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

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

[Out]

2/15*(6*c*e*x-5*b*e+16*c*d)*(c*x^2+b*x+a)^(3/2)/e^3/(e*x+d)^(3/2)-2/5*(c*x^2+b*x+a)^(5/2)/e/(e*x+d)^(5/2)-2/15
*(128*c^2*d^2+15*b^2*e^2-4*c*e*(-9*a*e+28*b*d)+16*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/e^5/(e*x+d)^(1/2)+2/
15*(128*c^2*d^2+23*b^2*e^2-4*c*e*(-9*a*e+32*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)*(-4*a*c+b^2)^(1/2)
*(e*x+d)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/e^6/(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)-2/15*(-b*e+2*c*d)*(128*c^2*d^2+15*b^2*e^2-4*c*e*(-17*a*e+32*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)*(-4*a*c+b^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/e^6/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.69, antiderivative size = 603, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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]

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*}

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Mathematica [C]  time = 13.44, 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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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \]

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

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]

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

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maple [B]  time = 0.19, size = 14453, normalized size = 23.97 \[ \text {output too large to display} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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