3.1965 \(\int \frac {(d+e x)^5}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\)
Optimal. Leaf size=231 \[ \frac {5 e^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{7/2} d^{7/2}}+\frac {5 e^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^3 d^3}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^4}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]
[Out]
-2/3*(e*x+d)^4/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+5/2*e^(3/2)*(-a*e^2+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^(7/2)/d^(7/2)-10/3*e*(e*x+d)^2/
c^2/d^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5*e^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3
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Rubi [A] time = 0.14, antiderivative size = 231, 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 =
{668, 640, 621, 206} \[ -\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {5 e^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^3 d^3}+\frac {5 e^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 c^{7/2} d^{7/2}}-\frac {2 (d+e x)^4}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]
[Out]
(-2*(d + e*x)^4)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (10*e*(d + e*x)^2)/(3*c^2*d^2*Sqrt[a*
d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*e^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c^3*d^3) + (5*e^(
3/2)*(c*d^2 - 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])])/(2*c^(7/2)*d^(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 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 640
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]
Rule 668
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(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] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
Rubi steps
\begin {align*} \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {(5 e) \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 c d}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (5 e^2\right ) \int \frac {d+e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c^2 d^2}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {\left (5 e^2 \left (c d^2-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}{2 c^3 d^3}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {\left (5 e^2 \left (c d^2-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 )}{c^3 d^3}\\ &=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {5 e^{3/2} \left (c d^2-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 )}{2 c^{7/2} d^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.06, size = 112, normalized size = 0.48 \[ -\frac {2 \left (c d^2-a e^2\right )^2 \sqrt {(d+e x) (a e+c d x)} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{3 c^3 d^3 (a e+c d x)^2 \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}} \]
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]
[Out]
(-2*(c*d^2 - a*e^2)^2*Sqrt[(a*e + c*d*x)*(d + e*x)]*Hypergeometric2F1[-5/2, -3/2, -1/2, (e*(a*e + c*d*x))/(-(c
*d^2) + a*e^2)])/(3*c^3*d^3*(a*e + c*d*x)^2*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])
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fricas [A] time = 3.84, size = 641, normalized size = 2.77 \[ \left [\frac {15 \, {\left (a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{2} - a^{2} c d e^{4}\right )} x\right )} \sqrt {\frac {e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \, {\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {e}{c d}}\right ) + 4 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - 2 \, {\left (7 \, c^{2} d^{3} e - 10 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{12 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{2} - a^{2} c d e^{4}\right )} x\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) - 2 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - 2 \, {\left (7 \, c^{2} d^{3} e - 10 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{6 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")
[Out]
[1/12*(15*(a^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4*e - a*c^2*d^2*e^3)*x^2 + 2*(a*c^2*d^3*e^2 - a^2*c*d*e^4)*x)*sqrt
(e/(c*d))*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 + 8*(c^2*d^3*e + a*c*d*e^3)*x + 4*(2*c^2*d
^2*e*x + c^2*d^3 + a*c*d*e^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d))) + 4*(3*c^2*d^2*e^2*x^
2 - 2*c^2*d^4 - 10*a*c*d^2*e^2 + 15*a^2*e^4 - 2*(7*c^2*d^3*e - 10*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^
2 + a*e^2)*x))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2), -1/6*(15*(a^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4
*e - a*c^2*d^2*e^3)*x^2 + 2*(a*c^2*d^3*e^2 - a^2*c*d*e^4)*x)*sqrt(-e/(c*d))*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(-e/(c*d))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x))
- 2*(3*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 10*a*c*d^2*e^2 + 15*a^2*e^4 - 2*(7*c^2*d^3*e - 10*a*c*d*e^3)*x)*sqrt(c*d
*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2)]
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giac [B] time = 0.77, size = 827, normalized size = 3.58 \[ \frac {{\left ({\left ({\left (\frac {3 \, {\left (c^{6} d^{10} e^{6} - 4 \, a c^{5} d^{8} e^{8} + 6 \, a^{2} c^{4} d^{6} e^{10} - 4 \, a^{3} c^{3} d^{4} e^{12} + a^{4} c^{2} d^{2} e^{14}\right )} x}{c^{7} d^{11} e^{2} - 4 \, a c^{6} d^{9} e^{4} + 6 \, a^{2} c^{5} d^{7} e^{6} - 4 \, a^{3} c^{4} d^{5} e^{8} + a^{4} c^{3} d^{3} e^{10}} - \frac {4 \, {\left (2 \, c^{6} d^{11} e^{5} - 13 \, a c^{5} d^{9} e^{7} + 32 \, a^{2} c^{4} d^{7} e^{9} - 38 \, a^{3} c^{3} d^{5} e^{11} + 22 \, a^{4} c^{2} d^{3} e^{13} - 5 \, a^{5} c d e^{15}\right )}}{c^{7} d^{11} e^{2} - 4 \, a c^{6} d^{9} e^{4} + 6 \, a^{2} c^{5} d^{7} e^{6} - 4 \, a^{3} c^{4} d^{5} e^{8} + a^{4} c^{3} d^{3} e^{10}}\right )} x - \frac {3 \, {\left (9 \, c^{6} d^{12} e^{4} - 46 \, a c^{5} d^{10} e^{6} + 89 \, a^{2} c^{4} d^{8} e^{8} - 76 \, a^{3} c^{3} d^{6} e^{10} + 19 \, a^{4} c^{2} d^{4} e^{12} + 10 \, a^{5} c d^{2} e^{14} - 5 \, a^{6} e^{16}\right )}}{c^{7} d^{11} e^{2} - 4 \, a c^{6} d^{9} e^{4} + 6 \, a^{2} c^{5} d^{7} e^{6} - 4 \, a^{3} c^{4} d^{5} e^{8} + a^{4} c^{3} d^{3} e^{10}}\right )} x - \frac {6 \, {\left (3 \, c^{6} d^{13} e^{3} - 12 \, a c^{5} d^{11} e^{5} + 13 \, a^{2} c^{4} d^{9} e^{7} + 8 \, a^{3} c^{3} d^{7} e^{9} - 27 \, a^{4} c^{2} d^{5} e^{11} + 20 \, a^{5} c d^{3} e^{13} - 5 \, a^{6} d e^{15}\right )}}{c^{7} d^{11} e^{2} - 4 \, a c^{6} d^{9} e^{4} + 6 \, a^{2} c^{5} d^{7} e^{6} - 4 \, a^{3} c^{4} d^{5} e^{8} + a^{4} c^{3} d^{3} e^{10}}\right )} x - \frac {2 \, c^{6} d^{14} e^{2} + 2 \, a c^{5} d^{12} e^{4} - 43 \, a^{2} c^{4} d^{10} e^{6} + 112 \, a^{3} c^{3} d^{8} e^{8} - 128 \, a^{4} c^{2} d^{6} e^{10} + 70 \, a^{5} c d^{4} e^{12} - 15 \, a^{6} d^{2} e^{14}}{c^{7} d^{11} e^{2} - 4 \, a c^{6} d^{9} e^{4} + 6 \, a^{2} c^{5} d^{7} e^{6} - 4 \, a^{3} c^{4} d^{5} e^{8} + a^{4} c^{3} d^{3} e^{10}}}{3 \, {\left (c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}}} - \frac {5 \, {\left (c d^{2} e^{2} - a e^{4}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{2 \, c^{4} d^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")
[Out]
1/3*((((3*(c^6*d^10*e^6 - 4*a*c^5*d^8*e^8 + 6*a^2*c^4*d^6*e^10 - 4*a^3*c^3*d^4*e^12 + a^4*c^2*d^2*e^14)*x/(c^7
*d^11*e^2 - 4*a*c^6*d^9*e^4 + 6*a^2*c^5*d^7*e^6 - 4*a^3*c^4*d^5*e^8 + a^4*c^3*d^3*e^10) - 4*(2*c^6*d^11*e^5 -
13*a*c^5*d^9*e^7 + 32*a^2*c^4*d^7*e^9 - 38*a^3*c^3*d^5*e^11 + 22*a^4*c^2*d^3*e^13 - 5*a^5*c*d*e^15)/(c^7*d^11*
e^2 - 4*a*c^6*d^9*e^4 + 6*a^2*c^5*d^7*e^6 - 4*a^3*c^4*d^5*e^8 + a^4*c^3*d^3*e^10))*x - 3*(9*c^6*d^12*e^4 - 46*
a*c^5*d^10*e^6 + 89*a^2*c^4*d^8*e^8 - 76*a^3*c^3*d^6*e^10 + 19*a^4*c^2*d^4*e^12 + 10*a^5*c*d^2*e^14 - 5*a^6*e^
16)/(c^7*d^11*e^2 - 4*a*c^6*d^9*e^4 + 6*a^2*c^5*d^7*e^6 - 4*a^3*c^4*d^5*e^8 + a^4*c^3*d^3*e^10))*x - 6*(3*c^6*
d^13*e^3 - 12*a*c^5*d^11*e^5 + 13*a^2*c^4*d^9*e^7 + 8*a^3*c^3*d^7*e^9 - 27*a^4*c^2*d^5*e^11 + 20*a^5*c*d^3*e^1
3 - 5*a^6*d*e^15)/(c^7*d^11*e^2 - 4*a*c^6*d^9*e^4 + 6*a^2*c^5*d^7*e^6 - 4*a^3*c^4*d^5*e^8 + a^4*c^3*d^3*e^10))
*x - (2*c^6*d^14*e^2 + 2*a*c^5*d^12*e^4 - 43*a^2*c^4*d^10*e^6 + 112*a^3*c^3*d^8*e^8 - 128*a^4*c^2*d^6*e^10 + 7
0*a^5*c*d^4*e^12 - 15*a^6*d^2*e^14)/(c^7*d^11*e^2 - 4*a*c^6*d^9*e^4 + 6*a^2*c^5*d^7*e^6 - 4*a^3*c^4*d^5*e^8 +
a^4*c^3*d^3*e^10))/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(3/2) - 5/2*(c*d^2*e^2 - a*e^4)*sqrt(c*d)*e^(-1/2)*
log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x))*c*d*e
- sqrt(c*d)*a*e^(5/2)))/(c^4*d^4)
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maple [B] time = 0.07, size = 3215, normalized size = 13.92 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^5/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2),x)
[Out]
5/4*e/c^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-15/32/c*d^3/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)+1/12*e^2
*d^5/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a+5/2*e^2*d^3/(-a^2*e^4+2*a*c*d^
2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x+5/4*e*d^4/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2
+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-5/6*e^3/c*x^3/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)+19/96*c*d^7/(-a^2*e^4+2*a*
c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)+19/12*e*c^2*d^8/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-5/2*e^2/c^2/d*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+73/12*e^9/c^2/
(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^4-97/12*e^5*d^4/(-a^2*e^4+2*a*c*d
^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2+11/4*e^4/c^3/d/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)
^(3/2)*a^2+5/2*e^2/c^2/d*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)-71/16*e^2/c^2*d/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a+5/96*e^8/c^5/d^5/(c*d*e*x^2+a*d*e+(
a*e^2+c*d^2)*x)^(3/2)*a^4-43/48*e^6/c^4/d^3/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^3-115/16*e/c*d^2*x/(c*d*
e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)-35/4*e^2/c*d*x^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)+e^4*x^4/c/d/(c*d*e
*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)-5/4*e^5/c^4/d^4/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2-19/16*e^3/c^2*x/
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a-5/2*e^6/c^2/d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e
^2+c*d^2)*x)^(1/2)*x*a^2+5/48*e^11/c^4/d^4/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^
(3/2)*x*a^5-61/48*e^9/c^3/d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x*a^4+5
/6*e^12/c^3/d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^5-61/6*e^10/c^2
/d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^4-43/24*e^5/c*d^2/(-a^2*e^4+
2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x*a^2+5/2*e^4/c*d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d
^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a+67/3*e^8/c*d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d
*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^3-11/6*e^4*c*d^5/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^
2)*x)^(1/2)*x*a-5/2*e^8/c^3/d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^3
+11/16*e^5/c^3/d^2*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^2+5/12*e^13/c^4/d^4/(-a^2*e^4+2*a*c*d^2*e^2-c^2
*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^6+19/6*e^2*c^2*d^7/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e
*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x-7/12*e^10/c^4/d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2
+c*d^2)*x)^(3/2)*a^5+73/96*e^8/c^3/d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*
a^4+19/48*e*c*d^6/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x+5/6*e^5/c^2/d^2*x
^3/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a+2/3*e^3*c*d^6/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e
+(a*e^2+c*d^2)*x)^(1/2)*a+5/2*e^4/c^3/d^3*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-5/2*e^4/c^3/d^3*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-5/4*e^9/c^4/d^
4/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^4-97/96*e^4/c*d^3/(-a^2*e^4+2*a*c
*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^2+5/96*e^12/c^5/d^5/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^
4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^6-5/4*e^6/c^3/d^3*x^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^2
-5/16*e^7/c^4/d^4*x/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^3+4*e^4/c^2/d*x^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)
*x)^(3/2)*a+67/24*e^7/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x*a^3-43/3*
e^6*d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^2-11/48*e^3*d^4/(-a^2*e
^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x*a+5/2*e^3/c*d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^
2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-5/2*e^7/c^3/d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a
*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^3+1/2*e^6/c^2*d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*
x)^(3/2)*a^3+4*e^7/c*d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^3-14/3*e
^11/c^3/d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^5
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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 \[ \int \frac {{\left (d+e\,x\right )}^5}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)
[Out]
int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{5}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**5/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)
[Out]
Integral((d + e*x)**5/((d + e*x)*(a*e + c*d*x))**(5/2), x)
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