3.5.50 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{x (d+e x)} \, dx\) [450]

Optimal. Leaf size=251 \[ \frac {\left (c d^2+5 a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e}-\frac {\left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 \sqrt {c} \sqrt {d} e^{3/2}}-a^{3/2} \sqrt {d} e^{3/2} \tanh ^{-1}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right ) \]

[Out]

-1/8*(-3*a^2*e^4-6*a*c*d^2*e^2+c^2*d^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))/e^(3/2)/c^(1/2)/d^(1/2)-a^(3/2)*e^(3/2)*arctanh(1/2*(2*a*d*e+(a*e^2+c*d^2)*x)/a
^(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))*d^(1/2)+1/4*(2*c*d*e*x+5*a*e^2+c*d^2)*(a*d*e+(
a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/e

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Rubi [A]
time = 0.17, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {863, 828, 857, 635, 212, 738} \begin {gather*} -a^{3/2} \sqrt {d} e^{3/2} \tanh ^{-1}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac {\left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\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 )}{8 \sqrt {c} \sqrt {d} e^{3/2}}+\frac {\left (5 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}}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c*d^2 + 5*a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e) - ((c^2*d^4 - 6*a*c*d^2*e^2
- 3*a^2*e^4)*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])])/(8*Sqrt[c]*Sqrt[d]*e^(3/2)) - a^(3/2)*Sqrt[d]*e^(3/2)*ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sq
rt[a]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^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 738

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

Rule 828

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

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 863

Int[((x_)^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + c*(
x/e))*(a + b*x + c*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b
*d*e + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||  !IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2
]))

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}}{x (d+e x)} \, dx &=\int \frac {(a e+c d x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{x} \, dx\\ &=\frac {\left (c d^2+5 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}}{4 e}-\frac {\int \frac {-4 a^2 c d^2 e^3+\frac {1}{2} c d \left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\right ) x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 c d e}\\ &=\frac {\left (c d^2+5 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}}{4 e}+\left (a^2 d e^2\right ) \int \frac {1}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx-\frac {\left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e}\\ &=\frac {\left (c d^2+5 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}}{4 e}-\left (2 a^2 d e^2\right ) \text {Subst}\left (\int \frac {1}{4 a d e-x^2} \, dx,x,\frac {2 a d e-\left (-c d^2-a e^2\right ) x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )-\frac {\left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\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 )}{4 e}\\ &=\frac {\left (c d^2+5 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}}{4 e}-\frac {\left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\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 )}{8 \sqrt {c} \sqrt {d} e^{3/2}}-a^{3/2} \sqrt {d} e^{3/2} \tanh ^{-1}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 232, normalized size = 0.92 \begin {gather*} \frac {\sqrt {a e+c d x} \sqrt {d+e x} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x} \sqrt {d+e x} \left (5 a e^2+c d (d+2 e x)\right )+\left (-c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )-8 a^{3/2} \sqrt {c} d e^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e} \sqrt {d+e x}}{\sqrt {d} \sqrt {a e+c d x}}\right )\right )}{4 \sqrt {c} \sqrt {d} e^{3/2} \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*(5*a*e^2 + c*d*(d +
2*e*x)) + (-(c^2*d^4) + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e +
c*d*x])] - 8*a^(3/2)*Sqrt[c]*d*e^3*ArcTanh[(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])/(Sqrt[d]*Sqrt[a*e + c*d*x])]))/(4*
Sqrt[c]*Sqrt[d]*e^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(606\) vs. \(2(213)=426\).
time = 0.08, size = 607, normalized size = 2.42

method	result	size
default	\(-\frac {\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3}+\frac {\left (a \,e^{2}-c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+a \,e^{2}-c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (a \,e^{2}-c \,d^{2}\right )^{2} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{2}}{d}+\frac {\frac {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}\right )^{\frac {3}{2}}}{3}+\frac {\left (a \,e^{2}+c \,d^{2}\right ) \left (\frac {\left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{4 c d e}+\frac {\left (4 a c \,d^{2} e^{2}-\left (a \,e^{2}+c \,d^{2}\right )^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{8 c d e \sqrt {c d e}}\right )}{2}+a d e \left (\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}+\frac {\left (a \,e^{2}+c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 \sqrt {c d e}}-\frac {a d e \ln \left (\frac {2 a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{x}\right )}{\sqrt {a d e}}\right )}{d}\)	\(607\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/x/(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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 8.89, size = 1303, normalized size = 5.19 \begin {gather*} \left [\frac {{\left (8 \, \sqrt {a d} a c d e^{\frac {7}{2}} \log \left (\frac {c^{2} d^{4} x^{2} + 8 \, a c d^{3} x e + a^{2} x^{2} e^{4} + 8 \, a^{2} d x e^{3} - 4 \, {\left (c d^{2} x + a x e^{2} + 2 \, a d e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {a d} e^{\frac {1}{2}} + 2 \, {\left (3 \, a c d^{2} x^{2} + 4 \, a^{2} d^{2}\right )} e^{2}}{x^{2}}\right ) - {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \sqrt {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 \, \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 {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 (2 \, c^{2} d^{2} x e^{2} + c^{2} d^{3} e + 5 \, a c d e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{16 \, c d}, \frac {{\left (4 \, \sqrt {a d} a c d e^{\frac {7}{2}} \log \left (\frac {c^{2} d^{4} x^{2} + 8 \, a c d^{3} x e + a^{2} x^{2} e^{4} + 8 \, a^{2} d x e^{3} - 4 \, {\left (c d^{2} x + a x e^{2} + 2 \, a d e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {a d} e^{\frac {1}{2}} + 2 \, {\left (3 \, a c d^{2} x^{2} + 4 \, a^{2} d^{2}\right )} e^{2}}{x^{2}}\right ) + {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \sqrt {-c d e} \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 {-c d e}}{2 \, {\left (c^{2} d^{3} x e + a c d x e^{3} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e^{2}\right )}}\right ) + 2 \, {\left (2 \, c^{2} d^{2} x e^{2} + c^{2} d^{3} e + 5 \, a c d e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{8 \, c d}, \frac {{\left (16 \, \sqrt {-a d e} a c d \arctan \left (\frac {{\left (c d^{2} x + a x e^{2} + 2 \, a d e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {-a d e}}{2 \, {\left (a c d^{3} x e + a^{2} d x e^{3} + {\left (a c d^{2} x^{2} + a^{2} d^{2}\right )} e^{2}\right )}}\right ) e^{3} - {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \sqrt {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 \, \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 {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 (2 \, c^{2} d^{2} x e^{2} + c^{2} d^{3} e + 5 \, a c d e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{16 \, c d}, \frac {{\left (8 \, \sqrt {-a d e} a c d \arctan \left (\frac {{\left (c d^{2} x + a x e^{2} + 2 \, a d e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {-a d e}}{2 \, {\left (a c d^{3} x e + a^{2} d x e^{3} + {\left (a c d^{2} x^{2} + a^{2} d^{2}\right )} e^{2}\right )}}\right ) e^{3} + {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} \sqrt {-c d e} \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 {-c d e}}{2 \, {\left (c^{2} d^{3} x e + a c d x e^{3} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e^{2}\right )}}\right ) + 2 \, {\left (2 \, c^{2} d^{2} x e^{2} + c^{2} d^{3} e + 5 \, a c d e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{8 \, c d}\right ] \end {gather*}

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)/x/(e*x+d),x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(a*d)*a*c*d*e^(7/2)*log((c^2*d^4*x^2 + 8*a*c*d^3*x*e + a^2*x^2*e^4 + 8*a^2*d*x*e^3 - 4*(c*d^2*x +
a*x*e^2 + 2*a*d*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(a*d)*e^(1/2) + 2*(3*a*c*d^2*x^2 + 4*a^2*d
^2)*e^2)/x^2) - (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(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*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(c*d)*e^(1/2)
+ 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)*e^2) + 4*(2*c^2*d^2*x*e^2 + c^2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d^2*x + a*x*e^2 +
(c*d*x^2 + a*d)*e))*e^(-2)/(c*d), 1/8*(4*sqrt(a*d)*a*c*d*e^(7/2)*log((c^2*d^4*x^2 + 8*a*c*d^3*x*e + a^2*x^2*e^
4 + 8*a^2*d*x*e^3 - 4*(c*d^2*x + a*x*e^2 + 2*a*d*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(a*d)*e^(1
/2) + 2*(3*a*c*d^2*x^2 + 4*a^2*d^2)*e^2)/x^2) + (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(-c*d*e)*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(-c*d*e)/(c^2*d^3*x*e + a*c*d*x*e^
3 + (c^2*d^2*x^2 + a*c*d^2)*e^2)) + 2*(2*c^2*d^2*x*e^2 + c^2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*
d*x^2 + a*d)*e))*e^(-2)/(c*d), 1/16*(16*sqrt(-a*d*e)*a*c*d*arctan(1/2*(c*d^2*x + a*x*e^2 + 2*a*d*e)*sqrt(c*d^2
*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(-a*d*e)/(a*c*d^3*x*e + a^2*d*x*e^3 + (a*c*d^2*x^2 + a^2*d^2)*e^2))*e^3
- (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(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*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(c*d)*e^(1/2) + 2*(4*c^2*d^2
*x^2 + 3*a*c*d^2)*e^2) + 4*(2*c^2*d^2*x*e^2 + c^2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d
)*e))*e^(-2)/(c*d), 1/8*(8*sqrt(-a*d*e)*a*c*d*arctan(1/2*(c*d^2*x + a*x*e^2 + 2*a*d*e)*sqrt(c*d^2*x + a*x*e^2
+ (c*d*x^2 + a*d)*e)*sqrt(-a*d*e)/(a*c*d^3*x*e + a^2*d*x*e^3 + (a*c*d^2*x^2 + a^2*d^2)*e^2))*e^3 + (c^2*d^4 -
6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(-c*d*e)*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(-c*d*e)/(c^2*d^3*x*e + a*c*d*x*e^3 + (c^2*d^2*x^2 + a*c*d^2)*e^2)) + 2*(2*c^2*d^2*x*e^2 + c^
2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-2)/(c*d)]

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

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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

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)/(x*(d + e*x)),x)

[Out]

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

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