3.3.82 \(\int \frac {x (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{d+e x} \, dx\)

Optimal. Leaf size=295 \[ -\frac {\left (3 a e^2+5 c d^2\right ) \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{5/2} d^{5/2} e^{7/2}}+\frac {\left (3 a e^2+5 c d^2\right ) \left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c^2 d^2 e^3}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{4 c d e (d+e x)}-\frac {1}{24} \left (\frac {3 a}{c d}+\frac {5 d}{e^2}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} \]

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Rubi [A]  time = 0.28, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {794, 664, 612, 621, 206} \begin {gather*} \frac {\left (3 a e^2+5 c d^2\right ) \left (c d^2-a e^2\right ) \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^2 d^2 e^3}-\frac {\left (3 a e^2+5 c d^2\right ) \left (c d^2-a e^2\right )^3 \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^{5/2} d^{5/2} e^{7/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{4 c d e (d+e x)}-\frac {1}{24} \left (\frac {3 a}{c d}+\frac {5 d}{e^2}\right ) \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[(x*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(d + e*x),x]

[Out]

((c*d^2 - a*e^2)*(5*c*d^2 + 3*a*e^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^2*d^2*e^3) - (((3*a)/(c*d) + (5*d)/e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/24 + (a*d*e + (c*
d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(4*c*d*e*(d + e*x)) - ((c*d^2 - a*e^2)^3*(5*c*d^2 + 3*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])])/(128*c^(5/2)*d^
(5/2)*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 664

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

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

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Mathematica [A]  time = 1.24, size = 276, normalized size = 0.94 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\sqrt {c} \sqrt {d} \sqrt {e} \left (-9 a^3 e^6+3 a^2 c d e^4 (3 d+2 e x)+a c^2 d^2 e^2 \left (-31 d^2+20 d e x+72 e^2 x^2\right )+c^3 d^3 \left (15 d^3-10 d^2 e x+8 d e^2 x^2+48 e^3 x^3\right )\right )-\frac {3 \sqrt {c d} \left (c d^2-a e^2\right )^{5/2} \left (3 a e^2+5 c d^2\right ) \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 )}{\sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}\right )}{192 c^{5/2} d^{5/2} e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[c]*Sqrt[d]*Sqrt[e]*(-9*a^3*e^6 + 3*a^2*c*d*e^4*(3*d + 2*e*x) + a*c^2*d^2*
e^2*(-31*d^2 + 20*d*e*x + 72*e^2*x^2) + c^3*d^3*(15*d^3 - 10*d^2*e*x + 8*d*e^2*x^2 + 48*e^3*x^3)) - (3*Sqrt[c*
d]*(c*d^2 - a*e^2)^(5/2)*(5*c*d^2 + 3*a*e^2)*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sq
rt[c*d^2 - a*e^2])])/(Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])))/(192*c^(5/2)*d^(5/2)*e^(7/2))

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IntegrateAlgebraic [F]  time = 180.77, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.46, size = 676, normalized size = 2.29 \begin {gather*} \left [-\frac {3 \, {\left (5 \, c^{4} d^{8} - 12 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} + 4 \, a^{3} c d^{2} e^{6} - 3 \, 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 - 31 \, a c^{3} d^{5} e^{3} + 9 \, a^{2} c^{2} d^{3} e^{5} - 9 \, a^{3} c d e^{7} + 8 \, {\left (c^{4} d^{5} e^{3} + 9 \, a c^{3} d^{3} e^{5}\right )} x^{2} - 2 \, {\left (5 \, c^{4} d^{6} e^{2} - 10 \, a c^{3} d^{4} e^{4} - 3 \, 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^{3} d^{3} e^{4}}, \frac {3 \, {\left (5 \, c^{4} d^{8} - 12 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} + 4 \, a^{3} c d^{2} e^{6} - 3 \, 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 - 31 \, a c^{3} d^{5} e^{3} + 9 \, a^{2} c^{2} d^{3} e^{5} - 9 \, a^{3} c d e^{7} + 8 \, {\left (c^{4} d^{5} e^{3} + 9 \, a c^{3} d^{3} e^{5}\right )} x^{2} - 2 \, {\left (5 \, c^{4} d^{6} e^{2} - 10 \, a c^{3} d^{4} e^{4} - 3 \, 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^{3} d^{3} e^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(3*(5*c^4*d^8 - 12*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^6 - 3*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 - 31*a*c^3
*d^5*e^3 + 9*a^2*c^2*d^3*e^5 - 9*a^3*c*d*e^7 + 8*(c^4*d^5*e^3 + 9*a*c^3*d^3*e^5)*x^2 - 2*(5*c^4*d^6*e^2 - 10*a
*c^3*d^4*e^4 - 3*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^3*d^3*e^4), 1/384*(3*(5*c
^4*d^8 - 12*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 + 4*a^3*c*d^2*e^6 - 3*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 - 31*a*c^3*d^5*e^3 + 9*a^2*c^2*d^3*e^5 - 9*a
^3*c*d*e^7 + 8*(c^4*d^5*e^3 + 9*a*c^3*d^3*e^5)*x^2 - 2*(5*c^4*d^6*e^2 - 10*a*c^3*d^4*e^4 - 3*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^3*d^3*e^4)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a sub
stitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perha
ps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing
0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution vari
able should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.W
arning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a
substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should pe
rhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replac
ing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution v
ariable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purge
d.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Error: Bad Argument Type

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maple [B]  time = 0.01, size = 1279, normalized size = 4.34

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x+1/8/c/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a+1/8/e^2*d*(c
*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)-3/32*e^2/c/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^2+3/16*d*(c*d*e
*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a-3/32/e^2*c*d^3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x-3/64*e^3/c^2/d^
2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^3+3/64*e/c*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2+3/64/e*d^2*
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-3/64/e^3*c*d^4*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+3/128*e^5/c^2
/d^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^4
-3/32*e^3/c*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1
/2)*a^3+9/64*e*d^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*
d*e)^(1/2)*a^2-3/32/e*c*d^4*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(
1/2))/(c*d*e)^(1/2)*a+3/128/e^3*c^2*d^6*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2
+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)-1/3*d/e^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)-1/4*d*a*((x+d/e)^2*c*d
*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-1/8*e*a^2/c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+1/16*e^3*a^3/c*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)
-3/16*d^2*e*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)+3/16*d^4/e*a*c*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)+1/4*d^3/e^2*c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+1/8*d^4/e
^3*c*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-1/16*d^6/e^3*c^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 [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(x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d),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 details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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