3.2476 \(\int \frac {(d+e x)^{7/2}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=659 \[ \frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {d+e x} \left (x (2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right )-\left (b^2 \left (9 c d^2 e-a e^3\right )\right )+8 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (5 a e^2+3 c d^2\right )\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (d+e x)^{5/2} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

-2/3*(e*x+d)^(5/2)*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)+2/3*(8*b*c*d*(3*a*e^2+c*d^2)-4*
a*c*e*(5*a*e^2+3*c*d^2)-b^2*(-a*e^3+9*c*d^2*e)+(-b*e+2*c*d)*(8*c^2*d^2-b^2*e^2-4*c*e*(-3*a*e+2*b*d))*x)*(e*x+d
)^(1/2)/c/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)-2/3*(-b*e+2*c*d)*(4*c^2*d^2-b^2*e^2-4*c*e*(-2*a*e+b*d))*EllipticE
(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),(-2*e*(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*
a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(e*x+d)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/c^2/(-4*a*c+b^2)^(3/2)/(c
*x^2+b*x+a)^(1/2)/(c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)+2/3*(a*e^2-b*d*e+c*d^2)*(16*c^2*d^2-b^2*e
^2-4*c*e*(-5*a*e+4*b*d))*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),(-2*e*(
-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*(c*(e
*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)/c^2/(-4*a*c+b^2)^(3/2)/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 659, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {738, 818, 843, 718, 424, 419} \[ \frac {2 \sqrt {d+e x} \left (x (2 c d-b e) \left (-4 c e (2 b d-3 a e)-b^2 e^2+8 c^2 d^2\right )+b^2 \left (-\left (9 c d^2 e-a e^3\right )\right )+8 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (5 a e^2+3 c d^2\right )\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 (d+e x)^{5/2} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^(5/2)*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (2*Sqrt[d + e
*x]*(8*b*c*d*(c*d^2 + 3*a*e^2) - 4*a*c*e*(3*c*d^2 + 5*a*e^2) - b^2*(9*c*d^2*e - a*e^3) + (2*c*d - b*e)*(8*c^2*
d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*a*e))*x))/(3*c*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]) - (2*Sqrt[2]*(2*c*d - b
*e)*(4*c^2*d^2 - b^2*e^2 - 4*c*e*(b*d - 2*a*e))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*Ell
ipticE[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*
d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c^2*(b^2 - 4*a*c)^(3/2)*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c]
)*e)]*Sqrt[a + b*x + c*x^2]) + (2*Sqrt[2]*(c*d^2 - b*d*e + a*e^2)*(16*c^2*d^2 - b^2*e^2 - 4*c*e*(4*b*d - 5*a*e
))*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*Ellipt
icF[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d -
(b + Sqrt[b^2 - 4*a*c])*e)])/(3*c^2*(b^2 - 4*a*c)^(3/2)*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], 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] && EqQ[m^2, 1/4]

Rule 738

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 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 843

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^{7/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^{5/2} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {(d+e x)^{3/2} \left (\frac {1}{2} \left (8 c d^2-9 b d e+10 a e^2\right )-\frac {1}{2} e (2 c d-b e) x\right )}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {2 (d+e x)^{5/2} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d \left (c d^2+3 a e^2\right )-4 a c e \left (3 c d^2+5 a e^2\right )-b^2 \left (9 c d^2 e-a e^3\right )+(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {4 \int \frac {\frac {1}{4} e \left (b^3 d e^2-4 a c e \left (c d^2+5 a e^2\right )+4 b c d \left (2 c d^2+5 a e^2\right )-b^2 \left (11 c d^2 e-a e^3\right )\right )+\frac {1}{2} e (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 c \left (b^2-4 a c\right )^2}\\ &=-\frac {2 (d+e x)^{5/2} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d \left (c d^2+3 a e^2\right )-4 a c e \left (3 c d^2+5 a e^2\right )-b^2 \left (9 c d^2 e-a e^3\right )+(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 c \left (b^2-4 a c\right )^2}-\frac {\left (4 \left (-\frac {1}{2} d e (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )+\frac {1}{4} e^2 \left (b^3 d e^2-4 a c e \left (c d^2+5 a e^2\right )+4 b c d \left (2 c d^2+5 a e^2\right )-b^2 \left (11 c d^2 e-a e^3\right )\right )\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 c \left (b^2-4 a c\right )^2 e}\\ &=-\frac {2 (d+e x)^{5/2} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d \left (c d^2+3 a e^2\right )-4 a c e \left (3 c d^2+5 a e^2\right )-b^2 \left (9 c d^2 e-a e^3\right )+(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {\left (2 \sqrt {2} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {a+b x+c x^2}}-\frac {\left (8 \sqrt {2} \left (-\frac {1}{2} d e (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )+\frac {1}{4} e^2 \left (b^3 d e^2-4 a c e \left (c d^2+5 a e^2\right )+4 b c d \left (2 c d^2+5 a e^2\right )-b^2 \left (11 c d^2 e-a e^3\right )\right )\right ) \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} e \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (d+e x)^{5/2} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d \left (c d^2+3 a e^2\right )-4 a c e \left (3 c d^2+5 a e^2\right )-b^2 \left (9 c d^2 e-a e^3\right )+(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) x\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {a+b x+c x^2}}+\frac {2 \sqrt {2} \left (c d^2-b d e+a e^2\right ) \left (16 c^2 d^2-16 b c d e-b^2 e^2+20 a c e^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 14.30, size = 5598, normalized size = 8.49 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

Result too large to show

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, {\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x + {\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \, {\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*sqrt(c*x^2 + b*x + a)*sqrt(e*x + d)/(c^3*x^6 + 3*b*c^2*x^5
+ 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)

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maple [B]  time = 0.37, size = 19258, normalized size = 29.22 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(7/2)/(c*x^2 + b*x + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^{7/2}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/(a + b*x + c*x^2)^(5/2),x)

[Out]

int((d + e*x)^(7/2)/(a + b*x + c*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(7/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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