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

Optimal. Leaf size=201 \[ \frac {\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{3/2} d^{3/2} e^{5/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}+\frac {1}{8} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \]

[Out]

1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/e+1/16*(-a*e^2+c*d^2)^3*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^(5/2)+1/8*(a/c/d-d/e^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)

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Rubi [A]  time = 0.12, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {664, 612, 621, 206} \[ \frac {\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{3/2} d^{3/2} e^{5/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}+\frac {1}{8} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a/(c*d) - d/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])/8 + (a*d*e + (c*d^
2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(3*e) + ((c*d^2 - a*e^2)^3*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])])/(16*c^(3/2)*d^(3/2)*e^(5/2))

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]

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

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Mathematica [A]  time = 0.67, size = 264, normalized size = 1.31 \[ \frac {\sqrt {c} \sqrt {d} \left (3 \left (c d^2-a e^2\right )^{7/2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )-\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d} (d+e x) \left (-3 a^3 e^5-a^2 c d e^3 (8 d+17 e x)+a c^2 d^2 e \left (3 d^2-10 d e x-22 e^2 x^2\right )+c^3 d^3 x \left (3 d^2-2 d e x-8 e^2 x^2\right )\right )\right )}{24 e^{5/2} (c d)^{5/2} \sqrt {(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[d]*(-(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(-3*a^3*e^5 - a^2*c*d*e^3*(8*d + 17*e*x) + a*c
^2*d^2*e*(3*d^2 - 10*d*e*x - 22*e^2*x^2) + c^3*d^3*x*(3*d^2 - 2*d*e*x - 8*e^2*x^2))) + 3*(c*d^2 - a*e^2)^(7/2)
*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])]))/(24*(c*d)^(5/2)*e^(5/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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fricas [A]  time = 1.26, size = 532, normalized size = 2.65 \[ \left [-\frac {3 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} - 3 \, c^{3} d^{5} e + 8 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5} + 2 \, {\left (c^{3} d^{4} e^{2} + 7 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{96 \, c^{2} d^{2} e^{3}}, -\frac {3 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) - 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} - 3 \, c^{3} d^{5} e + 8 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5} + 2 \, {\left (c^{3} d^{4} e^{2} + 7 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{48 \, c^{2} d^{2} e^{3}}\right ] \]

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

[Out]

[-1/96*(3*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*(8*c^3*d^3*e^3*x^2 - 3*c^3*d^5*e + 8*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5 + 2*
(c^3*d^4*e^2 + 7*a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e^3), -1/48*(3*(c^3*d
^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*(8*c^3*d^3*e^3*x^2 - 3*c^3*d^5*e + 8*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5 + 2*(c^3*d^4*e^2 + 7*a*c^2*d^2*e^4)*x)*
sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*e^3)]

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

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)^(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.Error: Bad Argument Type

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maple [B]  time = 0.01, size = 566, normalized size = 2.82 \[ -\frac {a^{3} e^{4} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{16 \sqrt {c d e}\, c d}+\frac {3 a^{2} d \,e^{2} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{16 \sqrt {c d e}}-\frac {3 a c \,d^{3} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{16 \sqrt {c d e}}+\frac {c^{2} d^{5} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{16 \sqrt {c d e}\, e^{2}}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, a e x}{4}-\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, c \,d^{2} x}{4 e}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, a^{2} e^{2}}{8 c d}-\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, c \,d^{3}}{8 e^{2}}+\frac {\left (\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e} \]

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

[Out]

1/3/e*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+1/4*e*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+1/
8*e^2*a^2/c/d*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-1/16*e^4*a^3/c/d*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*e^2*a^2*d*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*a*c*d
^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)-1/4/e*c*d^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-1/8/e^2*c*d^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2
)*(x+d/e))^(1/2)+1/16/e^2*c^2*d^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)

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maxima [B]  time = 0.50, size = 462, normalized size = 2.30 \[ \frac {3 \, a^{2} c d^{2} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{16 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} + \frac {c^{3} d^{6} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{16 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{4}} - \frac {3 \, a c^{2} d^{4} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{16 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{2}} - \frac {a^{3} e^{2} \log \left (2 \, c d x + \frac {c d^{2}}{e} + a e + 2 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} \sqrt {\frac {c d}{e}}\right )}{16 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} - \frac {\sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c d^{2} x}{4 \, e} + \frac {1}{4} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a e x - \frac {\sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c d^{3}}{8 \, e^{2}} + \frac {\sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{2} e^{2}}{8 \, c d} + \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}}}{3 \, e} \]

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)^(3/2)/(e*x+d),x, algorithm="maxima")

[Out]

3/16*a^2*c*d^2*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) + 1/16*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)*sqrt(c*d/e)
)/((c*d/e)^(3/2)*e^4) - 3/16*a*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^2) - 1/16*a^3*e^2*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) - 1/4*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c*d^2*x/e + 1/
4*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a*e*x - 1/8*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c*d^3/e^
2 + 1/8*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^2*e^2/(c*d) + 1/3*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e
)^(3/2)/e

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

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)^(3/2)/(d + e*x),x)

[Out]

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

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)**(3/2)/(e*x+d),x)

[Out]

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

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