### 3.2458 $$\int \frac{(a+b x+c x^2)^{5/2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=603 $-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (32 b d-17 a e)+15 b^2 e^2+128 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{a+b x+c x^2} \left (-4 c e (28 b d-9 a e)+15 b^2 e^2+16 c e x (2 c d-b e)+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (32 b d-9 a e)+23 b^2 e^2+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 e^6 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \left (a+b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}$

[Out]

(-2*(128*c^2*d^2 + 15*b^2*e^2 - 4*c*e*(28*b*d - 9*a*e) + 16*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(15*e^
5*Sqrt[d + e*x]) + (2*(16*c*d - 5*b*e + 6*c*e*x)*(a + b*x + c*x^2)^(3/2))/(15*e^3*(d + e*x)^(3/2)) - (2*(a + b
*x + c*x^2)^(5/2))/(5*e*(d + e*x)^(5/2)) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(128*c^2*d^2 + 23*b^2*e^2 - 4*c*e*(32*
b*d - 9*a*e))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[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)])/(
15*e^6*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]*Sqrt[b^2 -
4*a*c]*(2*c*d - b*e)*(128*c^2*d^2 + 15*b^2*e^2 - 4*c*e*(32*b*d - 17*a*e))*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqr
t[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[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)])/(15*c*e^6
*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.689455, antiderivative size = 603, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {732, 812, 843, 718, 424, 419} $-\frac{2 \sqrt{a+b x+c x^2} \left (-4 c e (28 b d-9 a e)+15 b^2 e^2+16 c e x (2 c d-b e)+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (32 b d-17 a e)+15 b^2 e^2+128 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (32 b d-9 a e)+23 b^2 e^2+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 e^6 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \left (a+b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(128*c^2*d^2 + 15*b^2*e^2 - 4*c*e*(28*b*d - 9*a*e) + 16*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(15*e^
5*Sqrt[d + e*x]) + (2*(16*c*d - 5*b*e + 6*c*e*x)*(a + b*x + c*x^2)^(3/2))/(15*e^3*(d + e*x)^(3/2)) - (2*(a + b
*x + c*x^2)^(5/2))/(5*e*(d + e*x)^(5/2)) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(128*c^2*d^2 + 23*b^2*e^2 - 4*c*e*(32*
b*d - 9*a*e))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[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)])/(
15*e^6*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]*Sqrt[b^2 -
4*a*c]*(2*c*d - b*e)*(128*c^2*d^2 + 15*b^2*e^2 - 4*c*e*(32*b*d - 17*a*e))*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqr
t[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[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)])/(15*c*e^6
*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 732

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 + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 812

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

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]

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 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 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)])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx &=-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\int \frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx}{e}\\ &=\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{2 \int \frac{\left (\frac{1}{2} \left (16 b c d-5 b^2 e-12 a c e\right )+8 c (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{(d+e x)^{3/2}} \, dx}{5 e^3}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{4 \int \frac{\frac{1}{4} \left (-112 b^2 c d e-64 a c^2 d e+15 b^3 e^2+4 b c \left (32 c d^2+17 a e^2\right )\right )+\frac{1}{2} c \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right ) x}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{15 e^5}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{15 e^6}+\frac{\left (2 c \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \, dx}{15 e^6}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\left (2 \sqrt{2} \sqrt{b^2-4 a c} \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 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 )}{15 e^6 \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 (2 \sqrt{2} \sqrt{b^2-4 a c} (2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\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 )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 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 )}{15 e^6 \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} \sqrt{b^2-4 a c} (2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\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 )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 13.2677, size = 8961, normalized size = 14.86 $\text{Result too large to show}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.378, size = 14453, normalized size = 24. \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt{c x^{2} + b x + a} \sqrt{e x + d}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out