3.26.42 \(\int \frac {(d+e x)^3}{(a+b x+c x^2)^{5/4}} \, dx\) [2542]

Optimal. Leaf size=662 \[ -\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {2 e \left (24 c^2 d^2+7 b^2 e^2-2 c e (9 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/4}}{3 c^2 \left (b^2-4 a c\right )}+\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{5/2} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)} \]

[Out]

-4*(e*x+d)^2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/4)+2/3*e*(24*c^2*d^2+7*b^2*e^2-2*c*e*(8*
a*e+9*b*d)+6*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(3/4)/c^2/(-4*a*c+b^2)+(-b*e+2*c*d)*(4*c^2*d^2+7*b^2*e^2-4*c*e*
(6*a*e+b*d))*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^(5/2)/(-4*a*c+b^2)^(3/2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c
+b^2)^(1/2))-1/2*(-b*e+2*c*d)*(4*c^2*d^2+7*b^2*e^2-4*c*e*(6*a*e+b*d))*(cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4
)*2^(1/2)/(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))*
EllipticE(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2*c^(1/2)*(c*x
^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1
/2))^2)^(1/2)/c^(11/4)/(-4*a*c+b^2)^(1/4)/(2*c*x+b)*2^(1/2)+1/4*(-b*e+2*c*d)*(4*c^2*d^2+7*b^2*e^2-4*c*e*(6*a*e
+b*d))*(cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c
*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*
a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(
1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))^2)^(1/2)/c^(11/4)/(-4*a*c+b^2)^(1/4)/(2*c*x+b)*2^(1/2)

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Rubi [A]
time = 0.49, antiderivative size = 662, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {752, 793, 637, 311, 226, 1210} \begin {gather*} \frac {\sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (6 a e+b d)+7 b^2 e^2+4 c^2 d^2\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac {\sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \left (-4 c e (6 a e+b d)+7 b^2 e^2+4 c^2 d^2\right ) E\left (2 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} \left (-2 c e (8 a e+9 b d)+7 b^2 e^2+6 c e x (2 c d-b e)+24 c^2 d^2\right )}{3 c^2 \left (b^2-4 a c\right )}+\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e) \left (-4 c e (6 a e+b d)+7 b^2 e^2+4 c^2 d^2\right )}{c^{5/2} \left (b^2-4 a c\right )^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}-\frac {4 (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + b*x + c*x^2)^(5/4),x]

[Out]

(-4*(d + e*x)^2*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (2*e*(24*c^2*d^2 +
7*b^2*e^2 - 2*c*e*(9*b*d + 8*a*e) + 6*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/4))/(3*c^2*(b^2 - 4*a*c)) + ((
2*c*d - b*e)*(4*c^2*d^2 + 7*b^2*e^2 - 4*c*e*(b*d + 6*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(c^(5/2)*(b^2
- 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - ((2*c*d - b*e)*(4*c^2*d^2 + 7*b^2*
e^2 - 4*c*e*(b*d + 6*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 -
4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[(Sqrt[2]*c^(1/4)*(a
+ b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(Sqrt[2]*c^(11/4)*(b^2 - 4*a*c)^(1/4)*(b + 2*c*x)) + ((2*c*d
- b*e)*(4*c^2*d^2 + 7*b^2*e^2 - 4*c*e*(b*d + 6*a*e))*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticF[2*
ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(2*Sqrt[2]*c^(11/4)*(b^2 - 4*a*c)
^(1/4)*(b + 2*c*x))

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 793

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p +
3))), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(
a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^{5/4}} \, dx &=-\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}-\frac {4 \int \frac {(d+e x) \left (\frac {1}{2} \left (-2 c d^2-3 b d e+8 a e^2\right )-\frac {5}{2} e (2 c d-b e) x\right )}{\sqrt [4]{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {2 e \left (24 c^2 d^2+7 b^2 e^2-2 c e (9 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/4}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left ((2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right )\right ) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{2 c^2 \left (b^2-4 a c\right )}\\ &=-\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {2 e \left (24 c^2 d^2+7 b^2 e^2-2 c e (9 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/4}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^2 \left (b^2-4 a c\right ) (b+2 c x)}\\ &=-\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {2 e \left (24 c^2 d^2+7 b^2 e^2-2 c e (9 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/4}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left ((2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{5/2} \sqrt {b^2-4 a c} (b+2 c x)}-\frac {\left ((2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {c} x^2}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{5/2} \sqrt {b^2-4 a c} (b+2 c x)}\\ &=-\frac {4 (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {2 e \left (24 c^2 d^2+7 b^2 e^2-2 c e (9 b d+8 a e)+6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/4}}{3 c^2 \left (b^2-4 a c\right )}+\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{5/2} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {(2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} c^{11/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.40, size = 247, normalized size = 0.37 \begin {gather*} \frac {-8 c \left (-7 b^3 e^3 x-b^2 e^2 (7 a e+c x (-18 d+e x))+2 b c \left (3 c d^2 (d-3 e x)+a e^2 (9 d+11 e x)\right )+4 c \left (4 a^2 e^3+3 c^2 d^3 x+a c e \left (-9 d^2-9 d e x+e^2 x^2\right )\right )\right )+3 \sqrt {2} (2 c d-b e) \left (4 c^2 d^2+7 b^2 e^2-4 c e (b d+6 a e)\right ) (b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{12 c^3 \left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(a + b*x + c*x^2)^(5/4),x]

[Out]

(-8*c*(-7*b^3*e^3*x - b^2*e^2*(7*a*e + c*x*(-18*d + e*x)) + 2*b*c*(3*c*d^2*(d - 3*e*x) + a*e^2*(9*d + 11*e*x))
+ 4*c*(4*a^2*e^3 + 3*c^2*d^3*x + a*c*e*(-9*d^2 - 9*d*e*x + e^2*x^2))) + 3*Sqrt[2]*(2*c*d - b*e)*(4*c^2*d^2 +
7*b^2*e^2 - 4*c*e*(b*d + 6*a*e))*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hypergeometric2F1[1/
4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(12*c^3*(b^2 - 4*a*c)*(a + x*(b + c*x))^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{3}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x+a)^(5/4),x)

[Out]

int((e*x+d)^3/(c*x^2+b*x+a)^(5/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((x*e + d)^3/(c*x^2 + b*x + a)^(5/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x+a)^(5/4),x, algorithm="fricas")

[Out]

integral((x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3)*(c*x^2 + b*x + a)^(3/4)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b
^2 + 2*a*c)*x^2 + a^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (a + b x + c x^{2}\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x+a)**(5/4),x)

[Out]

Integral((d + e*x)**3/(a + b*x + c*x**2)**(5/4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x+a)^(5/4),x, algorithm="giac")

[Out]

integrate((x*e + d)^3/(c*x^2 + b*x + a)^(5/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{{\left (c\,x^2+b\,x+a\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + b*x + c*x^2)^(5/4),x)

[Out]

int((d + e*x)^3/(a + b*x + c*x^2)^(5/4), x)

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