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

Optimal. Leaf size=274 \[ -\frac {3 \left (c d^2-a e^2\right )^5 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{256 c^{5/2} d^{5/2} e^{7/2}}+\frac {3 \left (c d^2-a e^2\right )^3 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{128 c^2 d^2 e^3}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e}+\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} \]

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Rubi [A]  time = 0.19, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {664, 612, 621, 206} \begin {gather*} \frac {3 \left (c d^2-a e^2\right )^3 \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}}{128 c^2 d^2 e^3}-\frac {3 \left (c d^2-a e^2\right )^5 \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 )}{256 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}}{5 e}+\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (a e^2+c d^2+2 c d e x\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[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x),x]

[Out]

(3*(c*d^2 - a*e^2)^3*(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])/(128*c^2*d^2*e^3
) + ((a/(c*d) - d/e^2)*(c*d^2 + a*e^2 + 2*c*d*e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/16 + (a*d*e
+ (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(5*e) - (3*(c*d^2 - a*e^2)^5*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqr
t[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(256*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]

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} \, dx &=\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e}-\frac {\left (2 c d^2 e-e \left (c d^2+a e^2\right )\right ) \int \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, dx}{2 e^2}\\ &=\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (c d^2+a e^2+2 c d e x\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}}{5 e}+\frac {\left (3 \left (c d^2-a e^2\right )^3\right ) \int \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{32 c d e^2}\\ &=\frac {3 \left (c d^2-a e^2\right )^3 \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}}{128 c^2 d^2 e^3}+\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (c d^2+a e^2+2 c d e x\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}}{5 e}-\frac {\left (3 \left (c d^2-a e^2\right )^5\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{256 c^2 d^2 e^3}\\ &=\frac {3 \left (c d^2-a e^2\right )^3 \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}}{128 c^2 d^2 e^3}+\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (c d^2+a e^2+2 c d e x\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}}{5 e}-\frac {\left (3 \left (c d^2-a e^2\right )^5\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 )}{128 c^2 d^2 e^3}\\ &=\frac {3 \left (c d^2-a e^2\right )^3 \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}}{128 c^2 d^2 e^3}+\frac {1}{16} \left (\frac {a}{c d}-\frac {d}{e^2}\right ) \left (c d^2+a e^2+2 c d e x\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}}{5 e}-\frac {3 \left (c d^2-a e^2\right )^5 \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 )}{256 c^{5/2} d^{5/2} e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.22, size = 384, normalized size = 1.40 \begin {gather*} \frac {\sqrt {c d} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d} (d+e x) \left (-15 a^5 e^9+5 a^4 c d e^7 (14 d-e x)+2 a^3 c^2 d^2 e^5 \left (64 d^2+268 d e x+129 e^2 x^2\right )+2 a^2 c^3 d^3 e^3 \left (-35 d^3+87 d^2 e x+489 d e^2 x^2+292 e^3 x^3\right )+a c^4 d^4 e \left (15 d^4-80 d^3 e x+54 d^2 e^2 x^2+688 d e^3 x^3+464 e^4 x^4\right )+c^5 d^5 x \left (15 d^4-10 d^3 e x+8 d^2 e^2 x^2+176 d e^3 x^3+128 e^4 x^4\right )\right )-15 \left (c d^2-a e^2\right )^{11/2} \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 )}{640 c^{7/2} d^{7/2} e^{7/2} \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

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),x]

[Out]

(Sqrt[c*d]*(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(-15*a^5*e^9 + 5*a^4*c*d*e^7*(14*d - e*x) + 2*a^3*c^2*
d^2*e^5*(64*d^2 + 268*d*e*x + 129*e^2*x^2) + 2*a^2*c^3*d^3*e^3*(-35*d^3 + 87*d^2*e*x + 489*d*e^2*x^2 + 292*e^3
*x^3) + c^5*d^5*x*(15*d^4 - 10*d^3*e*x + 8*d^2*e^2*x^2 + 176*d*e^3*x^3 + 128*e^4*x^4) + a*c^4*d^4*e*(15*d^4 -
80*d^3*e*x + 54*d^2*e^2*x^2 + 688*d*e^3*x^3 + 464*e^4*x^4)) - 15*(c*d^2 - a*e^2)^(11/2)*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])]))/(640*c^(7/2)*d^(7/2)*e^(7/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [F]  time = 180.04, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.46, size = 844, normalized size = 3.08 \begin {gather*} \left [\frac {15 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\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 (128 \, c^{5} d^{5} e^{5} x^{4} + 15 \, c^{5} d^{9} e - 70 \, a c^{4} d^{7} e^{3} + 128 \, a^{2} c^{3} d^{5} e^{5} + 70 \, a^{3} c^{2} d^{3} e^{7} - 15 \, a^{4} c d e^{9} + 16 \, {\left (11 \, c^{5} d^{6} e^{4} + 21 \, a c^{4} d^{4} e^{6}\right )} x^{3} + 8 \, {\left (c^{5} d^{7} e^{3} + 64 \, a c^{4} d^{5} e^{5} + 31 \, a^{2} c^{3} d^{3} e^{7}\right )} x^{2} - 2 \, {\left (5 \, c^{5} d^{8} e^{2} - 23 \, a c^{4} d^{6} e^{4} - 233 \, a^{2} c^{3} d^{4} e^{6} - 5 \, a^{3} c^{2} d^{2} e^{8}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{2560 \, c^{3} d^{3} e^{4}}, \frac {15 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\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 (128 \, c^{5} d^{5} e^{5} x^{4} + 15 \, c^{5} d^{9} e - 70 \, a c^{4} d^{7} e^{3} + 128 \, a^{2} c^{3} d^{5} e^{5} + 70 \, a^{3} c^{2} d^{3} e^{7} - 15 \, a^{4} c d e^{9} + 16 \, {\left (11 \, c^{5} d^{6} e^{4} + 21 \, a c^{4} d^{4} e^{6}\right )} x^{3} + 8 \, {\left (c^{5} d^{7} e^{3} + 64 \, a c^{4} d^{5} e^{5} + 31 \, a^{2} c^{3} d^{3} e^{7}\right )} x^{2} - 2 \, {\left (5 \, c^{5} d^{8} e^{2} - 23 \, a c^{4} d^{6} e^{4} - 233 \, a^{2} c^{3} d^{4} e^{6} - 5 \, a^{3} c^{2} d^{2} e^{8}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{1280 \, c^{3} d^{3} e^{4}}\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)^(5/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/2560*(15*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10
)*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*(128*c^5*d^5*e^5*x^4 + 15
*c^5*d^9*e - 70*a*c^4*d^7*e^3 + 128*a^2*c^3*d^5*e^5 + 70*a^3*c^2*d^3*e^7 - 15*a^4*c*d*e^9 + 16*(11*c^5*d^6*e^4
+ 21*a*c^4*d^4*e^6)*x^3 + 8*(c^5*d^7*e^3 + 64*a*c^4*d^5*e^5 + 31*a^2*c^3*d^3*e^7)*x^2 - 2*(5*c^5*d^8*e^2 - 23
*a*c^4*d^6*e^4 - 233*a^2*c^3*d^4*e^6 - 5*a^3*c^2*d^2*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3
*d^3*e^4), 1/1280*(15*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8
- a^5*e^10)*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)*sq
rt(-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*(128*c^5*d^5*e^5*x^4 + 15*c^5*d^9*
e - 70*a*c^4*d^7*e^3 + 128*a^2*c^3*d^5*e^5 + 70*a^3*c^2*d^3*e^7 - 15*a^4*c*d*e^9 + 16*(11*c^5*d^6*e^4 + 21*a*c
^4*d^4*e^6)*x^3 + 8*(c^5*d^7*e^3 + 64*a*c^4*d^5*e^5 + 31*a^2*c^3*d^3*e^7)*x^2 - 2*(5*c^5*d^8*e^2 - 23*a*c^4*d^
6*e^4 - 233*a^2*c^3*d^4*e^6 - 5*a^3*c^2*d^2*e^8)*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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(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.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 variable should perhaps be purged.Warning, rep
lacing 0 by ` u`, a substitution variable should perhaps be purged.Evaluation time: 0.45Error: Bad Argument Ty
pe

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maple [B]  time = 0.01, size = 1123, normalized size = 4.10

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),x)

[Out]

1/5/e*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(5/2)+15/256/e*a*c^2*d^6*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)-3/64*e^4*a^3/c/d*((x+d/e)^2*c*d*e+(
a*e^2-c*d^2)*(x+d/e))^(1/2)*x-9/64*a*c*d^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+3/256*e^7*a^5/c^2/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)-15/128*e*a^2*c*d^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)+1/8*e*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)*x+3/64*e^3*a^3/c*((x+d/e)^
2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-1/8/e*c*d^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)*x-1/16/e^2*c*d^
3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+3/128/e^3*c^2*d^6*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2
)-3/256/e^3*c^3*d^8*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)-3/64/e*a*c*d^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-15/256*e^5*a^4/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)+15/1
28*e^3*a^3*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)+3/64/e^2*c^2*d^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+9/64*e^2*a^2*d*((x+d/e)^2*
c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-3/128*e^5*a^4/c^2/d^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+1/16*
e^2*a^2/c/d*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)

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maxima [B]  time = 0.57, size = 915, normalized size = 3.34 \begin {gather*} -\frac {3 \, c^{4} d^{9} \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 )}{256 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{5}} + \frac {15 \, a c^{3} d^{7} \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 )}{256 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}} e^{3}} - \frac {15 \, a^{2} c^{2} d^{5} \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} + \frac {15 \, a^{3} c d^{3} 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 )}{128 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} - \frac {15 \, a^{4} d 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 )}{256 \, \left (\frac {c d}{e}\right )^{\frac {3}{2}}} + \frac {3 \, a^{5} e^{5} \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 )}{256 \, c d \left (\frac {c d}{e}\right )^{\frac {3}{2}}} - \frac {9}{64} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a c d^{3} x + \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c^{2} d^{5} x}{64 \, e^{2}} + \frac {9}{64} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{2} d e^{2} x - \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{3} e^{4} x}{64 \, c d} + \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} c^{2} d^{6}}{128 \, e^{3}} - \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a c d^{4}}{64 \, e} + \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{3} e^{3}}{64 \, c} - \frac {3 \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} a^{4} e^{5}}{128 \, c^{2} d^{2}} - \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} c d^{2} x}{8 \, e} + \frac {1}{8} \, {\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} a e x - \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} c d^{3}}{16 \, e^{2}} + \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {3}{2}} a^{2} e^{2}}{16 \, c d} + \frac {{\left (c d e x^{2} + c d^{2} x + a e^{2} x + a d e\right )}^{\frac {5}{2}}}{5 \, e} \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)^(5/2)/(e*x+d),x, algorithm="maxima")

[Out]

-3/256*c^4*d^9*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) + 15/256*a*c^3*d^7*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)*s
qrt(c*d/e))/((c*d/e)^(3/2)*e^3) - 15/128*a^2*c^2*d^5*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) + 15/128*a^3*c*d^3*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) - 15/256*a^4*d*e^3*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) + 3/256*a^5*e^5*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*(c*d/e)^(3/2)) - 9/64*sqrt(c*d*e*x
^2 + c*d^2*x + a*e^2*x + a*d*e)*a*c*d^3*x + 3/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c^2*d^5*x/e^2 + 9
/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^2*d*e^2*x - 3/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)
*a^3*e^4*x/(c*d) + 3/128*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*c^2*d^6/e^3 - 3/64*sqrt(c*d*e*x^2 + c*d^2
*x + a*e^2*x + a*d*e)*a*c*d^4/e + 3/64*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^3*e^3/c - 3/128*sqrt(c*d*
e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*a^4*e^5/(c^2*d^2) - 1/8*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^(3/2)*c*d^2
*x/e + 1/8*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^(3/2)*a*e*x - 1/16*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^
(3/2)*c*d^3/e^2 + 1/16*(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)^(3/2)*a^2*e^2/(c*d) + 1/5*(c*d*e*x^2 + c*d^2*x
+ a*e^2*x + a*d*e)^(5/2)/e

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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 )}^{5/2}}{d+e\,x} \,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)^(5/2)/(d + e*x),x)

[Out]

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

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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 {5}{2}}}{d + e x}\, 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)**(5/2)/(e*x+d),x)

[Out]

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

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