3.18.3 \(\int \frac {(d+e x)^{13/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=186 \[ -\frac {35 e^2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}}+\frac {35 e^2 \sqrt {d+e x} \left (c d^2-a e^2\right )}{4 c^4 d^4}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3} \]

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Rubi [A]  time = 0.14, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {626, 47, 50, 63, 208} \begin {gather*} \frac {35 e^2 \sqrt {d+e x} \left (c d^2-a e^2\right )}{4 c^4 d^4}-\frac {35 e^2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(35*e^2*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(4*c^4*d^4) + (35*e^2*(d + e*x)^(3/2))/(12*c^3*d^3) - (7*e*(d + e*x)^(5
/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(7/2)/(2*c*d*(a*e + c*d*x)^2) - (35*e^2*(c*d^2 - a*e^2)^(3/2)*ArcTa
nh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*c^(9/2)*d^(9/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 626

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{13/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac {(d+e x)^{7/2}}{(a e+c d x)^3} \, dx\\ &=-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {\left (35 e^2\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{8 c^2 d^2}\\ &=\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {\left (35 e^2 \left (c d^2-a e^2\right )\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{8 c^3 d^3}\\ &=\frac {35 e^2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{4 c^4 d^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {\left (35 e^2 \left (c d^2-a e^2\right )^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c^4 d^4}\\ &=\frac {35 e^2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{4 c^4 d^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}+\frac {\left (35 e \left (c d^2-a e^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 c^4 d^4}\\ &=\frac {35 e^2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{4 c^4 d^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 c^3 d^3}-\frac {7 e (d+e x)^{5/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{7/2}}{2 c d (a e+c d x)^2}-\frac {35 e^2 \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{9/2} d^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.33 \begin {gather*} \frac {2 e^2 (d+e x)^{9/2} \, _2F_1\left (3,\frac {9}{2};\frac {11}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{9 \left (a e^2-c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(2*e^2*(d + e*x)^(9/2)*Hypergeometric2F1[3, 9/2, 11/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(9*(-(c*d^2) +
a*e^2)^3)

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IntegrateAlgebraic [A]  time = 0.81, size = 279, normalized size = 1.50 \begin {gather*} \frac {e^2 \sqrt {d+e x} \left (-105 a^3 e^6+315 a^2 c d^2 e^4-175 a^2 c d e^4 (d+e x)-315 a c^2 d^4 e^2+350 a c^2 d^3 e^2 (d+e x)-56 a c^2 d^2 e^2 (d+e x)^2+105 c^3 d^6-175 c^3 d^5 (d+e x)+56 c^3 d^4 (d+e x)^2+8 c^3 d^3 (d+e x)^3\right )}{12 c^4 d^4 \left (-a e^2+c d^2-c d (d+e x)\right )^2}-\frac {35 e^2 \left (c d^2-a e^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e^2-c d^2}}{c d^2-a e^2}\right )}{4 c^{9/2} d^{9/2} \sqrt {a e^2-c d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(13/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(e^2*Sqrt[d + e*x]*(105*c^3*d^6 - 315*a*c^2*d^4*e^2 + 315*a^2*c*d^2*e^4 - 105*a^3*e^6 - 175*c^3*d^5*(d + e*x)
+ 350*a*c^2*d^3*e^2*(d + e*x) - 175*a^2*c*d*e^4*(d + e*x) + 56*c^3*d^4*(d + e*x)^2 - 56*a*c^2*d^2*e^2*(d + e*x
)^2 + 8*c^3*d^3*(d + e*x)^3))/(12*c^4*d^4*(c*d^2 - a*e^2 - c*d*(d + e*x))^2) - (35*e^2*(c*d^2 - a*e^2)^2*ArcTa
n[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(4*c^(9/2)*d^(9/2)*Sqrt[-(c*d^2) +
a*e^2])

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fricas [B]  time = 0.43, size = 638, normalized size = 3.43 \begin {gather*} \left [\frac {105 \, {\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} + {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \, {\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} - {\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}, -\frac {105 \, {\left (a^{2} c d^{2} e^{4} - a^{3} e^{6} + {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (8 \, c^{3} d^{3} e^{3} x^{3} - 6 \, c^{3} d^{6} - 21 \, a c^{2} d^{4} e^{2} + 140 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 8 \, {\left (10 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} - {\left (39 \, c^{3} d^{5} e - 238 \, a c^{2} d^{3} e^{3} + 175 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (c^{6} d^{6} x^{2} + 2 \, a c^{5} d^{5} e x + a^{2} c^{4} d^{4} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(a^2*c*d^2*e^4 - a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 2*(a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*s
qrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(
c*d*x + a*e)) + 2*(8*c^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a*c^2*d^4*e^2 + 140*a^2*c*d^2*e^4 - 105*a^3*e^6 + 8*(10*
c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 - (39*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c
^6*d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*e^2), -1/12*(105*(a^2*c*d^2*e^4 - a^3*e^6 + (c^3*d^4*e^2 - a*c^2*d^
2*e^4)*x^2 + 2*(a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d*sqrt(-(
c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (8*c^3*d^3*e^3*x^3 - 6*c^3*d^6 - 21*a*c^2*d^4*e^2 + 140*a^2*c*d^2*e^4
- 105*a^3*e^6 + 8*(10*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 - (39*c^3*d^5*e - 238*a*c^2*d^3*e^3 + 175*a^2*c*d*e^
5)*x)*sqrt(e*x + d))/(c^6*d^6*x^2 + 2*a*c^5*d^5*e*x + a^2*c^4*d^4*e^2)]

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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((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 449, normalized size = 2.41 \begin {gather*} -\frac {11 \sqrt {e x +d}\, a^{3} e^{8}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{4} d^{4}}+\frac {33 \sqrt {e x +d}\, a^{2} e^{6}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{3} d^{2}}-\frac {33 \sqrt {e x +d}\, a \,e^{4}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{2}}+\frac {11 \sqrt {e x +d}\, d^{2} e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c}-\frac {13 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{6}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{3} d^{3}}+\frac {13 \left (e x +d \right )^{\frac {3}{2}} a \,e^{4}}{2 \left (c d e x +a \,e^{2}\right )^{2} c^{2} d}-\frac {13 \left (e x +d \right )^{\frac {3}{2}} d \,e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c}+\frac {35 a^{2} e^{6} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{4 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{4} d^{4}}-\frac {35 a \,e^{4} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{3} d^{2}}+\frac {35 e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{4 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c^{2}}-\frac {6 \sqrt {e x +d}\, a \,e^{4}}{c^{4} d^{4}}+\frac {6 \sqrt {e x +d}\, e^{2}}{c^{3} d^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e^{2}}{3 c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(13/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^3,x)

[Out]

2/3*e^2*(e*x+d)^(3/2)/c^3/d^3-6*e^4/c^4/d^4*a*(e*x+d)^(1/2)+6*e^2/c^3/d^2*(e*x+d)^(1/2)-13/4*e^6/c^3/d^3/(c*d*
e*x+a*e^2)^2*(e*x+d)^(3/2)*a^2+13/2*e^4/c^2/d/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*a-13/4*e^2/c*d/(c*d*e*x+a*e^2)^2
*(e*x+d)^(3/2)-11/4*e^8/c^4/d^4/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^3+33/4*e^6/c^3/d^2/(c*d*e*x+a*e^2)^2*(e*x+d)
^(1/2)*a^2-33/4*e^4/c^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+11/4*e^2/c*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)+35/4*
e^6/c^4/d^4/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)*a^2-35/2*e^4/c^3/d^2
/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)*a+35/4*e^2/c^2/((a*e^2-c*d^2)*c
*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)

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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((e*x+d)^(13/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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 positive or negative?

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mupad [B]  time = 0.74, size = 319, normalized size = 1.72 \begin {gather*} \frac {2\,e^2\,{\left (d+e\,x\right )}^{3/2}}{3\,c^3\,d^3}-\frac {{\left (d+e\,x\right )}^{3/2}\,\left (\frac {13\,a^2\,c\,d\,e^6}{4}-\frac {13\,a\,c^2\,d^3\,e^4}{2}+\frac {13\,c^3\,d^5\,e^2}{4}\right )+\sqrt {d+e\,x}\,\left (\frac {11\,a^3\,e^8}{4}-\frac {33\,a^2\,c\,d^2\,e^6}{4}+\frac {33\,a\,c^2\,d^4\,e^4}{4}-\frac {11\,c^3\,d^6\,e^2}{4}\right )}{c^6\,d^8-\left (2\,c^6\,d^7-2\,a\,c^5\,d^5\,e^2\right )\,\left (d+e\,x\right )+c^6\,d^6\,{\left (d+e\,x\right )}^2-2\,a\,c^5\,d^6\,e^2+a^2\,c^4\,d^4\,e^4}+\frac {2\,e^2\,\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )\,\sqrt {d+e\,x}}{c^6\,d^6}+\frac {35\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^6-2\,a\,c\,d^2\,e^4+c^2\,d^4\,e^2}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{4\,c^{9/2}\,d^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(13/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

(2*e^2*(d + e*x)^(3/2))/(3*c^3*d^3) - ((d + e*x)^(3/2)*((13*c^3*d^5*e^2)/4 - (13*a*c^2*d^3*e^4)/2 + (13*a^2*c*
d*e^6)/4) + (d + e*x)^(1/2)*((11*a^3*e^8)/4 - (11*c^3*d^6*e^2)/4 + (33*a*c^2*d^4*e^4)/4 - (33*a^2*c*d^2*e^6)/4
))/(c^6*d^8 - (2*c^6*d^7 - 2*a*c^5*d^5*e^2)*(d + e*x) + c^6*d^6*(d + e*x)^2 - 2*a*c^5*d^6*e^2 + a^2*c^4*d^4*e^
4) + (2*e^2*(3*c^3*d^4 - 3*a*c^2*d^2*e^2)*(d + e*x)^(1/2))/(c^6*d^6) + (35*e^2*atan((c^(1/2)*d^(1/2)*e^2*(a*e^
2 - c*d^2)^(3/2)*(d + e*x)^(1/2))/(a^2*e^6 + c^2*d^4*e^2 - 2*a*c*d^2*e^4))*(a*e^2 - c*d^2)^(3/2))/(4*c^(9/2)*d
^(9/2))

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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((e*x+d)**(13/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

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