3.18.2 \(\int \frac {(d+e x)^{15/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=222 \[ -\frac {63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}+\frac {63 e^2 \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{4 c^5 d^5}+\frac {21 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{4 c^4 d^4}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3} \]

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Rubi [A]  time = 0.19, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {21 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{4 c^4 d^4}+\frac {63 e^2 \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{4 c^5 d^5}-\frac {63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(63*e^2*(c*d^2 - a*e^2)^2*Sqrt[d + e*x])/(4*c^5*d^5) + (21*e^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2))/(4*c^4*d^4) +
(63*e^2*(d + e*x)^(5/2))/(20*c^3*d^3) - (9*e*(d + e*x)^(7/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(9/2)/(2*c
*d*(a*e + c*d*x)^2) - (63*e^2*(c*d^2 - a*e^2)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2
]])/(4*c^(11/2)*d^(11/2))

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)^{15/2}}{\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)^{9/2}}{(a e+c d x)^3} \, dx\\ &=-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {(9 e) \int \frac {(d+e x)^{7/2}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {\left (63 e^2\right ) \int \frac {(d+e x)^{5/2}}{a e+c d x} \, dx}{8 c^2 d^2}\\ &=\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {\left (63 e^2 \left (c d^2-a e^2\right )\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{8 c^3 d^3}\\ &=\frac {21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {\left (63 e^2 \left (c d^2-a e^2\right )^2\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{8 c^4 d^4}\\ &=\frac {63 e^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{4 c^5 d^5}+\frac {21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {\left (63 e^2 \left (c d^2-a e^2\right )^3\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c^5 d^5}\\ &=\frac {63 e^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{4 c^5 d^5}+\frac {21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac {\left (63 e \left (c d^2-a e^2\right )^3\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^5 d^5}\\ &=\frac {63 e^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{4 c^5 d^5}+\frac {21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}-\frac {63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.27 \begin {gather*} \frac {2 e^2 (d+e x)^{11/2} \, _2F_1\left (3,\frac {11}{2};\frac {13}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{11 \left (a e^2-c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(2*e^2*(d + e*x)^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(11*(-(c*d^2)
+ a*e^2)^3)

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IntegrateAlgebraic [A]  time = 0.86, size = 367, normalized size = 1.65 \begin {gather*} \frac {e^2 \sqrt {d+e x} \left (315 a^4 e^8-1260 a^3 c d^2 e^6+525 a^3 c d e^6 (d+e x)+1890 a^2 c^2 d^4 e^4-1575 a^2 c^2 d^3 e^4 (d+e x)+168 a^2 c^2 d^2 e^4 (d+e x)^2-1260 a c^3 d^6 e^2+1575 a c^3 d^5 e^2 (d+e x)-336 a c^3 d^4 e^2 (d+e x)^2-24 a c^3 d^3 e^2 (d+e x)^3+315 c^4 d^8-525 c^4 d^7 (d+e x)+168 c^4 d^6 (d+e x)^2+24 c^4 d^5 (d+e x)^3+8 c^4 d^4 (d+e x)^4\right )}{20 c^5 d^5 \left (-a e^2+c d^2-c d (d+e x)\right )^2}-\frac {63 e^2 \left (c d^2-a e^2\right )^3 \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^{11/2} d^{11/2} \sqrt {a e^2-c d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(e^2*Sqrt[d + e*x]*(315*c^4*d^8 - 1260*a*c^3*d^6*e^2 + 1890*a^2*c^2*d^4*e^4 - 1260*a^3*c*d^2*e^6 + 315*a^4*e^8
- 525*c^4*d^7*(d + e*x) + 1575*a*c^3*d^5*e^2*(d + e*x) - 1575*a^2*c^2*d^3*e^4*(d + e*x) + 525*a^3*c*d*e^6*(d
+ e*x) + 168*c^4*d^6*(d + e*x)^2 - 336*a*c^3*d^4*e^2*(d + e*x)^2 + 168*a^2*c^2*d^2*e^4*(d + e*x)^2 + 24*c^4*d^
5*(d + e*x)^3 - 24*a*c^3*d^3*e^2*(d + e*x)^3 + 8*c^4*d^4*(d + e*x)^4))/(20*c^5*d^5*(c*d^2 - a*e^2 - c*d*(d + e
*x))^2) - (63*e^2*(c*d^2 - a*e^2)^3*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e
^2)])/(4*c^(11/2)*d^(11/2)*Sqrt[-(c*d^2) + a*e^2])

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fricas [B]  time = 0.44, size = 858, normalized size = 3.86 \begin {gather*} \left [\frac {315 \, {\left (a^{2} c^{2} d^{4} e^{4} - 2 \, a^{3} c d^{2} e^{6} + a^{4} e^{8} + {\left (c^{4} d^{6} e^{2} - 2 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} x^{2} + 2 \, {\left (a c^{3} d^{5} e^{3} - 2 \, a^{2} c^{2} d^{3} e^{5} + a^{3} c d e^{7}\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^{4} d^{4} e^{4} x^{4} - 10 \, c^{4} d^{8} - 45 \, a c^{3} d^{6} e^{2} + 483 \, a^{2} c^{2} d^{4} e^{4} - 735 \, a^{3} c d^{2} e^{6} + 315 \, a^{4} e^{8} + 8 \, {\left (7 \, c^{4} d^{5} e^{3} - 3 \, a c^{3} d^{3} e^{5}\right )} x^{3} + 24 \, {\left (12 \, c^{4} d^{6} e^{2} - 17 \, a c^{3} d^{4} e^{4} + 7 \, a^{2} c^{2} d^{2} e^{6}\right )} x^{2} - {\left (85 \, c^{4} d^{7} e - 831 \, a c^{3} d^{5} e^{3} + 1239 \, a^{2} c^{2} d^{3} e^{5} - 525 \, a^{3} c d e^{7}\right )} x\right )} \sqrt {e x + d}}{40 \, {\left (c^{7} d^{7} x^{2} + 2 \, a c^{6} d^{6} e x + a^{2} c^{5} d^{5} e^{2}\right )}}, -\frac {315 \, {\left (a^{2} c^{2} d^{4} e^{4} - 2 \, a^{3} c d^{2} e^{6} + a^{4} e^{8} + {\left (c^{4} d^{6} e^{2} - 2 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} x^{2} + 2 \, {\left (a c^{3} d^{5} e^{3} - 2 \, a^{2} c^{2} d^{3} e^{5} + a^{3} c d e^{7}\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^{4} d^{4} e^{4} x^{4} - 10 \, c^{4} d^{8} - 45 \, a c^{3} d^{6} e^{2} + 483 \, a^{2} c^{2} d^{4} e^{4} - 735 \, a^{3} c d^{2} e^{6} + 315 \, a^{4} e^{8} + 8 \, {\left (7 \, c^{4} d^{5} e^{3} - 3 \, a c^{3} d^{3} e^{5}\right )} x^{3} + 24 \, {\left (12 \, c^{4} d^{6} e^{2} - 17 \, a c^{3} d^{4} e^{4} + 7 \, a^{2} c^{2} d^{2} e^{6}\right )} x^{2} - {\left (85 \, c^{4} d^{7} e - 831 \, a c^{3} d^{5} e^{3} + 1239 \, a^{2} c^{2} d^{3} e^{5} - 525 \, a^{3} c d e^{7}\right )} x\right )} \sqrt {e x + d}}{20 \, {\left (c^{7} d^{7} x^{2} + 2 \, a c^{6} d^{6} e x + a^{2} c^{5} d^{5} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/40*(315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^
2 + 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e^5 + a^3*c*d*e^7)*x)*sqrt((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^4*d^4*e^4*x^4 - 10*c^4*d^8
- 45*a*c^3*d^6*e^2 + 483*a^2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^
5)*x^3 + 24*(12*c^4*d^6*e^2 - 17*a*c^3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85*c^4*d^7*e - 831*a*c^3*d^5*e^3 +
1239*a^2*c^2*d^3*e^5 - 525*a^3*c*d*e^7)*x)*sqrt(e*x + d))/(c^7*d^7*x^2 + 2*a*c^6*d^6*e*x + a^2*c^5*d^5*e^2), -
1/20*(315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^2
+ 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e^5 + a^3*c*d*e^7)*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^4*d^4*e^4*x^4 - 10*c^4*d^8 - 45*a*c^3*d^6*e^2 + 483*a^
2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^5)*x^3 + 24*(12*c^4*d^6*e^2
- 17*a*c^3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85*c^4*d^7*e - 831*a*c^3*d^5*e^3 + 1239*a^2*c^2*d^3*e^5 - 525*
a^3*c*d*e^7)*x)*sqrt(e*x + d))/(c^7*d^7*x^2 + 2*a*c^6*d^6*e*x + a^2*c^5*d^5*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)^(15/2)/(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 = 0.08, size = 635, normalized size = 2.86 \begin {gather*} \frac {15 \sqrt {e x +d}\, a^{4} e^{10}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{5} d^{5}}-\frac {15 \sqrt {e x +d}\, a^{3} e^{8}}{\left (c d e x +a \,e^{2}\right )^{2} c^{4} d^{3}}+\frac {45 \sqrt {e x +d}\, a^{2} e^{6}}{2 \left (c d e x +a \,e^{2}\right )^{2} c^{3} d}-\frac {15 \sqrt {e x +d}\, a d \,e^{4}}{\left (c d e x +a \,e^{2}\right )^{2} c^{2}}+\frac {15 \sqrt {e x +d}\, d^{3} e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} a^{3} e^{8}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{4} d^{4}}-\frac {51 \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^{2}}+\frac {51 \left (e x +d \right )^{\frac {3}{2}} a \,e^{4}}{4 \left (c d e x +a \,e^{2}\right )^{2} c^{2}}-\frac {17 \left (e x +d \right )^{\frac {3}{2}} d^{2} e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c}-\frac {63 a^{3} e^{8} \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^{5} d^{5}}+\frac {189 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^{3}}-\frac {189 a \,e^{4} \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^{3} d}+\frac {63 d \,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 {12 \sqrt {e x +d}\, a^{2} e^{6}}{c^{5} d^{5}}-\frac {24 \sqrt {e x +d}\, a \,e^{4}}{c^{4} d^{3}}+\frac {12 \sqrt {e x +d}\, e^{2}}{c^{3} d}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} a \,e^{4}}{c^{4} d^{4}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e^{2}}{c^{3} d^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} e^{2}}{5 c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(15/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^3,x)

[Out]

2/5*e^2*(e*x+d)^(5/2)/c^3/d^3-2*e^4/c^4/d^4*(e*x+d)^(3/2)*a+2*e^2/c^3/d^2*(e*x+d)^(3/2)+12*e^6/c^5/d^5*a^2*(e*
x+d)^(1/2)-24*e^4/c^4/d^3*a*(e*x+d)^(1/2)+12*e^2/c^3/d*(e*x+d)^(1/2)+17/4*e^8/c^4/d^4/(c*d*e*x+a*e^2)^2*(e*x+d
)^(3/2)*a^3-51/4*e^6/c^3/d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*a^2+51/4*e^4/c^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*
a-17/4*e^2/c*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)+15/4*e^10/c^5/d^5/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^4-15*e^8/
c^4/d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^3+45/2*e^6/c^3/d/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^2-15*e^4/c^2*d/(c
*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+15/4*e^2/c*d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)-63/4*e^8/c^5/d^5/((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^3+189/4*e^6/c^4/d^3/((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^2-189/4*e^4/c^3/d/((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+63/4*e^2/c^2*d/((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)^(15/2)/(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 = 0.73, size = 430, normalized size = 1.94 \begin {gather*} \frac {\sqrt {d+e\,x}\,\left (\frac {15\,a^4\,e^{10}}{4}-15\,a^3\,c\,d^2\,e^8+\frac {45\,a^2\,c^2\,d^4\,e^6}{2}-15\,a\,c^3\,d^6\,e^4+\frac {15\,c^4\,d^8\,e^2}{4}\right )-{\left (d+e\,x\right )}^{3/2}\,\left (-\frac {17\,a^3\,c\,d\,e^8}{4}+\frac {51\,a^2\,c^2\,d^3\,e^6}{4}-\frac {51\,a\,c^3\,d^5\,e^4}{4}+\frac {17\,c^4\,d^7\,e^2}{4}\right )}{c^7\,d^9-\left (2\,c^7\,d^8-2\,a\,c^6\,d^6\,e^2\right )\,\left (d+e\,x\right )+c^7\,d^7\,{\left (d+e\,x\right )}^2-2\,a\,c^6\,d^7\,e^2+a^2\,c^5\,d^5\,e^4}+\left (\frac {2\,e^2\,{\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )}^2}{c^9\,d^9}-\frac {6\,e^2\,{\left (a\,e^2-c\,d^2\right )}^2}{c^5\,d^5}\right )\,\sqrt {d+e\,x}+\frac {2\,e^2\,{\left (d+e\,x\right )}^{5/2}}{5\,c^3\,d^3}-\frac {63\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,{\left (a\,e^2-c\,d^2\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^8-3\,a^2\,c\,d^2\,e^6+3\,a\,c^2\,d^4\,e^4-c^3\,d^6\,e^2}\right )\,{\left (a\,e^2-c\,d^2\right )}^{5/2}}{4\,c^{11/2}\,d^{11/2}}+\frac {2\,e^2\,\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,c^6\,d^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(15/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

((d + e*x)^(1/2)*((15*a^4*e^10)/4 + (15*c^4*d^8*e^2)/4 - 15*a*c^3*d^6*e^4 - 15*a^3*c*d^2*e^8 + (45*a^2*c^2*d^4
*e^6)/2) - (d + e*x)^(3/2)*((17*c^4*d^7*e^2)/4 - (51*a*c^3*d^5*e^4)/4 + (51*a^2*c^2*d^3*e^6)/4 - (17*a^3*c*d*e
^8)/4))/(c^7*d^9 - (2*c^7*d^8 - 2*a*c^6*d^6*e^2)*(d + e*x) + c^7*d^7*(d + e*x)^2 - 2*a*c^6*d^7*e^2 + a^2*c^5*d
^5*e^4) + ((2*e^2*(3*c^3*d^4 - 3*a*c^2*d^2*e^2)^2)/(c^9*d^9) - (6*e^2*(a*e^2 - c*d^2)^2)/(c^5*d^5))*(d + e*x)^
(1/2) + (2*e^2*(d + e*x)^(5/2))/(5*c^3*d^3) - (63*e^2*atan((c^(1/2)*d^(1/2)*e^2*(a*e^2 - c*d^2)^(5/2)*(d + e*x
)^(1/2))/(a^3*e^8 - c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 - 3*a^2*c*d^2*e^6))*(a*e^2 - c*d^2)^(5/2))/(4*c^(11/2)*d^(11
/2)) + (2*e^2*(3*c^3*d^4 - 3*a*c^2*d^2*e^2)*(d + e*x)^(3/2))/(3*c^6*d^6)

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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)**(15/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

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