3*c*d*e^7 + 8*(c^4*d^5*e^3 + 17*a*c^3*d^3*e^5)*x^2 - 2*(5*c^4*d^6*e^2 - 18*a*c
^3*d^4*e^4 - 59*a^2*c^2*d^2*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e^4), 1/384*(15*(c^4
*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d*e*x^2
+ a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d
^3*e + a*c*d*e^3)*x)) + 2*(48*c^4*d^4*e^4*x^3 + 15*c^4*d^7*e - 55*a*c^3*d^5*e^3 + 73*a^2*c^2*d^3*e^5 + 15*a^3*
c*d*e^7 + 8*(c^4*d^5*e^3 + 17*a*c^3*d^3*e^5)*x^2 - 2*(5*c^4*d^6*e^2 - 18*a*c^3*d^4*e^4 - 59*a^2*c^2*d^2*e^6)*x
)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e^4)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 1455, normalized size = 5.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/e^2/(a*e^2-c*d^2)/(x+d/e)^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)-2/3/e*c*d/(a*e^2-c*d^2)*((x+d/e)
^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(5/2)-5/64/e^3*c^3*d^7/(a*e^2-c*d^2)*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(
1/2)+5/12/e*c^2*d^3/(a*e^2-c*d^2)*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)*x+5/24/e^2*c^2*d^4/(a*e^2-c*d^
2)*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)-5/12*e*c*d/(a*e^2-c*d^2)*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+
d/e))^(3/2)*x-5/32*e^3*d/(a*e^2-c*d^2)*a^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-25/128/e*c^3*d^7/(a*e
^2-c*d^2)*a*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)
)/(c*d*e)^(1/2)+5/32*e^4/(a*e^2-c*d^2)*a^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+15/32*c^2*d^4/(a*e^
2-c*d^2)*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-5/128*e^7/c/d/(a*e^2-c*d^2)*a^5*ln((1/2*a*e^2-1/2*c
*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)+25/64*e*c^2*d^5
/(a*e^2-c*d^2)*a^2*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e)
)^(1/2))/(c*d*e)^(1/2)+25/128*e^5*d/(a*e^2-c*d^2)*a^4*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x
+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)-25/64*e^3*c*d^3/(a*e^2-c*d^2)*a^3*ln((1/2*a*e^2-1/2*
c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)-5/24*e^2/(a*e^
2-c*d^2)*a^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)-15/32*e^2*c*d^2/(a*e^2-c*d^2)*a^2*((x+d/e)^2*c*d*e+
(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+5/64*e^5/c/d/(a*e^2-c*d^2)*a^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+5/
32/e*c^2*d^5/(a*e^2-c*d^2)*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-5/32/e^2*c^3*d^6/(a*e^2-c*d^2)*((x+
d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+5/128/e^3*c^4*d^9/(a*e^2-c*d^2)*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*
d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)

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maxima [B]  time = 1.31, size = 722, normalized size = 2.77 \[ -\frac {5 \, c^{4} d^{8} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{128 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{5}} + \frac {5 \, a c^{3} d^{6} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{32 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{3}} - \frac {15 \, a^{2} c^{2} d^{4} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{64 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e} + \frac {5 \, a^{3} c d^{2} e \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{32 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} - \frac {5 \, a^{4} e^{3} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{128 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} - \frac {5}{16} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a c d^{2} x + \frac {5 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c^{2} d^{4} x}{32 \, e^{2}} + \frac {5}{32} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{2} e^{2} x + \frac {5 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c^{2} d^{5}}{64 \, e^{3}} - \frac {5 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a c d^{3}}{64 \, e} - \frac {5}{64} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{2} d e + \frac {5 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{3} e^{3}}{64 \, c d} + \frac {5}{24} \, {\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} a - \frac {5 \, {\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} c d^{2}}{24 \, e^{2}} + \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {5}{2}}}{4 \, {\left (e^{2} x + d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/128*c^4*d^8*log(2*c*d*x + c*d^2/e + a*e + 2*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*sqrt(c*d/e))/((c*d/
e)^(3/2)*e^5) + 5/32*a*c^3*d^6*log(2*c*d*x + c*d^2/e + a*e + 2*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*sqr
t(c*d/e))/((c*d/e)^(3/2)*e^3) - 15/64*a^2*c^2*d^4*log(2*c*d*x + c*d^2/e + a*e + 2*sqrt(c*d*e*x^2 + c*d^2*x + a
*e^2*x + a*d*e)*sqrt(c*d/e))/((c*d/e)^(3/2)*e) + 5/32*a^3*c*d^2*e*log(2*c*d*x + c*d^2/e + a*e + 2*sqrt(c*d*e*x
^2 + c*d^2*x + a*e^2*x + a*d*e)*sqrt(c*d/e))/(c*d/e)^(3/2) - 5/128*a^4*e^3*log(2*c*d*x + c*d^2/e + a*e + 2*sqr
t(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*sqrt(c*d/e))/(c*d/e)^(3/2) - 5/16*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x
+ a*d*e)*a*c*d^2*x + 5/32*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c^2*d^4*x/e^2 + 5/32*sqrt(c*d*e*x^2 + c*
d^2*x + a*e^2*x + a*d*e)*a^2*e^2*x + 5/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c^2*d^5/e^3 - 5/64*sqrt(
c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a*c*d^3/e - 5/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^2*d*e +
5/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^3*e^3/(c*d) + 5/24*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^
(3/2)*a - 5/24*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^(3/2)*c*d^2/e^2 + 1/4*(c*d*e*x^2 + c*d^2*x + a*e^2*x +
a*d*e)^(5/2)/(e^2*x + d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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