3.5.48 \(\int \frac {x (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{d+e x} \, dx\) [448]

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

[Out]

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

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Rubi [A]
time = 0.17, 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 = {808, 678, 626, 635, 212} \begin {gather*} -\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} \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 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 626

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 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 678

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 808

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 ) \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 )}{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 = 0.55, size = 236, normalized size = 0.80 \begin {gather*} \frac {\sqrt {(a e+c d x) (d+e 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 \left (c d^2-a e^2\right )^3 \left (5 c d^2+3 a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{\sqrt {a e+c d x} \sqrt {d+e x}}\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*(c*d^2
- a*e^2)^3*(5*c*d^2 + 3*a*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c*d*x])])/(Sqrt[a*e
+ c*d*x]*Sqrt[d + e*x])))/(192*c^(5/2)*d^(5/2)*e^(7/2))

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Maple [A]
time = 0.08, size = 483, normalized size = 1.64

method	result	size
default	\(\frac {\frac {\left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) \left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}\right )^{\frac {3}{2}}}{8 c d e}+\frac {3 \left (4 a c \,d^{2} e^{2}-\left (a \,e^{2}+c \,d^{2}\right )^{2}\right ) \left (\frac {\left (2 c d e x +a \,e^{2}+c \,d^{2}\right ) \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{4 c d e}+\frac {\left (4 a c \,d^{2} e^{2}-\left (a \,e^{2}+c \,d^{2}\right )^{2}\right ) \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{8 c d e \sqrt {c d e}}\right )}{16 c d e}}{e}-\frac {d \left (\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3}+\frac {\left (a \,e^{2}-c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+a \,e^{2}-c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (a \,e^{2}-c \,d^{2}\right )^{2} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{2}\right )}{e^{2}}\)	\(483\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Fricas [A]
time = 2.58, size = 653, normalized size = 2.21 \begin {gather*} \left [-\frac {{\left (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^{\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 (10 \, c^{4} d^{6} x e^{2} - 15 \, c^{4} d^{7} e - 6 \, a^{2} c^{2} d^{2} x e^{6} + 9 \, a^{3} c d e^{7} - 9 \, {\left (8 \, a c^{3} d^{3} x^{2} + a^{2} c^{2} d^{3}\right )} e^{5} - 4 \, {\left (12 \, c^{4} d^{4} x^{3} + 5 \, a c^{3} d^{4} x\right )} e^{4} - {\left (8 \, c^{4} d^{5} x^{2} - 31 \, a c^{3} d^{5}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-4\right )}}{768 \, c^{3} d^{3}}, \frac {{\left (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^{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 (10 \, c^{4} d^{6} x e^{2} - 15 \, c^{4} d^{7} e - 6 \, a^{2} c^{2} d^{2} x e^{6} + 9 \, a^{3} c d e^{7} - 9 \, {\left (8 \, a c^{3} d^{3} x^{2} + a^{2} c^{2} d^{3}\right )} e^{5} - 4 \, {\left (12 \, c^{4} d^{4} x^{3} + 5 \, a c^{3} d^{4} x\right )} e^{4} - {\left (8 \, c^{4} d^{5} x^{2} - 31 \, a c^{3} d^{5}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-4\right )}}{384 \, c^{3} d^{3}}\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^(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*(10*c^4*d^6*x*e^2 - 15*c^4*d^7*
e - 6*a^2*c^2*d^2*x*e^6 + 9*a^3*c*d*e^7 - 9*(8*a*c^3*d^3*x^2 + a^2*c^2*d^3)*e^5 - 4*(12*c^4*d^4*x^3 + 5*a*c^3*
d^4*x)*e^4 - (8*c^4*d^5*x^2 - 31*a*c^3*d^5)*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-4)/(c^3*d^3)
, 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)*arcta
n(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*(10*c^4*d^6*x*e^2 - 15*c^4*d^7*e - 6*a^2*c^2*d^2*x*e^6 + 9*a^3*c*d
*e^7 - 9*(8*a*c^3*d^3*x^2 + a^2*c^2*d^3)*e^5 - 4*(12*c^4*d^4*x^3 + 5*a*c^3*d^4*x)*e^4 - (8*c^4*d^5*x^2 - 31*a*
c^3*d^5)*e^3)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-4)/(c^3*d^3)]

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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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Giac [A]
time = 2.50, size = 298, normalized size = 1.01 \begin {gather*} \frac {1}{192} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (2 \, {\left (4 \, {\left (6 \, c d x + \frac {{\left (c^{4} d^{5} e^{2} + 9 \, a c^{3} d^{3} e^{4}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x - \frac {{\left (5 \, c^{4} d^{6} e - 10 \, a c^{3} d^{4} e^{3} - 3 \, a^{2} c^{2} d^{2} e^{5}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x + \frac {{\left (15 \, c^{4} d^{7} - 31 \, a c^{3} d^{5} e^{2} + 9 \, a^{2} c^{2} d^{3} e^{4} - 9 \, a^{3} c d e^{6}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} + \frac {{\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 )} e^{\left (-\frac {7}{2}\right )} \log \left ({\left | -c d^{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 )} \sqrt {c d} e^{\frac {1}{2}} - a e^{2} \right |}\right )}{128 \, \sqrt {c d} c^{2} d^{2}} \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]

1/192*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*(6*c*d*x + (c^4*d^5*e^2 + 9*a*c^3*d^3*e^4)*e^(-3)/(c^3
*d^3))*x - (5*c^4*d^6*e - 10*a*c^3*d^4*e^3 - 3*a^2*c^2*d^2*e^5)*e^(-3)/(c^3*d^3))*x + (15*c^4*d^7 - 31*a*c^3*d
^5*e^2 + 9*a^2*c^2*d^3*e^4 - 9*a^3*c*d*e^6)*e^(-3)/(c^3*d^3)) + 1/128*(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)*e^(-7/2)*log(abs(-c*d^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))*sqrt(c*d)*e^(1/2) - a*e^2))/(sqrt(c*d)*c^2*d^2)

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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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