3.18.38 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{13/2}} \, dx\)

Optimal. Leaf size=236 \[ \frac {5 c^3 d^3 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt {c d^2-a e^2}}-\frac {5 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {662, 660, 205} \begin {gather*} -\frac {5 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}+\frac {5 c^3 d^3 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt {c d^2-a e^2}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}} \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)^(13/2),x]

[Out]

(-5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*e^3*(d + e*x)^(3/2)) - (5*c*d*(a*d*e + (c*d^2 + a*
e^2)*x + c*d*e*x^2)^(3/2))/(12*e^2*(d + e*x)^(7/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(3*e*(d +
e*x)^(11/2)) + (5*c^3*d^3*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sq
rt[d + e*x])])/(8*e^(7/2)*Sqrt[c*d^2 - a*e^2])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 660

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

Rule 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + 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] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + 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)^{13/2}} \, dx &=-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac {(5 c d) \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)^{9/2}} \, dx}{6 e}\\ &=-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac {\left (5 c^2 d^2\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{5/2}} \, dx}{8 e^2}\\ &=-\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac {\left (5 c^3 d^3\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 e^3}\\ &=-\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac {\left (5 c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{8 e^2}\\ &=-\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3 (d+e x)^{3/2}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac {5 c^3 d^3 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{8 e^{7/2} \sqrt {c d^2-a e^2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 220, normalized size = 0.93 \begin {gather*} \frac {15 c^3 d^3 (d+e x)^3 \sqrt {a e+c d x} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )-\sqrt {e} \sqrt {c d^2-a e^2} \left (8 a^3 e^5+2 a^2 c d e^3 (5 d+17 e x)+a c^2 d^2 e \left (15 d^2+50 d e x+59 e^2 x^2\right )+c^3 d^3 x \left (15 d^2+40 d e x+33 e^2 x^2\right )\right )}{24 e^{7/2} (d+e x)^{5/2} \sqrt {c d^2-a e^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)^(13/2),x]

[Out]

(-(Sqrt[e]*Sqrt[c*d^2 - a*e^2]*(8*a^3*e^5 + 2*a^2*c*d*e^3*(5*d + 17*e*x) + c^3*d^3*x*(15*d^2 + 40*d*e*x + 33*e
^2*x^2) + a*c^2*d^2*e*(15*d^2 + 50*d*e*x + 59*e^2*x^2))) + 15*c^3*d^3*Sqrt[a*e + c*d*x]*(d + e*x)^3*ArcTan[(Sq
rt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/(24*e^(7/2)*Sqrt[c*d^2 - a*e^2]*(d + e*x)^(5/2)*Sqrt[(a*e + c*d
*x)*(d + e*x)])

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IntegrateAlgebraic [A]  time = 36.71, size = 228, normalized size = 0.97 \begin {gather*} \frac {((d+e x) (a e+c d x))^{5/2} \left (\frac {c^3 d^3 \sqrt {a e+c d x} \left (15 a^2 e^4-30 a c d^2 e^2+40 c d^2 e (a e+c d x)-40 a e^3 (a e+c d x)+33 e^2 (a e+c d x)^2+15 c^2 d^4\right )}{24 e^3 \left (-e (a e+c d x)+a e^2-c d^2\right )^3}+\frac {5 c^3 d^3 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt {c d^2-a e^2}}\right )}{(d+e x)^{5/2} (a e+c d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(13/2),x]

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*((c^3*d^3*Sqrt[a*e + c*d*x]*(15*c^2*d^4 - 30*a*c*d^2*e^2 + 15*a^2*e^4 + 40*c*
d^2*e*(a*e + c*d*x) - 40*a*e^3*(a*e + c*d*x) + 33*e^2*(a*e + c*d*x)^2))/(24*e^3*(-(c*d^2) + a*e^2 - e*(a*e + c
*d*x))^3) + (5*c^3*d^3*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/(8*e^(7/2)*Sqrt[c*d^2 - a*e^2]
)))/((a*e + c*d*x)^(5/2)*(d + e*x)^(5/2))

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fricas [A]  time = 0.44, size = 837, normalized size = 3.55 \begin {gather*} \left [-\frac {15 \, {\left (c^{3} d^{3} e^{4} x^{4} + 4 \, c^{3} d^{4} e^{3} x^{3} + 6 \, c^{3} d^{5} e^{2} x^{2} + 4 \, c^{3} d^{6} e x + c^{3} d^{7}\right )} \sqrt {-c d^{2} e + a e^{3}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d^{2} e + a e^{3}} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (15 \, c^{3} d^{6} e - 5 \, a c^{2} d^{4} e^{3} - 2 \, a^{2} c d^{2} e^{5} - 8 \, a^{3} e^{7} + 33 \, {\left (c^{3} d^{4} e^{3} - a c^{2} d^{2} e^{5}\right )} x^{2} + 2 \, {\left (20 \, c^{3} d^{5} e^{2} - 7 \, a c^{2} d^{3} e^{4} - 13 \, a^{2} c d e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{48 \, {\left (c d^{6} e^{4} - a d^{4} e^{6} + {\left (c d^{2} e^{8} - a e^{10}\right )} x^{4} + 4 \, {\left (c d^{3} e^{7} - a d e^{9}\right )} x^{3} + 6 \, {\left (c d^{4} e^{6} - a d^{2} e^{8}\right )} x^{2} + 4 \, {\left (c d^{5} e^{5} - a d^{3} e^{7}\right )} x\right )}}, -\frac {15 \, {\left (c^{3} d^{3} e^{4} x^{4} + 4 \, c^{3} d^{4} e^{3} x^{3} + 6 \, c^{3} d^{5} e^{2} x^{2} + 4 \, c^{3} d^{6} e x + c^{3} d^{7}\right )} \sqrt {c d^{2} e - a e^{3}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d^{2} e - a e^{3}} \sqrt {e x + d}}{c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x}\right ) + {\left (15 \, c^{3} d^{6} e - 5 \, a c^{2} d^{4} e^{3} - 2 \, a^{2} c d^{2} e^{5} - 8 \, a^{3} e^{7} + 33 \, {\left (c^{3} d^{4} e^{3} - a c^{2} d^{2} e^{5}\right )} x^{2} + 2 \, {\left (20 \, c^{3} d^{5} e^{2} - 7 \, a c^{2} d^{3} e^{4} - 13 \, a^{2} c d e^{6}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{24 \, {\left (c d^{6} e^{4} - a d^{4} e^{6} + {\left (c d^{2} e^{8} - a e^{10}\right )} x^{4} + 4 \, {\left (c d^{3} e^{7} - a d e^{9}\right )} x^{3} + 6 \, {\left (c d^{4} e^{6} - a d^{2} e^{8}\right )} x^{2} + 4 \, {\left (c d^{5} e^{5} - a d^{3} e^{7}\right )} x\right )}}\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)^(13/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(c^3*d^3*e^4*x^4 + 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(-c*d^2*e +
a*e^3)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt
(-c*d^2*e + a*e^3)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(15*c^3*d^6*e - 5*a*c^2*d^4*e^3 - 2*a^2*c*d^2
*e^5 - 8*a^3*e^7 + 33*(c^3*d^4*e^3 - a*c^2*d^2*e^5)*x^2 + 2*(20*c^3*d^5*e^2 - 7*a*c^2*d^3*e^4 - 13*a^2*c*d*e^6
)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c*d^6*e^4 - a*d^4*e^6 + (c*d^2*e^8 - a*e^10)*
x^4 + 4*(c*d^3*e^7 - a*d*e^9)*x^3 + 6*(c*d^4*e^6 - a*d^2*e^8)*x^2 + 4*(c*d^5*e^5 - a*d^3*e^7)*x), -1/24*(15*(c
^3*d^3*e^4*x^4 + 4*c^3*d^4*e^3*x^3 + 6*c^3*d^5*e^2*x^2 + 4*c^3*d^6*e*x + c^3*d^7)*sqrt(c*d^2*e - a*e^3)*arctan
(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d^2*e - a*e^3)*sqrt(e*x + d)/(c*d*e^2*x^2 + a*d*e^2 + (c*d
^2*e + a*e^3)*x)) + (15*c^3*d^6*e - 5*a*c^2*d^4*e^3 - 2*a^2*c*d^2*e^5 - 8*a^3*e^7 + 33*(c^3*d^4*e^3 - a*c^2*d^
2*e^5)*x^2 + 2*(20*c^3*d^5*e^2 - 7*a*c^2*d^3*e^4 - 13*a^2*c*d*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)
*x)*sqrt(e*x + d))/(c*d^6*e^4 - a*d^4*e^6 + (c*d^2*e^8 - a*e^10)*x^4 + 4*(c*d^3*e^7 - a*d*e^9)*x^3 + 6*(c*d^4*
e^6 - a*d^2*e^8)*x^2 + 4*(c*d^5*e^5 - a*d^3*e^7)*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)^(13/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 443, normalized size = 1.88 \begin {gather*} -\frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 c^{3} d^{3} e^{3} x^{3} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+45 c^{3} d^{4} e^{2} x^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+45 c^{3} d^{5} e x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+15 c^{3} d^{6} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+33 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{2} e^{2} x^{2}+26 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a c d \,e^{3} x +40 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{3} e x +8 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a^{2} e^{4}+10 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a c \,d^{2} e^{2}+15 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, c^{2} d^{4}\right )}{24 \left (e x +d \right )^{\frac {7}{2}} \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, e^{3}} \end {gather*}

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)^(13/2),x)

[Out]

-1/24*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*x^3*c^3
*d^3*e^3+45*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*x^2*c^3*d^4*e^2+45*arctanh((c*d*x+a*e)^(1/2)/
((a*e^2-c*d^2)*e)^(1/2)*e)*x*c^3*d^5*e+15*arctanh((c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2)*e)*c^3*d^6+33*(c*d
*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^2*e^2*x^2+26*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*a*c*d*e^3*x
+40*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^3*e*x+8*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*e^4+
10*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d^2*e^2+15*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^4)
/(e*x+d)^(7/2)/(c*d*x+a*e)^(1/2)/e^3/((a*e^2-c*d^2)*e)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {13}{2}}}\,{d x} \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)^(13/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/(e*x + d)^(13/2), x)

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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}}{{\left (d+e\,x\right )}^{13/2}} \,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)^(13/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(13/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(13/2),x)

[Out]

Timed out

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